QA 471 R96 1893a CORNELL UNIVERSITY LIBRARIES Mathematics Library WhHe Hall CORNELL UNIVERSITY LIBRARY 924 059 551 501 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924059551501 Production Note Cornell University Library produced this volume to replace the irreparably deteriorated original. It was scanned using Xerox software and equipment at 600 dots per inch resolution and compressed prior to storage using CCITT Group 4 compression. The digital data were used to create Cornell's replacement volume on paper that meets the ANSI Standard Z39. 48-1984. The production of this volvime was supported in part by the Commission on Preservation and Access and the Xerox Corporation. 1991. fyvmll Wimvmiiri ^ihxM^ BOUGHT WITH THE INCOME FROM THE SAGE ENDOAVMENT FUND THE GIFT OF Hcnrg W. Sage 1891 MAmaiATIGS Cfrtrcnbon {pvtee ^ertee PURE GEOMETRY -R USSELL £onion HENRY FROWDE Oxford University Press Warehouse Amen Corner, E.C. MACMILLAN & CO., 112 FOURTH AVENUE AN ELEMEI^TAET TEEATISE ON PURE GEOMETRY WITH NUMEROUS EXAMPLES JOHN WELLESLEY RUSSELL, M.A. rOEHEBLT FELLOW OF UIBTOH COLLEGE UATHEHATICAL LECTHBEB OF BALLIOL AND ST. JOHN'S COLLEOEB, OXFOBD €>;rfoti AT THE CLAKENDON PKESS 1893 otb CLARENDON PRESS I B V MOaACB J PRINTER TO THE UNIVERSITY PREFACE In this treatise, the author has attempted to bring together all the well-known theorems and examples connected with Harmonics, Anharmonics, Involution, Projection (including Homology), and Reciprocation. In order- to avoid the difficulty of framing a general geometrical theory of Ima- ginary Points and Lines, the Principle of Continuity is appealed to. The properties of Circular Points and Circular Lines are then discussed, and applied to the theory of the Foci of Conies. The examples at the ends of the articles are intended to be solved by the help of the article to which they are appended. Among these examples will be found many interesting theorems which were not considered important enough to be included in the text. At the end of the book there is, besides, a large number of Miscellaneous Examples. Of these, the first pai-t is taken mainly from examination papers of the University of Oxford. Scattered throughout the book will be found examples taken &om that admirable collection of problems called Mathematieal Qtiestions and Solutions from the 'Educational Times.' For permission to make use of these, I am indebted to the kindness of the able editor, Mr. W. J. C. Miller, B.A., Registrar of the General Medical Council. vi Preface. The book has been read both in MS. and in proof by my old pupil, Mr. A. E. JoUiffe, B.A., Fellow of Corpus Christi College, and formerly Scholar of Balliol College, Oxford, whose valuable suggestions I have made free use of. To him I am also indebted for the second part of the Miscella- neous Examples. I am glad of this opportunity of ac- knowledging my great obligations to my former tutor, the late Professor H. J. S. Smith. My first lessons in Pure Geometry were learnt from his lectures ; and many of the proofs in this book are derived from the same source. I have assumed that the reader has passed through the ordinary curriculum in Geometry before attempting to read the present subject ; viz. Euclid, some Appendix to Euclid, and Geometrical Conies. I have not found it convenient to keep rigidly to any single notation. But, ordinarily, points have been denoted \sY A,B, C,..., lines by a, 6, c,..., and planes and conies by a, /3, y,.... The following abbreviations have been used — A straight line has been called a line, and a curved line has been called a curve. The point of intersection of two lines has been called the meet of the lines. The line joining two points has been called the join of the points. The meet of the lines AB and CB has been denoted by {AB; CB). To avoid the frequent use of the phrase ' with respect to ' or ' with regard to,' the word ' for ' has been substituted. The abbreviation 'r. h.' has sometimes been used for "rectangular hyperbola.' The single word ' director ' has been used to include the ' director circle ' of a central conic and the ' directrix ' of a parabola. Preface. vii The angle between the lines a and 6 has been denoted by Z db and the sine of this angle by sin ah. The length of the perpendicular from the point A on the line & has been denoted by (^, V). I have ventured to use the word ' mate ' to mean ' the point (or line) corresponding.' I have avoided using the word ' conjugate ' except in its legitimate sense in connection with the theory of polars. I shall be glad to receive, from any of my readers, correc- tions, or suggestions for the improvement of the book ; interesting theorems and examples which are not already included will also be welcomed. J. W. KUSSELL. February, 1893. CONTENTS CHAPTER I. FOBHULAE COSHEtHIHa SEOHTEBTS OF THE SAME LIBE. General properties of points and lines .... Menelaus's theorem. Oeva's theorem .... PAGE 2 7 CHAPTER II. . HABHOmC RAHOES AKD PENCILS. Polar of a point for two lines. Pole of a line for two points . i8 Locus of P, given i/OP = 2 a/OA -19 Complete quadrilateral. Complete quadrangle ... 19 CHAPTER III. HABMONIC PBOFERTLES OF A CIRCLE. Imaginary points and tangents Inverse points. Orthogonal circles Harmonic chord of two circles Salmon's theorem .... Self-conjugate triangle — polar circle Inscribed quadrangle — circumscribed quadrilateral 23 24 25 27 31 33 X Contents. CHAPTER IV. PROJECnOK. PAGE Line at infinity ■ • - 37 The eight vertices of two quadrangles ... . . 40 Homologous triangles ....•■ . . 4a CHAPTER V. HARHONIC PBOFERTIES OF A COKIC. Two lines or two points form a conic . . . . 47 Self-conjugate triangle 50 — of parabola, given a circumscribing triangle . . • 53 Centre — diameter— conjugate diameter — condition for circle — asymptotes — rectangular hyperbola — principal axes . • 53 CHAPTER VI. cabnot's theokeh. Newton's theorem — equations of conies ... . . 61 Hyperbola — conventional conjugate diameter — asymptotic pro- perties 63 Rectangular hyperbola — passes through orthocentre ... 65 CHAPTER VII. FOCI OP A COHIO. Coufocal conies. Focal projection 70 CHAPTER VIII. BECIPKOCATION. Focal properties of conies. Envelopes 88 Fencelet's theorem ... 94 CHAPTER IX. ADHARHOiriC OK CBOSS RATIO. Harmonic range 100 Cross ratio unchanged by projection ...... 103 To project the figure ABCB into the figure A'ffC'iy . . 103 Contents. xi PACE Homographic ranges and pencils 104 Ranges and pencils in perspective ....... 106 Triangle inscribed in one triangle and circtunseribed to another . 108 Projective ranges and pencils no CHAPTER X VAKISHINO POINTS OF TWO HOHOORAPHIC HAHBES. Locus of vertex of projection of two homologous triangles . 113 Formula fcts' + tc + mx' + n = o 114 Equation of a line and of a point 116 Common points of two homographic ranges . . .118 Ranges formed by the mates of a point in two homographic ranges . ... . ... . . . 119 Determine X, given .4X..ab tniES. Concentric circles have double contact 353 The circle about a triangle self-conjugate for a rectangular hyper- bola passes through the centre. Gaskin's theorem . . 353 Axes of conies through four concyclic points. Director circle . 353 Coaxal circles ^54 Foci of a conic. The circle about a triangle about a parabola passes through the focus. Gonfocal conies .... 355 xvi Contents. CHAPTER XXrX. PBOJXCmON, KEAX Am} rHAOIKABT. PACE Homologous triangles 264 Fole-Iocus touches sixteen conies a^S Common chords and common apexes a66 Harmonic envelope and harmonic locus of two conies . . . 267 Poncelet's theorem 268 Envelope of join of corresponding points of two homographic ranges on a conic. Conies having double contact. Envelope of last side of polygon inscribed in a conic, so that each side but one shall pass through a fixed point .... 269 CHAPTER XXX. OEHERALISATIOir BT PB0:rE(7n0Il. Generalisation by Seciprocation 279 CHAPTEK XXXI. HOHOLoay. Locus of vertex of projection of two figures in perspective . . 283 Coaxal figures and copolar figures ....... 285 Given a parallelogram, construct a parallel to a given line through a given point 288 MlSOEIXAITEOUS EzAUPLES , . 299 TEXT-BOOK OF PURE GEOMETRY. CHAPTEE I. FOEMULAE CONNECTING SEGMENTS OF THE SAME LINE. 1. One of the differences between Modern Geometry and the Geometry of Euclid is that a length in Modem Geometry has a sign as well as a magnitude. Lengths measured on a line in one direction are considered positive and those measured in the opposite direction are considered negative. Thus if AB, Le. the segment extending from A to B, be considered positive, then BA, i.e. the segment extending from B to A, must be considered negative. Also AB and BA diifer only in sign. Hence we obtain the first formula, viz. AB = -BA. Notice that by allowing lengths to have a sign as well as a magnitude, we are enabled to utilise the formulae of Algebra in geometrical investigations. In making use of Algebra it is generally best to reduce all the segments we employ to the same origin. This is done in the following way. 2 ^ ^ Take any segment AB on a line and also any origin 0. Then AB = OB— OA. This is obviously true in the above figure, and it is true for any figure. For OB- OA = OB+AO = AO+OB = AB; for AO+OB means that the point travels from A to and then from to B, and thus the point has gone from A to B. B 2 Formulae connecting Segments [ch. The fundamental formulae then are (i) AB=-BA; (2) AB=OB-OA. In the above discussion the lengths have been taken on a line. But this is not necessary; the lengths might have been taken on any curve. It is generally convenient to use an abridged form of the formula AB = OB—OA, viz. AB = I— a, where a= OA and 6 = OB. 2. A, B, C, D are any frntr collinear points ; show that AB. CD+AC. BB + AD. BC = o. Take A as origin, then CJ) = AD-AC= d—c, and so on. Hence AB. CD+AC. DB+AD.BG=b{d-c) + c{l>-d) + d{c-b) =bd-ic + cb-cd+dc—db=o. XiX. 1, A, B, C, D, are any Jive points in a plane; show that AOB.COD + AOC. BOB + AOD.BOC = o, where AOB denotes the area of the triangle AOB. Let a line meet OA, OB, OC, OB in A', B', C, £/. Then AOB = i. OA. OB Bin AOB. Hence the given relation is true if S { sin AOB . sin COD] = o, i. B. if 3 {sin^'OB'. sin COI/} = o. But p . A'B' = OA'. OB' sin A'OB', where p is the perpendicular from on A'B'CV. Hence the given relation is true if A'B'. CJ/ + A'C. VBf + A'jy. BfCf = o. Ex. 2. If OA, OB, OC, OD be any four lines meeting in apoint, show thai aiaAOB. sin COD + sin AOC . sin DOB + sin AOD . sin BOC = o. Ex. 3. Show also that cos AOB . sin COD + cos AOC. sin DOB + cos AOD . sin BOC = o, and cos^^OB. cos COB— cos ^OC. cosDOB— sin .4 02J. sin BOC = o. For Ex. 2 is true for OA' where A'OA is a right angle, and also for OA' and OD' where A'OA and D'OD are right angles. XiX. 4. From Ex. 2 deduce Ptolemy's Theorem amneeting four paints on a cirde. Take also on the circle. Then AB = s . B . sin AOB. "Ex.. 5. Show also that the relation of Ex. 2 halda if each angle involved he multiplied by Hie same quantity. For A0B=T0B-70A, if 07 be the initial line. Now take 70& = h . VOB, roc = h . roc, and so on. Then .4'Ofl' = A . AOB, &c. ; and the theorem is true for A'OB', &c. I.] of the same Line. 3 Ex. 6. Tf A, B, C be the angles 0/ a triangle and A', B', & be the angles which the sides BC, CA, AB make mth any line, then ainA .sin.4' + 8iiiB.8inB' + sm C. sin C= o. Draw parallels through any point. Ex. 7. OL, Oil, ON are any three lines through and PL, PM, PN make ecpjuU angles with OL, OM, ON in the same way, show that Pi . sin MON + par . sin if 01 + PN . sin LOM = o. 3. A, B, C are any three collinear points, and P is any other point ; show that PA\ BC+PB". CA + PC\ AB+BC. CA.AB = o. Drop the perpendicular PO from P on ABC. Then PA\ BC = {OA' + OP') BC = (a'' +p') (c - h). Hence '2{PA\BC)='2a'ic-i)+p^2{c-h)='2a'(c-h) = a^c—a'h + y^a—l''c + c'b—c''a= —{c — h)(a—c){b—a) ", =-BC.CA.AB. Ex. 1. If A, B, C be three coUinear points and a, b, c be the tangents from A, B, Cto a given circle, then a'. BC+ b'. CA + c\ AB + BC. CA. AB = o. Ex. 2. IfPbe any point on the base AB of the triangle ABC, then AP. CB'-BP. CA'= AB.(_CP'-AP. BP). Ex. 3. XfAjB, C, Dbe four points on a circle and P any point whatecer, show that £^BCD.AP'-A CDA.BP'+ABAB.CP'-AABC.BP'^ o, disregarding signs. Let AC, BD meet in inside the circle. Then A BCD oc BB.CO and BO. OD = CO.OA. Ex. 4. IfYA, YB, VC, YD be anyfowr lines through Y, then ainBYD.&iaCYD smCYD.siD.AYD sin AYD. sin BVD sinBYA.suiCVA siaCYB.aiaAYB sinAYCanBYC Draw a parallel to YD. Ex. 5. Jf A, B, Cbethe aivgUs of a triangle and A', B', C the angles which the sides BC, CA, AB make with any line, then sins', sin C sin C. sin .4' sin.4'. sins' _ sinB.sinC sinC.sin. in the same point ; similarly for AC. Ex. %. Ifad and iw meet an BD, then ai and cd meet on AC, and the above relation holds. Ex. 9. IfAB and CD meet in E and AD and BC in F, and if Edb cut AD in d and BCin b, and ifFac cut AB in a and CD in c, then Aa.Bb.Cc.Dd = aB.bC.cD. dA. Use the theorem sin A/an B = 0/6. Ex. 10. Beftoem ABCD, abed there holds also the fdOmring rdatvm sin 06 B. sin 6c C. sin cdi). sin (to .4 = sin £ 6c. sin Ccd. sin i) do. sin .4 a6. Ex. 11. If the lines AB, BC, CD, DA, which are not in the same plane, be met by any plane in a, b, c, d; then the relation of Ex. 7 holds ; and if this relation hM, the four points are in one plane. For the planes ABD, CBD, abed meet in a point, i. e. od and 6c meet on BD. Ex. 12. If the sides of the triangle ABC ahich is inscribed in a circle be cut by any transversal in D, E, F, show that the product of the tangents from D, E, F to the circle is numerically equal to AF. BD . CE. Ex. 13. Construct geometrically the ratio a/b-i-c/d. Ex. 14. The bisectors of the supplements of the angles of a triangle meet the opposite sides in coUinear points. Ex. 15. The bisectors of two angles of a triangle and the bisector of the sup- plement of the third angle meet the oppo^te sides in coUinear points. Ceva's Theorem. 8. If the lines joining any point to the vertices A, B, C of a triangle meet the opposite sides in D, E, F; then- AF.BD.CE=FB.BC.EA. To verify the sign of the formula. the point in which TO Formulae connecting Segments [ch. AI>, BE, CF meet must be either inside the triangle, in which case each of the ratios AF: FB and BD : DC and CE : EA is positive, or as at 0, or 0^, in which cases two of the ratios are negative and one positive. Hence the sign of the formula is correct. To prove the formula nimierically, we have AF:FB::AACF:AFCB:: AAOF-.AFOB :: l^ACF- LAOF : l^FCB- LFOB ::AAOC:ABOC. SimUarly BI):DC::ABOA:AAOC and CE:EA::hCOB:AAOB. Hence, multiplying, we see that the formula is true numerically. Conversely, if three points D, E, F, taken on the sides BC, CA, AB of a tricmgle, satisfy the relation AF.BD.CE = FB.DC. EA, then AB, BE, CF are concurrent. For, if not, let AB, BE cut in O ; and let CO cut AB in F'. Then since AB, BE, CF' are concuri'ent, we have AF'. BB.CE = F'B . BC. EA. But by hypothesis we have AF. BD.CE-FB.BG. EA. Dividing, we get AF': F'B -.-.AF: FB. Hence F and F' coincide, i.e. AB, BE, CF axe concurrent. Ex. L In the figure, show that OP OE OF _ AD* BE* CF ~ '■ Ex. 2. AO meets BC in D, BO meets CA in E, CO meets AB in F. GH is equal and parallel to BC and passes thrmgh A. BC meets 00 in L and RO in K. Similarly segments like KL are farmed on CA and AB. Show that the pro- dud of these segments is j^p ^j, pj, Ex. 3. Show thcA the necessary and sufficient condition that Aa, Bb, Cc should meet in a point is sin a AB . sin h BC . sin cCA = sin CA a . sin AB b . sin BC c. Ex. ^. If the lines Aa, Bb, Cc, Dd, . . . drawn through the vertices of a plane polygon ABCD .. . in the same plane meet in a point, then the continued product of sueh ratios as sin a AB : sin AB b is unity. I.] of the same Line. 1 1 Ex. 5. Iffhe lines joining a fined point to the opposite vertices of a polygon of an odd number of sides meet the sides AB, BC, CD, DE, ... in the points a, b, 0, d, . . ., show that the continued prod/uct ofsueh ratios as Aa/aB is unity. For Aa/aB = AO.aO sin AOa/aO.BO sin aOB. Ex. 6. The lines joining the centres of the escribed circles of a triangle to the middle points of the corresponding sides of the triangle are concurrent. Ex. 7. AO meets BC in P, BO meets CA in Q, CO meets AB in K ; PU meets QR in X, QU meets EP in Y, RU meets PQ in Z ; show that AX, BY, CZ are concurrent. Ex. 8. Through the vertices of a triangle are drawn paraMels to the reflexions of the opposite sides in any line ; show that these parallels meet in a point. For the angle between the reflexions of two lines is equal to the angle between the lines. Ex. 9. A', B', ff are the reflexions of A, B, C in a giren line through 0. OA', OB', Off meet BC, CA, AB in D, E, F. Show that D, E, F are coUinear. For BD-.BC-.-.BO sin BOA' : CO sin A'OC. Ex. 10. A cirde meets BC in D, Df, CA in E, E', and AB in F, F'. Show thai if AD, BE, GF meet in a point, so do ADf, BE', CF'. Ex. 11. A line meets BC, CA, AB in P, Q, R and AO, BO, CO in X, Y, Z. being any point. Show that QX. RY.PZ = RX.PY. QZ. Ex. 12. AA', BB', Cff meet in a point, show that the meets of BC. B'Cf, of CA, CA' and of AB, A'B' are coUinear ; and conversely, if the meets are coUinear, the joins are concurrent (See also IV. ii.) Let p. p' be the perpendiculars from A on A'B', A'ff and q, 3' those from Bon B'ff, B'A' and r-, r' those from C on ffA', ffBf. Then sinB'^'^ :sin^.4'0' •.-.p-.pf and y : r ■.-.AY-.CY, if .ICmeetX'tr in Y. Ex. 13. The linjes from the verHces of a triangle to the points of contact of any drck touching the sides of the triangle are concurrent. CHAPTER II. HAHMONIC RANGES AND PENCILS. 1. A range or row is a set of points on the same line, called the axis or base of the range. A pencil is a set of lines, called rays, passing through the same point, called the vertex or centre of the pencil. If A, B, A', B^ are collinear points such that AB : BA': : AB': A'B' or (which is the same thing) such that AB/BA'= -AB^IB^A', then {ABA'B^ is called a harmonic range. A, A' and B, B' are called harmonic pairs of points ; and A, A' are said to correspond and B, B' are said to correspond. Also A is said to be the fourth harmonic of A' (and A' oi A) for B and B'; so B is said to be the fourth harmonic of B' (and B' of B) for A and A'. Also AA' and JS5' are called hmmonic segments and are said to divide one another harmonically. The briefest and clearest way of stating the harmonic rela- tion is to say that {AA', BB') is harmonic. The relation may be stated in words thus — each pair of harmonic points divides the segment joining the other pair in the same ratio internally and externally. A B A' & For BA : AB'= -BA': A'B'. Ex. 1. I%e centres qf similitude of tim drdes divide the segment joining the centres of the circles harmonicaUy. Ez. 2. Ttie internal and external bisectors of the vertical angle of a- triangle cut the base harmonically^ CH. II.] Harmonic Ranges and Pencils. 1 3 Ex. 8. (BC, XX'\ {CA, TT'), {A3, ZZ') are harmonic ranges ; show that if AX, BY, CZ are concurrent, then X'Y'Z' are coUimar, and that if X'Y'Z' are cdlinear, then AX, BY, CZ are concurrent. Use the theorems of Ceva and Menelaus. 2. If [AA, BB^ he harmonic, then 211 211 AA'^AB'^AF' BB'~BA^ ba!' 211 211 I^I~2Zb ■'"Zb'' 'Wb~^a'^ Wa'' Taking any one of these formulae, say 2 _ I I choose A' as origin in the defining relation AB-.BA'-.-.AB^-.A'BT and use abridged notation. Then AB . A'B'= BA'. AB^ gives us (6-o)fc'=(-6)(6'-o) or hh'-ab' +bb' -ah = o, 211 or 266'= 06 + 06' or - =1 + Tj- a o o Conversely, if any one of these relations is true, then {AA', BB') is harmonic. For retracing our steps we see that AB:BA'::AB':A'B'. Ex. 1. ^ ^ bisect BB', then AB . AS = AA'. AS. Ex. 2. AX), BE, CF are the perpendiculars on BC, CA, AB, and {BC, DP'., {CA, EQ) and {AB, FB) are harmonic ; shme that PQR is the radical axis of the cireum-drde and the nine-point circle of ABC. Ex. 3. If {AA', BB') he harmonic, and P be any point on the line AB', show that R^ _PB Pff ''' ■ AA' ~ AB'^ AB'' Put PA'= PA-^AA', &e. Ex. 4. If EF divide both AA' and BB' harmonically, then AB . B'E . EA' = -A'B' . BE . EA. For we have i/EA - i/EB = i/EB' - i/EA'. If we call AA' the harmonic mean between AB and AB' and so on, Ex. I shows us that the G. il. between two lengths is equal to the G. M. between the A.M. andtheH.H. For A$ is the A. H. between AB and AB'. 14 Harmonic Ranges and Pencils. [ch. 3. If (AA', BB^ be harmonic, then aA^= aB. aB'; and conversely, if aA^= aB. aB', then {AA', BB') is harmonic, a being the middle point of A A'. For taking a as the origin in the defining relation AB:BA'::AB':A'B', we have {b-a) {b'-a') = (a'-b) {b'-a). But o'= —a, hence {b-a){b'+a) = {-a—b){b'—a), Le. bb' + ba-ab'-a^ + ab'—a^+bb'—ba= o, i.e. bb'r=a% Le. aA^=aB.aB'. The converse follows by retracing our steps. X!x. 1. Shom that the middle point qf either qftim harmonic segments is outside the other segment. Ex. 2. The chord of contact o/ tangents from A to any circle cute fhe diameter BSf through A in the fourth harmonic of A with rested to BB'. For 05* = OP" = OA.OA' by similar triangles, P being one of the points of contact and the centre. XiX. 3. Deduce a construction for the fourth harmonic <^ A' mth respect to BB' when A' is between B and B'. ^!z. 4. Deduce the connexion between the A, M., O. M., and M. M. of AB aiid ABf, I!x. 6. Deduce Oic formula a/AA'= i/AB+\/A&. We have AO.AA'= AI^= AB . AB'. Hence the result follows from a. AO = AB + AB^. Ex. 6. Do Ex. 4 and Ex. s, interchanging A and A'. Ex. 7. Show thai if {AA', BB') be harmonic and a bisect AA' and P bisect BB',then a . oB = (VI^ ± V2^)'. lix.8. Also AB'-.A'S':: BA: PA' ^ For 0A : 0A' = 0A' : PA . PA' = PA' : PBf. ^ Ex. 9. Oiven two segments AB, CD upon the same line, construct a segment XY which shaJX divide boOi AB and CD harmonically. Take any point P not on the given line. Through ABP and CDF construct circles cutting again in Q. Let PQ cut ABCD in 0. From draw tangents to the circles. With as centre and any one of these tangents as radius, describe a circle. This circle will cut the given line in the required points X and Y. For OJP= OT'=OP.OQ= OA.OB- OC. OD. 4. To find the relation between four harmonic points and a fifth point on the same line. Let {AA% BB') be harmonic, and take the fifth point P as origin. Then by definition AB/BA'= -AB'/B'A'. II] Harmonic Ranges and Pencils. 15 But AB = PB-PA = b-a, &c. Hence (6-ffl) ia'-h') + {a'-b] {b'-a) = o or 2aa'+2bb'={a + a'){b+b'), Le. 2.PA.PA'+2.PB. PB'= {PA + PA') {PB + PB'). Conversely, if this relation hold, (AA', BB') is harmonic. For reasoning backwards we deduce the relation AB/BA'= -AB'/B'A'. If (AA', BB') be harmonic, and a bisect AA' and )3 bisect BB', then PA.PA'+PB.PB'= 2.Pa.P^. For PA + PA'= 2 . Pa and PB+PB'= a . Pi3. Note that every relation of the second order connecting harmonic points must be identical with the relation of this article. Hence the following relations can be proved. Ex. 1. a . AB'. BA'= Z.AB. A'B' = AA'. BB'. Ex. 2. AB.AB' + A'B.A'B'= A' A'. Ex. 3. A'A^-^BfB^= {AB + A'Wy = 4.0^". Ex. 4. PA . A'B' + PA'. AB + PB. B'A + FB'. BA' = o. Ex.5. A&=a.aB.A^. Ex. 6. BA : BA' ■.:0B: A'P. Ex.7. PA. PA' -PIP + a. aB. pp. Ex. S.IfP and Q be arbitrary points, then PA . QB'. A'B + PA'. QB. AB' + PB.QA. B'A' + PB'. QA'. BA = o. Take P as origin and put QB' =b'-x, &c. Ex. 9. PB . PB'. AA' + PA\ A'0 + A'F'. $A = o. This is a relation of the third order, which vanishes when A coincides with A'. Hence we guess that it is the product of (a— a') into the harmonic relation. 5. J£ B, B' divide AA' in the same ratio internally and externally, then by definition {AA', BB') is a harmonic range. Now suppose this ratio is one of equality, then B becomes the internal bisector of the segment AA', i.e. B is the middle point of AA'; also B' becomes the external bisector of . the segment AA', Le. a point such that AB'= A'B', B' being outside AA'. But AB'/A'B'=^ {AA'+A'B')/A'B'= AA'/A'B'+ 1 ; and this can only be i when AA'= o or A'B'= co. Hence, 1 6 Harmonic Ranges and Pencils. [ch. assuming that A and A' do not coincide, we must have A'B'= oo , Le. B' must be at infinity. Also if .S' is at infinity, then AB' lA'S-=- 1 as above. Hence AB^= A'B', i.e. B' at infinity bisects A A' externally. Hence the two theorems — The point at infinity on any line bisects extei'nally every segment on this line. Every segment is divided Jiarmonically by its middle point and the point at infinity on the line, or, in other words, by its internal and external bisectors. 6. If any two points of a harmonic range coincide, then a third point coincides with them and the fourth may be anywhere on the line. Suppose AA' coincide. Then B lying between A and A' must coincide with them. So for BB', Suppose AB coincide. Then AB = o ; hence, from the defining relation AB . A'B'=: BA'. AB', we conclude that BA'= o or AB^= o, i.e. ABA' coincide or ABB'. So for AB', A'B, A'B'. Again, if ABA' coincide, then AB = o and BA'= o ; hence the relation AB.A'B'= BA'. AB' is satisfied wherever B' is. So for BA'B', &c. 7. A pencil of four concurrent rays is called a harmonic pencil if every transversal cuts it in a harmonic range. Harmonic pencUs exist for — If a pencil be obtained by joining any point to the points of a harmonic range, then every transversal cuts this pmcil in a har- monic range. Let (AA', BB') be a harmonic range and V any point. Join V to AA'BB", and let any transversal cut the joining lines in aa'bb'. Then ab:ha'= H^aVb: AbVa' = Va.Vb sin aVb-.Vb. Va' sin b Va'. „ ab ab' _ sin aF& sin aVb' Hence ^,-^^,- ^^f^' "^ ein b'Va' ' II.] Harmonic Ranges and Pencils. 17 Now aYV= AVB'; but for the transversal o^' we should have a F/3'= 1 80°-^ VB'. So in all cases aVb' is either equal to or supplemental to A VB'; hence in all cases sin aVb'= sin A VB'. So for the other angles. „ ah ab' ^nAVB sin^r^ ba' ■ h'a'~ sin BVA' ' sinB'VA' AB AB' ^ . ., = pX'"^ n'A' "y similar reasonmg = — I by deilnition. Hence ab/ba'-i- ab'/b'a'= — i ; henee {aa', 66') is a har- monic range. We denote the pencil subtended by ABA'B' at V by V{ABA'B') ; and we may briefly state the above theorem thus — if (AA', BB^) is a harmonic range, then V(AA', BB') is a harmonic pencil (or more briefly still — is harmonic). Ex. 1. If B bisect AA' and Vn be drawn parallel to AA', then the pencil Y {AA', Bn) is harmonic. For (AA', B fl) is harmonic, fl being the point at infinity on AA'. Ex. Z. If a transversal be drawn parallel to the ray VB of the harmonic pencil V (AA', BB') meeting the other rays in aVa', then V bisects aa'. For b is at infinity. C i8 Harmonic Ranges and Pencils. [ch. Ex. 3. The internal and external bisectors of an angle form with the rays of the angle a harmonic peticil. Draw a parallel to one of the bisectors ; or use Eu. VI. 3 and A. Ex. 4. 2/ a pair <>f correspmding rays of a harmonic pencil be perpendicular, they are the bisectors of the angles between the other pair of rays. Ex. 5. JfV (AA', BB') be harmonic, prove that 2 cot AVA' = cot ArB + cot AYS'. Take a transversal perpendicular to VA. Ex. 6. Also ifVa bisect the angle AVA', then tan^ aVA'= tan a VB . tan a VB'. Take a transversal perpendicular to Fa. Hx-I. zain AVB'. Bin BVA'= a sin AVB. sin A'VB' = Bin AVA'. sin BVB'. .„ _ sin PVA' sin PVB sin PFB' Ex. 8. 2 . = + Bin AVA' Bin AVB sin AYS' where VP is an arbitrary line through V. Also dediKe Ex. 5, 8. The polar of a point for two lines BA and BC is the fourth harmonic of BO for BA and BC. The pole of a line LM for two points A, B is the fourth harmonic of the meet of LM and AB for A and B. If through there be drawn the transversal OPQ cutting BA in P and BC in Q, then tJie locus of B, the fourth harmonic of for P and Q, is the polar of for BA and BC. For the pencil B (OPEQ) is harmonic. If the two lines BA, BC be parallel, i.e. if £ be at infinity, the theory still holds, if we consider B to be the limit of a finite point. To construct the polar of for ilA, HC where fl is at infinity, draw any transversal OPQ meeting ii4 in P and iiC in Q, and take B so that (OPQB) is harmonic, and through E draw a parallel ilB to il A and ilC; then ii.R is the polar of for the parallels ilA, SIC. Ex. 1. The polars of any point for the three pairs of sides of a triangle meet the opposite sides in three coUinear points. Let AO, BO, CO meet the opposite sides in P, Q, R, and let the polars of meet these sides in f, ^, B'. Then BP/PC = -BP'/P'C, and so on Now use the theorems of Menelaus and Ceva. Ex. 2. Thepolea of any line for the pairs qf vertices qfa triangle conned con- currentty with the opposite vertices. II.] Harmonic Ranges and Pencils. 19 Ex. 3. The poles of any line for the pairs of points BC and CA are collinear with the meet of the line and AB. Ex. 4. The polars of Ofor BA, BC and for CB, CA meet on AO. 9. Through a given point is drawn a line meeting two fixed lines in P and Q, and on OPQ is taken tlie point B such that i/OB = 1/OP+ i/OQ ; find the locus ofE. Take the polar w of for the two given lines, and let OPQ meet this line in R. Then we know that 2/0R= i/OP+j/OQ. Now draw parallel to n and half-way between and n the line n' cutting OPQ in R'. Then 0B'= OB/2, i.e. 2 /OR = I /OR'. Hence \/0B'= i/OP+i/OQ; hence n' is the required locus. Ex. 1. A transversal through the fixed point meets fixed lines in A.B.C,. .. and on OA is taken a point P such that i/OP = 1/0.4 + I /OB + 1/OC+ . . .; find the locus of P. Replace i/OA + i/OB by i/OL, and so on. Ex. 2. A transversal through the fixed point meets fixed lines in A, B,C,. . .; find the direction of the transversal when Z 1/0.4 is (i) o maximum (ii) a minimum. Perpendicular and parallel to the locus of P. Ex. 3. A transversal through the fixed ptnnt meets fixed lines inA,B,C,... and on OA is taken a point P such that i/OP = a/OA + b/OB + c/OC+ . . . , where a, b, c, . . . are any muMip'iers ; find the locus of P. Also find the direction of the transversal when "ia/OA is (i) a max., (ii) a min. Whatever a, 6, c, . . . are, we can, by taking the integer*: large enough, make ka, kb, kc,... all integers. Hence k/OP = 070.4 + V/OB + c^/OC+ . . . where k, a', V, cf, . . . are all integers. Now by Ex. i find the locus of Q such that i/OQ = {i/OA + ... a' times) + (i/OB + . . . V times) + . . . and draw a parallel through P to the locus of Q such that OP = k. OQ. This parallel is the required locus. 10. A complete quadrilateral is formed by four lines called the sides which meet in six points called the vertices of the quadrilateral. These six points can be joined by three other lines called the diagonals. The diagonals are also called the harmonic lines of the quadrilateral and the harmonic linesform the sides of the harmonic triangle. These names are derived c 2 20 Harmonic Ranges and Pencils. [ch. from the following property — called the harmonic property of a complete guadrilMeral. Each diagonal of a complete qiiadrilateral is divided har- monically by the other two diagonals. Let the four sides of the complete quadrilateral meet in the three pairs of opposite vertices AA', BB', CC. Then AA',BB',CC', or /3y, ya, a/3 are the harmonic lines. We have to show that the ranges (AA', §y), [BB', ya), (CC, ap) are harmonic. To prove that (AA', /3y) is harmonic consider the triangle whose vertices are AA' and any other of the vertices, say AA'G. Since ByB" are collinear, we have ^Pp CB.A'y.AB'= CB'.Ay.A'B. Also since AB, A'B', Cfi are concurrent, we have CB . A'P .AB'= - CB'. A p . A'B. Dividing we get A'y/Ay= -A'^/A^; hence (AA', j8y) is harmonic. Similarly (BB', ya) and (CC, a/3) are har- monic. 11. Using a ruler only, construct the fourth harmonic of a given point for two given points. To construct the fourth harmonic of y for B and B'. On any line through y take two points A and A'. Let A'B, AB' cut in G and AB, A'B' in C. Then CC cuts BB' in the required point a. For BB' is a diagonal of the complete quadrilateral formed by AB, AB', A'B, A'B'; hence (BB', ya) is harmonic. Ex. 1. AO, BO, CO meet BC, CA, AB in F, Q, R ; QS, SP, PQ meet BO, CA, AB in X, Y, Z. Shmo that {BC, PX), {CA, QT), {AB, RZ) are harmonic ranges, and thai XYZ are collinear. Ex. 2. 1/a transcersal meet BC, CA, AB in X, Y, Z, and the join uf A to the meet of BY and CZ eui BC in P; skou> Oiat {BC, PX) is harmonic, and, thai Pit three lines farmed like AP are cortcurrent. II.] Harmonic Ranges and Pencils. 21 12. A compkU quadrangle, is formed by four points called the vertices which are joined by six lines called the sides of the quadrangle. These six lines meet in three other points called fhe harmonic points of the quadrangle ; and the har- monic points are the vertices of the harmonic triangle. Some writers give the name diagonal-points to the harmonic points. The following is the harmonic property of a complete quadrangle. The angle at each harmonic point is divided harmonically by the joins to the other harmonic points. Let ^5CZ) be the v four points form- \*^^^^^^~~-~lr — — 7^ "- ing the quadrangle. \ "^^^^^^^^-4^ ' — ^ Then U, r, TTare V=*=='^^^^^^/V^ the harmonic points ° \ / / of the quadrangle ; \ // and we have to show \ Z' that the pencils c U{AI), VW), V{BA, WU), W{CD, UV) are harmonic. To show that the pencil W{CB, UY) is harmonic, it is sufficient to show that the range {LM, UV) is harmonic, L being the meet of ^C and UV, and M of BD and UV. Con- sider the triangle formed by UV and any vertex, say UVC. Then because BDM are coUinear, we have CB.VM. UI)=CB. UM.VB. Also because UB, VD, CL are concurrent, we have CB.VL.UI)=-CI).UL.VB. Hence dividing we get VM / UM = —VL / UL. Hence {UV, LM) is harmonic, i. e. W(CD, UV) is harmonic. Similarly U{AB, VW) and V{BA, WU) are harmonic. 13. Using a ruler only, construct the fourth harmonic of a given line for two given lines. To construct the fourth harmonic of VU for VA and VB. 22 Harmonic Ranges and Pencils. Through any point Z7 on YJJ draw any two lines YAB and YBQ, cutting F^ in ^ and B, and YB in B and G. Then if ^(7 and BJ) meet in PT, FW is the required line. For Z7, F, TF are the harmonic points of the quadrangle A,B,C, D. Hence Y{BA, WU) is harmonic. She 1. Through one of the harmonic points of a comply guculrangle is drawn the line pardttel to the join of the other fuxt harmonic points ; s}uiw that two of tht segments cut off between opposite sides of the quadrangle are bisected at the harmonic point. Ex. 2. Through V, one of the harmonic points of a quadrangle, is drawn a line parallel to one side and meeting the opposite side in P and the join of the other liarmonic points in Q, show that YP = PQ. Ex. 3. In the figure of the quadriMeraZ in $ lo, show that Aa, A'a, B0, B'fi, Cy, ffyform the six sides of a quadrangle. We have to show that the six lines pass three by three through four points. Consider aA. BB', yC Since A'ffC are collinear and {fiy, AA') is harmonic, o^, OB', yCf are concurrent. Similarly oA, &B, fC are concurrent, also aA', HB, yff, and also aA', BB', yC. Ex. 4. In the figure of the quadrangle in § I2, the sides of the triangle XJVW »«■«< the sides in six new points which are the lertices of a quadrilaterai. CHAPTER III. HAEMOKIC PEOPEKTIES OF A CIBCLB. 1. Every line meets a circle in two points, real, coincident or imaginary. For take any line I cutting a circle in the points A and B. Now move I parallel to itself away from the centre of the circle. Then A and B approach, and ultimately coincide when I touches the circle. But when I moves stUl further from the centre, the points A and B become invisible ; yet, for the sake of continuity, we say that they still exist, but are invisible or imaginary. (See also XXVII.) 2. From every point can be drawn to a circle two tangents, real, coincident or imaginary. For take any point T outside the circle, and let TP and TQ be the tangents from T to the circle. Now let T approach the centre of the circle along OT. Then TP and TQ approach, and ultimately coincide when T reaches the circumference. But when T moves still further towards 0, the tangents TP and TQ become invisible ; yet, for the sake of continuity, we say that they still exist, but are invisible or imaginary. (See also XXVII.) 3. Two points which divide any diameter of a circle har- monically are said to be inverse points for this circle. If be the centre and r the radius of the circle, then inverse points B, B' must lie on the same radius of the circle and be such that OB . 0B'= r". 24 Harmonic Properties of a Circle. [ch. Ex. 1. The inverse of any point at infinity for a circle is the centre of tlie drde ; and conversely, the inverse of the centre is any point at infinity. "Ex.. 2. Evury tim points and their inverses for a cirde lie on a circle. Ex. 3. Given a pair of inverse points for a drde, the circle must be oru of a certain system of coaxal circles. Ex. 4. Iffourpainti (AA', BB') he hannonic, so are the four inverse points {oaf, W)far any drde. r» r' r'. AB For Oa = — , Oh = -; hence ab=-^j-^^. Ex. 5. If BB' be a pair of inverse poitUs on the diameter AA' of a drde, and ifPbe any point on the drde ; then PA, PA' bisect the angle BPB', and the ratio PB : PB' is independent of the position of P. Ex. e. Also if perpendiculars to AA' at AA' BB' meet any tangent to the drde in aa' bV, show that Oa and Oa' bisect the angle hOV, bang the centre, and that the ratio Ob -.01/ is independent of the position of the tangent. 4. Two circles are said to be orthogonal when the tangents to the circles at each point of intersection are at right angles. Ex. 1. If tam drcUs are orthogonal at one of their meets, they are orthogonal at the other, Ex. 2. If the orthogonal drdes a and ft xchose centres are A and B meet in F, show that AP touches and BP Umclies a. Ex. S. The radii of two drdes are a and b and the distance between thdr centres is S ; show that the necessary and sufficient condition that the drdes should be orthogonal is o' + 6' = S". 5. Every circle which passes through a pair of points inverse for a circle is orthogonal to this circle ; and conversely, every circle orthogonal to a circle cuts every diameter of this circle in a pair of inverse points. First, let the circle y pass through the inverse points BB' ai the circle CO. Let P be one of the meets of ') is harmonic. Ex. 3. The UcMS of the middle points of PP' and QQf is the coaxai circle whose centre bisects AB. For the locus of X and Y is the auxiliary circle. Also each meet of the circles is on the locus ; for the tangent to either circle at a meet is divided harmonically. "Ex.. 4. If R he any point on a circle, A and B fixed points on a diameter arul equidistant from the centre, the envelope of a line which cuts harmonically the two circles with A, B as centres and AB, BR as radii is independent of the position of R on the circle. Its foci are A and B. Also 26'= AR'' + BR'-AB' = a OiJ' + a OA'-,iOA'. Hence b" = OIC-OA', which is constant. 7. Through a point U is drawn a variable chord PP' of a circle and on PP' is taken the point B stick that (UR, PP'} is harmonic ; to show that the locus ofB is a line. Take the centre of the given circle a>. Let OU cut o) in AA'. From any position of R drop a perpendicular RU' to UO. On iJ?7 as diameter describe the circle /3 pas- sing through U'. Now since {BU, PP') is har- monic, PP' are inverse points for 13. Hence to and ^ are orthogonal. Hence UU' are inverse for m. Hence Z7' is a fixed point. Hence the locus of iZ is a fixed line, viz. the perpendicular to OU through the inverse of U for the given circle. The locus of B is called the polar of U for the circle. We may briefly define the polar of a point for a circle as the locus of the fourth harmonics of the point for the circle. Also if RU' ia given, ?7is called its pole for the circle, and U and BU' are said to be pole amd polar for the circle. 8. IfUhe outside the circle, the polar of U for the circle is ■ the chord of contact of tangents from U to the circle. III.] Harmonic Properties of a Circle. 27 For take the chord ZZPP' very near the tangent JJT. Then when TV coincide, iJ, being between them, coincides with them ; i.e. one posi- tion of B, is at T. So another position of JS is atT'. Hence rT' is the polar. The polar of the centre of the circle is the line at infinity. (See IV. 3.) For if U coincide with 0, then PP' is bisected at U. Hence B is at infinity. The pole of the line at infinity for a circle is the centre of the circle. For if ii be always at infinity, PP' is always bisected at Z7, i.e. Z7is the centre of the circle. The polar of a point on the circle is the tangent at the point. For suppose ZJto approach A, then since OU.OU'=- OA", we see that U' also approaches A. Hence when tT is at ^, U' is at ^ ; and the polar of Z7, being the perpendicular to OU through Z7', is the tangent at U. Similarly, the pole of a tangent to a circle is the point of contact. 9. Salmon's theo- rem. — If P and Q be any two points and if PM be the perpendicular from P on the polar of Q for any circle, and if QN be the perpendicular from Q on the polar ofP fw the same circle, then OP/PM = OQ/QN, being tlie centre of the circle. From P drop PX perpendicular to OQ and from Q drop ^r perpendicular to OP. Then P' being the inverse point 28 Harmonic Properties of a Circle. [ch. of P, and C the inverse point of Q, we have Also since the angles at X and Y are right, we have or. op= ox.oq, . ■. OP/OQ= Oq/OP'= OX/OY={OQ'-OX)MOP'-OT) =Xq/YP'=PM/QN. Hence OP/PM = OQ/QJV: We may enunciate this theorem more briefly thus — If p, q be thepdlars ofP, Qfor a circle whose centre is 0, then OP/{P,q) = OQ/(Q,p). Ex. Xf a, b, p be the polars of the points A, B, P for a circle centre is 0, show that (f.o) . U,jp) _{0,a,) (a,a) {B,p) (0,&) For OA . (0, a)= OB. (0, b) = r". {A,b) 10. If the polar of P pass thronigh Q, then the polar of Q passes through P. If the polar of P pass through Q, then, P' being the inverse of P, P'Q is per- pendicular to OP. Take (^ the inverse of Q. Then OP.OP'=OQ.Oqf. Hence PP'QQ' are concyclic. Hence OQ'P = OP'Q is a right angle. Hence P(^ is the polar of Q, Le. the polar of Q passes through P. The points P and Q are called conjugate points for the circle. We may define two conjugate points for a circle to be such that the polar of each for the circle passes through the other. Note that if PQ cut the circle in real points BR', then, since the polar of P passes through Q, we see that {PQ, BR') is harmonic ; and hence the polar of Q passes through P. The pole of the join of P and Q is tM meet of the polars of PandQ. III.] Harmonic Properties of a Circle. 29 For if the polars of P and Q, meet in iZ, then, since the polars of P and Q pass through iJ, therefore the polar of i? passes through P and Q. 11. On emery line, there is an infinite number of pairs of con- jiigate points for a given circle ; and each of these pairs is har- monic with the pair of points in which the line meets the circle. On the line take any point P, and let the polar of P meet the line in P'. Then P and P' are conjugate points ; for the polar of P passes through P'. Also if PP' meet the circle in BBf, then (PP', BR') is harmonic ; for P' is on the polar of P. Conversely, every two points which are harmonic with a pair ofpoinis on a circle are conjtigate for the circle. 12. If the line p contain the pole of the line q, tlien q con- tains the pole of p. Let P be the pole of p and Q of q. We are given that p contains Q, Le. that the polar of P passes through Q. Hence the polar of Q passes through P, i.e. q passes through P, i.e. q contains the pole of p. The lines p and q are called conjtigate lines for the circle. We may define two conjugate lines for a circle to be such that each contains the pole of the other. Through every point can he draim an infinite number of pairs of lines which are conjugate for the circle, and each of these is harmonic with the pair of tangents from the point. For take any line p through the given point U and join U to the pole Pofp. Then j9 and UP are conjugate lines, for UP contains the pole oip. Draw the tangents UT and UT' from U, and let the polar TT' of Umeet p in P'. TT' meets UPinP since U is on the 30 Harmonic Properties of a Circle. [ch. polar of P. Now the range (PP', TT') is harmonic, for P' is on the polar of P; hence the pencil TJ{1P'P', TT') is har- monic, i.e. the conjugate lines p and UP are harmonic with the tangents from Z7. Conversely, every pair of lines which are harmonic with the pair of tangents frcym a point to a circle are conjugate for the circle. For let UQ and TJQ' be harmonic with the tangents UT, UT' from U. Let UQ and UQ' cut the polar TT' of ZJin P and P'. Since U{QQ', TT') is harmonic, hence {PP', TT') is harmonic. Hence UP' is the polar of P ; for the polar of P passes through P' since {PP', TT') is harmonic, and passes through U since P is on the polar of Z7. Hence since the pole of UP' lies on UP, we see that UP and UP' are conju- gate lines. Ex. L Find the loom of all the points conjugate to a given poinU IjX. 2. AU the lines conjjigate to a given line are concurrent. Ex. 3. WTien two points are conjugate, so are their polars ; and mhen two lines are conjugate, so are their poles. £x. 4. A point can he found conjugate to each of two given points ; and a line can be found conjugate to each of two given lines. SiX. 6. ]f the circle a be orthogonal to the circle $, then the ends of any diameter of a are conjugate for /3. XjX. 6. The circle on the segment PQ joining any pair of conjugate points for a circle as diameter is orthogonal to the given circle. For PO cuts the new circle in the inverse of P. Ex. 7. XfE'C be the polnr of A, CfA' ofB and A'Bf ofC; Own BC is the polar of A', CA of B' and ABofC. Ex. 8. Reciprocal triangles are homologous. Tliat is, if A is the pole of £'0', B of C'A', C of A'B', then AA', BB', CC meet in a point. This follows from sin BAA' : sin A'AC : : {A', c') : {A', V) and OA' : {A', c')::OC : (C, a'). (See also XIV. 3.) Ex. 9. X/P and Q be a pair of conjugate points for a circle to which they are external, then (i) P(^ is equal to the sum of the squares of the tangents from P and Q ; (ii) PQ is twice the tangent from the middle point of PQ ; (iii) PU. UQ is equal to the square of the tangent from U, V being the foot of the perpendicular from the centre of the circle on PQ ; (iv) the circle on PQ as diameter is orthogonal to the given circle. Take C the middle point of PQ and B the pole of PQ. III.] Harmonic Properties of a Circle. 31 Then JJQ meets OP perpendicularly in Y, say. Hence PQ- = 0P»+0Q=-20r. OP = 0I^-¥0(i^-2r' (i) = 2C0" + 2CQ=-2J'2 (ii) and PV. UQ = VR.VO = VCf-r' (iii). (iv) follows at once from (ii), or because the circle on PQ as diameter passes through T. Ex. 10. M and N are the projections of a point P on a circle on turn perpen- dicular diameters, Q is thepok o/MN/or the circle, and U and Vare the projections of Q on the diameters. Show that W touches the circle. UV is the polar of P. 13. Pairs of conjugate lines at the centre of a circle are called pairs of conjugate diameters of the circle. Every ^air of conjugate diameters of a circle is orthogonal. Take any diameter AA' of a circle whose centre is 0. The diameter conjugate to AA' is the line through conjugate to AA', i.e. is the join of to the pole of AA'. But the tangents at A and A' meet at infinity in Q., say. Hence OQ. is the conjugate diameter ; hence the diameter conjugate to A A' is parallel to the tangent at A, i.e. is perpendicular to AA'. £jX. 1. Tfte pole of a diameter is the point at infinity on any line perpendicular to the diameter ; and the polar of any point CI at infinity is the diameter perpen- dicular to any line through CL. "Ex.. 2. Any two points at infinity which S!ubtend a right angle at the centre are conjugate. 14. A triangle is said to be self-conjugate for a circle when every two vertices and every two sides are conjugate for the circle. Such a triangle is clearly such that each side is the polar of the opposite vertex. Hence the other names — self-recipro- cal or self-polar. Self-conjugate triangles exist. For on the polar of any point A take any point B. Then the polar of B passes through A and meets the polar of A in C say. Then ABC is a self-conjugate triangle. For BC is the polar of A, CA is the polar of B ; hence C, the meet of BC &nd CA, is the pole ofAB. Hence AB are conjugate points, and BC, AC are conjugate lines. So for other pairs. 32 Harmonic Properties of a Circle. [ch. Ex. The triangle formed by the line at infinity and any hoo perpendicylar diameters of a circle is self-conjugate for the circle. 15. There is only one circle for which a given triangle is self- conjugate; and this is real only when the triangle is obtuse- angled. Suppose the triangle ABC is self-conjugate for the circle whose centre is 0. Then since A is the pole of BC, it fol- lows that OA is perpendicular to BC; so OB is perpendicular to CA, and OC to AB. Hence is the orthocentre of ABC. Let OA meet BC in A', OB meet CA in B' and OC meet AB in C. Then the square of the radius of the circle must be equal to OA . OA' and to OB . OB' and to OC . OC; and this is possible if is the orthocentre, for then these pro- ducts axe equal. Now describe a circle (called the polar circle of the triangle) with the orthocentre as centre and with radius p, such that p'=0A . 0A'=0B . 0B'=0C. OC. Then the triangle ABC is self-conjugate for this circle. For BC, being drawn through the inverse point A' of A perpendicular to OA, is the polar of ^ ; so for CA and AB. Also this circle is imaginary if the triangle is acute-angled ; for then is inside the triangle and hence p^ (= OA . OA') is negative. Ex. L Describe a circle to cut Oie three sides of a given triatigle harmonically. When is this circle real ? Ex. 2. In any triangle the circles on the sides as diameters are orthogonal to the polar circle. Ex. 3. If any three points X, Y, Z be talten on the sides BC, CA, AB of a triangle, the circles on AX, BY, CZ as diameters are orthogonal to the polar circle. Ex. 4. The circle on each of the diagonals of a guadriUUeral as diameter is orthogonal to the polar circle of each of the four triangles formed by the sides of the quadrilateral. Ex. 6. Hence the two sets of circles are coaxal. Hence the middle points of the three diagoruils of a quadrilateral are coUinear; and the four ortho- centres qf the four triangles formed by the sides of a quadrilateral are coUinear, Ex. 6. Every circle cutting two of the circles on the three diagonals of a quadri- lateral orthogonally, cuts the third also orthogoruUly. For it cuts two circles of a coaxal s^rstem orthogonally. III.] Harmonic Pr(^erties of a Circle. 33 16. Thi harmonic triangle of a quadrangle inscribed in a circle is self-conjugate for the circle. Let UVW be the harmonic triangle of the quadrangle ABGB inscribed in a circle. Then UVW is self-conjugate for the circle. Let UV meet AC in. L and BD in M. Then since V{WU, BA) is harmonic, hence {WL, AG) and {WM, BD) are harmonic. Hence L and M lie on the polar of W, i.e. UV is the polar of W. Similarly VW is the polar of U, and WU of V. 17. With the rukr only, to construct the polar of a given point for a given circle. To construct the polar of V for the given circle, draw through V any two chords AD and BC of the circle. Let BA, CD meet in U, and AC, BD meet in W. Then by the above theorem WU is the polar of V. Ex. Through U one of ihe harmonic points of a quadrangk inscribed in a drde is drawn a chard catting the circle in oo', and the pairs of opposite sides in hV, a/; show that if one of the segments aa', W, of is bisected at U, ihe others are cUso bisected at U. Let the transversal cut the opposite side of the harmonic triangle in X, then UX divides each segment harmonically. 18. The three diagonals of a quadrilateral circumscribing a circle form a triangle self-conjugate for the circle. Let the three diagonals AA', BB', CC of the quadrilateral 34 Harmonic Properties of a Circle. [ch. BA, AS, B^A', A'B circumscribing the circle form the tri- angle a^y. Then afiy (the harmonic triangle of the quadri- lateral) is self-conju- ^^^ gate for the circle. \ ^^^^'a,,^-^~~ ""---- harmonic, hence C (/3y, \^^-^'^^f-%J '« ^^') and C'{^y, AA') °\'( i^~ \/^~ ^^^ harmonic, i.e. Cy \1 ) L is the fourth harmonic "Nv 1 yj of a^ for the tangents V T / from C, and G'y is the \ 1 / fourth harmonic of afi \ 1/ for the tangents from jj; C Hence aji is con- jugate to Cy and to C'y, i.e. the pole of a/3 lies on Cy and on Cy. Hence y is the pole of a^. Similarly a is the pole of /3y, and y3 of ya. 19. With the ruler only, to construct the pole of a given line for a given circle. This may be done by the above theorem ; but better by finding by § 1 7 the meet of the polars of two points on the given Une. £!z. The tioo lines joining the opposite meets 0/ common tangents of turn cirdes which are not centres oj similitude cut the line of centres in the limiting poiTits. For these points are two vertices of a self-conjugate triangle with respect to both circles. 20. Hie harmonic triangle of a quadrilateral circumscribed to a circle coincides with the harmonic triangle of the inscribed quadrangle formed by the points of contact. In the figure of § 18, let B'A, AB, BA', A'B' touch the circle in a, b, c, d. Comparing with the figure of § i6, we see that we have to prove that ac and bd meet in y, that ba and cd meet in a, and that cb and da meet in /3. Now ba is the polar of A and cd is the polar of A'; hence ba and cd meet in the pole of AA', i.e. in the pole of /3y, i.e. ba and cd pass through a. Similarly ac and bd pass through y, and cb and da pass through j3. III.] Harmonic Properties of a Circle. 35 The theorem is sometimes erroneously stated thus — Ofthx two quadrilaterals formed ty four tangents to a circle and the points of contact, the four internal diagonals are concurrent and form a harmonic pencil, and the two external diagonals are col- linear and divide one another harmonically. The former part follows from y being a harmonic point of the quadrangle. The latter part follows from /3a being a harmonic line of the quadrilateral. Ez. L If the whole figure be symmetrical for AA' and if the angle ABA' be right, show that ac, bd bisect t}te angles between AA' and BB'. By elementary geometry each of the angles at 7 is 45°. Ex. 2. AA' meets ab in P and cd in P, and so on. Show that the six points PP'QQfBR' lie three by three on four lines. CHAPTER IV. PEOJECTION. 1. Given a figure <}> in one plane ir consisting of points A, B, C, ... and lines Z, w, «,..., we can construct another figure ' consisting of corresponding points A', B', C,... and lines V, m', n',... in the following way. Take any point F (called the vertex of jprojedion) and any plane it' (called the plane of projection). Then A', B', C, ... and V, m', n', ... are the points and lines in which the plane of projection meets the lines and planes joining the vertex of projection to A, B, C,-. and I, m, n,.... Each of the figures and <^' is called the projection of the other ; and they are said to be in projection. Also each of the points A and A' is said to be the projec- tion of the other ; so for the points B and f , C and C, &c., and for the lines I and {', m and mf, n and n', &c. The line in which the planes of the figures and ^' meet may be called the axis of projection. When the vertex of projection is at infinity we get what is called parallel projection ; in this case all the lines AA', BB^, GC, ... are parallel. A particular case of parallel pro- jection is orthogonal projection. The lines AA', BB', CC, ... are called the rays of the projection ; and projection is sometimes called radial projec- tion to distinguish it from orthogonal projection. Figures in projection are also said to be in perspective in different planes ; and then the vertex of projection is called the centre of perfective, and the axis of projection is called the axis of perfective, and each figure is called Projection. 37 the perspective or picture of the other. Note that figures may also be in perspective in the same plane. (See XXXI.) Some writers use the term conical projection or central projection or central perspective for radial projection. 2. The projection of the join of two points A, B is the join of the projections A', B' of the points A, B. The projection of the meet of the two lines I, m is the meet of the projections V, m' of the lines I, m. The projection of any point on the axis of projection is the point itself. Every line and its projection meet on the axis of projection. The proofs of these four theorems are obvious. The projection of a tangent to a curve y at a point A is the tangent at A' {the projection of A) to the curve ■/ {the pro- jection ofy). For when the chord AB of y becomes the tangent at A to y hy B moving up to A, the chord A'B' of y becomes the tangent at A' to / by J5' moving up to A'. The projection of a meet {i.e. a common point) of two curves is a meet of the projections of the curves. The projection of a common tangent to two curves is a common tangent to the projections of the curves. The proofs of these theorems are obvious. 3. The plane through the vertex of projection parallel to the plane of one of two figures in projection meets the plane of the other figure in a line called the vanishing line of this plane. Each vanishing line is parallel to the axis of projection. For the axis of projection and the vanishing line in the plane w are the meets of it with i/ and with the plane through V parallel to it'. Every point at infinity in a plane lies on a single line {called the line at infinity). Let A be the point at infinity on any line I in the plane ti. Through any point V not in the plane draw a plane p parallel to the given plane. Then p passes through A ; for 38 Projection. [CH. p, being parallel to the plane of I, meets I at infinity. Similarly /> passes through every point at infinity in ir. Also every point of intersection of ir and p is at infinity on •77. Hence the points at infinity on ir are the points of inter- section of the two planes ir and p. And as two planes when not parallel meet in a line, we may say for the sake of con- tinuity that two parallel planes also meet in a line. Hence the points at infinity in a plane lie on a line. The vanishing line in one plane is the projection of the line at infinity in the other plane. ■ For the plane joining V to the vanishing line is parallel to the other plane. To project a given line to infinity. With any vertex of projection, project on to any plane parallel to the plane containing the given line and the vertex of projection. Then the projection of the given line will be the intersection of these two parallel planes and will therefore be entirely at infinity. 4. The vanishing point of a line is the point in which the line meets the vanishing line of its own plane. The angle between the projections of any two lines I and m is the angle which the vanishing points of I and m subtend at the vertex of projection. Let I and m meet in A, and let I meet the vanishing line i in B and let m meet i in C We have to show that the pro- jection of the angle BAG is equal to BVC, V being the vertex of projection. Now the plane of projection i/ is parallel to the plane BVC. Also A'B' is the meet of the plane A VB and n-'. Hence A'B' and VB (being the meets of the plane A VB with the two parallel planes it' and BVC) are parallel. Similarly A'C and VC are parallel. Hence Z B'A'C'= Z BVC. IV.] Projection. 39 Ex. AU angles whose bounding lines hare the same vanishing points are projected into eguai angles. 5. To project any two given angles into angles of given mag- nitudes and at the same time any given line to infinity. Let the given angles ABC, BEF meet the line which is to be projected to infinity in AC, BF. Then since A, C are the vanishing points of the lines BA, BC, hence the angle A'B'C is equal to ^ FC ; so Z B'E'F'= L BYF. Hence to construct F draw on ^C a segment of a circle containing an angle equal to the given angle A'B'C, and on BF and on the same side of it as before describe a segment of a circle con- taining an angle equal to the given angle B'E'F'. Let these segments meet in F. Kotate F about ACBF out of the plane of the paper. Then if we project with vertex F on to a plane parallel to the plane TAGBF, the problem is solved. For the line ABCF will go to infinity. Also ABC vrill be projected into an angle equal to A VC, i.e. into an angle of the required size. So for BEF. The segments may meet in two real points or in one or in none. Hence there may be two real solutions of the problem or one or none. Ex. In the exceptional case when the vanishing line is parallel to one of the Unes of one of the angles, give a construction for the vertex of projection. Let A be at infinity. Through C draw a line making with CF the supplement of A'B'C This wdll meet the segment on DF in V. 6. Given a line I and a triangle ABC, to project I to infinity and each of the angles A, B, C into an angle of given size. Suppose we have to project A, B, C into angles equal to a, 13, y, where of course a-l-/3 + y=i8o°. Let I cut BC, CA, AB in P, Q, B. Of the points P, Q, B let Q be the point which lies between the other two. On BQ describe a seg- ment of a circle containing an angle equal to a. On QP and on the same side of I describe a segment of a circle contain- ing an angle equal to y. These two segments meet in Q ; hence they meet again in another point, F say. For if the supplements of the segments meet in F, then BVQ+QYP= 180° — a + 180° — y = 180° -t- /3> 180°, which is impossible. 40 Projection. [ch. Now rotate F about I out of the plane of the paper. With Y as vertex of projection, project on to any plane parallel to TPq ; and let A'^C be the projection of ABC. We have to prove that A'=a, JS'=/3, C'=y. Through JB draw a parallel to YF meeting YQ, in Z. Then JJ7X=a, YXR=y, and Z^F=/3. Also A'l^ is parallel to FK, £'C' is parallel to FP and therefore to i?X, and C^' is parallel to YQ. Hence the sides of the triangles A'B^G and YUX are parallel. Hence the angles are equal ; i.e. A'-=^a, B'=fi, C=y. 7. To project any triangle into a triangle with given angles and sides and any line to infinity. Project as above the given triangle ABC into A'B'C in which LA'=La', LB'- Ih', IC'= Ic', a'Vc: being the triangle into which ABC is to be projected. On YA' take a point P such that FP :F^': : Vcf: B'C. Through P draw a plane parallel to A'B'C cutting YB' in Q and YC in E. Then by similar triangles FP : VA': : QB : B'C; hence QB = b'c'. So BP = c'a', PQ = a'V. Hence PQB is super- posable to a'b'c' and in projection with ABC. Hence we can project any triangle into an equilateral triangle of any size and any line to infinity. Ex. L Project any fcur given points into the angular points of a sqtmre of given size. Let ABCD (II. 12) be the given points. Project ZTV to infinity and the angles VAU, LWM into right angles. Then in the projected figure AB and CD are parallel, and also AD and EC. Also BAD is a right angle and also AWD. Hence the figure is a square. We can change its size as before. The construction is always real since the semicircles on LM and UV must meet since Lit and VV overlap. Ex. 2. Project any tioo homologous triangles (see § 11) simultaneously into equilateral triangles. Is the construction always real 1 Ex. 3. Project any three angles into right angles. Let the legs of the angles A and B meet in L and M, and let LM cut the legs of C in DE ; then on LM and DS describe semicircles. Ex. 4. If tiBO quadrangles have the same harmonic points, then the eight vertices lie on a conic ; as u particular case, if any three of the points are collinmr, the eight vertices lie on two lines. Project one of the sides UV of the harmonic triangle to infinity, and the angles UAV and t7.ilTinto right angles, and the angle LWM into a right angle. The quadrangles are now a square and a rectangle with parallel sides and the same centre ; hence the vertices by symmetry IV.] Projection. 41 lie on a conic whose axes are parallel to the sides. If however 3' is on £i), clearly this conic degenerates into the common diagonals ; so if JS' is on BA, the conic degenerates into BA and CD, and if B' is on BC into BC and AT). (See also XII. 7.) 8. In projecting from one plane to another, there are in each plane two points such that every a/ngle at either of them is pro- jected into an equal angle. Let the given planes be it and ir'. Draw the planes a and fi bisecting the angles between the planes n and ir'. Through the vertex of projection V draw a line perpendicular to o cutting the planes tt and ir' in E, E', and a line through V perpendicular to ^ cutting the planes ir and ir' in F, F'. Then every angle at E will be projected into an equal angle at E', and every angle at F will be projected into an equal angle at F'. The figure is a section of the solid figure by a plane through F perpendicular to the planes w and ■n'. Let this plane meet the axis of pro- jection in K, and let the legs of any angle at ^ in IT meet the axis of projection in L, M. Then the angle LEM projects into the angle LE'M. But EE=E'K by construction and Z EEL = I E'EL = 90°. the figure EELM is Hence the angle LEM Hence superposable to the figure E'KLM. is equal to the angle LE'M, i.e. any angle at E is projected into an equal angle at E'. So any angle at F is projected into an equal angle at F'. 9. The projection of a harmonic range is a harmonic range. For if A'B'C'iy be the projection of the harmonic range ABCB, then V and the lines AB, A'B> lie in one plane. Hence by IL 7. The projection of a harmonic pencil is a harmonic pencil. 42 Projection. [ch. Draw any line cutting the rays of the harmonic pencil U(ABCD) in a, b, c, d. Let U' (A'B'C'I)') be the projection of the pencil U{ABCI)), and a', b', c', d' the projections of a, b, c, d. Then a being on TJA, a' is on TJ'A', and so on ; hence U' {A'B'C'I/) is harmonic, if {a'b'c'd') is harmonic. And {a'b'c'd') is harmonic, since {abed) is harmonic. 10. To prove by Projeetion the harmonic properly of a com- plete quadrangle. In the figure of II. 12, suppose we wish to prove that V{BA,WV) is harmonic. Project CD to infinity. Then YA WB is a parallelogram and U is the point at infinity on BA. Let VW cut BA in 0. Then in the new figure y{BA, WU) is harmonic, for {BA, OU) is harmonic since BO = OA and Z7 is at infinity. It follows that V{BA, WU) is harmonic in the given figure. So IJ{AI),YW) and W{CD, UV) can be proved to be harmonic. Ex. Praoe by Projection the harmonic property of a complete guadrUatereU. Homologous Triangles. 11. Two triangles ABC, A'B'C are said to be homologoiis (or in perspective) when AA', BB', CC meet in a point (called the centre of homology or centre of perspective) and also {BC; B'C'), {CA ; C'A'), {AB; A'Bf) lie on a line (called the axis of homology or the axis of perspective). IV.] Projection. 43 If two triangles in the samepJme he copolar, they are coaxal ; and if coaxal, they are copolar. (i) Let the two triangles ABC, A'B'C be copolar, i.e. let AA', BB', CO' meet in the point ; then they are coaxal, i.e. {BC; B'C), {CA ; C'A'), {AB; A'B') He on a line. Call these three points X, Y, Z. Then we have to show / f ^ ^^ / IM \ "\ ^\ '^^"-^-^ B\ / 1 1 \ /" CN \; 1 1 I , 1 \ 1 1 k / that YZ passes through X Project YZ to infinity. Then in the new figure AA', BB', CC meet in a point ; also AB is parallel to A'B' and AC to A'C. Hence OB:OB'::OA:OA'::OC:OC'. 44 Projection. [ch. And since OB : OB' : : OC : OC, BC is parallel to B'C, i.e. X is at infinity, i.e. X lies on YZ, i.e. XYZ are collinear. Hence in the original figure XYZ are collinear, i.e. the tri- angles are coaxal. (ii) Let the triangles be coaxal, i. e. let {BC; B'C), (CA ; G'A'), {AB ; A'B') be collinear ; then they are copolar, i.e. AA', BB', CC meet in a point. Project XYZ to infinity. Then in the new figure BC is paraUel to B'C, CA to CA', aad AB to J.'.B'. Let AA' and BB' meet in 0. Then OB : OB' : : AB : A'B' -.-.BC: B'C; and Z OBC = Z OB'C. Hence the triangles OBC and OB'C are similar. Hence LBOC= /-B'OC. Hence CC passes through 0. Hence .4 J^', J5^, CC meet in a point. Hence AA', BB', CC meet in a point in the original figure. 12. If the triangles are not in one plane, the proofs are simpler. If two triangles he copolar, they are coaxal. (Use the same figure as before, but remember that now the triangles are in diiferent planes.) Since AB, A'B' lie in the plane OAA'BB', hence AB, A'B meet in a point on the meet of the planes ABC, A'B'C. Similarly {CA ; CA'), {AB ; A'Bi) lie on this line, i.e. the triangles are coaxal. If two triangles be coaxal, they are copolar. The three planes BCXB'C, CAYC'A', ABZA'B' meet in a point ; hence their meets AA', BB', CC pass through this point, i.e. the triangles are copolar. I!z. 1. Hence {by taking the angle beticeen the planes evanescent) deduce that coaxal triangles in the same plane are copolar ; and {by a ' reducUo ad absurdum' proof) that copolar triangles are coaaxtl. Ex. 2. Iftim triangles ABC, A'B'C in the same plane be such that AA', BB', CC meet in a point ; and if on any line through not in the plane be taken txm points V, V ; show that VA, V'A' meet in a point A", and YB, V'B' in a point B", and VC, V'C in a point C ; and that the three triangles ABC, A'B'C, A"B"C' are such that corresponding sides meet in threes at three points on the same line, vis. the meet of the given plane and the plane ()f the triangU A"B"C'. For the triangles AA'A", BB'B" are coaxal (and not in the same plane) ; hence they are copolar. This gives us another proof that triangles in the same plane which are copolar are also coaxal. IV.] Projection. 45 "Ex.. 3. The sides BC, B'ff of two triangles in the same plane meet in X, and, CA, GA' meet in T, and AB, A'B' meet in Z ; and X, T, Z are coUinear. The lines joining A, B, Cto any vertex V not in the plane ABC cut any plane through X, Y, Z but not through V in A", B", G". Show that A' A", B'B", CC" meet in 1* point V such that AA', BB', CC meet in the point where YV cuts the plane of the triangles. For B"A" passes through Z. This giyes us another proof that triangles in the same plane which are coaxal are also copolar. Ex. 4. If three triangles ABC, A'B'C, A"B"C", which are homologous in pairs, be such that BC, B'C, Bf'ff' are coruMrrent and CA, CA', C'A" and AB, A'B', A"B" ; then the three centres of homology of the triangles taken in pairs are coUinear. For the triangles AA'A", BB'B" are copolar and therefore coaxal. Ex. 5. If three triangles ABC, A'B'C, A"B"C' be such thai AA'A", BB'B", CCC are concurrent lines ; then the axes of homology of the triangles taken in pairs are concurrent. For the triangles whose sides are AB, A'^, A"B" and AC, A'C, A"C' are coaxal and therefore copolar. Ex. 6. If the paints A', Bf, C lie on the lines BC, CA, AB, and if AA', BB', CC meet in a point, show that the meets of BC, B'C, of CA, CA' and of AB, A'B' lie on a line which bisects the lines drawn from A, B, C to BC, CA, AB parallel to B'C, CA', A'B'. The line is the axis of homology of the two triangles. Let AB, A'B' meet in Z, and BC, B'C In X. Bisect AL (parallel to B'C) in 0. It is sufficient to prove that AZ . BX .LO^-ZB.XL. OA. But LO = OA ; and AZ -.BZ = AC : CB = LX : XB. Ex. 7. The triangles ABC, A'B'C are coaxal; if {BC; B'C) be X, (CA; CA') be Y, {AB ; A'B') be Z, {BC ; B'C) be X', {CA' ; CA) be Y', (aB', A'B) be Z's then XY'Z', X'YZ', X'Y'Z are lines. CHAPTEE V. HARMONIC PEOPEETTES OP A CONIC. 1. We define a conic section or briefly a conic as the pro- jection of a circle, or in other words, as the plane section of a cone on a circular base. The plane of projection may be called the plane of section. From the definition of a conic it immediately follows that— Every line meets a conic in two points, real, coincident, or imaginary. From every point can be drawn to a conic two tangents, real, coincident, or imaginary. For these properties are true for a circle, and therefore for a conic by projection. 2. There are three kinds of conies according as the vanish- ing line meets the circle, touches the circle, or does not meet the circle, or more properly according as the vanishing line meets the circle in real, coincident, or imaginary points. If the vanishing line meet the circle in two points P and Q, then, V being the vertex of projection, the plane of section is parallel to the plane VPQ, and therefore cuts the cone on both sides of F. Hence we get a conic consisting of two detached portions, extending to infinity in opposite directions, called a hyperbola. K the vanishing line touch the circle, and TT' be the tangent, then the plane of section, being parallel to the plane VTT' which touches the cone, cuts the cone on one side only Harmonic Properties of a Conic. 47 of F. Hence we get a conic consisting of one portion ex- tending to infinity, called a parabola. If the vanishing line does not meet the circle, the plane of section is parallel to a plane through F which does not meet the cone except at the vertex, and therefore cuts the cone in a single closed oval curve, called an ellipse. Since the line at infinity is the projection of the vanishing line, it follows that the line at infinity meets a hyperbola in two points, touches a parabola and does not meet an ellipse, in other words, the line at infinity meets a hyperbola in two real points, a parabola in two coincident points, and an ellipse in two imaginary points, or, again, a hyperbola has two real points at infinity, a parabola two coincident points, and an ellipse two imaginary points. 3. A pair of straight lines is a conic. For let the cutting plane be taken through the vertex, so as to cut the cone in two lines. Then these lines are a section of the cone, i.e. a conic. But properties of a pair of lines cannot be directly obtained by projection from a circle. For let the cutting plane meet the circle in the points P and Q. Then the projection of every point on the circle except P and Q is at the vertex, whilst the projection of P is any point on the line VP and the projection of Q is any point on the line VQ. Now if we take any point B' on one of the lines YP and VQ, its pro- jection is P or Q unless R' is at the vertex and then its pro- jection is some point on the rest of the circle. To get over this difficulty we take a section of the cone parallel to the section through the vertex. Then however near the vertex this plane is, the theorem is true for the hyperbolic section ; hence the theorem is true in the limit when the section passes through the vertex and the hyper- bola becomes a pair of lines. 4. A pair of points is a conic. This follows by Eeciprocation. (See VIII.) For the re- 48 Harmonic Properties of a Conic. [ch. ciprocal of two lines is two points and the reciprocal of a conic is a conic. Hence two points is a conic. Clearly however we cannot obtain two points by the section of a circular cone. 5. As in the case of the circle we define the, polar of a point for a conic as the locus of the fourth harmonics of the point for the conic. The polar of a point for a conic is a line. Through the given point U draw a chord PP' of the conic and on this chord take the point B, such that {PP', TIB) is harmonic. We have to show that the locus of i2 is a line. Now by hypothesis the conic is the projection of a circle. Suppose the range {PP', TIB) is the projection of ijpp', wr) in the figure of the circle. Then since {PP', UR) is harmonic, so is {j^p', ur). Hence r is on the locus of the fourth har- monics of u for the circle ; hence the locus of r is a line. Hence by projection the locus of ii is a line. As in the case of the circle, if the line w is the polar of U for a conic, then XT is defined to be the pole of u for the conic ; and U and u are said to be pole and polar for the conic. We have proved above implicitly that The projection of a pole a/nd polar for a circle is a pole and polar for the conic which is the projection of the circle. The following theorems now follow at once by projection. IfPhe outside the conic, the polar ofP is the chord of contact of tangents from P. If P he on the conic, the polar of Pis the tangent at P, and the pole of a tangent is the point of contact. Note that a point is said to be inside or outside a conic according as the tangents from the point are imaginary or real, Le. according as the polar of the point meets the curve in imaginary or real points. When the point is on the conic, its polar, viz. the tangent, meets the curve in coincident points and the tangents from the point coincide with the tangent at the point. v.] Harmonic Properties of a Conic. 49 Ex. 1. P<)isa chord of a conic through the fixed point V, and u is the polar of V ; show *Aaf (P, m) - ' + (Q, m) - ' is constant. viz. = 2 . (17, m)-' by similar triangles. Ex. 2. If further a be any line, show that (P,a ) (Q^ ^ (t^o) (P,«) (g,«) \U,u)' Take the meet of PQ and a as origin. Ex. 3. From any point on the line u, tangents p and q are drawn to a conic, and V is the pole ofu, and A is any point; show that {rJ,p) {V,q) ■{u,u) Take U on the range UA as origin. 6. Since a pole and polar project into a pole and polar, the whole theory of conjugate points and conjugate lines for a conic follows at once by projection from the theory of conjugate points and conjugate lines for a circle. Hence all the theorems enunciated in III. 10-12 for a circle follow for a conic by projection. Ex. X.Ifa series of conies be drawn touching two given lines at given points, the polar of every point on the chord of contact is the same for all. Let the conies touch TL and TM at L and M. The polar of P on LM passes through T the pole of LM and passes through the fourth har- monic of P for LM. Ex. 2. Tfie pole of any line through T is the same for aM. Ex. 3. TP, TQ tmmh a conic at P and Q, and on PQ is taken the point V such that TV bisects the angle PTQ, and through V is drawn any chord EVBf of the conic ; show that TU also bisects the angle BUB'. Draw TU' perpendicular to TU ; then TU' is the polar of U. Hence (ZD, BB') is harmonic, Z being on TU'. Ex. 4. A is a fixed point, P is a point on the polar of A' for a given conic. The tangents from P meet a fixed line in Q, B. AB, PQ meet in Jl; and AQ, PB in Y. Shaw that XY is a fixed line. Viz. the fourth harmonic of BB for BP and BA, B being the meet of QB with the polar ot A. Ex. 5. The polar of any point taken on either of two conjugate lines for a conic meets t?te lines and the conic in pairs of harmonic points. For if P be the point, its polar meets tUe other line in the pole of the line on which P is. Ex. 6. A, B, C are three points on a conic and CT is the tangent at C ; if C(TD, AB) he harmonic, show that CD passes through the pole of AB. Ex. 7. TP, TQ touch a conic at P, Q ; the tangent at R meets PQin N,PT in L, QT in M ; show that (LM, BJT) is harmonic. Ex. 8. A and B are two fixed points ; a line through A cuts a fixed conic E 50 Harmonic Properties of a Conic. [ch. in C and D, BD cuts the polar of A in F, and BC cuts the polar in E ; show that DE and CF meet in a fixed point. Viz. the fourth harmonic of B for A and the meet of AB with the polar of A, Tix. 0. Through U, the mid-point of a chord AB of a conic is drawn any chord PQ. The tangents at P and Q cut AB in L and M. Prove that AL = BM. If R be the pole of PQ, then RClis the polar of U, n being the point at infinity upon AB. Hence VL = UM. 'Ex. 10. The tangents TP, TP' to a conic are cut by the tangent at Q (which is parallel to the chord of contact PP') in L, L'; show that LQ = QL', Ex. 11. Through the point IT is drawn the chord PQ of a conic and UT is drawn perpendicular to the polar of V; show that UT bisects the angle PYQ or its supplement. 7. The theory of self-conjugate triangles for a conic follows at once hy projection from a circle, since the theory involves only the theory of poles and polars. Of the three vertices of a self-conjugate triangle two are outside and one inside the conic. Let ZTVW be the vertices of the given triangle. Then if Z7 is outside, VW, being the polar of Z7, cuts the conic. Also V, W form a harmonic pair with the meets of VW with the conic ; hence F or W is outside the conic. If Z7 is inside, VW does not cut the conic, and hence V and W are both outside the conic. Hz. 1. Of the three sides of a self-conjugate Mangle two meet the conic and one does not. X!x. 2. The joins of n points on a conic meet again in three times as many paints as there are combinations ofn things taken four together, cmd of these meets one-third lie wiOiin and two-thirds without the conic. Sz. 3. Show that one vertex of a triangle self-conjugate for a given conic is arbitrary, that the second vertex may be taken anywiiere on the polar of the first, and that the third vertex is tlien known. Ex. 4. Show that one side may be taken arbitrarily and complete the construc- tion. 8. The harmonic points of a quadrangle inscribed in a conic form a triangle which is self-conjugate for the conic. The harmonic lines of a quadrilateral circumscribed to a conic firm a triangle which is self-conjugate for the conic. If a quadrilateral be circumscribed to a conic, the harmonic tria/ngU of this quadrilateral coincides with the harmonic triangle of the inscribed quadrangle formed by the points of contact. v.] Harmonic Properties of a Conic. 5 1 For these propositions are true for the circle, and they follow for the conic by projection. So also — the guadrcmgle construction for the polar of a point applies to a conic. Through a given point P draw a pair of tangents to a conic. By the quadrangle construction obtain the polar of P for the conic, and join P to the points where this polar cuts the conic. The joining lines are the tangents from P to the conic. Ex. 1. A, B, C, D are four points on a conic ; AB, CD meet in E, and AC BD meet in H, and the tangents at A and B meet in G ; show that E, G, H are coUinear. Ex. 2. A system of conies touch AB and AC at B and C. D is a fixed point and BD, CD meet one of the conies in P, Q. Show that PQ meets BC in a fixed point. Viz. the pole of AD. Ex. 3. Through the fixed point A is drawn the variable chord PQ of a conic, and the chords PV, QV pass through the fixed point B. Show that UV passes through a fixed point. Viz. the fourth harmonic of A, for B and the polar of B. Ex. 4. PP', QQ' are chords of a conic through C, and A and Bare the points of contact of tangents from C. Show that a conic which toucfies tfiefour lines PQ, P'Q', P'Q, PQ^ and passes through B, touches BC at B. For AB is the polar of C for the new conic. Ex. 5. The lines AB, BC, CD, DA touch a conic at a, b, c, d, and AB and CD are parallel. If ac, bd meet at E, and AD, BC meet at F, show that FE bisects AB and CD. For iiAB and CD meet at n, then FE and FR are conjugate lines. Ex. 6. Through one of the vertices V of a triangle WW self-conjugate for a conic are drawn a pair of chords of the conic harmonic with VV and VW. S)iow that the lines joining the ends of these chords all pass through YarW. Through V draw the chord PQ, and join Q to V. 9. If one point on a conic he given and also a triangle self- conjugate for the conic, tlien three other points are known. Let A be the given point and UVW the given self-con- jugate triangle. Let UA cut WV in L. Then the other point I) in which UA cuts the conic is known since (UALB) is harmonic. Similarly the points C and B where VA and WA cut the conic are known. £ 2 52 Harmonic Properties of a Conic. [ch. The four points A, B, C, Z> form an inscribed quadrangle of which TJTW is the harmonic triangj£i By construction {UALB) is harmonic ; hence W{UAYI)) is harmonic. Similarly WiUAVG) is harmonic. Hence WB and WC coincide, i.e. WB passes through G. Similarly TIB passes through G. Hence the pole of UW is the meet of AC and BB. But the pole of UW is F. Hence BB passes through F. Ex. 1. Show that if one tangent of a conic be given and also a self-conjugate triangle, then three other tangents are known ; and that the four tangents together form a circumscribed quadrilateral of which the given triangle is the harmonic triangle. Ex. 2. If two sides of a triangle inscribed in a conic pass through two vertices of a triangle sAf-conjugate for the conic, then the third side wiU pass through the third vertex. 10. Properties peculiar to the parabola follow from the fact that the line at infinity touches the parabola. The lines TQ, TQ' touch a parabola at Q, Q', and TV bisects QQ' in Y and meets the curve in P ; show that TP = PY. Take the point at infinity ai on QQ'. Then since to lies on the polar of T, hence the polar of to passes through T. Since (toF, QQ;) is harmonic, hence the polar of a> passes through F. Hence TY is the polar of oj. Now suppose the line at infinity to touch the parabola in X2. Then u is v.] Harmonic Properties of a Conic. 53 on the polar of i2, viz. the line at infinity ; hence TY passes through 12. Also P and 12 being points on the curve, there- fore {TV, PI2) is harmonic; hence TP = PY. For clearness the figure is drawn of which the above figure is the projection. In this case, as in other cases, the theorem might have been proved directly by projection. Ex. 1. The line half-way between a point and its polar for a parabola touches the parabola. Ex. 2. The lines joining the middle points oj the sides of " triangle sdf- amjugatefor a parabola touch the parabola. Ex. 3. The nine-point circle of a triangle self-conjugate for a parabola passes Oirough the focus. Ex. 4. Through the vertices of a triangle circumscribing a parabola are drawn lines parallel to the opposite sides ; show that these lines form a triangle self- conjugate for the parabola. Being the harmonic triangle of the circumscribing quadrilateral formed by the sides of the triangle and the line at infinity. Ex. 5. No two tangents of a parabola can be parallel. For if possible let them meet at ai on the line at infinity ; then three tangents are drawn from at infinity. Also since {u>V, QQ') is harmonic, Y is on the polar of (0. Hence the locus required is the polar of Fo>) is harmonic, i.e. QF=FQ'. Hence PP' bisects every chord parallel to Z)D'. So DD' bisects every chord par- allel to FP'- Ex. 1. A pair of conjugate diameters form vnth the line at iinflnily a self- conjugate triangle. Ex. 2. In the hyperMa one and only one of a pair of conjugate diameters cuts the curve in real points. Ex. 3. The polar of a point is parallel to the diameter conjugate to the diameter containing the point. Ex. 4. The tangents at the end of a diameter are parallel to the conjugate diameter. Ex. 5. The line joining any point to the middle point of its chord of contact passes through the centre. Ex. 6. The sides of a parallelogram inscribed in a conic areparallel to apair of conjugate diameters ; and the diagonals meet at the centre. Ex. 7. The diagonals of a parallelogram drcumscribirtg a conic are conjugate diameters ; and the points of contact are the vertices of a parallelogram whose sides are parallel to the above diagonals. Ex, 8. A tangent cuts two parallel tangents in P and Q, show that CP and CQ are conjugate diameters. For, reflecting the figure in the centre C, this reduces to Ex. 7. 14. If each diameter of a conic he perpendicular to its con- jugate diameter, the conic is a circle. Take any two points P, Q on the conic. Bisect PQ in F and join CV. Then CV is the diameter bisecting chords parallel to PQ, L e. CV and PQ are parallel to conjugate diameters. Hence CV and PQ are perpendicular. Also PV = VQ. Hence CP= CQ. Hence all radii of the conic are equal, i.e. the conic is a circle. 15. The asymptotes of a conic are the tangents from the centre. They are clearly the joins of the centre to the 56 Harmonic Properties of a Conic. [ch. points at infinity on the conic. In the hyperbola they are real and distinct, in the parabola they coincide with the line at infinity, and in the ellipse they are imaginary. The asymptotes are harmonic with every pair of conjugate diameters. For the tangents from any point are harmonic with any pair of conjugate lines through the point. Any line cuts off equal lengths between a hyperbola and its asymptotes. Let a line cut the hy- perbola in Q, Q' and its asymptotes in B, Bf; then BQ = Q'B'. On BB' take the point at infinity w and bisect Q^ in V. Then since (Q^, Fw) is harmonic, the polar of is at infinity, its polar passes through C. Hence CV is the polar of o). Hence CV and Ca> are conjugate lines. And CB, CB' are the tangents from C. Hence C{BB', Vai) is harmonic. Hence {BB^, V(o) is harmonic. Hence BV= VB'. But QV=VQ'. Hence BQ=Q'B'. The proof appUes whether we take QQ^ to cut the same branch in two points or (as in the case of qq) to cut different branches of the hyperbola. The intercept made by any tangent between the asymptotes is bisected at the point of contact. For let Q and Q' coincide ; then BQ = QB'. Ex. 1. Given the asymptotes and one point an. a hyperbola, construct any number of points on the curve. Ex. 2. Given tlie asifmptotes and one tangent of a hyperbola, construct any number of points and tangents of the curre. Ex, 3. Two of the diagonals of u quadrilateral formed by too tangents of a hyperbola and the asymptotes are parallel to the chord joining the points of contact of the tangents. Consider the harmonic triangle of the quadrangle formed by the points of contact and the points at infinity on the hyperbola. Ex. ^ If a hyperbfUa be drawn through tao opposite vertices of a parallelogram v.] Harmonic Properties of a Conic. 57 ■with its asymptotes parallel to the sides, show that the centre lies on the join of the other vertices. 16. A rectangular hyperbola is defined to be a hyperbola whose asymptotes are perpendicular. Conjugate diameters of a rectangular hyperbola are equally inclined to the asymptotes. For they form a harmonic pencil with the asymptotes, which are perpendicular. Ex. The lines joining the ends of any diameter of a rectangular hyperbola to any point on the carve are equally inclined to the asymptotes. 17. A principal axis of a conic is a diameter which bisects chords perpendicular to itself. All conies have a pair of principal axes; but one of the prin- cipal axes of a parabola is at infinity. Consider first the hyperbola. Then the asymptotes are real and distinct. Now the bisectors of the angles between the asymptotes are harmonic with the asymptotes and are therefore conjugate diameters. But the bisectors are also perpendicular. Hence they are a pair of conjugate diameters at right angles. Each of the bisectors is therefore a prin- cipal axis ; for each bisects chords parallel to the other, Le. perpendicular to itself. Consider next the parabola. We might say that here the asymptotes are coincident with the line at infinity ; and the bisectors of the angles between a pair of coincident hnes are the line with which they coincide and a perpendicular to it. Hence the principal axes of a parabola are the line at in- finity and another line called the axis of the parabola. Or thus — All the diameters of a parabola are parallel. Draw chords perpendicular to a diameter, then the diameter bisecting these chords is perpendicular to them and is called the axis of the parabola. The other principal axis (like the diameter conjugate to any of the other parallel diameters) is the line at infinity. Consider last the ellipse. Here the asymptotes are ima- ginary and this method fails. But it will be proved under 58 Harmonic Properties of a Conic, Involution that there is always a pair of conjugate diameters of any conic at right angles. Hence the ellipse also has a pair of principal axes. (See XIX. 4.) An axis cuts the conic at right angles. For the tangent at the end of an axis is the limit of a bisected chord. A central conic is symmetrical for each axis. For the principal axis AL bisects chords perpendicular to itself. Let PMP' be such a chord. Then P' is clearly the reflexion of J* in AL, i.e. the conic is symmetrical for AL. The same proof shows that A jpardbola is symmetrical for its axis. Ex. 1. The tangent at P meets the axis CA in T and PN is ike perpendicular on CA 1 shou! thcU CN. CT - CA''. For PN is the polar of T. Tix. 2. PQ, PR toudi a conic at Q, B. PM is drawn perpmdicular to either axis. Show that PM bisects the angle QMR. CHAPTEE VI. CAHNOTS THEOREM. 1. The sides BC, CA, AB of a triangle cut a conic in the points A^A^, BiBi, C, C^, show tliat ACj . AC, . BA, . BA^ . CB, . CB^ = AB^ . AB^ . BC, .BC^.CA,. CA^. By definition a conio is the projection of a circle. Let the points ABCA^A^... be the pro- jectionsof^'^C^/^j'... in the figure of the circle. Now in the circle we have A'C,'. A'C,'. B'A,'. B'A,'. CB(. C'B/ = A'B,'. A'B,'. B'C.'.B'C^. OA^. CA.; for A'G^. A'C;= A'B,'. A'B^, and so on. Let Fbe the vertex of pro- jection. AC,_ AAVC , ■^^®° BC," ABVC, ^ AV.C,V. Bin AVC , BV. C^V. sin BVC„^ _AV sinAVG^ ~ BV' sin BVC^ and so for each ratio. 6o Camot's Theorem. [ch. AC, .AC^ ... _ sin AVC,. Bin A VC^... Hence ^^^^^^^ - sin ^F^^. sin ^F^^... where each segment is replaced by the sine of the corre- sponding angle. Also the last expression sin^'FC' sin^'FC;... ^ ,^. , A'C,'. A'C,' ... = sinA'rB,'.BinA'rB,'... ' ""*^ ^^'' "l""^' A'B.'.A'B,'... by exactly the same reasoning as before, and this has been proved equal to unity. Hence AC, . AC, . BA, . BAj . CB^ . CB^ = AB,^ . AB^ . BC, . BC, . CA, . CA^. Ex. 1. The sides AB, BC, CD, . . . qf a polygon meet a come in A^ A,. B^ B^, C,C,,, , , ; show that AAt.AAj.BBi.BB^.CCi.CC,... ^BAi.BA,. CB^. CB^. SCi.DC,.. . Ex. 2. By taking Um conic to be a line and the line at infinity, deduce Menelaus's theorem from Camot's theorem. Ex. 3. If a conic txmch the sides of the iiiangle ABC in A,, B^, C^; then AAi , BBi, CC^ are concurrent. For AB^". CAi'. BC{' = AC^. BA^. CB^ ; and we cannot have AB, . CA, . BC^ = + AC^ . BA^ . CBi , for then Ai B^ C, would cut the conic in three points. Ex. 4. If the vertex A in Catinot's theorem be on the conic, show that the ratio AC.^ : AB^ must be replaced by Bin TAG : sin TAB, AT being the tangent at A. For Bi C, is ultimately the tangent at A. Ex. 5. What does Camot's theorem reduce to when A, B, and C are on the curve ? Ex. 6. If through fixed points A, B we draw the chords AB^ Bj, BA^ Ai of a conic meeting in the variable point C, then the ratio BAi . BAi . CBi . CBi 4- .^-Bi • -^B^ . CAy. CA^ is constant. Ex. 7. Deduce the corresponding theorem when Bis at infinity. Ex. 8. A, B, C are three points on a conic ; the tangents at ABC meet in GHK; points DEF are taken an BC, CA, AB such that AD, BE, CF are concurrent : shmo that GD, SB, KF are concurrent. For sin D6B/ainDGC = DB/DC-i-BG/CG. Now use two forms of Ceya's theorem. Ex. Q. AC touches a conic at A, AB meets it again in C„ and BC meets it in Ai, Ai ! if the circle of curvature at A meet AB in C, show that AC. CAi . CA^ . BCy . BA = AC,, . BA^ . BA, . CA'. Consider the circle of curvature as the limit of the circle through B^ B^ Vg . Ex, 10. IfAi Ai be parallel to the tangent at A, this reduces to Ad. BCi .BA = AC^. BAi . BA.^ . Ex. 11. Deduce the expression s CD' -V CPfor the central chard of curvature. VI.] Carnoi's Theorem. 6i Ex. 12. A conic aiis the sides BC, CA, AB of a triangle in Pj P, , Q1Q2, 5i Ri ; BQi and CR^ meet in X, AP^ and CR^ in Y. and AP^ and BQ^ in Z ; show that AX, BY, CZ are comMrrent. Z. If, on tJie sides BC, CA, AB of a triangle, the pairs of points AjA^, B^B^, Cfi^ be taken, such that AC^.AC.,.BA,. BA, . CB, . CB, = AB, . ABj. BC, . BC^. CA, . CA^, then the six points A,, A^, B„ B.^, C,, C.^ lie on a conic. Through the five points (XXIV. 2) A„ A^, B„ B^, C, draw a conic. If this conic does not pass through C^, let AB cut the conic again in y.^. Then we have AC,.Ay^.BA,. BA^ . CB, . CB^ = AB, . AB, .BC,.By,. CA, . CA^. Dividing the given relation by this relation we have ACJAy, = BC,/BY,. Hence C^ and y^ coincide. Hence the six points A,, A2, B,, S,, C„ C, lie on a conic. Ex. 1. If from any two points the vertices of a triangle he projected upon the opposite sides, the six projections lie on a conic. Ex. 2. The parallels through any point to the sides of a triangle meet the sides in sir points on a conic. Ex. S. ffa conic which has two sides of a triangle as asymptotes touch the third side, the point of contact bisects the side. Ex. 4. A conic can be drawn to touch the three sides of " triangle at their middle points. 3. Newton's theorem — If two chords of a conic UPQ, TJLM he drawn in given directions through a variable point U, show that the ratio of UP. UQ to TIL . UM is independent of the position of U. Let U'P'Q', U'L'M' be an- other position of the chords UPQ, ULM. Then PQ is par- allel to P'Q' and LM to L'M'. Let PQ, P'Q' meet at infinity in o)', and LM, L'M' at infinity in 0). Apply Camot's theorem to the triangle a>'U'7. Then (o'Q'. co'P'. U'L'. U'M'. YQ. VP =w'Q. w'P. VL'. VM'. Vq. U'P'. 62 Camoi's Theorem. [ch. From Q drop the perpendicular QX on QV. Then a)'Q'/a)'Q = (a)'Z+ZQ')/<-'^ = i +XQ'VZ = i. So (i>'I'= lil'P. Hence CT'i'. C'Jf'. FQ. FP= 7i'. FJIf'. CT'Q'. CT'P' i. e. ZJ'P'. U'Q'^U'L'. U'M'= VP. VQ^VL'. VM' In exactly the same way the triangle wUV gives us VP. VQ-^VL'. VM'= UP. UQ-i-UL. UM. Hence UP. UQ-^UL. UM= U'P'. U'Q^^U'L'. U'M', i. e. UP. UQ-i-UL. UM is independent of the position of U. Ex. If the tangents from T to the conic touch at P and Q, show that TP:TQ::CP' •.CQ', where CP", CQ' are the semi-diameters paraUel to TP, TQ. Take 17 at T and C successively. 4. In a parabola QV= 4 . SP. PV. Besides QVQ^ draw a second double ordinate qvg" of the dia- meter PV. Now PV meets the parabola again at Q,, a point at infinity. Also by Newton's theorem we have VQ . VQ ' __ vq . v^ VP. vn,~ vP.vn' But Fi2 = vQ.. Hence VQ • yQ^-^ VP=vq. v^^vP, i. e. QV^ -r- PV is constant. To obtain the value of this constant take qc^ through the focus S. Then by Geometrical Conies ^: GP\ VI.] Camot's Theorem. 63 6. In a hyperbola QV : PV. VP' : Besides QVQ' draw a second double ordinate qvg^ of the diameter PCP' Then by Newton's theo- rem VQ.VQ^-.VP. VP' •.-.vq.vg' -.vP.vP', i.e. QV^-.PV. FP' is con- stant. To obtain the value of this constant, take V at C, and let D be the position of Q. Then QY^= CD' and PV. rP'= PC. CP'= CP'- Henee QV : PV. VP' : : CD' : CP\ the formula required. But this is not the formula given in books on Geometrical Conies ; for in the above formula either P or D is imaginary, since, of two conjugate diameters of a hyperbola, one only meets the curve in real points. Take P real and J) imagin- ary. Then CD^ is negative, otherwise D would be real. On CD take the point d, such that Cd'= —CD\ Then d is real, for Cd' is positive. Then QV' : PV. VP'::- Cd' : CP% i.e. QV'iPV.P'V:: Cd':CP', which is the formula given in books on Geometrical Conies, the d here replacing the I> of the books. We may call CD the true and Cd the conventional semi- diameter conjugate to CP. It is sometimes convenient to employ the symbol D for the conventional point d when the meaning is clear from the context. Note that the locus of d is the so-called conjugate hyper- bola. The theorems of § 5 and § 6 may also be obtained directly from Camot's theorem by using the triangle contained by DC, VC, VQ. 7. If the diameter conjugate to PCP' meet the curve in 64 Camofs Theorem. [ch. the imaginary points D and If, and if the tangent at P meet an asymptote in T, then CI>'=—PT^, i. e. PT is equal to the conventional CD and parallel to it. In the figure of § 6 let BQQ' be parallel to the tangent at P, and let 12 be the point at infinity on the asymptote CB. Then by Newton's theorem BQ . BQ'-i-BD.''= rq . rq'-i- ri2^ rqq' being parallel to BQQ'. But BS1 = ril. Hence BQ.BQ'=rq.rq'. Now take B at T, then BQ . BQ' = TP\ Again, take r at C, then rq.r^^-rq^=- GD\ Hence TP^ = -CD' = Gd\ Hence TP = Cd, i. e. TP represents Cd in magnitude and direction. Notice that we have incidentally proved the theorem — If a chord Q<^ of a hyperbola drawn in a fixed direction cut one of the asymptotes in B, then BQ . BQ' is constant and the same whichever asymptote is taken. For BQ. BQ'= TP^= T'P^= B'Q. RQ'. It follows that BQ . QB' and BQ'. Q'B' are constant and equal. For BQ'=QB'. Also BQ.BQ'=B'Q.B'q = Cd^, Cd being parallel to BQQfB'. Ex. 1. Pd is parallel to an asymptote. For by reflexion in C we get the complete parallelogram TT'tt^, and clearly Pd is parallel to fT'. Ex. 2. Given in magnitude and position a pair of conjugate diameters qf a hyperbola, construct the asymptotes. Ex. 3. Through any point Ron an asymptote of a hyperbola is drawn a line paraUel to the reed diameter P'CP cutting the curve in QQ', show that RQ.B(^ = -CP'. Ex. ^ If the same line cut the other asymptote in E', show that QR . QR'= (fR . (^Rf= CP°. Ex. 6. Given a pair of conjugate diameters of a hyperbola in magnitude and position, construct the axes in magnitude and position. Use Ex. a and Ex. 4. Ex. 6. The tangent at Q to a hyperbola meets a diameter CD or Cd {which meets the curve in imaginary points) in T, and the parallel through Q to the conjugate diameter CP meets CD in V, show thai CV. CT = Ctf = —CcP. For (Dff, TV) is harmonic, and C bisects DD'. Since CV. CT is nega- tive, V and T are on opposite sides of C. Of the above harmonic range notice that DD' are imaginary points and TV real points. VI.] Camot's Theorem. 65 Ex. 7. Given a pair of conjugate diameters of a hyperbola in position and a tangent and its point of contact, construct the axes in magnitiide and position. CV. CF = CI" giTes the lengths of the diameters. Ex. 8. The polar of d is d'T. Consider the chords intercepted on dT and dP. Ex. 9. ]f through a variable point V a chord PUQ be drawn in » fixed direction, and also a chord VB paraMel to one of the asymptotes in the case of the hyperbola, or parallel to the axis in the case of the parabola, then VP.UQ — VR is constant. 8. In a rectangular hyperbola, conjugate diameters are equal and equally inclined to the asymptotes. Also diameters which are perpendicular are equal. Since conjugate diameteis are harmonic with the asymptotes which are perpendicular, they are equally inclined to the asymptotes. Again CD = FT. But in the r. h. TCT' is a right angle and TP = PT'. Hence CP = PT, hence CP= CD. Draw CQ perpendicular to CP. Then IBCq = ^o''-lBCP=lAGP=lBGI>, since IPCT^IBCT; hence CQ is the reflexion of CD in CB. Hence CQ = CD. Hence CQ = CP. Similarly CD is equal to the semidiameter perpendicular to it. Notice that the true formulae are CP" = —C]y= — CQ' ; so that if a diameter meet a r. h, in real points, the perpendi- cular diameter meets the curve in imaginary points. Ex. The perpendicular chords LM, L'M' ofar.h. meet in U, show that UL.UM = - UL'. Ulf. For VL . XJM : TJL'. UM' as the squares of the parallel diameters, i.e. asCP": -CO". 9. Every rectangular hyperbola which circumscribes a triangle passes through the orthocentre. Let ABC be the triangle and P its orthocentre. Suppose a r. h. through ABC cuts the perpendicular AD in Q. Then from the r. h. we have DQ.DA=-DB DC. And from p 66 Camot's Theorem. Elementary Geometry we have DP . DA = — BB . DC. Hence DQ=DP, i.e. Q coincides with P, i. e. the r. h. passes through the orthocentre. For the converse see XXI. 9. Ex. 1. Ufa triangle PQR whidi is right angled at Q be inscribed in a r. h., Ok tangent atQis Vie perpendicular frmn Q on PR. XjZ. 2. J/ a r, k. circumscribe a triangle, the triangle formed by the feet of the perpendicuiars from the vertices on the opposite sides is self-conjugate far the r. h. Being the harmonic triangle of ABCP. CHAPTER VII. FOCI OF A CONIC. 1. A focus of a conic is a point at which every two conju- gate lines are perpendicular. A directrix of a conic is the polar of one of the focL The polar of a focus is called the corresponding directrix. From the definition of a focus it at once follows that every two perpendicular lines through a focus are conjtigate. The theory of the foci of a conic is given in Chapter XXVIII. It is there shown that — Every conic has fottr foci. All the foci are inside the curve. The foci lie, two 'by two, on the principal axes ; the pair SS' on one axis (called the focal axis) are real, and those FF' on the other axis are imaginary ; also SS' are equidistant from the centre on opposite sides, and so are FF'. One real focus of a parabola is at infinity on the axis of the parabola. All the foci of a circle coincide mth the centre. Note that the focal axis is the major axis in an ellipse, the transverse axis in a hyperbola, and the axis in a parabola. Ex. 1. Tangents at the ends of a focal chord meet on the directrix. Ex. % If OS meet the directrix in Z, then SZ is perpendicular to the polar of T. Being perpendicular to the polar of Z. Ex. 3. PSQ is a focal chord of a conic. UOV is any chard of the conic through the middle point of PQ. Parallels throitgh U, V to PQ meet the directrix corresponding to S in M, N. Show that PQ bisects the angle MSN. Let the polar of (which is parallel to PQ) meet the directrix in R ; then SR and SP are conjugate lines. F 2 68 Foci of a Conic. [ch. 2. If from any point P on a conic, a perpendAcMlar PM be drawn to the directrix which corresponds to a focus S, then SP -T- PM is constant. Take any two points P and P' on the conic. Let the tangents at P and P' meet in T. Let PP' meet the correspond- ing directrix in K, and STinB. From P and P' drop the perpendiculars PM and P'M' on the directrix. Now SE and ST are conjugate lines at the focus ; for the polar of K, which lies on PP' and on the directrix, is TS. Hence SK is perpendicular to ST. Also {KPBP') is harmonic, since E is the pole of ST. Hence S(EPBP') is harmonic. Hence SE and ST, being perpendicular, are the bisectors of the angle PSP'. Now since SE bisects the angle PSP' (externally in the figure), we have SP : SP': : PE : P'E : : PM : P'M'. Hence SP-.PM:: SP': P'M'; in other words, SP:PM is constant. In the parabola, SP= PM. For let SA be the axis. Then SA meets the parabola again at infinity, at il, say. Hence (XASQ) is harmonic, since XZ is the polar of S. Hence SA = AX. But SP : PM :: SA: AX, for A is on the parabola. Hence SP = PM. In the ellipse, SP < PM. Since a focus is an internal point, S must lie between A and A'. X A S £ Let A be the vertex between S and X. Then since A' is & point on the ellipse, we have SP : PM : : SA': A'X. But SA'< A'X, hence SP < PM. In the hyperbola, SP>PM. Since the focus is an internal point, 8 must lie outside the segment AA'. ^ AX A^ VII.] Foci of a Conic. 69 As before SP:PM:: SA': A'X > i. The corresponding property in tlie circle is that the radius is constant. For the focus is the centre. Hence the directrix is the line at infinity. Hence PM = P'M'. Hence SP=SP', i.e. CP= CP'. Ex. 1. Any two tangents to a conic subtend at a focus angles which are either eqitai or supplementary. Ex. 2. Show that it is not true conversely that ' if any two tangents to a conic subtend at a point on an axis angles which are equai or supplementary, then this point is a focus.' The foot of the perpendicular from T on the axis is such a point. 3. Assuming from Chapter XXVIII that a conic has a real focus, we have just shown that this focus possesses the SP : PM property by which a focus is defined in books on Geometrical Conies. This opens up to us all the proofs given in such books. It will be assumed that these proofs are known to the reader ; and the results will be quoted when convenient. Properties of Conies which can be best treated by the methods of Geometrical Conies will be usually omitted from this treatise. 4. In any conic, the semi-latus rectwm is equal to the harmonic mean between the segments of any focal chord. Let the focal chord FSP cut the directrix in K. Then {KPSP'} is harmonic since S is the pole of XK. Hence 2 (KS)-' = (EP)-' + (EP'r'- But KP : KS : KP': iPMiSX: P'M' ■.iSP.SL.SP', for SP:PM::SL: LU::SL : SX. Hence 2{SL)-' = (SP)-' + {SP')-\ Ex. L If The the pole of the focal chord PQ of a parabola, s/joio that PQ oc ST'- Ex. 2. A focal chord of a central conic it proportional to the square of the parallel diameter. 70 Foci of a Conic. [ch. 5. If the tangent at P meet the tangents at the vertices AA' of the focal axis in UV, then UU' subtends a right angle at S and S'. Also if US, U'S' cut in E, and US', U'S cut in F, then EF is the normal at P. For since A U and PU subtend equal angles at S and since A'U' and PU' subtend equal angles at S, it follows that USU' is a right angle. Similarly UU' subtends a right angle atS'. Again, F is the orthoeentre of the triangle UEU'. Hence EF is at right angles to UU'. Let PU cut the axis in T and draw the ordinate PN. Then (TUPU') = (TANA') is harmonic. Also if EF cut UU' in P', then since UU' is a harmonic side of the quadrilateral SF, FS', S'E, ES, we have {TUP'U') harmonic. Hence P' and P coincide, i. e. EF passes through P, Hence EF is the normal at P. Ex. 1. I/a circle through the foci cut the tangent at the vertex A in U, V and the tangent at the vertex A' in V, V, show that the diagonals of the rectangle UV'V'V touch the conic. Ex. 2. Given the focal axis AA' in magnitude and posiHon and one tangent, construct tJiefod. 6. If the tangent at a point P of a central conic cut the focal axis in T, and if the normal at P cut the same axis in G, then CG.CT=CS\ For since the tangent and normal bisect the angle SPS', it follows that P (SS', TG) is harmonic ; hence CG.CT=CS\ ISx. 1. Given the axes in position and one tangent and its point of contact, construct the foci. Ex. 2. In the parabola, S bisects GT. For S' is at infinity. £x. 3. Given the axis of a parabola in position and one tangent and its point of contact, construct the focus. Confocal Conies. 7. Confocal conies (or briefly confocals) are conies which have the same foci. If one of the given foci is at infinity, we have confocal parabolas, which may also be defined as parabolas having the same focus and the same axis. VII.] Foci of a Conic. 71 Two confocals can he drawn through any point, one an ellipse and one a hyperbola, and these cut at right angles. Join the given point P to the foci S, S', and draw the bisectors PL and PL' of the angle SPS'. Since both foci are finite, the conic must be an ellipse or a hyperbola. If it be an ellipse, then Q being any point on the ellipse, SQ + S'Q = SP+S'P; so that one and only one ellipse can be drawn through P with S and S' as foci. Similarly one and only one hyperbola can be drawn. And the two conies cut at right angles, for PL and PL' are theu- tangents at P. If one focus is at infinity, the ellipses and hyperbolas become parabolas, and we get the theorem — Of the system of parabolas which have the same focus and the same axis, two pass through any point and these are orthogonal. This can be easily proved directly. 8. One confocal and one only can be drawn to touch a given line. Take responding side of the triangle formed by the polars of the vertices for any conic lie on a line. 9. If through the point of contact of two conies which touch, any chord be drawn, the tangents at its ends will meet on the common chord of the two conies. 10. If on a common chord of two conies, any two points be taken, and from these, tangents be drawn to the conies, 8o Reciprocation. [CH. the diagonals of the quadrilateral so formed will pass through one or other of the meets of the common tangents of the conies. 1 1 . If a and j3 be two conies having each double contact with the conic y, the chords of contact of a and /3 with y and their common chords with each other meet in a point. 12. Jf a, p, Y be three conies, having each double contact with the conic tr, and if o and /3 both touch y, the line join- ing the points of contact will pass through a meet of the common tangents of a and j3. Point Beciprocation. 8. If the base conic is a circle (the most common case), the reciprocation is generally called point reciprocation, the centre of the base circle is called the origin of reciprocation, and the radius k of the base circle is called the radius of re- ciprocation. The reason of the name point reciprocation is that the value of k is usually of no importance. By recipro- cation is meant point reciprocation unless the contrary is stated or implied in the context. In point reciprocation, the angle between two lines is equal to the angle subtended by the reciprocal, points at the origin of re- ciprocation. V Let p and g be the lines, and P and Q the reciprocals of p and q. Let be the origin of reciprocation. Then P being VIII.] Reciprocation. 8i the pole of j) for a circle whose centre is 0, OP is perpen- dicular to p. So OQ is perpendicular to q. Hence FOQ, is equal to the angle between p and g. In point reciprocation, the angle between a line p and the line joining the origin of reciprocation to a point Q, is equal to the angle between the line q and OP, P and q being the reciprocals ofp and Q. This follows at once, as before, from the above figure. In point reciprocation, ifP he the reciprocal of p and if be the origin of reciprocation, then OP is inversely proportional to the perpendicular from on p. For OP. OP, = OP. {0, p) = k\ 9. The reciprocal of a figure for a given point and a given radius Ic may be obtained without considering a circle at all. To obtain the reciprocal of P — on OP take a point P, , such that OP. OP, = k'; and through Pj draw a perpendicular p to OP. To obtain the reciprocal of p — drop the perpen- dicular OP, from top, and on OP, take the point P, such that OP. OP, = k\ Instead of taking OP. OP, = ¥, we may take OP. OP, = -Af", i.e. we may take P and P, on opposite sides of 0. This is called negative reciprocation, and is equivalent to reciprocating for an imaginary circle whose radius is k v —i. Ex. L The reciprocal of the origin of reciprocation is the line at infinity ; and conversely, the reciprocal of the line at infinity is the origin. For the polar of the centre of the base circle is the line at infinity ; and conversely. Ex. 2. The reciprocal of a line through the origin is a point at infinity ; and Ex. 3. Reciprocate a quadrangle into a paraBelogram. Take at one of the harmonic points. Ex. 4. IJte reciprocal of the meet of OP and m is the line through i{ parallel top. Ex. 5. IfP and Q be points on a curve smh that PQ passes through 0, then in the reciprocal for 0, p and q are parallel tangents. Ex. 6. The reciprocal far of the foot of the perpendicular from on pis the line through P perpendicular to OP. Q 82 Reciprocation. [CH. Hx. 7. 7%e reciprocoZ o/a triangle far its arthocentre is a triangle having the same arthocentre. Ex. 8. On the sides, BC, CA, AS of a triangle are taken points P, Q, B such that the angles POA, QOB, ROC are right, being a fixed point ; show that PQR are coUinear, Reciprocating for 0, we have to prove that the three perpendiculars from the vertices on the opposite sides meet in a point. Ex. 9. Ttte reciprocal of the curve p =f(r)for the origin is Ic'/r =f(k''/p). Let h be the tangent at A to the given curve. Then B is on the reciprocal curve and a touches it. Hence. p = (0, 6) = k'/OB = k^/r', and r = OA = k' (0, a) = Ic'/p'. Beciprocatkm of a conic into a circle. 10. The reciprocal of a circle, taking a circle with centre as base conic, is a conic having a focus at 0. Let U be the centre of the given circle a. Take u the re- ciprocal of U, i. e. the polar of U for the base circle F whose centre is 0. Let p be any tangent to a touching at T. Take P the reciprocal of p. Draw the perpendicular PM from P to u. Then since p is the polar of P and u the polar of Uior T, we have by Salmon's theorem (III. 9) OP/{P, u) = OU/{U, p), i.e. OP/PM=OU/UT. viii.] Reciprocation. 83 Hence OP/PM is constant, i. e. the locus of P is a conic with as focus. But the reciprocal of a for F is the locus of the poles for F of the tangents to o, i. e. is the locus of P. Hence the reciprocal of a circle a for the circle F whose centre is is a conic a' having a focus at 0. Briefly, the reciprocal of a circle for a point Ois a conic having a focus at 0. Since e = OP/PM = OU/UT, we see that the reciprocal of a circle for a circle whose centre is 0, is an ellipse, parabola or hyperbola according as 0U< = > UT, i.e. according as Ois inside, on or outside the given circle. This is a particular case of a general theorem. (See § 21.) Let OU = S, UT = B, and let k be the radius of the base circla Then e = b/R. Also OX.OU= K^. Hence 1c' /S = OX = a/e - ae. Hence a = FR/{Ii!' - 8"). Sx. Show (hat the semi-latus rectujn I = k'^/R. This follows from I = a (i— e^) ; or directly by noticing that an end of the latus rectum through O reciprocates into a tangent of a parallel to OV. Notice that I is independent of S, i. e. of the relative positions of the circles. 11. Conversely, the reciprocal of a conic, taUng any circle whose centre is at afoms as lose conk, is a drck. Let be the given focus, and XZ or u the corresponding directrix. Take any point P on the conic a', and let p be its reciprocal, L e. the polar of P for the base circle F whose centre is at 0. Draw the perpendicular PM from P to u. Take the reciprocal U of u. Draw the perpendicular UT from Utop. Then since p is the polar of P and u the polar of U for the conic F, we have by Salmon's theorem 0U/UT=0P/PM=e. Hence OU/ UT is constant. Also U" is a fixed point ; hence UT is of constant length. Hence the perpendicular from U on p is constant, Le. p envelopes a fixed circle a. But the reciprocal of .a' for F is the envelope of the polars for F G 2 34 Reciprocation. [ch. of the pointB on a'. Hence the reciprocal of the conic a' for a circle F whose centre is at one of the foci of the conic is a circle a. Briefly, the. reciprocal of a conk for one of its foci is a circle. Ex. 1. Tlie envelope of the polar far a of the centre of a circle which touches two given circles a and B is a circle. Ex. 2. Deduce a construction for the centre of a circle touching three given circles. Ex. 3. Given four points A, B, C, D, show that, with D as focus, one conic can be drawn touching BC, CA, AB, and four conies through ABC. Show also thfU, if ABB he a right angle, a conic, with focus at D, can he found to touch the Jive conies. In a right-angled triangle the nine-point circle touches the circum- circle. Ex. 4. Of the above four cmics, the sum of the latera reda of three is equal to the latus rectum of the fourth. Ex. 5. The reciprocals of equal circles are conies having equal parameters. Ex. 6. Reciprocate far the orthocentre of ABC the theorem — ^ If DEF he the feet of the perpendiculars from A, B, C on BC, CA, AB, then the radius of the circle about ABC is double the radius of the circle about I)EF.' Ex. 7. Four conies a, 0, y, ^ + --^ = cb-AD^AD'BG~^ AG. DB-DG . AB-CB . AD ~ CB.AD {c-a){b-d)-{c-d){h-a)-{b-c)(d-a) ~ CB.AD H 2 lOO Anharmonic or Cross Ratio. [ch. _ 06 — cii— a& + arf—c 6 + eg + d&—tto—6d + 6a+cd— ac _ CB.AB = o. ., , AC DB AD EC AB CD Also ^■>^-- = cB'AD-DC-AB-BD-AC = -'- We have now shown that the three fundamental cross ratios A, fx, v are connected by the above four relations. Two of these are independent and give |u, v in terms of A. The other two can be derived from these. Hence given any one cross ratio of four points, the other 23 can be cal- culated. liz. 1. Giien ^ + -=i, /» + - = !, shoio that v + - = I and Kuy = —i. \ Ex. 2. Given X + - = i, X/jy = — I, show that ¥■ K + - = I ana, u + - = i. K V Ex. 3. If (,AB, CD) — I, show that either A and B coincide, or C and D ; and conversely, if A and B coincide, or C and S. then {AB, CD) = i. Ex. 4. Xf ^wo points of a range of four points coincide, each of the cross ratios is equal to o, i, or co ; and no cross ratio can equal o or i or oo unless two points coincide. Ex. 5. Shaw that no real range can be found of which all the cross ratios are equcd. Ex. 6. Of the three X, n, v, two are positive and one negative. Ex. 7. If any cross ratio of the range ABCD is equal to the cwresponding cross ratio of the range A'BtfSf , then every two corresponding cross ratios of the ranges are equal. For if X = X', then ft = f/ and v — v'. Two such ranges are said to be homographic, and we denote the fact by the equation {ABCD} = {A'B'C'I/). Ex. 8. If {ABB'tX} = {A'B'BCf) and (ABB'D) = {A'ffBI/), show tliat {BB'CD) = {B'BCD'). Divide {BB', AC) = {B'B, A'C) by {BB', AD) = {B'B, A'l/). 5. If (AC, BD) be harmonic, then (AC, BD) = - 1. .^ AB AD , AB AD ^"^ BC=-DC'^''"'^BC-^DC=-'- If (AC, BD) be harmonic, then (AC, BD) = (AC, DB) ; and conversely, if (AC, BD) = (AC, DB), then either (AC, BD) is harmonic or two points coincide. hence ^ . -^ = ± i, i. e. {AC, BD) = ± i. IX.] Ankarmonic or Cross Ratio. loi For if {AC, BB) = {AC, BB), ,. ab dc ad bc ^ ^ab dc." ***'" bc-ab=bc-ab'^'''', — tan" and cos" is the angle at which the cirdes m AC and BD as diameters intersect. AC DB _ sin APC sin DPB '"' "" CB ' AD~ sin CPB ' ain APD' and IAPC=-, ^DPB=--, iCPB=-". Then (AB'C'B') = [ABCB) by hypothesis = {AB'C'B") by projection. Hence {AB'C'B') = {AB'C'B"), i.e. D' and B" coincide, i.e. the join BB' of any pair of corresponding points passes through 0. Ex. 1. If A be the meet of two corresponding rays of two homographic pencils, then any two transversals through A will cut the pencils in ranges in perspective. Ex. 2. If a cross ratio of the range ABCD be equal to the corresponding cross ratio of the range A'B'(fI/, show that every two corresponding cross ratios are equal. (See also § 4, Ex. 7.) Place the two ranges so that A and A' coincide and that the lines AB and A'B' do not coincide. Then, as above, the ranges are in per- spective ; and hence every cross ratio is equal to the corresponding cross ratio. Ex. 3. If {ABGS) = {A'B'Clf) and (ABCE) = {A'B'CTE') and so on, then (ABCSE,..) and {A'B'Cl/E'...) are homographic ranges. Ex.4. // {Ur, AA') = {VV, BB') = {Vr, CC) =.•-, slum that {ABC. .) = {A'E'(f.. .). For {Ur, AB) = {VV, A'B'). Ex. 5. IfP he a variable point on the line joining two fixed points A, B, and P' a variable point on the line joining the fixed points A', Bf, such that AP/BP -i- A'PjB'P' is constant, then P and P generate homographic ranges. For if C be a position of P and C of f, we have AC/BC ~ AP/BP = A'C/PKy ^ A'P'/PfP', i.e. {ABCP) = {A'B'CP). Ex. 6. If VA, VB, YP and V'A', V'B', Y'P' be such that sin AVP/sin BVP -=- sin A'V'P'/aia BiY'P' is constant, then YP and Y'P' generate homographic pencils. Ex. 7. Also if tan .4rP/tan A'VP" be constant. Take AYB and A'Y'B' right angles. Ex. 8. If AP.B^P'-h- BP be constant, then, P and P' generate homographic ranges. For AP.B^P'^BP.n'P'= AC.B'C^BC.Cl'C, hence (AB, CP) = (fl'B', CP'). io8 Anharmonic or Cross Ratio. [ch. Bx. 9. If Ok triat^gle ABC be circumscribed to the triangle LMN ; show that on infinite number of triangles can be drawn which are inscribed in the triangle LMN and at the same time dreumscribed to the triangle ABC. Take any point R on LM ; let AR cut NL in Q, and let BR cut KM in P. It will be sufficient to prove that PQ passes through C. Let BC cut KM in X, let CA cut LN in Y, and let AB cut ML in Z. Then {KMPX) = iJ (A'JlfPJr) = (ZMRL) = A {ZMRL) = {KYQL). Hence the ranges {KMPA') and {KYQL) are in perspective. Hence MY, PQ, XL meet in a point, i.e. PQ passes through C. Hence PQJK is inscribed in LMK and circumscribed to ABC. Ex. 10. Six points A, B, C, D. E, F are taken, SMcA that AB, FC, ED meet in a point G, and also FA, EB, DC in H; show that BC, AD, FE also meet iti a point. Let BE and CF meet in P, CF and AD in R, and AD and BE in Q. Then (^BPQE) = G{BPQE) = {ARQD) = H{ARQD) = (FRPC) = (CPRF). Ex. 11. AO meets BC in D. BO meets AC in E, CO meets AB in F. JC, Y, Z are taken such that (AD, OX) = (BE, OY) = (CF, OZj = — I ; show that the triangle XYZ circmnscribes the triangle ABC. For (AD, OX) = {EB, OT). Ex. 12. The paints A and B move on fixed lines ihrorugh 0, and U and V are fixed paints cdlinear urith ; if VA and VB meet on a fixed line, show that AB passes through a fixed point. Take several positions of the point A, viz. AiA^A, Join .4, U cutting the given line in C, , and join C,Y cutting OB in B, . Similarly construct C^C^... and B^B, Then (.A,A,A,...)==V(A,A^A,...) = (C,C,C,...) = r(C^C2C,...) = (BiB,B,...). Hence the ranges (A^A^A, ..,) and (B^BiB,...) are homographie. Al.so when A is at O, B is also at 0. Hence the ranges are in perspective. Hence .^iB,, A^B^, A^B^, ... meet in a point, i.e. .4B passes through ;i fixed point. Ex. 13. If the points A,B,Cmore on fixed lines through 0, and AB turn about a fixed point P, and BC turn about a fixed point Q, show that CA turns about a fixed point. Ex. 14. If the vertices of a pdlygon move an fixed concurrent lines, and aM but one of tlie sides pass through fixed points, this side a7ui every diagonal will pass through a fixed point. 16. 1/ two homographie pencils at different vertices have the ray joining tlie vertices as a ray corresponding to itself in tJie two pencils, then the pencils are in perspective. Let the two homographie pencils be V{V'ABC...) and F' (VA'B'C...). Let VA cut V'A' in o. Let VB cut V'B' in /3. Let a/3 cut VV in v. If a/3 does not cut VC and V'C in the same point, let a/3 cut FC in y and V'C in /. Now F ( V'ABC. ..)=¥'{ VA'B'C. . . ). Hence (va^y) = (va^y), IX.] Anharmonic or Cross Ratio. 109 by considering the sections of these pencils by a^. Hence y and y' coincide. Hence YC, V'C meet on a/3. So eveiy pair of corresponding rays meet on a/3. Hence the pencils are in perspective. Ex. 1. I/(ABCI)...) and (A'B'Cfl/...') be tuoo homographic ranges, and any two points V, V be taken on AA', show that the meets of VB and V'B', of VC and V'C, of VD and Vl/, ^c, aU lie an a line. Ex. 2. If AB pass through a fixed point V, and A and B move on fixed lines meeting in 0, and if V, W be fixed points eoUinear with 0, show that the locus oftlie meet ofAVand BW is a line. Let AV and BW cut in P. Take several positions A^A^... of A, BiB,... ofB, P,Pj... ofP. Then V{pP,P^...) = {OA,A,...) = V{OAiA^...) = {OB^B,...) = W[OP,P.,...). Now the pencils ViOP^P.^...) and W{OPiP.i...) have a common ray; hence they are in perspective. Hence all the meets (VPi\ WA)i (FPj ; WPi), ... lie on a line. Ex. 3. Show that the meet of UV arul OB, and the meet of VW and OA lie on the locus. Ex. 4c. If A and B move on fixed lines through 0, and AB, BP, and APpass through fixed coUinear points U, V, W, show that the locus of P is a line through 0. Ex. 5. If each side of a polygon pass through one of a set of collinear points whilst all but one of its vertices slide on fixed liTies, then will the remaining vertex and every meet of two sides describe a line. 17. If (ABC...) and (A'B'C..) he two homographic ranges on different lines, then the meet of AB" and A'B, of BC and B'C, and generally ofPQ' and P'Q, where PP', QQ are amy two pairs of corresponding points, all lie on a line (called the homo- graphic axis). Let the two lines meet in a point which we shall call no Anharmonic or Cross Ratio. [ch. X or Y', according as we consider it to belong to the range (^5C...) or to {A'B'G'...). Take the points X' and F cor- responding to the point Z (= Y') in the two ranges. Then every cross meet such as (PQ'; P'Q) lies on X'Y. (See figure of § 1 2.) For by hypothesis {X.YABC. . . ) = {X'YA'B'C. . . ). Hence A'(XYABC...) = A {X'TA'B'C'...) ; and these two pencils have the common ray AA'; hence they are in per- spective ; hence {A'X; AX'), {A'Y ; AT), {A'B ; AB'\ ... all lie on aline. But {A'X ; AX') is X', and (A'Y; AT) is Y. Hence {A'B ; AB') lies on the fixed line X'Y; i.e. every cross meet lies on a fixed line, for AA', BB' are any two pairs of corresponding points. 18. By Reciprocation, or by a similar proof, we show that ifV{ABCD...)and V {A'B'G' B'...) U homographk pencils, then all the cross joins such as the join of {VB ; Y'C) with (V'B' ; VC) pass through a fixed point (called the Iwmographic pole). Ex. 1. 1/ A, B, C be any three points on a line, and A', B', C' be any three points on another line, show that the meets of AB' and A'B, of AC and A'C, and of BC ancl B'C, are cdllmear. Consider JC {= Y') as above. Ex. 2. When two ranges are in perspective, the axis of homography is the polar of the centre of perspective for the lines of the ranges. Projective ranges and pencils. 19. If range a is in perspective with range P, and range ^ vsdth range y, and range y with range 6, and so on ; then each of the ranges a,l3,y,b... is said to be projective with every other. If pencil a is in perspective with pencil p, and pencil 3 with pencil y, and pencil y with pencil 8, and so on ; then each of the pencils a, /3, y, 6... is said to be projective with every other. Projective ranges are homographic. For the range a is homographic with the range /3, being in perspective with it ; so /3 with y, y with 6, and so on ; hence each is homographic with every other. Projective pencils are homographic. IX.] Ankarmonic or Cross Ratio. 1 1 1 For the pencil a is homographic with the pencil /3, being in perspective with it ; and so on. Homographic ranges are projective. For they can be put in perspective with the same range on the homographic axis. nomographic pencils are projective. For they can be put in perspective with the same pencil at the homographic pole. A range and a pencil are said to be projective, when the range is projective with a section of the pencil. Hence a range and a pcmil which are projective are homo- graphic ; and a range and a pencil tvhich are Iwmographic arc projective. CHAPTER X. VANISHING POINTS OF TWO HOMOGEAPHIC EANGES. 1. The points corresponding to the two points at infinity in two homographic ranges are called the vanishing points. To construct the vanishing points. Let the ranges be {Q.IABC...)= {J'D.'A'BV ...), where ii and il' are the points at infinity, and I and J' are the vanishing points. First, suppose the ranges to be on different lines. Through A' draw A'ui parallel to AB (and therefore passing through X2) cutting the homographic axis in w. Then A a> will cut A'B' in the vanishing point J'. Similarly I can be constructed. Second, suppose the ranges to be on the same line. Join ilABC... to any point V, not on the line ; and let the joining lines cut any other line in odbc... Then Vo is parallel to AA'. By using the homographic axis of the two homographic ranges abc... and A'B'C'..., find the point J' in A'B'C... corresponding to o in abc Then J"' is the vanishing point belonging to the range A'B'C For (QABC.) = (oabc.) = (J' A'B'C...). Similarly I can be constructed. 2. In two homographic ranges (ABCP...) and {A'B'CP'...), on the same or different lines, if I correspond to the point Q.' at infinity in the range {A'B'...), and J' correspond to the point i2 at infinity in the range (AB...), then IP.J'P' is the same what- ever corresponding points P and P' are taken. Vanishing Points of two Homographic Ranges. 1 1 3 For we have (J 12 ABGP. . . ) = {^'J'A'B'C'F . . .) ; hence (^P, ISl) = {A'P', Of J'), i.e. AI/IP -=- A il/Q-P = A'n'/D.'P'^ A'J'/J'P'. But AQ,/ap=-i and 4'i27I2T'= -i. .-. AI/IP = J'P'/A'J', . : IP. J'P'= I A . J' A', which is constant. Conversely, if IP. J'P' he constant, then P and P' generate ranges which are homographic, and I and J' are the points cor- responding to the points at infinity in the ranges. For let A and A' be any two positions of P and P', then IP. J'P'=IA . J' A'. Hence retracing the above steps, we get (AP, 112) = {AT', n.'J'). Hence P and P' are cor- responding points in the ranges determined by Alil and A'il'J', and I and J' correspond to i2' and i2 in these ranges. Ex. 1. If through the centre of perspective of the two ranges (ABC, . . ) and {ASCf ...), there be drawn a parallel to AB' meeting AB in I and a parallel to AB meeting AB' in J', prove geometrically that lA . J' A = IB . J'B' = ... = 10. J'O. Deduce theformvla IP . J'P' for any two homographic ranges. Ex. 2. If OP. OP' be constant, being the meet of the lines on which P and P' lie, show that P and P' generate hmnographvi ranges. Ex. 3. If I and J' be the vanishing points of the homographic ranges {ABCP...) = {A'B'G'P'...), shoie that (o) AP:AI:: A'P': J'P'; (6) AP/BP -:- A'P'/B'P' = AI/BI. Ex. 4. Show also that AP.J'P'-i- A'P" is independent of the position of P. For (An, PQ) = (A' J', PV). Ex. 5. If 0, A, B be fixed points on the fixed line OAB, and 0, A', B' be fixed points on the line OA'Pl which may have any direction in space, show that the meet of A A' and BB' describes a ^here. Through the meet V of AA' and BB' draw ri parallel to A'B'. Then 7 is a fixed point, for (lOAB) =r{IOAB) ={n'OA'B'). Again, through Fdraw VJ' parallel to AB. Then 7* is a fixed point on OA', i.e. OJ' is known, i.e. IV is known. Ex. 6. If one of two capolar triangles be rotated about the axis of homology, show thai the centre of homology describes a circle, whose centre is on the axis. Viz. the meet of the spheres determined by AB, A'P! and by AC, A'Cf, whose centres are on the axis. 3. Take any two origins U and V on the lines of the ranges. Then IP =UP—UI = x—a, say; and J'P'= V'P'- VJ' =iif-a', say. I 114 Vanishing Points of two [ch. Hence we get (x—a) {of—a') = constant, or xjf —a'x—a3f + aa'= constant, a relation of the form to;' + lx+ mxf + w = o. Hence the distances x and x" of corresponding points in two homographic ranges from any fixed points on the lines of the ranges are connected by a relation of the form kxx' + lx+mx' + n = o, where k, I, m, n are constants. Conversely, if the distances he connected by this relation, the points generate homographic ranges. For if kxx^ + lx+mx'+n = o, then Tc{x + -){^+-) = --n, or IP . J'P' = constant, where m/h = lU and l/Jc = J'V. The above relation assumes a neat form if we take W to coincide with 17. For then IP . J'P'= lU. J' U', .: xx'-a'x-ax = o, or a/x+a',x'= i, or UI/UP+U'J'/U'P'=i. ^.IfP and P' be connected by the relation Ix+mx' + n = o, P and P' generate homographic ranges in which the vanishing points are at infinity ; and conversely, the corresponding points of two homographic ranges whose vanishing points are at infinity, are connected by a relation of the form Ix + mx' + n = o. (The reasoning employed in the general case does not apply here because I and J' are at infinity, and hence we cannot start with the equation IP .J'P' = constant.) If lx-\-mxf + n = o, then x'= fix + y (say). Hence P'Q'= TQ- V'P'=^-af (say)= P(^-x) = p . PQ. Hence the two lines are divided proportionally by the two sets of points, which therefore form homographic ranges. Also putting a; = 30 , we get a;' = 30 ; hence O and li' are corresponding points, i. e. I and J' are at infinity. Conversely, if I and J are at infinity, then (AB, Pil) = {A'B', P'Q.'}. X.] Homographic Ranges. 115 „ AP A'P' x-a 3f-o! rLence =: or = PB P'B' h-x b'-af .: x{l'—a') + af{a-'b) + a'h-aV=o, which is of the form lx+ mx' + w = o. Or, we may consider the equation Ix+mx' +n = o as the limit of the relation Icxaf + lx+ mx'+n = o when k decreases indefinitely. Since the latter equation determines two homo- graphic ranges however small k is, we may assume this to be true in the limit when ft = o. Two homographic ranges in which the vanishing points are at infinity may be called similar homographic ranges. Ex. 1. 1/ (^ClIAB...) = (J'n'A'B'...), and AB = A'B'; show that AB = AI + A'J', and AI = -B'J'. Ex. 2. Through the vertex V of the parallelogram VIOJ' is drawn a line cutting 01 in, A and OJ' in A', show that 01/ OA + OJ'/OA' = i. Ex. 3. Find the values of the constants in the relation xaf + lx + mxf +n — o. The relation is i + l/x^ + m/x + n/xx' = o. Put X = ; .: I = —3f= —Y'J'; so m = -UI. Again, put a; = o and x^^VU'; .: n.= VV'. UI. Hence UP.V'P'-r'J'.UP-UI .V'P' + V'V'.UI = o. Another yalue of n is UV. VJ'. These values come also at once from IP.J'F'=nj.J'U' or IV.J'r'. Ex. 4. Deduce iheformvla when the vanishing points are at infinity. Dividing by VI and putting UI = x , we get c.UP+r'P'= VV (where c is the limit of V'J'/UI), or tc + ma/ + m = o. Ex. 5. Show thai the formula Ix + mx* + » = o can be vjritten VP/UV+r'F'/V'V= I. Put P = U and P' =V' successively. Ex. 6. Show that iy properly choosing V, the generai relation can be thrown into the form xx' + 1 {x—xf) + n = o. Ex. 7. Show that corresponding points PP' of two homographic ranges on the same line are connected by a relation of the form UP.r'F' + y.FP' + S = o, provided VV = VV. Show also that y - TO = VJ' and 8 = m.uv'= Vj'.r'v. 5. The following are geometrical applications. Ex. 1. If 0, C, A. A', V, V be fixed points of which OAA'O' are colUnear, and if points P and P' be taken on AU and A'V' such that u . DP/AP + P . VP'/A'P' = 7, I 2 ii6 Vanishing Points of two [ch. where a, P,y are constanis, shovo that the locus of the meet qf OP and OfV is a line. Reducing the given relation to any origins on ^17 and A'V, it is clearly of the form xi^ + lx + na/ + n = o. Hence P and P' generate homographic ranges. Also putting P = A, we get P'=A'. Hence in the two homographic pencils 0{P...) and 0' (P'...), OC is a common ray. Hence the locus is a line. Ex. 2. The same is true if any one of the following relations hold — (i) a . UP/AP + P/A'P' = 7, V being at infinity ; (ii) a/AP + P/A'P' =7, U and V being at infinity ; (iii) a . VP/AP + p . V'P' = 7, A' being at infinity ; (iv) a.VP + e . Y'P =7, A and A' being at infinity. Ex. 3. //" 7 = o in any of these relations, the locus passes through the meet of OU atid aV- XjX. 4. Obtain the Cartesian equation of a line, vis. Ax + By + C = o. Consider the pencils at the points at infinity on the axes. Ex. 5. ff Pit, PW drawn in given directions from P meet given lines OM aiirf OiS^ in M and Jf so that a.PM+fi. Pit = 7, show that P moves on a line. For Pit and Pit are proportional to the x and y of Ex. 4. Ex. e. If 0, U, V be fixed points, and if points P and P' be taken on OU and or' such that „ . UP/ OP + . VP'/OP' = 7, tiien PP passes through a fixed point. Ex. 7. The same is true if any one of the following relations hold — a/OP + e. VP'/OP' =7, V being at infinity ; a/ OP + P/OP' — y, V and F* being at infinity ; a.UP+P. T'P' =7, being at infinity. Ex. Q. If y = o, the point is on UV. Ex. 0. Zf p, g, r, the perpendiculars from A, B, Con a line, be connected by the relation K.p + fi.g + v.r == o, then the line passes through a fixed point. Divide by p and use Ex. 6. 6. If P and P' be connected by a relation ■which can be reduced to the form Icxxf -\-lx + m3f+n-= o, we have proved that P and P' generate homographic ranges. The following converse is very important, viz. Any relation which can be reduced to the form kxaf+hf+mx' + n — o is true of every pair of corresponding points of two homographic ranges, provided it is true of three pairs of corresponding points. Let the two homographic ranges be (ABCB...) and {A'B'C'D'...). Suppose the above relation (in which x= UP and x^= V'P') is satisfied when P is at ^ and P' at A', and X.] Homographic Ranges. 117 when P is at 5 and P' at S, and when P is at C and P' at C. Then it will be satisfied when P is at Z> and P' at B', D and D' being any other two corresponding points of the ranges. Tor if not, suppose that when P is at D, the above relation gives E' as the position of P'. Then since the given relation determines two homographic ranges, we have {ABCD) = {A'B'C'E') ; but {ABCD) = {A'B'C'B') by hypothesis. Hence D' and E' coincide, i.e. the given relation is true for every pair of corresponding points of the two ranges. Ex. 1. If the point P on the line AB and the point P on the line B'Cf be connected by the relation A. . AP/BP + ft . C'P'/B'P' = I, show that P and P' gmerate homographic ranges, and that B and B' are cor- responding points in these ranges. Find also the valvss of \ and /i. Prove also conversely, that if {ABCP) = (A'B'C'P') then the relation holds. Taking any origins we get \(x-a) (a/-6')+M(a^-0 (?:-b) = (x-b) (nf -V), which ia of the form feca/ + lx + mxf + n = o. Hence P and P' generate homographic ranges. Take P at B, then x = b, .: \ {b—a) (x'-b') = u, :. d = V. Hence f is at B , i.e. B and B' correspond. Again, let P be at C when P' ia at C Put a/= d. :. \ (c-ffl) = (c-b), .: X = BC/AC. Let P' be at A' when P is at A. Put a; = o. .-. II (a'-O = a'-V, .: n = B'A'/CA'. Conversely, if {ABCP) = {A'B'C'P'), the relation BC AP B'A' C^ _ AC' BP * C^' ' W¥' is true ; for it is of the form kci/ + te + rrad + k = o, and it is satisfied by {A, A'), {B, B') and (C, C). Ex. 2. Treat the following relations in the same way — (o) \/BP + n . aP'/B'P' = I ; (6) \.AP/IP + ii.C'P'= i; (c) \/IP + ix.C'P'= IS (d) AP.B'P' + \.CP + n.CP'=AC.B'Ci (e) IP.B'P' + K.CP + ii = o. Results— (a) X = BC, ,1 = B'J'/CJ'; {b) \ = IC/AC, n = i/CA'; (c) \ = IC, /i = i/CJ'; {d) K --^ -B'J', fi AI; {e) K = -B'J', 11=-- -IC.B'C. ii8 Vanishing Points of two [ch. Sx. 3. Shtm that the foUauKng eipuitions are satisfied by every ftoo homo- graphic ranges. , , AP B'F' AF B'V , AP P!P' AP B'pf Each equation is of the required form, and X, /«, v can be determined so that the equation shall be satisfied by any three pairs of points. Ex. 4. Deduce in 'Ex. 3 rfe/inite formidae for (ABCP) = (,A'B'CP'), i.e. determine the vaiues of K, ft, v. Ex. 5. V {ABCD) = (A'B'Ciy), prove that ,. AB.CD AC.DB AD. BO ^"^ -^'Bi-^-^^*-l^^°' C ieing an arbitrary point on the line A'B'. Take D and 1/ as variable points. Ex, e. Ifthepenca ViABCD) be hmwgraphic toith the range {A'B'CI/), show that sin ArB.aiaCTD ain AVC . sin DVB sin AVD. sin BVC _ a7W ■*" A^ * A^ ~ "• Use Ex. 5 {a). Ex. 7. SItou) that VP, V'P generate homographie pencils if sin^ rP sin BVC sin C'V'P' sinB'V'A' ^"' sin BTP ' sin^FC '*' SnWWF ' sin C'V'A' ~ '' or (6) \ cot Srp + /I cot S'7''P'= i, or (c) X tan^rP+/j tan CF'P'= I. Ex. 8. If YP and V'P' generate two homographie pencils, and AVP = 9 and B'Y'P' = V, VA and V'B' being any initial lines, shaie that tan 9. tanfl' + X tanS + ntanfl' + K = o; and conversely, if this relation be satisfied, then VP and V'P' generate homo- graphic penctts. Take transversals perpendicular to the initial lines, then tan 8 oc X and tanS'oc a/. Common points of two homographie ranges on the same line. 7. Suppose corresponding points in two ranges on the same line to be connected by the relation k.UP. V'P' + 1. UP+m .V'P' +n = o. For the origins U and V we can take the same point on the line, called U or V' according as it is considered to belong to one or the other range. The equation becomes h.UP. UP' +1. UP+m. UP'+n = o. X.] Homographic Ranges. 119 Now if P correspond to itself, P must coincide with P'. Hence the equation giving the self-corresponding or common points of the two ranges is k.UP^ + {l+m) UP+n = o. Hence every two homographic ranges on the same line have two common points, real, coincident, or imaginary. A graphic construction of the common points will be found in XVI. 6. Ex. 1. I/B andF be the common points of the homographic ranges (ABC.) and {A'B'Cf. . .), show that (_EFAA') = [BFBB') = (EFCC) =■... For {EF, AB) -= (JSf, A'B'), .: (,EF, AA') = {EF, BB'). Ex. 2. Jf (EFABC. . .) = (EFA'B'C. . .) = (EFA"B"C"...) = . . . , then {EFAA'A". . .) = {EFBB'B". . .) = {EFCCC. ..) = .... Ex. 3. If {EF, PP') be constant, then PF' generate homographic ranges of which EF are the common points, Ex. 4. Xf ABC..., A'B'C... be homographic ranges on the same line, and if P', Q be the points cffrresponding to the point P ( = Q') according as it is con- sidered to belong to the first range or the second, show that P*, Q generate homo* graphic ranges whose common points are the same as those of the given ranges. The range generated by P" is homographic with the range generated by P, i.e. by Q', and this is homographic with the range generated by Q. Hence range f = range Q. Again, suppose P is a common point of the given ranges ; then P' coincides with P, i.e. P' coincides with Q'; hence P. coincides with Q, i.e. F' coincides with Q, i.e. P is a common point of the derived ranges. Ex. 5. If .Z be the fourth harmonic of P for P' and Q, then PX is divided harmonically by the common points. Let the given homography be defined by PA . PA' + 1 . PA + m . PA' + n = o. Put ^ = P and A'= P>, .: PP* -= -n/m. Put ^ ^ ft ^'= 0'= P, .: PQ = -n/l, .: 2/PX= i/PP'+i/PQ = -(l + m)/n. Now E and F are given by x' + Q + m) x + n - o, .: i/PE + i/PF = -{1 + m)/n ^ a/PX, .-. (PX, EF) is harmonic. Ex. 6. Construct the fourth harmonic of a given point for the (unknmon) common poirUs of two given homographic ranges. Ex. 7. Show thai {EF, gP') = {EF, AA')' in Ex. 4. For {EF, AA')'= {EF, PP') . {EF, QC/) where P=-Q'. This gives us another proof of Ex. 4, using Ex. 3. Ex. 8. If ABA'B' be given coUinear points, find a point X in the same line, such that the compound ratio AX.A'X -r- BX .B'X may he a given quantity. 1 20 Vanishing Points of two [ch. X is one of the common points of the homographic ranges deter- mined by AP/BP-=r BfP'/A'P'= the given quantity. £z. 9. Determine the point X, given the value of AX . A'X -r BX. 8. If one of the common points of two homographic ranges {ABC.) and (A'B'C..) on the same line be at infinity, then the points ABC... divide the line in the same ratios as the points A'WC ; and conversely. For if {AB, CO.) = {A'B', CO). AC ^_4^G^ ^ ^^^" CB' AQ.~ C'B' ' A'Ll ' But Q.B^A il = - 1 = ilB:^A'Q. ; .-. AC:CB::A'G':C'B!; and similarly for any other pair of segments ; i.e. the line is divided similarly by the two sets of points. Conversely, if the line be divided similarly by the two sets of points. Since AC:CB:: A'C: C'B', we have, retracing our steps, {AB, CO) = {A'B', C'il). So {DB, Ci2) = {B'B', C'ii), and so on. Hence {aABC.) = {0. A'B'C'...), i.e. {ABC.) and {A'B'C'...) are two homographic ranges with a common point at infinity. Or thus — Let the homography be given by Icxxf+lx+mnf+n = o. The common points are given by Jcaf+{l+m)x+n = o. If one of the common points be at infinity, then A; = o, i.e. the homography is given hy lx + wx' + n = o, i.e. the ranges are similar. Conversely, if the ranges are similar, then Ix+mx^+n = o, i.e. ft = o, i.e. one of the common points is at infinity. Ex. 1. If in two homographic ranges on different lines the points at infinity correspond, the ranges are similar ; and conversely. Ex. 2. If one of the common points of two homographic ranges on the same line he at infinity, the other, E, is given by EA : EA': : BA : B'A'. X.] Homographic Ranges. 121 Ex. 3. Shaw oJso that B is the meet with AA' of the radical axis of any two circles through AB' and A'B. Ex. 4. If AB/A'B' = BC/B'C = ... = — i, show that one common point is at infinity, and that the other bisects all the segments AA', SB', CC, Ex. 5. If each of the common points be at infinity, then all segments joining corresponding points are equal : and conversely. For if F be at infinity, the ranges are divided proportionally, hence AB/A'B' = EB/BB' = i, for E is also at infinity. Conversely, if AB = A'B', BC = B'C, ..., the ranges are divided pr%)ortionally ; hence F is at infinity. And E is given by EB/EB' = i, hence E is also at infinity. Or thus. In this case the quadratic kx' + (l + m)x + n = o has both roots infinite. Hence ft = o and i + m = o. Hence the homography is given by I (x—x') + n = o, i.e. x—x'= constant, i.e. AA' is constant. And conversely, if AA' is constant, then ft = o and i + m = o. Hence both common points are at infinity. Common rays of two homographic pencils having the same vertex. 9. In amy two JiomograpMc pencils having the same vertex, two rays exist, each of which corresponds to itself. Let the pencils be V{ABC...)-= V{A'B'C' ...). Suppose a line to cut the pencils in the ranges (afcc.) = (a'fcV...), a being on VA, and so on. Then if VA and VA' coincide, a and a' will coincide. Hence if e and / be the self-corre- sponding points of the ranges (aftc ...) and (a'h'c' ...), Ve and Yf&ve the self-corresponding or common rays of the pencils F(^J5C...)and V{A'B'G'...). Ex. 1. If VP and VP be a pair of corresponding lines in two homographic pencils whose common lines are VE and YF, show that sin EYP/ain FVP -i- sin EYF'/sin. FVP' is constant. Ex. 2. Find a point on u given line through which shall pass a pair oj corresponding lines of two given homographic pencils. Either of the common points of the homographic ranges determined on the line by the pencils. Ex. 3. If YA, Y'A' generate homographic pencils at Y and Y', show that in two positions YA is parallel to Y'A'; and that any transversal in either of these directions is cut by the two pencils proportionally. For vrithout altering the directions of the rays, superpose Y' on Y. Ex.4. Two given homographic pencils Y(abc..^ and Y' {a'Vcf ...) meet a line in the points ABC... and A'BIQ ... ; determine the position of the line so that AB = A'B', BC = BI0, CD = CI/, by construction. Hence OB — (fEf. Hence AB = A'&, and so on. Hence there are two such lines, one parallel to each of the lines Vy, V'. I!x. 6. Given any two homographic pencils, one can be moved parallel to itself so as to be in perspectrce with the other. 10. If I, J' correspond to the points at infinity in two homographic ranges on the same line, and Msect IJ', and 0' be the jmnt corresponding to 0, then the common points E, F are given ly 01^"= OF^ = or. ocy. For (012, IE) = (lyr, q:e\ where i2 or fl' is the point at infinity upon the line, or E£l _ O'il' EJ' ■'■ iLi ' OE~Wj' ' We' But E^-^ia = I and 0'12'-- 12 V= - i, .-. OI.) OP.OP'-OI.PP' + OI.OO'=os (c) OP' + IP.PP' + OI.O(y=o. Ex. A.IfE and F coincide, P and P' are connected by the relation UP. VP'-UJ'. VP-UI. VP'+Vd'^ o. For putting P = P' in the general relation UP . UP' - VJ'. UP-UI.UP' + UU'. UI = o, and noticing that -^.UO = UI+ UJ', we get PP--a . DO . !7P+ UV. VI = o. And this is a perfect square, hence UU'. UI = UOf. Ex. 5. If E and F coincide, show that P and P' are connected by the relatim OP.OP'= 01. PP'. It is of the required form, and is satisfied by (I, Jl'), and by {E, E) and [F, F) since E and F coincide with 0. Ex. Q.IfE and F coincide, show also that (a) (OP)-> + (OP')-'= (0^)-' + (0.4')-'; (6) OP. OA/AP = OP'. OA'/A'P>; (c) 01^ = PI. PP'. Ex. 7. Any two ranges wh^se comtnon points coincide, can be placed in perspective with two ranges whose corresportding segments are equal. For join the two ranges to any point V and consider the ranges on any line parallel to VO. 11. If the common points be imaginary, then the ranges {ABC.) , =!> (do, cd). sm dpb sm cp smapb ^ 126 Anharmonic Properties of [ch. by projection. Hence ABCD subtend the same cross ratio at every point P on the conic. The cross ratio subtended by the points {AB, CD) on a conic at any point on the conic is called the (ross ratio of the points (AB, CD) on the conic. Notice that, making P coincide with A, the cross ratio of (AB, CD) is equal to A {AB, CD) = A {TB, CD), where AT is the tangent to the conic at A. Ex. 1. Show that in a circle the pencils p (abed) and p' (abed) are super- posable in all cases. This gives another proof of § 2. Ex. 2. A tangent to an ellipse meets tlie auxiliary circle in ZZ'j show that tlw cross ratio 0/ the four points (AA', ZZ') oji the circle is (i— e) -j- (i +«). Consider the pencil at the point opposite to Z'. Ex. 3. Prove that the cross raUo [AB, CD) of the four points A, B, C, Ban a cirde is AC/CB -r AD/DB, AC being ike Imgth of the line joining A to C. For sin APC =AC-i-aR. Ex. 4. Conjugate lines far a conic meet the conic in four points which subtend a harmonic pencil at every point an the conic. Consider the pencil at one of the four points. Such points are called harmonic points on the conic, Ex. 5. If AA', BB' be pairs of harmonic points on a conic, show that AA' and BB are conjuyate lines for the conic. Ex. 6. The chorda AB, CD of a conic are conjugate, and ACB is a right angle ; through D is drawn the chord DP meeting AB in Q; show tliat CA, CB are the bisectors P(7-=-sin BPC. sin APD. But PA . PB sin APB = a . AB, and so on. Hence a .y . AB . DC-^fi.b. BC . AD = P{AC, BD) is constant, i.e. ay = & . /3 . 6. Ex. 1. If the perpendiculars let faUfrom any point on a conic on the sides of an inscribed polygon of an even number of sides be called i, 2, 3, ,,,, a n, show that 1 . 3 . 5 . . . , {an— i) -i- s . 4 . 6 sn is constant. Suppose the theorem holda for 2 n — 2 sides. Then 1.3.5 (2TO-3) = fc. 2.4.6 (2n-4)i. And by the above theorem {an—i)x = k'{an—a)(sn). Multiplying, I. 3. 5. ... (2W — i)=J,-". :4. 4.6 an. Hence by Induction. Ex. 2. The product of the perpendiculars from any point on a conic on the sides of any inscribed polygon varies as the product of the perpendiculars on the tangents at the vertices. Make the alternate sides in Ex. i of zero length. Sx. 3. If the conic be a circle, the products are equal, in the theorem aiui in Ex. I and Ex. a. (See Ex. 3, § a.) Sx. 4. The product of the perpendiculars from any point on a conic on two fixed tangents is proportioruU to the square of the perpendicular on the chord of contact. Ex. 5. The product of the perpendiculars Jrom any point on a hyperbola on two fixed lines parallel to the asymptotes is proportional to the perpendicular on the intercept on the curve. For a.7-j-/3.S = a', y-f- /3'. S' and a = o'. Ex. 6. Theproduct of the perpendiculars from any point on a parabola on two fixed diameters is proportional to the perpendicular on the intercept on the curve. 128 Anharmonic Properties of [ch. 4. Any number of fixed jooints on a conic subtend homograpMc pencils at variable points on the conic. Let the fixed points be A, B, C, B,... and take two other points P, Q on the conic ; we have to prove that P{ABCB ...) = Q{ABCB ...). This follows at once from the fact that P{ABCB) = Q(ABCB), where ABCB are any four of the fixed points. Ex. 1. P, U, V are points on a hyperhda, P being variable ; show that the lines PV and PV intercept on either asymptote a constant length. Instead of the asymptote consider at first a chord LU of the conic, and let PU, PV cut LM injp andy. Then (p) = U{P) = r{P) = {pf). And the common points of the homographic ranges (p) and (y) are seen, by taking P at i and U, to be L and M. Hence in the given case the common points coincide at infinity ; hence pff is constant. Ex. 2. Through a fixed point are drawn lines parallel to the rays of the pencils subtended at two points on a parabola by the other points on the parabola ; show that corresponding lines cut off on a fixed diameter a constant length. Join the ranges determined on the line at infinity to the fixed point and proceed as above. Ex. 3. The fixed line DA meets a fixed conic in A, and EB touches at a fixed point B. A point is taken on the conic. Through A is drawn a variable line meeting the conic again in P and EB in Q, OP ineets DA in U and OQ meets DA in V. Find the position of when VV is of constant UngOi. First take EB to be a chord BC. Then (to = 0{TJ) = 0{P) = A (P) = (0) = 0(SI) = (D- And the common points are where OB and OC meet DA. In the given case therefore these coincide. And they must be at infinity. Hence OB is parallel to DA. 5. The locus of the meets of corresponding rays of two homo- graphic pencils, at different vertices and not in perspective, is a conic which passes through the vertices. Let the pencils be {PQR . . .) and 7 (PQR ...). Then we have to prove that the locus of the points PQR ... is a conic through and V. Since the pencils are not in perspective, corresponding to the ray VO in the V pencil, we shall have some ray OU, say, in the pencD which does not coincide with VO. Draw any cii-cle touching OU at 0. Let this circle cut OF in V, OP in P', and so on. Now V {OPQ . . .) = by hypothesis {UPQ ...) = o{UP'q...) = o{OP'qr...) = y'iop'q...) XI.] Points on a Conic. 129 from the circle. Hence the two pencils Y{OPAB. Ex. 2. The locus of the points of contact of parallel tangents to a system of confocal conies is a rectangular hyperbola through the foci. Prove this, and obtain the reciprocal property of coaxal circles. 10. Converse of Pappus's theorem. If a point move so that its perpendicular distances a, fi, y, 6 from four fixed lines AB, BC, CD, DA are connected by the relation a.y = h.fi.h. then the loms of P is a conic through ABCD. 134 Ankarmonic Properties of Points on a Conic. ^ a.y AB DC. . . tt For — — ; • -rfryr — -— IS coDstant. Hence, reasoning as in p . 8 xsC . A.D § 3, we see that P{AC, BD) is constant. Hx. 1. Gnen hm pairs of lines which are conjugate for a cirde, the locus of the centre of the circle is a rectangular hyperbola. Let AB, CD be conjugate, and also BC, AD. Assume to be a position of the centre. From drop OP perpendicular to DC to meet AB in F'. Then P' is the pole of CD, hence OP.OP = (radius')". So if OQ, per- pendicular to AD, meet BC in C, we have OP.OP'= OQ.OQ'. Also OP = y, OP'xa, OQ = S,OQ'cce. Hence u.7 oc^. S. Hence the locus of is a conic through ABCD. Also the orthocentre of ADC gives OP.0P'=^ OQ. OQ'. Hence the conic is a r. h. Ex. 2. The /ocMS of the foci of conies inscribed in a parallelogram is a r. h. circumscribing the parallelogram. Here 0.7 = ^.!. 11. The projection of a conic is a conic. We have to prove that any projection of a conic can be placed in perspective with a circle. Now every projection of a conic is such that all the points on it subtend homographic pencils at two points on it ; for this is true in the conic which was projected and is a projective property. Hence the projec- tion is the locus of the meets of two homographic pencils and is therefore a conic. CHAPTER XII. ANHARMONIC PHOPEETIES OF TANGENTS OF A CONIC. 1. Four fixed tangents of a conic cut any variable fifth tangent of the conic in a constant cross ratio. Consider first the circle of which the conic is the projec- tion. Let the iixed tangents of the conic be the projections of the tangents at ABCD of the circle, and let the variable tangent of the conic be the projection of the variable tangent at P of the circle. Let the tangent at A cut the tangent at P in a, and so on. Then if be the centre of the circle, Oa is per- pendiculartoPJ.. Hence the pencils {abed) and P{ABCD) are superposable and therefore homographic. But P {ABCD) is independent of the position of P on the circle. Hence {abed), i. e. {abed), is independent of the position of the variable tangent of the circle. Hence the proposition is true for a circle ; and being a projective theorem, it follows at once for the conic by projection. The constant cross ratio {ab, cd) determined on a variable tangent by four fixed tangents is called a cross ratio of the four tangents. 136 Anhannonic Properties of [ch. Notice that the point where a tangent cuts itself is its point of contact ; for as two tangents approach, their meet approaches the point of contact of each. Similarly any number of tangents of a conic determine on two other tangents of the conic two ranges which are homographic. Notice that we have in the above proof incidentally shown that the range determined on any tangent of a conic by several otlier tangents of the conic is homographic with the pencil sub- tended at any point on the conic by the points of contact of the other tangents. Ex. L SAow that the angle aOb is the same for every position 0/ the variable tangent. Thia gives us another proof of the proposition of § i. Ex. 2. A variable tangent of a conic meets at Q and Q' the tangents at the ends P, P 0/ aftxxd diameter of the conic ; show that PQ . Pf^ = CD', CD being the semi'-diame/er amyvgaie to CP. For P and P are the vanishing points of the ranges determined by and C on the tangents at P and P. Hence PQ . PQ^ is constant. To get the constant in the ellipse, take QQ' parallel to PP. To get the constant in the hyperbola, take an asymptote as QQ'. Then PQ = PQ' = CD. Ex. 3. If the joins of the ends PP of a diameter to a point on the conic cut the tangents at P and P in Q and Q', show that PQ . P'Q' - 4 . CD". Ex. 4. IfR and B' be the meeti of these joins and BV , them, OR. CR' = CD', and R and R' are conjugate points. Ex. 5. A variable tangent to a conic meets the adjacent sides AB, BC of the parallelogram ABCD circumsiribed to the conic in P and Q ; show that AP . CQ is constant. Ex. 6. A variable tangent cuts the asymptotes of a hyperbola in T and P; show that CT . CP is constant, C being the centre. Ex. 7. Deduce the equation of a hyperbola referred to its asymptotes, viz. xy i= constant. Ex. 8. B and C are the points of contact of tangents from A to a conic. A variable tangent meets AB in P and AC in Q. Show that the locus qf {BQ ; CP) is a conic touching the given conic at B and C. For B {S) = (0) = (P) = C(JJ). Also when P approaches B, R ap- proaches B. Ex. 9. Tfie hco pairs of tangents from a pair of conjugate points meet any tangent in two pairs of harmonic points. Such pairs of tangents are called harmonic pairs of tangents. Ex. 10. If AA', BP be pairs of harmonic points an a conic, show that the four tangents at ABA'B' cut any fifth tangent in a harmonic range. Ex. 11. On a fixed tangent of a conic are taken two fixed points AB and also two variable points QR, such that {AB, QR) = — i ; show that the locus of the meet of the other tangents from Q and R is the join of the points cf contact of the other Umgenis from A and B. XII.] Tangents of a Conic. 137 2. If AB, BC, CD, DA touch a conic, and p, q, r, s be the perpendiculars from A, B, C, D on a variable tangent of the conic, then p .r — k. q. s. Let two variable tangents cut BC in P, P' and AD in Q, Q'. Then (BC, PP') = {AD, Q<^). BP P^_AQ ^D •'* PC' BP'~ QD ' A^' .-. BP.QD-^PC.AQ is constant. ^ , BP q . AQ p But ^=- and p=-, . *. p.r-i- q. s is constant. Ex. 1. Extend the theorem to a a n-sided circumscribed polygon. Hx. 2. Dediux a theorem concerning a n-sided circumscribed polygon. Ex. 3. I/tht conic be a circle, show that p.r -i- q. s is equal to OA. OC-T- OB. OB, being the centre. For sin AOQ = sin BOP. Ex. 4. If the conic be a parabola, then p.r = q. s. For taking the line at infinity as tangent, k = p'.r' -i- q'.s^= i. Ex. 5. Show that for any conic the k ofp .r = k.q.sisthe cross ratio of the four tangents divided by the cross ratio of the pencil formed by four lines drawn parallel to them through any vertex. 138 Anharmonic Properties of [ch. Let P<) meet AB in M and CD in iV. Then JlfQ -T- sin Jlf^Q = A(i -j- sin ^ilf(} ; and so on. Hence the ratio of cross ratios corresponding to (JlfJf, (HP) is Aq.VC-^QB.BP =p.T -^ i.s. Ex. 6. Tht lines AB BC, CD, DA tmtch a conic ; one tangent meets AB, CD in M, N and another tangent meeti AD, BC in P, Q ; show that AM.BQ. CN. DP= AP. BM . CQ.DN. Ex. 7. The sides BC, CA, AB of a triangle touch a conic at P, Q, R; s)u)W that if I be any tangent (i) (pt).{A,t)a:{B,t).{C,i); (ii) (R.i).(Q,t)€C(A,ty. 3. Deduce, from tlie theorem a. y = fc. /3. 8 of XL 3, the theorem p.r = k.q.s Vy Eedprocation. Call the sides of the inscribed figure in XI. 3 a, h, c, d; and let the reciprocals of a, h, c, d be the points A, B, G, B of a four-sided figure circumscribing a conic ; then p, the reciprocal of P, touches this conic. The given theorem a . y = A . /3 . 8 asserts that {P,a).{P,c)^{P,b).{P,d) is constant. But by Salmon's theorem OP/{P, a) = OA/(A, p), and so on. Hence, dividing by OP^, we see that (A,p) (C^^(B^ {I>,p) OA ' OC ' OB ' OD is constant. Now is a fixed point, hence {A,p).(C,p)^{B,p).{B,p) is constant, i.e. p.r-i- q.s is constant. Ex. L Giten any fixed point and any conic, two lines s and h can be found such that OP' -— {P, s) . (P, k) is constant, P being u variable point on the conic. Viz. the lines corresponding to the foci of a reciprocal of the conic for 0. Ex. 2. AA', BB' CC are the three pairs of opposite vertices of a quadrilateral circumscribed to a pardbola whose focus is S; show that SA . SA' = SB.SB''=SC. SC. Take the four-sided figure whose vertices are AB'A'B. Then p.r = q.s. Hence in the reciprocal circle we have SA . SA'.a. y = SB. SB'. /9 . 8. But fc = I in the circle. Hence SA . SA' = SB . SB', =SC.SC' similarly. XII.] Tangents of a Conic. 1 39 Ex.3. IfiTit tangents at ABCB ... to a circle meet in LMN ..., then, t being any tangent and the centre of the circle, II {A, t) : II (i, t) :: II OA: II OL, II demiting a product. For II (r, a) = II (T, I) in a circle. 4. TJie lines joining corresponding points of two homographk ranges which are on different axes and not in perspective touch a conic which touches the axes. Let the ranges be {PQR...) and (JP'Q'R ...) on the axes OP and OP'. Since they are not in perspective, the point which corresponds in the range (P'Q'^...) to will be some point 0' not coinciding with 0. Draw any circle touching OP' at 0', and from and P' draw the second tangents to this circle, meeting in p. Then the range (P) = range (P') by hypothesis = range (j?) from the circle. Hence the ranges (P) and {p) are homographic. Also when P' coincides with O', both P and p coincide with 0. Hence the ranges are in per- spective. Now rotate the figure of the circle out of the original plane about the axis 00'. Then the ranges (P) and [p) are still in perspective. Hence all the lines Pp, Qq, Rr, ... meet in a point, say V. Hence, taking F as vertex of projection, p projects into P, and therefore the line P'p into the line P'P. Hence, since P'p in all positions touches a circle, P'P in all positions touches the projection of a circle, i. e. a conic. Also, since the circle touches Op and OP', the conic touches the projections of these lines, viz. OP and OP'. Notice that if the ranges be in perspective the envelope of PP' degenerates into the centre of perspective and the meet of the axes. 5. One, and only one, conic can he drawn touching five given 140 Ankarmonic Properties of [ch. The envelope of a line which cuts four given lines vn a given cross ratio is a conic touching the given lines. These propositions can be proved like the reciprocal pro- positions in XI. 6 or they may be deduced from these by Beciprocation. 6. Every two conies have four common tangents. Two conies cannot have more than four common tangents ; for if they had five, they would coincide. Also we see that two equal ellipses laid across one another have four common tangents. Hence we conclude that if two conies have not apparently four common tangents, some of the tangents are imaginary, or coincident. (See also XXVII. 4.) Touching fmr given lines can be drawn an infinite number of conies. For we can draw a conic touching the four given lines and any fifth line. All the conies which touch four given lines have a common self-conjugate triangle, viz. the harmonic triangle of the quadrilateral formed by the common tangents. Ex. L Given two hoTnograpkic ranges ABC... and A'B'C... on different lines ; show thai two points can be found at each of which the segments AA', BB', CCf ... subtend the same angle. Viz. the foci of the touching conic. Ex. 2. There are also turn paints at which AA', BB', CC... subtend angles having the same bisectors. Let the enveloped conic touch the lines in P and Q. The required points are the meets of PQ with the director ; as may be shown by reciprocating for one of these meets. Ex. 3. The vertices A, B, C of a triangle lie on the fixed lines MN, NL, LM, and the sides BA, AC pass through the fixed points W and V ; show that tlie envelope ofBC is a amic totuhing the five hnes LM, LN, YW, NV, MW. Ex. 4. AU but one of the sides of a polygon pass through fixed points and each vertex moves on a fixed line; find the envelope of the remaining side. Ex. 5. From the variable point situated on a fixed line are drawn the lines OA, OB, OC to the fixed points ABC, meeting BC, CA, AB in X, Y, Z ; BC, YZ meet in X', CA, ZX meet in Y', and AB, XY meet in Z'. Show that the line X'Y'Z' envelopes a conic which touches each side of the triangle at the fourth harmonic of tite fixed line for tite side. By a previous example X'Y'Z' are coUinear. Also {0) =A{0) = (JT) = (JT'J since {BC, XX') is harmonic = (Y') similarly. XII.] Tangents of a Conic. 141 Hence X'Y' envelopes a conic. Let the locus of meet BC in P. Then when coincides with P, X coincides with P, X' coincides with P' where [PP, BC) = - i, Z and T coincide with C, and Z and Z' coincide with B. Hence BC touches at P'. Ex. 6. Eeciprocate Oie prmious example. Ex. 7. JVie vertices BC of a triangle lie an given lines and the vertex A lies on a amic on which also lie fixed points VW through which the sides CA, AB pass. Show that the envelope of BC is a conic touching the given lines. Ex. 8. The side BC of a triangle touches a conic, and the vertices B and C move on fixed tangents of this conic, whilst the sides AB, AC pass' through fixed points ; show that the locus of A is a conic through the fixed points. Ex. B.Ife {ab, cd) mean the cross ratio determined on the line e by the lines a, b, c, d ; show that e (ab, cd) . c (ab, de) . d (db, ec) = i, where a, b, c, d, e are any five lines. Estimate the cross ratios on any tangent to the conic touching abede. Ex. 10. S?u)W thai the problem —' To find a line on which five given lines, no three of which are concurrent, shall determine a range homographic with a given range ' — has four solutions. Ex. 11. Given in position two pairs of conjugate diameters of a conic and a tangent, construct the conic. Construct the parallel tangent (which is equidistant from the centre). Let these tangents cut a pair of conjugate diameters in LL' and MM'. Then LM and L'M' also touch the conic. Proceeding simi- larly with the other pair, we hare seven tangents. Ex. 12. Given in position a pair of conjugate diameters and two tangents, construct the conic. Ex. 13. Prove the converse q/' § 2. 7. If two quadrangles have the same harmonic points, then their eight vertices lie on a conic ; as a particular case, if any three of the vertices are collinear, the eight vertices lie on two lines. Let ABCB, A'B'G'iy be the two given quadrangles, and UVW the common harmonic triangle. If no three of the eight vertices lie on a line, we can draw a conic through any five, say A', B', C', D' and A. Then from the inscribed quadrangle A'B'C'B' we see that UVW is a self-conjugate triangle for this conic. Also by hypo- thesis UVW is the harmonic triangle of the quadrangle ABCD. Hence (see figure of V. 9) B is such that ( WANS) is harmonic ; hence B is on the conic, for A is on the conic, and W is the pole of UV; similarly C and I) are on the conic. 142 Anhannonic Properties of [ch. Hence ABCDA'B^C'iy lie on a conic. If three of the vertices lie on a line, say ACD', then since B'l)' passes through F we see that JB' also lies on AG. Again, BD and also A'C form with AC or B'D' a pair har- monic with VU and VW. Hence BD and A'C coincide. Hence the eight vertices lie on two lines, i.e. on a conic. Ex. 1. Prove thai bvo quadrilaterals which have the same harmonic triangle are such thai 'the eight sides touch a conic (which may be two points). Ex. 2. A conic can be drawn through the eight points qf contact of two conies inscribed in tlie same quadrilaiercU. Ex. 3. The eight tangents at the four meets of any two amies touch the tame conic. 8. Any number of tangents of a parabola determine on two other tangents of the parabola two ranges which are similar. Let the two ranges be {FQB...) and (P'Q'iJ'...). Let 12 and 12' be the two points at infinity upon the lines PQ and P'^. Then since the line at infinity touches the parabola, the line i2i2' is a tangent. Hence the two ranges (SIPQM...) and (H'P'Q'B'. ..) are homographic ; also the points at infinity QSl' correspond. Hence the ranges are similar. Conversely, the lines joining corresponding points of two similar ranges which are on different axes and not in perspective touch a parabola which touches the axes. For if the ranges iPQR...) and {P'QfK...) are similar, the ranges {HPQR ...) and (QfP'Q'Bf. . . ) are homographic. Hence the lines i2i2', PP', Q(^, ... all touch a conic which touches PQ and P'Qf. And this conic is a parabola since Oil' touches it. Ex. L One and ordy one parabola can be draum touching four given lines. Ex. 2. The envelope of a line which ciUs three given lines in a constant ratio is a paraJmla. Ex. 3. Every two parabolas ?iave three finite common tangents. Ex. 4. Touching three given lines can be drawn an irifinite number of parabolas. Ex. 5. TP, TP' touch a parabola at P and P', and cut a third tangent in Q, (/; Shaw that QP : TP :: T^fe92...) = o(r,r,...) [corresponding rays being perpendicular] = (P,Pj...). Hence PQ, i.e. PY, enTelopes a conic touching P^ P, and Q, 4i ie. the polar of and the line at infinity. This parabola touches the tangent at B, a foot of a normal from ; for if OY be OR, then PY is the tangent at R. Ex. 9. If instead of being perpendicular to the variable line, PY make a given angle with it ; show that PY envelopes a parabola, which touches the polar of 0, and also touehes the tangents at the points where the tangents make the above angle with Hie radii from 0. Ex. 10. If the given angle be the angle between the polar of and the conjugate diameter, the envelope reduces to a point; and tbelocus of Y is a circle. For when P is at infinity, Q coincides with it. Ex. 11. ff through every point on a line, there be drawn the chord of a conic which is bisected at this point, the envelope of these chords is a parabola which touches the line. Consider the pencil of diameters. Ex. 12. Through points PQ... on the line I are drawn the lines PI*, QQ^, ... parallel to the polars of P, Q,... for a conic; show that PP", QQ',,.. touch a parabola which touches I. L 146 Poles and Polars. Reciprocation, [ch. Ex. 13. The reciprocoUs of the four points A, B, P, Q are the four lines ' Q> C be Pi ^) -K, p', q, ^. Then on a fixed line p is taken a variable point B, and from M are drawn the tangents q, c[ to the reciprocal conic, and the line p' is taken such that (pp', qc[) is harmonic. "We are given that p' always passes through E, and we have to prove that E is the pole of p. But this is obvious, for p and p' are conjugate in all positions oip', since (^pp', qc[) = — i. Hence ^' always passes through the pole of j3, i.e. E is the pole of jj. XIII.] Poles and Polars. Reciprocation. 147 Ex. 1. The reciprocal of a triangle self-conjugate for a conic is a triangle self- conjugate for the reciprocal conic. Ex. 2. A triangle self-conjugate far the hase conic reciprocates into itself. Ex. 3. A conic, its reciprocal, and the base conic have a common self-conjugate triangle. "Viz. the common self-conjugate triangle of ther given conic and the base conic. 4. Given any two conies, a base conk can he found for which they are reciprocal. Of the two given conies a and ^3, let P be a common point, g a common tangent, and UVW the common self-con- jugate triangle. Describe by XXV. 12 the conic F for which UVW is a self-conjugate triangle and P is the pole of q. Then F is the required base conic. For let o! be the reciprocal of a for F. Then since P is on a, its reciprocal g touches a'. Again, since q touches a, its reciprocal P is on a'. Also since UVW is self-conjugate for a and F, it is self-conjugate for a'. Hence a, a' and /3 pass through P, touch q, and have UVW as a self-conjugate triangle. Now by V. 9 to be given a point and a self-conjugate triangle is equivalent to being given four points. Hence a, a' and ^ pass through the same four points and touch the same line. But by XXI. 3, Ex. 4, two, and only two, conies can satisfy these conditions. Hence a! coincides with a or ji. Now if the meets of the conies are distinct, a' cannot coin- cide with a. For let q touch a at B. Then, by XI. 7, Ex. 6 and 7, a and F have double contact, and PB passes through the common pole A of the chord of contact BC. Now A is the pole of BC for a and F. Hence A must be ?7 or 7 or W. Hence PB passes through U" or F or W. Hence PB is a common chord of a and ^, i.e. i2 is a common point ; which is impossible unless a and fi touch. Hence a' does not coincide with a. Hence a' coincides with /3. Hence o and /3 are reciprocal for F. If two or more of the common points of a and )3 coincide, this may be taken as the limit of a case when no two coin- cide ; and the proposition still holds. L 2 148 Poles and Polars. Reciprocation. Note that there are four base conies. For we may take any one of the four common tangents as the reciprocal of P. Then as the conies are reciprocal, each of the common points ■will have, as polar, one of the common tangents. The above construction is imaginary unless the conies have a real common point and also a real common tangent. Ex. The cross ratio 0/ the four ccmnum points qf two amies for one of the conies is egwd to the ooss ratio of the four eammon tangents for the other conic. 5. 'Reciprocate — a segment divided in a given ratio. Let .^C be divided in B. Let I be the line AB and i the line at infinity, and let ii be the meet of I and i. The reci- procals of the points ABGil on the line I are the lines abcta through the point L. Also the reciprocal of i is the centre of the base conic. Hence AB -*• BC= —{AC, BQ) of the given range of points = —{ac, 6 a)) of the reciprocal pencil, where oj is the join of L to 0. As a particular case the middle point of a segment AC reel- procates into the fourth harmonic for a and c of the join ofac to the centre of the base conic. Ex. Seciprocate the theorem — ' The locus of the centres of conies inscribed in a given quadrilateral is a line which bisects each of the three diagonals.' CHAPTEE XIV. PHOPEETIES OF TWO TEIANGLES. 1. If the vertices of two triangles lie on a conic, the sides touch a conic ; and conversely. Let the vertices ABC, A'B'C of the two triangles lie on a conic. Let AB, AC meet B'C in L, M ; let A'B', A'C meet BC in L', M'. Then {C'LMB') = A {C'BCB^ = A' (C'BCB') = {M'BCL'). Hence the six lines CM', LB, MC, B'L', B'C, BC touch a conic ; i.e. C'A', AB, AC, B'A', B'C, BC touch a conic ; i.e. the sides of the triangles touch a conic. Let the sides touch a conic. Then A (C'BCB') = {C'LMB') = {M'BCL') = A'{C'BCB'). Hence the six points C, B, C, B', A, .d' lie on a conic ; i.e. the vertices lie on a conic. IiX. 1. If turn triangles circumscribe the same conic, then a conic drawn through fifx of Hie vortices teillpass through the sixth also. Ex. 2. ^f two triangles he inscribed in the same conic, then a conic drawn to touch five of the sides mil touch the sixth also. Ex. 3. Iftu>o conies be such that one triangle can be drawn which is circum- scribed to one conic cmd inscribed in the other, then an infinite number qf such triangles can be drawn. For suppose ABC to be circumscribed to and inscribud in y. Draw any tangent to /3 cutting 7 in S' and C From ff and C draw the other tangents to meeting in J.'. Then, since ABC, A'B/Cf are circumscribed to 0, the vertices ABCA'B'ff lie on one conic ; hence A' lies on 7. Hence A'B'Cf satisfies the required conditions. 1 50 Properties of two Triangles. [ch. Ex. 4. IJ BC 6e (he points of contact of tangents from A, and B'C' he the points of contact of tangents from A' to a conic ; show that the triangles ABC, A'B'C are inscriptible in a conic, and circumscriptible to a comic. Let AB, AC cut B'C in L, M ; let A'B', A'Cf cut BC in i', M'. Then (LB'CM) of poles = A'{BL'M'C) of polars. Hence (LB' CM) = {BL'M'C). Hence the triangles are circumscriptible, and therefore inscriptible. Ex. 5. If Obe t)ie centre of the conic circumscribing ABC, A'B'Cf {of Ex. 4), and if BC and B'C meet in I), show that DO bisects AA'. For D is the pole of AA' for the new conic as well as for the given conic. Ex. 6. A conic is drawn through a fixed point A and through the points of contact B, C of tangents from A to a circle, so as to touch the circle at a variable point P. Show that the curvatures of all the conies at the points P are equal. In Ex. 4 let A'B'C coincide in P. Then the circle of curvature of the conic at P is the circum-circle of A'B'C, whose radius is one- half of that of the given circle. Ex. 7. Through a point Oona conic is drawm a line cutting the conic in p and tlie sides of an inscribed triangle in a, b, c ; show that (abq)} is constant. Draw another line a'b'c'p' and consider the triangles ABC, Opp'. 2. If two triangles le self-conjugate for a conic, the six vertices lie on a conic, and the six sides touch a conic ; conversely, if the six vertices lie on a conic, or if the six sides touch a conic, the triangles are sdf-conjugate for a conic. In the figure of § i, let AUG, A'B'C be self-conjugate for a conic. Then the polar of C is A'B', the polar of L where B'C and AB meet is A'C, the polar of M where B'C and AC meet is A'B, and the polar of B' is A'C Hence (fl'LMB') = A' (B'CBC) = (L'CBM') = (M'BCL'). Hence the six sides CM', LB, MC, B'L', B'C, BC touch a conic ; and hence the six vertices lie on a conic. If the two triangles are inscriptible in a conic y, describe by XXV. 12 a conic a such that ABCis self-conjugate for a, and that A' is the pole of B'C for a. Let the polar of B' for o cut B'C in C". Then ABC and A'B'C" are self-con- jugate for a ; hence ABCA'B'C" lie on a conic. But this conic is y, for the points ABCA'B' lie on both conies. Hence B'C' culs y in three points unless C and C" coincide. Hence C and C" coincide. Hence ABC, A'B'C are self- conjugate for a conic, viz. for the conic a. If the two triangles are circumscribed to a conic, they are also inscribed in a conic, and the above proof applies. XIV.] Properties of two Triangles. 151 Ex. 1. Ifima triangles be self-conjugate for a conic a, then a conic j8 drawn to touch five of the sides mil touch the sixth also, and a conic y drawn to pass through five of the vertices will pass through the sixth also; and 7 and are reciprocal for a. Ex. 2. Through the centre of a conic and the vertices of a triangle self-conjugate for the cotiic can be drawn a hyperbola with its asymptotes parallel to any pair of conjugate diameters of the conic. For, adding the line at infinity, we have two self-conjugate triangles. Ex. 3. If two conies be such that one triangle can be circumscribed to one conic which is self -conjugate for tJie other conic, then an infinite number of such triangles can be drawn. Let ABC be the given triangle touching conic and self-conjugate for conic a. Take any tangent B'O of /3, and take its pole A' for a ; draw from A' one tangent A'S to ^, and take C, the pole oiA'B' for a. Then, since ABC, A'B'Cf are self-conjugate for a, the sides touch a conic. But five sides touch /3 ; hence the sixth side OA' touches 0. Hence A'BfCf satisfies the required conditions. Ex. 4. If two conies be such that one triangle can be inscribed in one conic which is self-conjugate for the other conic, then an infinite number of such triangles can be drawn. Ex. 5. An infinite number of triangles can be described having the same cir- cumscribing, nine-point, and polar circles as a given triangle. For the nine-point circle is given when the circum-circle and the polar circle are given, being half the circum-circle, taking the ortho- centre as centre of similitude. Ex. 6. Gaskin's theorem. The circum-circle of any triaiigle self-conjugate for a conic is orthogonal to the director circle of the conic. (See also XXIII. 5, Ex. 9.) Let the two circles meet in T. Let the polar PP" of 2" for the conic meet the circum-circle in QQ'. Then, as in Ex. 4, since T is the pole of QQ', it follows that TQQ" is a self-conjugate triangle for the conic. Hence QQ' are conjugate points for the conic ; hence if CT meet PP' in V, we ha«e VQ. rQ' = rP', for V bisects PP'. Also PTf is a right angle. Hence VQ . VQ' = VT' ; i.e. CT touches the circum-circle. Hence the circles are orthogonal. Ex. 7. Tim conies $ and a are such that triangles can be circumscribed to which are self-conjugate for a ; find the locus of the point from which the pairs of tangents to a and H are harmonic. From P, any point on the locus, draw tangents PT and PT to 0. These tangents are conjugate for a, for they are harmonic for the tangents to a. Hence the pole of PT, viz. Q, lies on PT', and the pole of PP, viz. B, lies on PT. Hence the triangle PQR is self-conjugate for a. Let ABC be a triangle self-conjugate for a and circumscribed to 0. Then since the two triangles ABC, PQR are self-conjugate for the same conic, their sides touch a conic, i. e. QS always touches 0. Hence P, the pole of QB for a, always lies on the reciprocal of for a. Ex. 8. If two conies 7 and a are such that triangles can be inscribed in 7 which are self-conjugate for a, find the envelope of a line which cuts a and 7 in pairs of hairrumic points. 152 Properties of two Triangles. [ch. Ex. 8. XfQ and S be the poinfs of contact of the tangents from P to any conic a, and any conic 7 he dravm to pass through P and to touch QR at Q, then triangles can be inscribed in 7 which are seif-conjugate for a. For PQQ is such a triangle, QQ being QB. Ex. 10. If Q and B be the points of contact of the tangents from P to any conic a, and any conic P be drawn to touch PQ at P and to touch QB, then triangles can be circumscribed to which are self-conjugate for a. For PQQ is such a triangle, QQ being QB. Ex. 11. Jf triangles can be circumscribed to P which are self -conjugate for a, then triangles can be inscribed in a which are self-conjugate for ; and cot^- For we can reciprocate a into 0. Ex. 12. The triangle ABC is inscribed in the conic a, and the triangle DEF is self-conjugate for a. Show that a conic fi can be found such that DEF is circum- scribed to P and ABC is self-conjugate for 0. Viz. that conic inscribed in DEF for which A is the pole of BC. Ex. 13. The centre of the circle circumscribing a triangle which is silf-conjugate for a parabola is on the directrix. Consider the triangle 00.0.' where OCl, 00,' are the tangents to the parabola from the centre of the circle. Ex. 14. The conic a is drawn touching the lines PQ, PB at Q, B ; the conic is drawn touching the lines QP, QB at P, B; show that (i) triangles can be inscribed in a which are self-conjugate for 0, (ii) triangles can be inscribed in which are self-conjugate for a, (iii) triangles can be circumscribed to a which are self-conjugate for 0, (iv) triangles can be circumscribed to which are self- conjugate for a, (t) triangles can be inscribed in a. and circumscribed to 0, (vi) triangles can be inscribed in and circumscribed to a. On BP and BQ take L, L' consecutive to JJ ; on PB, QB take U, W conseeutire to P, Q ; on QP, PQ take N, N' consecutive to Q, P. Then consider the triangles (i) QBL, (ii; PBL', (iii) QPM, (iv) PQM', (v) BQN, (vi) BPN'. Ex. 15. If a triangle can be drawn inscribed ina and circumscribed to and also a triangle self-conjugate for a and circumscribed to 0, then the conies a and are related as in Ex. 14. At B, one of the meets of a and 0, draw BQ touching and meeting a again in Q ; draw the tangent at Q, and on it take N consecutive to Q. Then by the first datum QN touches 0, at P say. Then by the second datum QB is the polar of P for a, i. e. PR touches a at R. Similarly many other converses of Ex. 14 can be proved. Ex. 16. The centre of a circle touching the sides of a triangle self-conjugate for a rectangular hyperbola is on the r. h. For triangles can be inscribed in the r. h. which are self-conjugate for the circle. Now one triangle self-conjugate for the circle is Ono', and two of its vertices nn' lie (at infinity) on the r. h. ; hence 0, the centre of the circle, lies on the r. h. Ex. 17. Given u triangle self-conjugaie for a r.h,, we know four points on the r. h. Viz, the centres of the touching circles. XIV.] Properties of two Triangles. 153 Ex. 18. Qiven a sdf-conjvgale triangle of a conic and a point on the director, sftoM) that Jour tangents are knoum, viz. the directrices of the four conies which can le drawn to circumscribe the triangle and to have the point as corresportding focus. Beciprocate for the point. Sx. 19. The necessary and sufficient condition that triangles can he dreum- scribed to a circle which are self-conjugate for a r. h. is that the centre of the circle shall he on the r. h. Ex. 20. An instance of Ex. 14 is a rectangular hyperbola which passes through the vertices of a triangle and also through the centre of a circle toiKhing the sides. This follows from Ex. 15 and Ex. 19. Ex. 21. If two conies $ and 7 be so situated that one triangle can be circum- scribed to P so as to be iTucribed in 7, then an infinite number of such triangles can he draimi, and all of these mil be self-conjugate for a third conic a ; also the two conies and 7 are reciprocal for a. The first part has been proved. To prove the third part, notice that ABC, A'S'ff are self-conjugate for a conic a. Define 7 by ABCA'B' ; then since the polars of these points for a, viz. BC, CA, AB, SC, Of A' touch /3, it follows that fi is the reciprocal of 7 for a. Again, take any point A" on 7, and let J!" be one of the points in which the polar of A" for a (which touches /3) cuts 7. Let the polar of if" for a (which touclies 3 and passes through A"^ cut the polar of A" in 0". Then the triangle A"B"C" is self-conjugate for a. Hence, since two sides touch 3 and two vertices are on 7, it is circumscribed to P and inscribed in 7. Ex. 22. Prove by this article that * The ortfiocentre of a triangle iriscribed in a rectangular hyperbola lies on the r. h.* The given triangle and the triangle formed by the orthocentre and the points at infinity on the r. h. are self-conjugate for the polar circle. 3. The two triangles ABC, A'B'C are said to be reciprocal for a conic if A be the pole of B'C, B of CA', C of A'B', A' of BC, B' of CA and C of AB for the conic. Two triangles which are reciprocal for a conic are homologous ; and conversely, if two triangles be homologous they are reciprocal for a conic. Let the triangles ABC, A'B'C be reciprocal for a conic ; then they are homologous. For let BC and B'C meet in U, and let AA' meet BC in L and B'C in L'. Then the polar of B is A'C, the polar of C is A'B', the polar of U where BC and B'C meet is A' A, the polar of L where BC and A' A meet is A'U. Hence (LBCU) of poles = A' {UCB'L'). Hence (LBCU) = (L'B'CU) ; hence the ranges (LBCU) and {L'B'CU) are in perspective. Hence LL', BB', CC 1 54 Properties of two Triangles. [CH. meet in a point, i. e. the triangles ABC, A'^C' are homo- logous. Let the triangles ABC, A'B'C be homologous, then they are reciprocal for a conic. For let BG and A'C meet in M.. By XXV. 1 2 describe a conic such that the triangle A'BM is self-conjugate for it, and that A is the pole of B'C. Then A' is the pole of BC, B is the pole of A'C, and A is the pole of B'C. Hence C is the pole of AB. Now let the polar of C cut C'B' in B". Then the triangles ABC and A'B"C' are reciprocal and therefore homologous. Hence AA', BB", CC meet in a point. But AA', BB', CC meet in a point. Hence B' and B" coincide, i. e. the triangles ABC, A'B'C are reciprocal for the above conic. Given a triangle ABC and a conic a, we can describe the reciprocal triangle A'B'C, and then determine the centre and axis s of perspective of the triangles ABC, A'B'C. It is convenient to call the pole and s the polar of the triangle ABC for the conic a. Ex. 1. If two triangles be reciprocal for a conic, show that the centre of homology of the triangles is the pole of the axis of homology for this conic. Ex. 2. BC, CA, AB meet any conic in XX', TY', ZZ', and the conic meets AX again in L, AX' in V, BY in M, BY' in W , CZ in N, CZ' in N'. Show that UJ, MM', NN', meet BC, CA, AB on a line. Viz. on the axis of homology of ABC and its reciprocal for the conic. XIV.] Properties of two Triangles. 155 Ex. 3. Any triangle inscribed in a conic and the triangle firmed by the tangents at the vertices are homologous. Ex. 4. Hesse's theorem. If the opposite vertices AA' and the apposite vertices BB' of a complete quadrilateral be conjugaie for the same conic, then the opposite vertices CC^ are also conjugatefor this conic. ^See also XX. i, Ex. ii.) Let the triangle reciprocal to the triangle ABC for the conic be PQR. Then QR passes through A', since A and A' are conjugate. So RP passes through B'. Hence PQ passes through C, ; for the triangles ABC and PQR are homologous. Hence C and C are conjugate. Ex. 5. If two pairs of opposite sides of a complete (faadrangle be conjugatefor the same conic, then the third pair is also conjugatefor this conic. Ex. 6. The points PP', QQf, BR' divide harmonicaily the diagonals AA', BB', CC of a quadrilateral ; show that the six points P, F', Q, Q', R, M' lie on a conic. CHAPTEE XV. pascal's theobem and beianchon's theorem. Pascal's Theorem. 1. The meets of opposite sides of a hexagon {six-point) inscribed in a conic are collinear. Let the six points be A, B, C, D, E, F. Let the opposite sides AB, BE meet in M, and the opposite sides BC, EF meet in N. Let AF meet MB in G, and let CB meet NF in H. Then we have to show that MN, FG, HB are con- current. This is true i£{EMGB) = (ENFH), for the ranges, having a common point, will be in perspective ; i.e. if A (EBFB) = G(EBFB\ which is true. Hence the meet M of AB, BE, the meet N of BC, EF, and the meet L of CB, FA are collinear. Pascal's Theorem and Brianckon's Theorem. 157 Conversely, if the meets of opposite sides of a hexagon {six- pomt) are collinear, the six vertices lie on a conic. For if LMN are collinear, we have {EMGD) = (ENFH). Hence A (EBFB) = C{EBFD). Hence A, B, C, B, E, Flie on the same conic. The line LMN is called the Pascal line of the six-point ABCDEF. Observe that for every different order of the points A, B, C, JD, E, F we get a different Pascal line. Notice that if two consecutive points, e.g. JSand C, coincide, the side BG becomes the tangent at B or C. Ex. 1. If AD, BE, CF meet in a point, the Pascal line is the polar of this point. Ex. 2. The triangles ABC, A'B'C are homologous. BC meets A'B' in Y and A'ff in Zf, CA meets B'C in Z and B'A' in X', and AB meets CA' in X and CB'.in Y'. Show that BY.BZ'.CZ.CX'.AX.AY'= CY. CZ'.AZ.AX'.BX.BY'. For XY'ZX'TZ' lie on a conic. Ex. 3. In every hexagon inscribed in a conic, the two triangles formed by taking aUemate sides are homologous. Ex. 4. Six points on a conic determine 60 hexagons inscribed in the amie. Ex. 5. The 60 Pascal lines bdonging to six given points on a conic intersect three by three. Let the homologous triangles of any one hexagon be STYZ, X'Y'Z'. Then XX', TT', ZZ' meet in a point. Also XX' is the Pascal line of CDEBAF, TT' oiABCFED, ZZ' of BCDAFE. Ex. 6. Two triangles are inscribed in a conic. The sides of the one meet the sides of the other in nine points. Show that any join of two of these nine points is a Pascal line of the six vertices of the triangles, unless it is one of the sides of the triangles. Ex. 7. ABC is a ttiangle inscribed in a circle. P is any point on this circle. A perpendicular at P to PA meets BC in D, to PB meets CA in B, and to PC meets AB in F. Shaw that DBF is a line passing through the centre of the circle. Call the centre of the circle 0. Let PD, PE meet the circle in A', B'. Then AA'PB'BC proves that ODE are concurrent. Ex. 8. Reciprocate Ex. 7, (i)for t?te circle itself, (ii) for any circle. Ex. 9. IfAOA', BOB', COC^, POP" be chords of a conic, show that the meets qfPA, B'C, ofPB, CA', of PC, A'B', of P'A', BC, of PB/, CA and ofP'C, AB aii lie on the same line through 0. Use (BCCTP'A'A), {ffCCPAA'), (^BAPP'CB'). Ex. 10. Taking the conic as a circle and as its centre, deduce by recipro- cating far P the theorem — The orthocentre of a triangle about a parttMa is on the directrix. 158 Pascal's Theorem and [ch. Ex. 11. A, B, C, B, E are any five points. BA, BCmeet in A' ; AS, CD meet in : BC, DE meet in C ; CD, EA meet in Hf ; DE, AB meet in E' ; and AD, BC meet in F. Show that FBf tauthes the conic through A'B'C'D'E'. Ex. 12. AA', BB', C(f are the diagonals of a complete quadrilateral, A'B'C being colUnear points. AO meets BC in M, CO meets AB in L, LM meets B'C in N and AC in P. If PB and ON meet in R, show that R is the remaining intersection of the conies OBB'AA' and OBB'CC, and that OR is the tangent at to the conic OCOAA'. Consider the hexagons 0RBA'B!A, ORBffB'C, and OOCA'CA. Ex. 13. ABC, A'B'Cf are coaxal triangles ; AC and A'B' meet in P, AB and A'C meet in Q ; show that BCB'ffPQ are on a conic. Ex. 14. The chord QQ' of a conic is parallel to the tangent at P, and the chard PP" is parallel to the tangent at Q ; show that PQ and P'Q' are parallel. Consider PPP'Q'QQ. Ex. 15. The tangents at the vertices of a triangle inscribed in a conic meet the opposite sides in three collinear points. Ex. 16. PQ, PR are chmds of a parabola. PR meets the diameter through Q in V, and PQ meets the diameter through R in XJ j show that UV is parallel to Ike tangent at P. Consider PPRnciQ, where Jl is the point at infinity on the parabola. Ex. 17. Deduce by Reciprocation a property of a cirde. 2. Since Pascal's theorem is true for a hyperbola however near the hyperbola approaches two lines, it is true for two lines, the six points being situated in any manner on the two lines. But each case may be proved as in § i. Ex. 1. If any four-sided figure be divided into two others by a line, the three meets of the internal diagonals are coUinear. Let the four-sided figure ABCD be divided into two others ABFE, EFCD. Now apply Pascal's theorem to ACEBDF. Ex. 2. P, Q, R are fixed points on the sides MN, NL, LM of a triangle. A is taken on MN, AQ meets LM in B, BP meets NL in C, CR meets MN in A', A'Q meets LM in Pi , B'P meets NL in C ; show that ffA passes through R. Consider the hexagon BQCA'PB'. Ex. 3. On the fixed lines LM, MN, NL are taken the fixed points C, A, B. On BC is taken the variable point P ; NP meets CA in Q, and MP meets BA in R. Sluyw ttiat RLQ are coUinear. Consider ACMPNB. 3. IfOQ and OR be the tangents of a conic at Q and R, and ifP he any point on the conic, then PQ and PR cut any line through in points which are conjugate for the conic. Let PQ and PR cut any line through in J' and G. Let FR and GQ meet in U. Consider the six-point PQQUBB. XV.] Brianchoris Theorem. 159 Then since the meets of opposite sides are collinear, the six points lie on a conic. But five points lie on the given conic ; hence the sixth point TJ also lies on the given conic. Hence F and G are two harmonic points of the inscribed quadrangle PQUB. Hence F and G- are conjugate points. Conversely, if any two conjugate points lying on a line through he joined to the points of contact of the tangents from 0, tJien tlie joining lines meet on the conic. Let F and G be conjugate points on a line through 0. Join FQ cutting the conic again in P, and join PR cutting FG in G'. Then F and G' are conjugate, and also F and G. Hence G' coincides yrith Cr ; L e. FQ and GR meet on the conic. So FR and GQ meet on the conic. Ex. 1. IfPV be amjugate points for a central conic, and QQ' be the ends of the diameter which bisects chords parallel to PP" ; show that PQ, P'Ql cut on the conic, and so do P(y, P'Q. Ex. 2. JF/'-R and R' be conjugate points lying on a diameter of a hyperbola, sliow that parallels to the asymptotes through R and R' cut again on the curve. Sx. 3. T?ie diameter bisecting the chord QC/ of a parabola cuts the curve in P, and RR' are points on this diameter equidistant from P; show that the other lines joining QQ^RR^ meet triangles formed by taking aitemate vertices are homologous. Sz. 2. Six tangents to a conic determine 6o hexagons circumscribed to the conic. £iX. 3. The 6o Brianehon points belonging to six given tangents to a conic are coUinear three by three. Beciprocate. £z. 4. The hexagon formed by tite six lines in order obtained by joining aUemate pairs of vertices of a Brianchon hexagon is a Pascal hexagon. For the triangles are coaxal. XV.] Brianckon's Theorem. i6i Hz. 6. 'Rtd'pnmie Ex. 4. Ex. 6. Thne angles have coUinear vertices. Show thai their six legs intersect in twelve other points which can be divided in four ways into a Pascal hexagon and a Brianchon hexagon. Ex. 7. If two triangles he the reciprocals of one another for a conic a, the meets of non-corresponding sides lie on a conic 0, and the joins of non-correspond- ing vertices touch a conic 7 ; and i8 and 7 are reciprocals for a, Jf one iriartgle he inscribed in the other^ the three conies coincide. Ex. 8. Steiner's theorem. The orthocentre of a triangle circumscribing a parabola is on fite directrix. Let ABC be the triangle. Through Z, the meet of BC and the direc- trix, draw the other tangent Zn where n is at infinity. Through Z', the meet of CA and the directrix, draw the other tangent Z' Df where n' is at infinity. From the circumscribing six-side ABZnn'Z'A we conclude that ZZ', Bd' and An meet in a point. Now ZZ' is the directrix ; Sn' is a parallel through B to Z'n', i.e. Bn' is the perpen- dicular from B on CA; so .4 O is the perpendicular from A on BC. Hence these two perpendiculars meet on the directrix ; i.e. the ortho- centre is on the directrix. Ex. 9. The orihocentres of the four triangles formed by taking three out of four given lines are coUinear. Ex. 10. ABCDEA is a pentagon circmnscribing a parabola ; show that the parallel through A to CD^ and the paraUel through B to DE meet on CE. Ex. 11. ABCBA is a quadrilateral circumscribing a parabola; show that the parallel through A to CD and the parallel through C to DA meet on the diameter through B. Ex. 12. The lines AB, BC, CD, DA touch a conic in L, M, N, B ; shmo that AC, BD, LN, MR are concurrent. Consider ALBCNDA and ABMCDEA. Ex 13. The lines BC, CA, AB touch a conic oil, M, N; show that AL, BE, CN are ccmcurrent Ex. 14. The line CB'A touches a conic in P, ACB touches in F', B'CA' touches in Q and CBA' in Q'. Show that A'P, AQ meet on CC, and so do A'P, AQf. Ex. 15. If tu>o triangles be inscribed in a conic, their sides touch a conic. Consider the Pascal hexagon ABffA'BfC, and the Brianchon hexagon BC, CA, A'C, CBf, BfA', AB. Ex. 16. If turn triangles be circumscribed to a conic, their vertices lie on a conic. Ex. 17. IfAB and AC touch a conic at B and C, and A'B' and A'C touch the same conic at B! and ff, then ABCA'B'ff lie on a conic and the six sides touch a conic. The proof is like that of Ex. 15. 5. If OQ and OB he the tangents of a conic at Q and R, and if any tangent meet OQ, OB in K, L ; then the joins ofK and L to any point E on QB are conjugate lines. Let LE cut OQ in M, and let KE cut OB in N. Consider the six-side EL, LB, BN, NM, MQ, QK. Since ML, QB, M 1 62 Pascal's Theorem and Brianchons Theorem. KN meet in a point, the six sides touch a conic. But five sides touch the given conic ; hence the sixth side MN also touches the given conic. Hence ML, KN, being two har- monic lines of the circumscribed quadrilateral KLNM, are conjugate lines. Conversely, if through any point E on QR any two conjugate lines be draivn cutting OQ in M, K and OR in L, N, then MN and KL touch the conic. For if KL does not touch, let KL' touch. Then JEL and EL' are both conjugate to EK. Hence L and L' coincide. Hence KL touches ; so MN touches. £z. 1. Parallel to a diameter of a conic are drawn a pair of conjugate lines ; shmo that the diagonals of the parallelogram formed by these lines and the tangents at the ends of the diameter touch the conic. SiX. 2. Ttm paraUel lines which are conjugate for a hyperlola meet the c^mptotes in points such that the other lines joining them touch the curve. Ex. 3. Jfthe tangents of a parabola at P and Q cut in T, and on the diameter through P there be taken any point R ; show that BT is conjugate to the parallel through R to the tangent at Q. Ex. 4. TJirough a point on the chord of contact PQ of the tangents from T to a parabola are drawn parallels to TP and TQ meeting TQ and TP in R and U; show that RU touches the parabola. CHAPTEE XVI. HOMOGEAPHIC EANGES ON A CONIC. 1. Two systems of points ABC... and A'B'C ... on a conic are said to be homographic ranges on the conic when the pencils P(ABC...) and Q (A'B'C...) are homographic, P and Q being points on the conic. Hence two ranges on a conic which are homographic subtend, at any points on the conic, pencils which are homographic. To construct homographic ranges on a conic, take two homographic pencils at points P and Q on the conic ; the rays of these pencils will determine on the conic two homo- graphic ranges. Given one of these pencils, three rays of the other pencil may be taken arbitrarily. Hence given a range of points on a conic, in constructing a homographic range on the conic, three points may be taken arbitrarily. 2. If (ABC. . . ) and (A'B'C. .■)le two Jiomographic ranges on a conic, then the meet ofAB' and A'B, of BC and B'G, and generally ofPQ' and P'Q, where PP', QQ' are any two pairs of corresponding points, all lie on a line (called the homographic axis). First consider all the meets which belong to A and A'. These all lie on a line. For A(A'B'C...) = A'(ABC.). Hence aU the meets (AB' ; A'B), (AC; A'C),(AB';A'D), ...lie on an axis. So all the meets which belong to B and 5' lie on an axis. So for CC, BU, .... M 2 164 Homographic Ratiges on a Conic. [ch. We have now to prove that all these axes are the same. The inscribed six-point AB'CA'BC shows that the meets {AB'; A'B), {B'C; BC), {CA'; C'A) are collinear. Now {AB'; A'B) and {CA'; C'A.) determine the axis of ^^'; so (AB'; A'B) and {B'C; BC) determine the axis of BB'. Hence the axes of AA' and BB' coincide ; i.e. every two axes, and therefore all the axes, coincide. Hence all the cross meets {PQ' ; P'Q) lie on the same line. 3. Given three pairs of corresponding points ABC, A' B'C of two homographic ranges on a conic, to construct the point D' corresponding to D. The meets {AB'; A'B) and {AC; A'C) give the homo- graphic axis ; and we knowthat{jiZ)'; A'D)is on the homo- graphic axis. Hence the construction — Let A'B cut the homographic axis in d, join A&, cutting the conic again in the required point 2)'. 4. Two homographic ranges on a conic have two common points, vie. the points where the homographic axis cuts the conic. Let the homographic axis cut the conic in X and T. To get the point X' corresponding to X, we join .4' to X cutting XT in X and then join AX cutting the conic again in X'. Hence X' is X. So Y' is T. And there can be no common point other than X and T. For if Z>and 2J' coincide, then each coincides with 8. Hence D, D' and 8 must be at X or T. 5. Reciprocally, two homographic sets of tangents to a conic can be formed by dividing two tangents homographicdUy in ABC... and A'B'C...; then the second tangents from ABC... unit form a set of tangents homographic with the second tangents from A'B'C... For any tangent will cut the two sets in homographic ranges. Again, all the cross joins mil pass through a point called tlic homographic pole ; and the tangents from the homographic pole XVI.] Homographic Ranges on a Conic. 165 wiU he the sdf-corresponding lines in tlie two sets of homographic tcmgents. This follows by Eeciproeation from the previous articles. Ex. 1. The points of contact of two homographic sets of tangents are homo- graphic ranges; and conversely, the tangents at points of tvjo homographic ranges on a conic farm homographic sets of tangents. Ex. 2. If 0(/ be fixed points an a conic and AA' variable points on the conic, such that {00', AA') is constant; show that A and A' generate homo- graphic ranges on the conic of which and 0' are the common points. Ex. 3. If the lines joining a fixed point P on a conic to the corresponding points AA' of two homographic ranges on the conic cut the homographic axis in aa', show that aa' generate homographic ranges, and that the ranges obtained by varying P are identical. For (Jry, aa') is constant and independent of the position of P on the conic. Ex. 4. A conic is drawn through the common points of two homographic ranges AB..., A'B' ... on the same line. P is any point on the conic, and PA, PA' cut the conic again in a, a'. Show that aa' generate homographic ranges on the conic, and thai the ranges obtained by varying P are identical. Ex. 6. Beeiprocate Examples 3 arul 4. Ex. 6. The pencils A {PQB...) and A' (PQR...) are homographic. A line meets AP in p, A'P in pf, and so on. Show that there are tivo positions of the line such that pp' = qg' = rr' •= Viz. the asymptotes of the conic through AA'PQB. . . . Ex. 7. The joins of corresponding points of two homographic ranges on a conic touch a conic having double contact with the given conic at tite common points of the given ranges. Let AA' cut T7 in L, the tangent at X in a, and the tangent at Y in a' ; let BB' cut X7 in M, the tangent at X in b, and the tangent at Yin v. LetAB',A'Be\itJYmK. Then {ALA'a) = jr(,ALA'a) = {AYA'^) = {BYB'JTj [since X, Y are the common points] = Y{BYB'X) = (Bb'B'M) = {BfMBV). Now Aff, LM and A'B meet in K. Hence aV passes through K. So a'b passes through K. Hence J[Y, aV, a'b are concurrent. Hence, by Brianchon, a conic touching the conic at X and at Y and touching ^1.4' will also touch BB', and similarly Cff, etc. (See also XXIX. 10.) 6. Given a conic and a ruler, construct the common points of two homographic ranges on the same line. Let the ranges be ABC . ■ and A'B'C . ■ . Take any point p on the conic, and let pA, pA', pB, pB', ... cut the conic again in a, a', b, 6', ... . The ranges abc... and a'b'c'... on the conic are homographic ; for (dbc.) =p{abc...) = {ABC.) = (A'B'C...) = p {A'B'C'...) = {a'b'c'...). 1 66 Homographic Ranges on a Conic. Now determine the homographic axis of the ranges (aibc...^ and (a'ft'c'...) by connecting the cross meets (flV ; a'h), etc. ; and let this axis cut the conic in x and y. Then iipx and py cut AB in X and Y, X and Y are the common points of the ranges ABG... a,nd A'B'C... . For {XYABC.) =p (XYABC.) = (xydbc.) = (ajya'ftV...) = p (xya'b'c'...) = (XYA'B'C...) ; i.e. XY correspond to themselves in the ranges ABC... and A'B'C... &iven a conic and a ruler, construct the common rays of two homographic pencils having the same vertex. Join the vertex to the common points of the ranges deter- mined by the pencils on any line. CHAPTEE XVII. EANGES IN INVOLUTION. 1. Ip we take pairs of corresponding points, viz. AA', BB', CC, DD', EE', ... on a line, such that a cross ratio of any four of these points (say AB', C'E) is equal to the corre- sponding cross ratio of the corresponding points (viz. A'D, CE'), then the pairs of points AA', BB', CC, ... are said to be in involution or to form an involution range. Or more briefly — If the ranges {AA'BB'CC'...) and (A'AB'BC'C ...) are homographic, then the pairs of points AA', BB', CC, . . . are in involution. To avoid the use of the vague word ' conjugate ' let us call each of a pair of corresponding points, AA' say, the mate of the other, so that A is the mate of A' and A' is the mate of A. Let us call AA' together a pair of the involution. There is no good notation for involution. The notation we have used above implies that A and B are related to one another in a way in which A and B' are not related ; and this is not true. If we use the notation AB, CD, EF, ... for pairs of points in involution, this objection disappears ; but there is now nothing to tell us that A and B are corre- sponding points. 2. The following is the fundamental proposition in the subject and enables us to recognise a range in involution. If two homographic ranges, viz. {AA'BCD ...)and {A'AB'C'D'. ..), be such that to one point A corresponds the same point, vie. A', 1 68 Ranges in Involution. [ch. whichever txmge A is supposed to belong to, the same is true of every other point, and the pairs of corresponding points A A' , BB', CC, Biy, ... are in involution. We have to prove that {AA'BB'CC'BU...) = {A'AB'BC'Ciri)...), given that (AA'BCB ...) = (A'AB'C'B'. . .). Now if P be considered to belong to the first range, its mate P' in the second range is determined by the equation {AA'BP) = (A'AB'P'). Let P be ff, then the mate P' of B' is given by the equation (AA'BB') = {A'AB'P'). Now we have identically {AA'BB') = (A'AB'B). Hence P' is B. Hence B has the same mate, viz. B', whichever range it is considered to belong to. Again, we may consider the homography to be determined by the equation {AA'CP) = (A'AC'P') ; hence, as before, Ohas the same mate in both ranges. Similarly every point has the same mate in both ranges, i.e. {AA'BB'CC'...) = (A'AB'BC'C...). The commonest case of this proposition is — If {AA'BC) = (A'AB'C) ; then AA', BB', CC are in involution. Two pairs of points determine an involution. For the pairs of points PP' which satisfy the relation (AA'BP) = {A'AB'P^ are in involution. Ex. 1. If [CB, AA') and {CfBf, AA') be harmonic, then {AA', BB', C(f) are in involution. Ex. 2. ^ (JCA, A'BT) = {AB, A'C) = -i, then {AA', BBf, CCT) is an involution. Ex. 3. If {AA', EC) = {BBT, CA) = {CC, AB) = -i, show that (A'A, BTCT) = {BfB, CA') = {CC, A'Bf) = -i, and that {AA', Bff, ffC), {BB', A(f, A'C) and {CC, AB', A'B) are involu- tions. Project the range so that A goes to infinity. Ex. 4. If {AB, XX*) = {CB, ZX'), where A, B, C, D are fixed points m the same line, Oien X and X' generate homographic ranges. For {AB, XX') = {DC, X'X), hence {AD, BC, XX') is an involution. Hence {ADBX) = {DACX'). Ex. 6. ABC and A'BfC are homologous triangles. BC and B'C meet in X, XVII.] Ranges in Involution. 169 CA and CA' meet in Y, and AB and A'B' meet in Z. OAA', OBB', OCC meet the line 2:YZ in JC , Y', Z'. Show that {XX', YY' , ZZ') is an involution. For {XX'Y'Z') = 0(XABG) = A (XOBO) = (XX'ZY) = {X'XYZ). 3. To construct mth the ruler only the mate of a given point in a given involution. Let the involution be determined by the two pairs AA', BB'. Take any vertex F, and let VA, VA', YB, VB', &c. cut any line in a, a', b, V, &c. Then the ranges AA'BB"... and a'db'b... are homographic ; iQi{a'ab'b...)={A'AB'B...) by projection tlirough 7 =(^^'£5'...) by invo- lution. Construct the homographic axis A;* of these ranges. We observe that F is on A/ix, being the cross meet (Aa ; A'a'). Take any point X on AA'. Let Xa' cut A^i in f. Let A^ cut aa' in x'. Let Fa/ cut AA' in X'. Then X' is the mate of X in the given involution. For (XAA'BB'...) = (x'a'ab'b...) by the homographic axis = (X'A'AB'B...) by projection through F. Hence (XA A' BB'... ) = (X'A'AB'B...). Hence X' is the mate of Z in the involution. 4. If A A', BB', CC be three pairs of points in involution, the foUomng relations are true, vie. AB'. BC. CA' = - A'B. B'C . CA, AB'. BC. C'A'= - A'B. B'C. CA, AB . B'C. CA'= - A'B'. BC. CA, AB.B'C. CA'=-A'B'. BC. CA. Take any one of the relations, viz. AB.B'C. CA'= - A'B'. BC. CA. This is true i£ AB/BC^AC/i = -A'B'/B'C-^A'C/i, i.e. if AB, BC^AC/CC = A'B'/B'C-^A'C/CC, i.e. if (AC, BC) = (A'C, B'C). And this is true ; hence the relation in question is true. 170 Ranges in Involution. [ch. Similarly the other relations can be proved. Conversely, if any one of these relations be true, then AA', BR, CC are in involution. For suppose AB . B'C. CA'= - A'B'. BC . C'A ; then as above (AC, BC')= {A'C, B'C) ; hence AA', BB', CC are in involution. Remark that, given one of these relations, the othei-s follow at once. For in the definition of involution, there is no distinction made between two corresponding points. Hence in any relation connecting the points, we may inter- change A and A', or B and B', or C and C, or we may make any of these interchanges simultaneously. To obtain the second relation from the first, we inter- change G and C, to obtain the third we interchange B and Bf, to obtain the fourth we interchange B and 5' and C and C simultaneously. Ex. 1. If {AA', BB', CC) be in involution, then {A'A, BC) . {B'B, CA) . {(TC. AB) = - 1. Ex. 2. Circles of a coaxal system whose centres are A, B, C touch the sides of a triangle LMN in P, Q, R, and circles of the same system whose centres are A' , B', C pass through the vertices of the triangle; if PQR be a line, then (AA', BB', C(f) is an invcAviiffn, For LS' : LQ' : : A'C : A'B. 5. If AA', BB', CC, ...be in involution, and if any fixed pair of cot responding points UU' be taken as origins, and ifPP' be any variable pair of corresponding points, then UP. UP'^U'P. U'P' is constant. It will be sufficient to prove that UP. UP'^ U'P. U'P'= UA . UA'-r- U'A . U'A', where AA' is a fixed pair of corresponding points. This is true iSPU/UA-^PU'/U'A = P'U'/U'A'^P'U/UA', i.e. if {PA, UU') = (P'A', U'U). And this is true ; hence the relation in question is true. Particular cases of this theorem are — AB . AB'^ A'B . A'B'= AC.AC-i- A'C . A'C, CA . CA'^ CA . C'A'= CD . CZ/h- C'B . CI/. XVII.] Ranges in Involution. 171 Conversely, if TJV he fixed points, and ifPP' be variable points such that UP. UP'-^ U'P. U'P' is constant ; then PP' generate an involution in which UU' are corresponding points. For take any point A and let A' be the position of P' when P is at A. Then UP. UP'-r- U'P. U'P'= UA . UA'^ U'A . U'A'; hence (PA, UU') = (P'A', U'U), i. e. P and P' are corre- sponding points in the involution determined by the two pairs ^4', UU'. 6. In an involution range, if any two of the segments AA', BB', . . . bounded by corresponding points overlap, then every two overlap ; and if any two do not overlap, then no two overlap. For suppose AA' and BB" overlap, then any two others CC and BB' overlap, A B A> B' AB.AB' AC. AC For A'B.A'B'~A'C.A'C' But since A A' and BB' overlap, the sign of AB . AB'^ A'B . A'B' is — . Hence the sign of AC . AC'-^ A'C . A'C is — . Hence AA' and CC overlap ; for if AA' and CC do not overlap, the sign of this expression is + , as we see from the figures — A A' C C A C C A' We have shown that if AA' and BB" overlap, then A A' and CC overlap. Hence, since CC and AA' overlap, it follows that CC and BB^ overlap, i.e. that every two such segments overlap. Conversely, suppose A A' and BB' do not overlap, then CC and BIZ do not overlap ; for if they do overlap, by the first part of the proof it follows that AA' and BB' overlap. 7. The centre of an involution range is the point correspond- ing to the point at infinity. IfObe the centre of the involution of which P and P' are a 172 Ranges in Involution. [ch. pair of corresponding points, then OP . OP' is constant ; and, conversely, if a pair of points P and P' he taken on a line, such that OP . OP' is constant, then P and P' generate an involution range of which is the centre. Let be the centre of the involution range {AA', BB', PP', ...). Then Of being the point at infinity upon the line, we have by definition (OH'AA'BB'PP') = (QfOA'AB'BP'P) ; .: {OQ.',AP) = {QfO,A'P'); .: OAIAQ!^ OP/PQf- QfA'/A'0-^Q.'P'/P'0, and J.fl'=Pii' and il'A'=U.'P'; .: OP. 0P'= OA . OA', -which is constant. Conversely, if OP . OP' be constant, let A' be the position of P' when P is at A. Then we have OP ■ 0P'= OA . OA'. Hence by writing the above steps backward we get (OQ.'AP) = {a'OA'P'), where il' is the point at infinity on the line. Hence P and P' are a pair of corresponding points in the involution determined by (Oil', AA'), i.e. P and P' generate an involu- tion of which is the centre. Ex. 1. JfObe the centre of the invdidion (AA, BB', CCf, ...), sham Vtat AB . AB'-^ A'B . A'BT = AO -k- A'O. To prove this, make the relation projective by introducing infinite segments in such a manner that the same letters occur on each side of the relation. We get AB . AB' ^ A'B . A'B' = AO .Aa'-i- A'O. A'n', and this is a particular case of the theorem AB.ABf-i-A'B.A'Bf = AG . AC-=r A'C .A'(f. XiZ. 2. Shrno that OA : OB : : AB' : BA' ; and deduce three other relations by interchanging corresponding points^ Ex. 3. If a bisect AA' arid bisect BB', show that (a) a.AO.afi = AB . AB' ; (6)4.00.0^ = AB.AB'-^A'B.A'BT ; [c) a . AA' .a0 = AB. AS -A'B . A'B'. For if be origin, then aa! = W. Ex. ^.IfB bisect CC and R' be the mate of R, then RC = RRf. BO. Ex. 6. Any two homographic ranges, whether on the same line or rtot, can be placed in two ways so as to be in invdufion. Viz. by placing / on J' and placing A and A' on the same or opposite sides of I. XVII.] Ranges in Involution. 173 Sx. 6. Of the two involutions one is overlapping and the other not. Ex. 7. Any line through the radical centre of three circles cuts them in a range in invdtttion. 8. A point on the line of an involution range which coincides with its mate is called a double point (or focus) of the involution. An involution range has two, amd only two, double points ; and the segment joining the double points is bisected by the centre and divides the segment joining any pair of corresponding points If AA', PP' be two pairs of corresponding points of an involution whose centre is 0, we have seen that OP.OP'= OA.OA'. Suppose P and P' coincid e in E. Then OE^^ OA . OA', hence OE = ± -/OA . OA'. Hence there are two double points, Ea,nd Fsaj, which are equidistant from 0. Also, since 0E^= 0F^= OA . OA' and O bisects EF, it follows that {AA', EF) is harmonic, Le. EF divides the segment joining any two corresponding points harmonically. Notice that tlie centre is always real, being the mate of the point at infinity. But the double points will be imaginary when OA . OA' is negative, i.e. when lies between A and A'. The dmhU points cannot coincide, for then each coincides with 0, in which case OA . 0A'= 0E''= o ; i. e. ^ or A' coincides with 0, and A' or A is anywhere, i. e. half the points are at and half are indeterminate, i e. the involu- tion is nugatory. 9. The dmible points of am, overlapping involution are im,aginary and those of a non-overlapping involution are real. Take the centre of the involution. Then OA . 0A'= OB. 0B'= ■■■= OE'=OFK Now in an overlapping involution Oil' and AA' over- lap, Le. lies between A and A'. Hence OA . OA' is nega- tive, i.e. OE' and OF' are negative, i.e. E and F are imaginary. 174 Ranges in Involution. [ch. Similarly in a non-overlapping involution, OE^ and OJ" are positive, i.e. JE and T are real. An overlapping involution is sometimes called a nRgaiiw involution and a non-overlapping involution is called aposiUve involution. Ex. \. If E and F divide harmonically the segments AA', Bff, CC, ... , show that {AA', BEl , CCf,...) is an involution. Bisect the segment EF at 0. Then OA.OA'= OB.0B'= .■■ = OFf. Ex. 2. IfE and F he the doublepoints qf{AA', BB', CC,...), show that AB . AB'-^ A'B . A'Br= AF?-h-A'F?. _ _ ,, AB.A^ AE.AF Ex. 3. Also ■ = • A'B. A'B' A'E.A'F Ex. 4. Also AB'. BE . EA' = -A'B ■ B'E . EA. Ex. 5. Also EF'. ae' = AB. Aff. A'B . A'ST. For £f = 2 . 05 = 2e if (jo'=e'. Ex. 6. Also 4.a$.aE = \a/AB . ABf + ^/A'B . A'fff. Ex. 7. If the segments AA', BB',... joining corresponding points have the same middle point, show that AA', BB',... form an involution; and find the centre and double points. a' the point at infinity and E the middle point are harmonic with every segment AA'. Hence a', B are the double points and ft' is the centre. Ex. 8. If AA', BB' be pairs of points in an involution, one of whose double points is cU infinity, then AB = —A'B'. For E the other double point must bisect AA' and BB'. Ex. 9. If any two segments AA', BB' joining corresponding points in an involution have the same middle point, then all such segments have the same middle point. For the other double point must be n'. Ex. 10. IfAF.AP=A'P.A'P. show that the points P arid P farm an involution in which A and A' are corresponding points ; and find the centre and double points. Ex. IL If any transversal through V {the internal vertex qf the harmonic triangle of a quadrilateral circumscribing a conic) cut the sides in AA', BB' and the conic in PP ; show thai (AA', BB', PP') is an involution, the dmiile points being V and the meet of VW with the transversal. Ex. 12. Through a given point draw a line meeting two conies (or two pairs of lines) in points AA', BB' such that {OAA'BPI) = (OA'AB'B). Join to the meet of the polars of 0. Ex. 13. If ABC..., A'B(f... be two homographic ranges on the same Une, and if the mates of P {= Qf) be P' and Q, we know that the ranges A and A' and the ranges P" and Q hare the same common points (E, F say) ; show that P has the same fourth harmonic for P'Q and for EF. (See X. 7. Ex. 4.) XVII.] Ranges in Involution. 175 We have only to prove that P (= Q') is one of the double points of the involution determined by P'Q, BF. Now (PQSfj = {PcyEF) from the first homography = {PPEF) = {PP'FE). Hence PQ, EF, PP are in involution, i.e. P is a double point of the involution. Ex. 14. With the same data, P and the fourth harmonic of P for P'Q generate an involuHofn. 10. A system of coaxal circles is cut by any transversal m pairs of points in involution. For if the transversal cut the circles in AA', BB , CC, ... and the radical axis in 0, then OA . 0A'= OB. 0B'= OC. 0C'=... . Hence AA', BB', CC, ... form an involution of which is the centre. Ex. 1. Give a geometrical construction for the double points of the involution determined an a line by a system of coaxal circles. Ex. 2. A line touches two circles in A and A' and cuts a coaxal cirde in B and B' ; show that (AA', BB') is harmonic. Ex. 3. Of the involution determine by a system of coaxal circles on the line of centres, the limiting points are the double points. Ex. ^. If a line meet three circles in three pairs of points in involution, then either the circles are coaxal or the line passes through their radical centre. Ex. 6. If each of the sides of a triangle meet three circles in pairs of points in involution, the circles are coaxal. Sx. 6. The three cirdes drawn through a given point V, one coaxal with the circles a and 0, one coaxal unth the circles and y, and one coaxal vnth the circles 7 and a, are coaxal. Let two of the circles cut again in V, and consider the involution on rv. Ex. 7. Two circles a and are drawn having the radical axis p with the circle 7, and S and t are drawn huving the radical axis q with 7 ; show that the meets ofaS and of 0( are concydic. Consider the involution on the radical axis of a and S. IL IfEF be the double points of an involution of which A A' and BB' are any two pairs of corresponding points, then {AB', A'B, EF) are in involution, and so are {AB, A'B', EF). For {AB", A'B, EF) are in involution if (ABEF) = {B'A'FE), i.e. = (A'B'EF) ; and this is true, for E corresponds to itself and so does F. Similarly {AB, A'B", EF) are in involution. 176 Ranges in Involution. [ch. Sx. 1. Prme the fdllovnng construction for the double points of the involuUan A A', Sff, CC,... viz. — Take any point P and let the circles through PAS' and PA'B cut in Q ; so let the circles through PAB and PA'B/ cut in R ; then the circle through PQR cuts AA' in the required double points. For if the circle through PQR cut AA' in EF, then from the radical axis PQ we see that {AB', A'B, EF) are in involution ; hence (ABEF) = (B'A'FE) = (A'B'EF). So from the radical axis PR, we get {AB'EF) = (A'BEP). Hence {ABEFB') = {A'B'EFB). Hence EF are the double points of the involution detei-mined by AA' and BB'. Ex. 2. If E and F be the limiling points of the circles on the coUinear segments AA', BB' as diameters, show thai the circles on AB, A'B', and EF as diameters are coaxal. Ex. 3. IfE,F be the common points of the two homographic ranges {ABC...') and {A'B^(f ...), then AB', A'B, EF are in involution. Ex. 4. Prove the foUowing construction for the common points of the two hmiographic ranges (ABC.) and (A'B'C..) — Take any point P and let the circles PAEf and PA'B cut in Q, and let the circles PAC and PA'C cut in R; then the circle PQR wiU cut AA' in the required points. Ex. 5. Oiven two pairs of points AA', Bff of two homographic ranges and one common point, construct the other. 12. If A A', BB', CC ie pairs of points in involution, and if P, Q, B be the middle points of AA', BB', CC, show that A'A\ QB+B'B'. BP+ CC. PQ+ 4PQ. QB.BP = o ; and if U he any point on the same line, then {ATP + A'W) QR+(BIP+B'ZP) BP+{CIP+C'V')PQ = -4PQ. QB.BP. Take the centre of the involution as origin and use abridged notation ; then if OA'=ai, and so on, A'A^= (a—aif= o' + Oi"— 200,= (a+a,)^ — 4aai. But a + Oi= 2p and QB = r—q, and aai= 66,= cCi= A, say; .-. A'A\QB = {4P^-4\){r-q); .: 2{A'A\ QB) = 4 Sj>'(r-2)-4\S(r-2) = -4(q-p)(r-q) (p-r) = -4PQ. QB.BP. Again, if x be the distance of U from the origin AIP= {x-a}\ XVII.] Ranges in Involution. 177 Hence 2 {(^p-« + ^'U-')QiJ} = S{[2ar'-2aj(o+aJ + aHa.''] (r-g)} = 2ar'2(r-2)-a;Sj)(r-g) + 2 {a' + a,^— 2aa,+ 2A}(r— g) = - 4Pe . Q5 . i?P by the former part. Ex. 1. With Oie same notation, show that AB . AB/AC. Aff = PQ/PR. Ex. 2. Also if E he a double point, then A'A^/PE-B'B'/QE = 4PQ. Ex. 3. Also, X being any point on the satne line, XA . XA'. QR + XB. XS .RP+XC. XC. PQ = o. Ex. 4. Also XA . XA' . EF-^ XE'. FP+XF' .PE = 0. Ex. 5. Also XA . XA'-XB . XBf + aPQ .XO = 0. Ex. e. Also RC.PQ = RA. RA'. QR + RB. RBf. RP. Ex. 7. Given two coUinear segments AA', BBI, determine a point C such that CA.CA'-.CB.CB' ■.•.\: i. Determine the point R from the relation RP : RQ : : \ : 1. Through any point V draw the two circles VAA', YBB cutting again in V . Draw the circle through VV, having its centre on the perpen- dicular to AA' through R. This circle will cut AA' in the required points (see Ex. i). 13. Take any point Y on the line of the involution. Then OA = 7A- VO = x-r, say ; so 0A'= af-r. .'. OA. 0A'= constant gives {x- r) (x'—r) = constant Hence the distances of pairs of points in an involution from any point on the line satisfy the relation Icxxf + Z (a;+ a;') + m = o, where Tc, I and n are constants. Conversely, if this relation be satisfied, the pairs of points form an involution. For kxx^+l{x+x^) + n = o can be thrown into the form {x—r) {x' — r) = constant ; which is the same as OA . 0A'= constant. Or thus. If {AA', BB', CC, ...) be in involution, then {AA'BB'CC', ...) is homographic with {A'AB'BC'C, ...). Hence corresponding points in the two ranges are connected by a relation of the form xx' +lx+ muf +n= o. Also, as th ere is no distinction in an involution between A and A', we must N 1 78 Ranges in Involution. [ch. have I = m. Conversely, if aa/ + Z (a; + a;') + w = o, ^ and A' generate homographic ranges in which A and A' are inter- changeable. Hence A and A' generate an involution. Ex. L Shaw that P, P* determine an involution if AP. B'P' + K . AP+ li . B'P' + v = o, provided X— ^ = AB'. Ex. 2. Show Oiat P, P' determine an involulion if B.AP.BP' = AB.PP'; owd that A and B are the double points. Ex. 3. Show that P, P' determine an involution ifAP + B'P'= v ; and that one douNe point is at infinity. Find also the second double paint. 14. If {AA', BB", CO') he pairs of points of an involution, then ^^.-BB'+^^^.B'A+^.AB^o. We have to prove that CA . BB'. C'B'. C'B+ CB.B'A. C'A'. C'B + CB'.AB.C'A'.C'B'=o. This relation is of the first order in A and in .4'. Consider the points X, X' connected by the relation CX . BB". C'B'. CB+CB. B'X . C'X'. C'B + CB'. XB . C'X'. C'B'= o. Beducing to any origin, this relation assumes the form xsf + lx+mx'+n = o. Hence X and X' generate homographic ranges. Now the relation is satisfied by X = (7 and X'= C, and hyX = B and X'= B', and by Z = JB' and X'= B. Hence the homography is that determined by {CBB') - ifl'B'B), Le. is the given involution. Hence the above relation be- tween X and X' is satisfied by any corresponding pair of points of the involution. Hence the relation is satisfied if we replace X, X' by A, A'. T, 1 c. ... . ./ AS.A'C AB'.A'C Ex. L Show that AA' = — — — + — -7-= Joii Jy(j „ „ ., AB.A'C AA'.B'Cr ^^■^■^^ -a^Tmb^^a^Tmg-'- Bx.3. .ibo ^,.Bff-^^.GA~AB. XVII.] Ranges in Involution. 179 ^^.5. Also AB^-^ ^^. Ex. 6. Also, P being any point on the same line, §L^.BB'.FA'^^.B'A.PB'^^.AB.FB^O. Ex.7, ^feo ^.BC.PA' + -^.CA.PB'=AB.PC. Of A (/i» N 2 CHAPTER XVIII. PENCILS IN INVOLUTION. 1. The pencU of lines VA, VA', VB, VB', VC, YG', ... is said to form a pencil in involution if y{AA'BBrCG' ...)=Y{A'AB:BG'G...). Any transversal cuts an involution pencil in an inmlution range ; and, conversely, the pencil joining any involution range to any point is in involution. Let a transversal cut an involution pencil in the pairs of points J.-d.', BB', CC' Then, since V{AA'BB'CC'...) = V{A'AB'BC'C...), we have V(AA'BC) = ViA'AB'C) ; hence (AA'BG) = (A'AB'G'). Hence G, G' are a pair in the involution determined by the pairs AA', BB'. Similarly for any other pair of points in which the transversal is cut by a pair of lines of the invo- lution pencil. Conversely, ii {AA'BB'GG' ...) = {A'A]ffBG'G...), we have {AA'BG) = (A'AB'G') ; hence V{AA'BG) = Y{A'AB'G'). Hence YG, YG' are a pair of rays in the involution pencil determined by Y(AA', BB'). So for any other pair of corresponding rays. Ex. 1. 1/ V he any point on the homographic axis of the two homographic ranges (ABC.) = {A'Wff ...) on different lines; show that r{AA',Blf,CC,...) is an invdlviion pencil. Let jr'r be the mates of the point X(= r') where AB and A'B' meet. Then 7 ia on X'y. Hence r(J[J:'ABC...) = 7{ZTABC...) Pencils in Involution. i8i = ViX'TA'B'ff...) by homography = r(^X^'5'C...). Hence Y{XX', AA', BB', ...) is an involution. Hx. 2. Reciprocate Ex. i. Ex. 3. Two homographic pencils hare their vertices at infinity. Shau> that any line through their homographic pole determines an involution of which the pole is the centre. Ex. 4. Any tvoo homographic pencils can 6e placed in two ways so as to be in involution. Let the pencils be V {ABC ...) = V (A'B'C...). First, superpose the pencils so that V is on V and VA on V'A'. This can be done in two ways. Let FX ( = V'JP) be the other common line of the two pencils r(ABC...) = r(,AB'C...). Then in the original figure AVJ:= A'V'Z'. Second, place V on V and VA on Y'.X' and YJC on V'A'. The two pencils are now in involution ; for VA ( = V'X') has the same mate, viz. V'A' ( = YX) whichever pencil it is supposed to belong to. If the vertices are at infinity, place the pencils so that all the rays are parallel. Let any line now cut them in the homographic ranges (abc...) = (o'6'c'...). Now slide {a'h'c'...) along (o6c...) until the two ranges are in involution (either by Ex. 5. of XVII. 7, or by a construc- tion similar to the above). 2. A pencil of rays in involution has two double rays (i.e. rays each of which coincides with its corresponding ray), and the double rays divide harmonically the angle between every pair of rays. Let any transversal cut the pencil in the involution (AA', BB', CC, ...), and let E, F be the double points of this involution. Then the ray corresponding to VE in the pencil is clearly YE itself. Hence YE is a double ray. So VF is a double ray. Also (AEA'F) is a harmonic range ; hence Y (AEA'F) is a harmonic pencil. Similarly VE, YF divide each of the angles BVB', CYC, ... harmonically. There is nothing in an involution pencil which is analo- gous to the centre of an involution range. In fact the point at infinity in the range AA', BB^, CC, ... w^ill project into a finite point on another transversal, and will project into the mate of this finite point. If, however, Y is at infinity, i.e. if the rays of the pencil are parallel, then all sections of the pencil are similar, and there is a central ray which is the locus of the centres of all the involution ranges determined on transversals. Ex. 1. Xfthe angles AVA', BYB', CYC,... be divided harmonically by the tame pair 0/ lines, the pencil Y {AA', BB', CC,...) is in involution. i82 Pencils in Involution. [ch. Ex. 2. i^ (Ae anglea he bisected by the same line, then the pencU is in involution. Ex. 3. V the douile rays of a pencil in involution be perpendicular, they bisect all the angles bounded by corresponding rays. Ex. 4. If two angles AVA', BYB' hounded by corresponding rays of a pencil in involution have the same bisectors, thm all such angles have the same bisectors. Ex. 5. Find the locus of a point at which every segment (AB) of an in- volution subtends the same angle as the corresponding segment {A'B'). The circle on EF as diameter. Ex. 6. Through any point are drawn chords AA', BBf, CC, ... of a conic 1 show that AA', BB', C(f subtend an invduMon pencil at any point of the polar of 0. Ex. 7. Reciprocate Ex. 6. Ex. 8. If ABA' B/ he four points on a conic, and if through any paint on the external side UW of the harmonic triangle qf ABA'B' tliere he drawn two tangents OT and OP to the conic ; show that {AA', BB', XT') is a pencil in involution, the imMe lines being OU and 07, Ex. 9. Through a fixed point is drawn a variable line to cut the sides of a given triangle in A'B'C ; find the locus of the poirU P such that {PW,A'0') =-i. Now B (AC, BT) = - 1, .-. BB' and BP generate an involution .-. B(P) = B(^) = {B') = 0(,P), . : the locus is a conic through B and 0. 3. 1/ AVA',BVB', CrC, ... le all right angles, then the pencil V{AA', BB', CC, ...) is in involution. We have to show that Y{AA'BB'CC'...) = Y{A'AB'BC'C...). Produce AVio a, BY to b, and so on. Then if we place YA on VA', YA' will fall on Ya, and 80 on. Hence the two pencils Y{AA'BB'...) and Y{A'aB'b...) are superposable and therefore homographic. But Y{A'aB'b...) is homographic with Y(A'AB'B...) ; hence Y(AA'BB'...) and Y{A'AB'B...) are homographic. Otherwise : — From the vertex F drop the perpendicular YO on any transversal AA'BB"... . Then, since AY A' is a right angle, we have YO' = AO . OA'. Hence OA .0A'= OB . 0B'= OC. 0C'= - • XVIII.] Pencils in Involution. 183 Hence {AA', BB', CC, ...) is an involution range. Hence V{AA', BB', CC, ...) is an involution pencil. Ex. 1/ through the centre of an overlapping involution (AA', BB', ...), there he drawn VO perpendicular to AA' and smh that Y(?= AO . OA', then any four points of the involution subtend at Y a pencil superposable to that subtended by their mates. 4. In every involution pencil, there is one pair of correspond- ing rays which is orthogonal ; and if nuyre than one pair he orthogonal, then every pair is orthogonal. (See also XX. 6.) Take any transversal cutting the pencU in the involution (AA', BB', CC, ...). Through the vertex Fdraw the circles VAA', VBB' cutting again in V. Let VV cut AA' in 0. Then OA.OA'=OV.OY'=OB.OB'. Hence is the centre of the involution. Hence OC. 0C'= OA . 0A'= OV. OV. Hence the four points V, V, C, C are concyclic. In this way, we prove that all the circles VAA', VBB', VCC, ... pass through V. Also every circle through VV ■ cuts AA' in a pair of points PP' of the involution ; for OP. 0P'= OV. 0V'= OA . OA'. Let the line bisecting VV at right angles cut AA' in Q. Describe a circle with Q as centre and with QF as radius, cutting AA' in TF'. Then P, P' are a pair in the involution, and PVP' is a right angle. This construction fails in only one case, viz. when VV is perpendicular to AA'. In this case, the orthogonal pair are W and the perpendicular to VV through V. Also if two pairs are orthogonal, every pair is orthogonal. For suppose AVA', BVB' are right angles. Then the centres of the circles AVA' and BVB' are on AA'. Hence AA' bisects W orthogonally. Hence the centres of all the circles AVA', BVB', CVC, ... are on AA'. Hence all the angles AVA', BVB', CVC, ... are right angles. Ex. 1. Show that a giren line VX through the vertex alicays biiects one of the angles AVA', BVB',... of an involution ; and if it bised two of the angles, it bisects aU. IHscitss the case when YX is perpendicular to one of the double rays. Take AA' perpendicular to YX, and take the centre of the circle onKX 184 Pencils in Involution. [ch. Sx. 2. Slum that the pencils r (AA', Eff, Cff, ...)and V {A' A, B'B, CC, ...) 0/ S 4 are superposable. For lAVBf is equal to lA'V'B or its supplement. XiX. 3. Oiven two homographic pencils, we can always find in the first pencil rays VA, VB, and in the second pencil corresponding rays V'A', V'B', such tfcai both A VB and A' VBf are right angles. Can more than one such pair exist 1 Hz. 4. {AA',BBf ,CQf,..^isanin'miutian. Show t?iat the circles PAA', PBB',PCC,..., where P is any point, are coaxal. Ex. 5. Deduce a construction for the mate 0/ a given point in flie involuliom. 'Ex.. 6. Also given AA' and BB', and the middle paint of Cff, construct Cand ff. Ex. 7. Given two segments AA', BB' qf an involution, construct geometrically the centre 0. Ex. 8. Given a segment AA' of an involution and the centre 0, construct the mate of C. Ex. 9. Oiven two involutions (AA', BB', ...) and (aa', 66', ...)(mthe same Une : find two points which correspond to one another in both involutions. Ex. 10. If any two circles be drawn through AA' and BB', their radical axis passes through 0. Ex. 11. If A, A' generate an involution range, and QA be perpendicular to PA and QA' to PA', show that if P be a fixed point, then Q generates a line. For the locus of the centres of the coaxal circles PAA' is a line. 5. Every overlapping pencil in involution can ie projected into an orthogonal involution. Let any transversal cut the pencil in the overlapping involution of points (AA', BB', CC, ...). On AA', BB' as diametei-s describe circles. Since AA', BB' overlap, the circles will cut in two real points U and U'. Now, since in the pencil in involution U{AA', BB', CC, ...) two pairs of rays, viz. UA, UA' and UB, UB', are orthogonal, hence every pair is orthogonal. Eotate U about AA' out of the plane of the paper. With any vertex W on the line joining U to the vertex V of the given pencil, project the given pencil on to any plane parallel to the plane UAA'. Then VA projects into a line parallel to UA, and VA' projects into a line parallel to UA'; hence AVA' projects into a right angle; similarly BVB', CVC, ... project into right angles. xvin.] Pencils in Involution. 185 XjX. 1. (AA'.BB", CC, ...') is an involution. Shou} that the circles on AA', SB", CC, ...as diameters are coaxal. X]z. 2. Show also that AA', BB', CC, ... subtend right angles at two points in the plane. When are tliese points real ? 6. IfP, Q, B 6e the fourth harmonics of the point Xfor tlie segments AA', BB', CC of an involution range, tJien PA^ QB^ QB^ BP BC^ PQ^ PQ.QB.BP _ XA"- ' XP"^ XW' Xq^ XC ' XB"^ XP.XQ.XB~°' Join the points to any vertex V ; and cut the pencil so formed by a transversal aa', bV, cc', p, q, r, x. Then since the given relation is homogeneous in each point, as in proving Carnot's Theorem, we see that the relation is also true of the projections aa', &c. of the given points. Now take aa' parallel to TX. Then a; is at infinity. Hence xc^.xp xa^.xp xa^.xp xV.xq xc-.xr xp . xq. xr Hence the given relation is true if pa^. qr+ qW. rp + rc^.pq+pq . qr . rp = o. But now p, q, r bisect aa', hb', cc'; hence this relation is true by XVII. 12. If in addition to the above notation P^, Qj, i?, bisect AA', BB', CC, show that in Examples i-6 AB.AB' _^ AC. AC ^PQ , XQ ■ XB.XB' ■ XC.JTC ~ PR ' JTR' Ex. 2. AB.AB'-r-AC.AC=PQ. XQ^ ^ PR . Xi?,. For JTQ . XQi = XB . XB', &c. TA YA' YB . YB' ^^•3-S^'-««-^^^xb:xF'-^^-^« YC . YC XC. XC ^ ' Ex. 4. YA . YA'. ^+YB. YB'. ^ + YC. YC. -^ .Axj -^Vl ^-/»i wltere Y is any point on the same line. YA.YA'.^+YB.YB' QR.XP RP.XQ P Q.XR ■ XA.XA' ■*■ XB.XB' * XG.XC ' Take Y at infinity. .„ „ QR RP PQ ^■^■XP,^XQ.*XRr°- 1 86 Pencils in Involution. Ex. 7. IJ {CP, AA') = {CQ, BS') = (PP', BB^ = (W, AA') = (CC,P'' + J . tan JTFP + J . tan A'VP' + n = o, where I and n are constants. Ex. 5. IfVA', VB', rC be three bisectors of the angles BVC, CVA. AVB {either three external, or one external and tmo internal), then V{AA',BB',CC,...) is an involulion. Ex. 6. y VX and VT be fixed lines, and VP, VP' be variable linen satisfying the condition Bin XVP 4- sin YVP = -sin XVP'-i-ain YVP', then VP, VP' generate an involution. CHAPTEE XrX. INVOLUTION OF CONJUGATE POINTS AND LINES. 1. The pairs of points on a line which are conjugate for a conic form an involution. Let I be the line and L its pole. Let A A', BB', CC, ... be the pairs of conjugate points on I. Then the polar of A which lies on I passes through L. Also the polar of A by hypothesis passes through A'. Hence LA' is the polar of .4. So LA is the polar oiA', and so on. Hence {AA'BB'CC ...) of poles = L{A'AB'BC'G...) of polars = {A'AB'BC'C). Hence {AA', BB', CC, ...) form an involution. The double points of the involution of conjugate points on a line are the meets of the line and the conic. For these meets are harmonic with every pair of conjugate points on the line. This affords another proof that conjugate points on a line generate an involution. 2. The pairs of lines through a point which are conjugate for a conic form an involution. Let L be the porat and I its polar. Let LA, LA'; LB, LB'; LC, LC; ... be the pairs of conjugate lines, the points AA', BBf, CC, ... being on I. Then the pole of LA which passes through L is on I. But the pole of LA by hypothesis lies on LA'. Hence the pole of LA is A'; so the pole of LA' is A, and so on. Hence L {AA'BB'CC...) of polars = {A'AB'BC'C.) of poles = L {A'AB'BC'C..). Hence L{AA', BB', CC, ...) form an involution. 1 88 Involution of [ch. The double lines of the involution of conjugate lines through a point are the tangents from the point. For these tangents are harmonic with every pair of con- jugate lines through the point. This alfords another proof that conjugate lines through a point generate an involution. Ex. 1. Through every point can he drawn a pair of lines which are conjugate fiir a given conic and also perpendicular. Ex. 2. Xffioo pairs of conjugate lines at a point are perpendiaiiar, ail pairs are perpendicular. Ex. 3. Given that I is the polar of L, and given that ABC is a self-amjugate triangle, construct the tangents from L. 3. An important case of conjugate lines is conjugate diameters, i. e. conjugate lines at the centre. The double lines of the involution of conjugate diameters are the tan- gents from the centre, i.e. the asymptotes. Ex. 1. Conjugate diameters of a hyperbola do not overlapt and conjugate diameters of an tUipse do overlap. Ex. 2. Throiigh the ends P and J) of conjugate semi-diameters CPj CD of a conic are drawn parallels, meeting the conic again in Q and E ; show that CQ, €E are also conjugate diameters. For if CX bisect FQ and BE, and CT be the diameter conjugate to CX, then CX, CY are the double lines of the involution C{FQ, DE). Hence C (XF, PD, QE) is an involution. Ex. 3. The joins of the ends of two pairs qf conjugate diameters PF', DIf and Q(/, EE' are parallel four by four. Ex. 4. Two of the chords joining the ends of diameters parallel to a pair of tangents are parallel to the chard of contact. Ex. 5. The conjugate diameters CQ, CE cut the tangent at P in S, R'j slune tluURP.PR'= CD'. For P is the centre of the involution determined by the variable conjugate diameters CQ, CE on the tangent at P. Also in the hyperbola the double points are on the asymptotes. Hence RP.PR'-=-Pr' = CI>\ In the ellipse the diagonals of the quadrilateral of tangents at P, P', D, ly give a case of CQ, CE. Hence RP.PBf= Clf. Ex. 6. Parallel tangents of a conic cut the tangent at P in R, Rf; shmn tltat RP.PR'= CD\ Ex. 7. The conjugate diameters CQ, CE cut the tangents at the end of tlie diameter PP" in H, R'; show that PR . P'R' = CD'. XIX.] Conjugate Points and Lines. 189 4. Defining the principal axes of a central conic as a pair of conjugate diameters which are at right angles, we can prove that — The principal axes of a conic are always real. For by XVIII. 4, one real pair of rays of an involution pencil is always orthogonal. A central conic {unless it be a circle) has only one pair of principal axes. For by XVIII. 4, if two pairs of rays of the involution pencil are orthogonal, then every pair is orthogonal, i.e. the conic is a circle. £z. Given the centre of a conic and a self-conjugate triangle, construct the axes and asymptotes. Let be the centre and ABC the triangle. Through draw OA', OB', 0(J parallel to BC, CA, AB. The asymptotes are the double lines, and the axes the orthogonal pair of the involution {AA', BB', CC). 5. The feet of the normals which can be drawn from any point to a central conic are the meets of the given conic, and a certain rectangular hyperbola which has its asymptotes parallel to the axes of the given conic, and which passes through the centre of the given conic, and through the given point. Let be the given point. Take any diameter CP, and let the perpendicular OY on CP cut the conjugate diameter CB in Q. Then, taking several positions of P, &c., C(ft&-) = CID,1),...) = C(P,P, ...) = C{T,Y,...)= 0(Y,T,...)= 0{Q,Q,...). Hence the locus of Q is a conic through C and 0. This conic is a rectangular hyperbola with its asjrmptotes parallel to the axes, as we see by making CP coincide with CA and CB in succession. Now let B be the foot of a normal from to the given conic, then B is on the above rectangular hyperbola ; for, drawing CP perpendicular to OB, or meets CD, i.e. CB, in jB. liZ. 1. The same conic is the locus of points Q sucft that the perpendicular from Q on the polar of Q passes through 0. For QO, being perpendicular to the polar, is perpendicular to the diameter conjugate to CQ. igo Involution of [ch. XiZ. 2. The norma!Ls at the four points where a conic is cut by a rectangular hyperbola which passes through the centre and has its asymptotes parallel to the axes, are concurrent at a point on the rectangular hyperbola. For let the normal at one of the meets R cut the hyperbola again in 0. Ux. 3. Eight lines can be draum from a given point to meet a given central conic at a given angle. Ex. 4. Deduce the corresponding theorems in the case qf a parabola. Here the centre n is on the curve ; hence one of the meets is the point n. Rejecting this, we see that three normals and six obliques can be drawn from any point to a parabola. Ex. 5. If OL, OM, ON, OR be conairrerU normals to a conic, the iangerUs at L, M, N, R touch a parabola which also touches ike axes of the conic and the polar of Of or the conic. Reciprocate for the given conic. Ex. 6. is on the directrix of the parabola. Reciprocate for 0. 6, A common chord of two conies is the line joining two common points of the conies. On a common chord of two conies the involution of conjugate points is the same for each conic, the double points being the common points. Conversely, if two conies have on a line the sam^ involution of conjugate points, this line is a common chord, the double points of the involution being tJie common points of the two conies. Two common chords of two conies which do not cut on the conies may be called a pair of common chords. We know that a pair of common chords meet in a vertex of a common self-conjugate triangle of the two conies. Conversely, every point which has the same polar for two conies is the meet of a pair of common chords. Let E be the point. Join E to any common point A of the two conies. Let EA cut the polar of iJ in P and the conies again in B and B'. Then (EP, AB) is harmonic, and also (EP, AB'). Hence B and B' coincide, ie. EA passes through a second common point. So EC passes through Z>. Hence two conies have only one common self-conjugate tri- angle ; for if VV TF' be a self-conjugate triangle, and UVW the harmonic triangle belonging to the meets of the conies, then IT coincides with U, V, or W, and so on. (See also XXV. 12.) XIX.] Conjugate Points and Lines. 191 If, however, the two conies touch at two points the above proof breaks down, and there is an infinite number of common self-conjugate triangles. Let the conies touch the lines OP and OQ at P and Q. On PQ take any two points VW such that {PQ, VW) = -1. Then OVW ia clearly a common self-conjugate triangle. Notice that if two conies have three-point or four-point con- tact, the common self-conjugate triangle coincides with the point of contact. Ex. L The common chords which pass through one of the vertices of the comTnon seJf-conjiigate triangle of two conies are a pair in the involution deter- mined by the pairs of tangents from this point. UV, VW being the double lines. XjX. 2. Reciprocate Ex. i. Ex. 3. The conic y touches the conic a at the two points L and it, and touches the conic at the two points N and R. Show that two common chords of a and meet at the intersection of LM and NR. 7. A common apex of two conies is the meet of two com- mon tangents of the conies. At a common apex of two conies the involution of conjugate lines is the same for each conic, the double lines being the com- mon tangents. Conversely, if two conies have at a point the same involution of conjugate lines, the point is a common apex, the double lines of the involution being the common tangents of the two conies. The join of a pair of common apexes of two conies has the same pole for both conies. Conversely, if a line have the same pole for two conies, this line is the join of a pair of common apexes of the conies. These results follow by Keciprocation. 8. Since two conies have only one common self-conjugate triangle, it follows that the harmonic triangle of the quadrangle of common points coincides with the harmonic triangle of the quadrilateral of common tangents. Let UVW be the harmonic triangle of the quadrangle formed by the common points a, b, c, d, and let A A', BB', CC be the opposite vertices of the quadrilateral formed by the common tangents of the two conies. Then AA' being a side 192 Involution of [ch. of the common self-conjugate triangle, must coincide with VT, YW, or WTJ, say with YW, as in the figure. So B^ coincides with YYTJ, and GG' with TJY. The polars of any common apex of two conies for the two conies pass through the meet of two common chords of the conies. Take the common apex B. Now JS is on WU, the polar of Y. Hence the polar of B for either conic passes through Y, the meet of the common chords ad, be. The common chords ad, he are said to belong to the com- mon apex B. So to every common apex belong two common chords. Similarly, the poks of any commcus is I and another line. Ex. 6. The polars of any two points for amies of a four-point system form two homographic pencils. For jr(iafiV...) = Y'{LMN...). Ex. 7. The pencil of tangents at one of the four common points of a system of four-point conies is homographic with that at any other of the four points. 3. Taking I at infinity we deduce the following theorem — The locus of the centres of a system of conies circumscribing a given quadrangle is a conic which passes through the harmonic points of the quadrangle, through the middle points of the six sides of the quadrangle, and through the common conjugate points for the system on the line at infinity. The following is a direct proof of this proposition. Let ABCD be the given quadrangle, and the centre of one of the circumscribing conies. Join to the middle points m, n, r, s of the sides AB, BC, CD, DA ; and draw Om', On', Or', Os' parallel to AB, BC, CD, DA. Then since Om bisects a chord parallel to Om', Om and Om' are conjugate diameters. So On, On', and Or, Or', and Os, Os' are conjugate diameters. Hence {mm', nn', rr', ss') is an involution. Hence (mnrs) = 0{m'n'r^^). But the rays of (m'n'r's') are in fixed directions. Hence [rrmrs) is constant. Hence the locus of is a conic through the four points m, n, r, s. Now define this locus by five of the centres, then the locus passes through the middle point of the side AB. Similarly the locus passes through the middle point of every side. The locus also passes through the harmonic points of the quadrangles ; for these are the centres of the three pairs of lines which can be drawn through the four points. The locus also passes through the common conjugate points on the line at infinity ; for these, being the double points of the involution in which the line at infinity cuts the conies, are the points of contact of the conies which can be drawn through the four points to touch the line at in- XXII.] Pole-locus and Centre-locus. 213 finity, L e. are the centres of the two parabolas which can be drawn through the four points. Notice that the centre-locus by the former proof also passes through the conjugate point for the system of every point at infinity. If the quadrangle is re-entrant, it is easy to see that the sides of the quadrangle cut the line at infinity in an over- lapping involution. Hence the common conjugate points at infinity are imaginary, and the centre-locus is an ellipse. So if the quadrangle is not re-entrant, the centre-locus is a hyperbola. Ex. 1. Given four fovnts on a conic, and a given paint on the ceTitre-locus as centre, construct the asymptotes. Ex. 2. Five points ABODE are taken. Show that the five conies which bisect the sides of the five quadrangles BCDE, ACSE, ABDE, ABCE and ABCI) meet in a point, Ex. S.Ifa pair of oj^posite sides of the quadrangle be parallel, the centre- locus is a pair of lines. Ex. A. Xf a pair of sides, not opposite, be parallel, the centrc'locus is a parabola. Ex. 5. 1/ two pairs of sides, not opposite, be parallel, the centre-locus is a line {and the line at infinity). Ex. 6. A variable line cuts off from two given conies lengths which are bisected at the same point P. Show that the locus of P is the centre-locus belonging to the meets of the conies. Ex. 7. Thepolars of any point on the centre-locus for conies of the system are parcUlel. Ex. 8. The asymptotes of any conic of the system are parallel to a pair of conjugate diameters of the centre-locus. Let be the centre of that conic of the system which meets the line at infinity in jjp'. Now the centre-locus meets the line at infinity in the double points e,/ of the inTolution {pp', ...). Hence {pp',ef) = -i. Hence Z (pp', ef) = —i where Z is the centre of the centre-locus. But Ze, Zf are the asymptotes of the centre-locus. Hence Zp, Zpf are conjugate diameters of the centre-locus. And Zp, Zp' are parallel to Op, Op', which are the asymptotes of the conic whose centre is 0. Ex. 9. If one of the four-point conies be a circle, the centre-locus is a rect- angular hyperbola. For the common conjugate points at infinity, being conjugate for a circle, subtend a right angle at any finite point, i.e. the asymptotes of the centre-locus are perpendicular. Ex. 10. The axes of every conic circumscribed to a cyclic quadrangle are in the same directions. 214 Pole-locus and Centre-locus. Ex. 11. The locus of the centres of reOangiHar ftHperMiM circamscrxbing a given triangle is the nine-point circle. E!z. 12. If two of the fbur-point conies he rectangular hyperbolas, the centre- locus is a circle. Ex. 13. The nine-point circles of the four triangles formed by four points meet in a point. Ex. 14. Given three points A, B, Con a circle, and the ends P, Q of a dia- meter s show that me centres of the rectangular hyperbolas BCPQ, CAPQ, ABPQ lie on the nine-point eirde of ABC. The centre of the r. h. BCPQ is the middle point of BC, for the tan- gents at B and C are perpendicular to PQ. Ex. 15. The locus of the centres of all conies through the vertices of a triangle and its centroid is the maximum inscribed ellipse. 4. To find the centre of the centre-locus. Since ms and nr are parallel to BD, and since mn and sr are parallel to AC, hence mnrs is a parallelogram. Also the centre of any conic cii-cumscribing a parallelogram is the meet of the diagonals. Hence the required centre is the meet Z of mr and sn. Similarly Z is on the join of the middle points of AC and £D, Note that Z is the centre of mass of equal masses at A, B, C, B. Ex. 1. Several conies have three-point contact at A and pass through B. Show that the centres of the conies lie on a conic whose centre is such that 3 . ^0 = OB. Ex. 2. The six fourth harmonies of the ends of the six sides of a quadrangle for the meets with any transversal lie on a conic ; and the lines joining opposite pairs of these points meet in a point. Project the transversal to infinity. CHAPTER XXIII. INVOLUTION OF A QUADEILATEEAL. 1. The three pairs of lines which join any point to the (^osite vertices of a qimdrilateral are in involution. Let be the point, and AA', BB', CO' the opposite vertices of the quadrilateral. Let OA cut A':ff in a and A'B in H. Then (AA'BG) = (HA'BC) = A {HA'BC) = {GA'C'B')= 0(AA'C'B'). Hence {AA'BG) = (A'AB'C). Hence (AA', BB', CC) is an involution. Ex. L Frme the theorem by considering the section of the guadram/le OABC by A'B'. Ex. 2. Seduce a construction for the mate Off of OC in the invdlvtion determined by (AA', BB'). Ex. 3. Deduce the property of the harmonic points of a quadrangle. Ex. 4. If arty point he joined to the vertices of a triangle and to the meets of any line with the sides of the triangle, thepencU so farmed is in involution. Ex. 6. If any point be joined to the vertices ABC of a triangle, and if OA' OB', Off be drawn paraM to BC, CA, AB, then {AA', BB', Off) is an invo- lution. Ex. 6. If any point be joined to the vertices ABC of a triangle, and A'B^ff be points on the sides of the triangle, such that (AA', BB', Off) is an involu- tion ; then A'B'ff are collinear. Ex. 7. The perpendiculars through to OA, OB, OC meet BC, CA, AB in collinear points. Ex. 8. The six radical axes of four circles through the same point form an involution. 2i6 Involution of a Quadrilateral. [ch. Ex. 9. The orthogonal projectums of the vertices of a guadrilateral on any line are in involution. Ex. 10. If AA', BB', CC be the vertices themsdves, thenAB.AB'-T-AC.AC' = A'B . A'Bf ^A'C.A'C. For the ratios ABf ■¥ AC, &o., are not altered by orthogonal projec- tion. Ex. 11. Also ifP, Q, n bisect AA', BB', CC, thmAB.AB'^AC.AC = PQ-^PIi. For FQR are collinear. Ex. 12. An infinite number of pairs of lines can be found which divide the diagmals of a quadrilateral harmonically. The pair of lines through any point are the double lines of the involution {AA', BB', CC). Involution of four-tangent Conies. 2. The pair of tangents from any point to a conic and the pairs of lines joining this point to the oj^osite vertices of any quadrilateral circumscribing the conic are four pairs of lines in involution. c Let AA', BB', CC be the vertices of the quadrikleral. Let OP, OP' be the tangents from the point 0. Let the meets {OP; AB), {OP'; AB), {OP; A'B'), {OP'; A'B') be called L, M, N, B. Then 0{PP'AB) = {LMAB) = {NBB'A') = 0{PP'B'A'). Hence {PP'AB) = {P'PA'B'). Hence (PP', AA', BB') is an involution. Hence OB, OB' belong to the involution deter- mined by 0{PP', AA'). Similarly OC, OC belong to this involution. Hence (PP', AA', BB', CC) is an involution. XXIII.] Involution of a Quadrilateral. 217 3. The system of conies which can he dravm to touch four given lines is such that the pairs of tangents from any point to conies of the system form an involution. For the tangents OP, OP' to any conic of the system belong to the invohition {AA, BB', CC), determined by the opposite vertices of the given quadrilateral of tangents. Note that we have above given an independent proof that (AA', BB', CC) is an involution. For touching the four given lines and any other line OP we can draw a conic. Note also that we should expect OA, OA' ; OB, OB' ; OC, OC to belong to the involution {PP', QQ', ..-) of tangents. For each pair of opposite vertices may be considered to be a conic which touches the four lines ; and OA, OA' are the tangents from to the conic {A, A'). 4. If two sides BA and AB' coincide, we get the theorem — If a triangle BA'B' be circumscribed to a conic, amd if A be the point of contact of BB'; then the tangents from are a pair in the involution {AA', BB'). If the sides CB and C'B coincide and also the sides CB' and C'B', we get the theorem — If a conic touch the lines CB, CB' at B aiid B', then the tangents from are a pair in the involution 0{CC, BB") of which OCis a double line. If the sides BA, AB' and B'A' coincide, we get the theorem — If a system of conies have three-point contact mth the line BB' at B' and touch a line through B, then the tangents from form an involution of which OB, OB' are a pair. For three-point contact and three-tangent contact are equivalent. If all four sides coincide, we get the theorem — The tangents from Oto a system of conies having four-point contact at a point B' form an involution of which OB' is a double line. Ex. 1. The pencil formed by the pairs of tangents from any point to two circles and the joins of the point to the centres qf similitude is in involution. Ex. 2. If the line joining the centres of similitude SS' of two circles cut the circles in AA', BBf ; then AA', BB", SS' are in involution. Ex. 3. If VP, VQ be the tangents from any point Y to a conic, and if i, a. 2 1 8 Involution of a Quadrilateral. [ch. 3, 4 ie fke successive rertices of a four-point figure circumscribed to the conic, show that sinPn .sinPr3_ sinPr2.sinPr4 sin QV I . sin QV3 ~ sin QVa . sin QV4 Ex. 4. Extend the theorem to any sn-poini circumscribed polygon. Ex. 5. Through every point can be draum a pair of lines which are amjygate far every conic of a four-tangent system. Viz. the double lines of the involution of tangents. Ex. 6. Through ecery point can be drawn a pair of lines to divide the diagonals of a given guadrilaterai harmonically ; and these meet any insaribed conic in harmonic points. For they are the common conjugate lines through the point. Ex. 7. Through every point can be draum two conies of a four-tangent system ; and the tangents to these conies at the point are the common conjugate lines at the point. Draw a conic a of the system to touch OE, one of the double lines of the involution of tangents at 0. Then, since OE is a double line, OE is the other tangent from to a. Hence a passes through 0. Ex. 8. The tangents at one qf the intersections of turn conies inscribed in the same quadrilateral are harmonic with the lines joining the poirU to any tun opposite vertices of the quadrSateral. Ex. 9. ABC is a triangle and a given point. Through 0, and parallel to the sides BC, CA, AB, are drawn the lines OX, OT, OZ ; show that the double lines of the involution {,XA, TB, ZC) are the tangents at to the tuio parabolas which can be inscribed in ABC so as to pass through 0. Ex. 10. P, Q, B are the points of contact of the lines BC, CA, AB with a conic, and OT, OP are the tangents from any point ; show that {BC, PA, XT') and (RQ, AA, TI') are involutions. Ex. 11. If OP, OQ be a pair in the involution obtained by joining to the three pairs of opposite vertices of a quadrilateral, the lines OP, OQ and the sides of the quadrilateral touch a conic. Ex. 12. TTiree vertices of a four-point figure circumscribed to a conic lie on three fixed lines through a point; show that the fourth vertex lies on a fourth fixed line Oirough the same point. Ex. 13. Extend the theorem to any sn-point figure. Ex. 14. By taking the two sides coincident which pass through the anth vertex, deduce a simple solution of the problem — ' Circumscribe to a given conic a polygon of an^i vertices, each vertex to lie on one of a set of sn-i fixed concurrent lines.' Ex. 15. Show that the problem — ' To circumscribe to a given conic a polygon of an vertices, each veiiex to lie on one (f a set of an fixed concurrent lines ' — is either indeterminate or impossible. 5. The three circles on the diagonals of any quadrilateral as diameters are coaxal. The three middle points of the diagonals of a quadrilateral lie on a line (called the diameter of the quadrilaterai). XXIII.] Involution of a Quadrilateral. 219 Tht directors of a system of conies touching the sides of a quadrilateral are coaxal, and three circles of the coaxal system are the three circles on the diagonals as diameters. The centres of a system of conies touching the sides of a quad- rilateral lie on a line which also contains the middle points of the diagonals of the quadrilateral. Let AA', BB', CC be the opposite vertices of the quadri- lateral. Let the circles on A A' and BB^ as diameters meet in and 0'. Then in the involution pencil {AA', BB', CC), since AOA' and BOB' are right angles, COG' is a right angle. Hence the circle on CC as diameter passes through ; and similarly through C Hence the circles on AA', BB', CC as diameters are coaxal. Hence their centres, viz. the middle points of AA', BB', CC, are collinear. Again, the tangents OP, OF' from to any conic touching the sides of the quadiilateral belong to the involution 0(AA', BB', CC). Hence POP' is a right angle. Hence the director of this conic passes through ; and similarly through 0'. Hence this director, and similarly all the directors, belong to the above coaxal system. But the centre of a conic is the same as the centre of its director. Hence the centres of the conies lie on a line, viz. the line of centres of the coaxal system of circles. The locus of centres is the diameter of the quadrilateral ; for three circles of the system are the circles on A A', BB', CC as diameters. The radical axis of the coaxal system of directors is the directrix of the parabola of the system of conies. For the directrix is the limit of a director, and the radical axis is the limit of a coaxal, when each becomes a line. The limiting points of the coaxal system of directors are the centres of the rectangular hyperbolas of the system of conies. For when the coaxal becomes a point, the director becomes a point, and the conic becomes a rectangular hyperbola, the director being the centre of the r. h. Note that the director of a conic which consists of two points is the circle on the segment joining the points as 220 Involution of a Quadrilateral. [ch. diameter, and the centre of the conic is the point half-way between the points. Ex. 1. The directors of all conks touching turn given lines OP, OQ at P, Q are coaxal, the axis being the radical axis of the point and the circle on PQ as diameter. Ex. 2. T?ie polar circle of a triangle circumscribing a conic is orthogonal to the director circle. Let ABC be the triangle. Take any fourth tangent A'B'C. Then the circle on AA' as diameter passes through the foot D of the perpen- dicular from A on BC. Now A and D are inverse for the polar circle. Hence the polar circle is orthogonal to the circle on AA', and similarly to the circles on BBf, CC; and hence to the director, for this belongs to the same coaxal system. Ex. 3. The locus of the centre of a rectangular hyperbola which touches a given triangle is the polar drcU of the triangle. For the polar circle cuts orthogonally the director circle which is the centre in a r. h. Ex. 4. If the vxM-point drde of a triangle circumscribing a r. h. pass through the centie of the r. h. ; show that the centre aiso lies on the circum-cirde, and that the centre of the circum-cirde lies an the r, h. The centre lies on the nine-point circle and on the polar circle and therefore on the circum-cirde, as the three circles are coaxal. Let the asymptotes meet the circum-circle in P, Q. Then ABC, OPQ are inscribed in the same conic, hence PQ touches the r. h. Hence the point of contact is the centre of the circle. Ex. 6. The diameters of the five quadrilaterals which can be formed by five given lines are concurrent. Prove this, and deduce a construction for the centre , E, no three of which are coUinear, subtend a pencil komographic with a given pencil. Take BI/ and BE' so that B {ABCffE') shall be homographic with the given pencil. Draw a conic through ABC to touch BB^ at B. Con- struct the point F in which D^ cuts this conic, and construct the point P in which FE cuts this conic. P is the required point. "XXIV.] Constructions of the First Degree. 225 4. As an example of cases in which some of the given points are at infinity, let us construct a conic, given one asymptote, the direction of the other asymptote, and two other points. Let I be the given asymptote, and m any line in the direction of the other asymptote, and A and B the two given points. We may take C and B to be the points at infinity on I, and E to be the point at infinity on m. Then M is the point at infinity on AB. Hence the construction is — To find where any line AG through A cuts the conic again, let AG and I meet in L, and let a parallel through L to AB cut a parallel through B to Z in N. Then a parallel through ^ to m cuts AG irv the required point F. XiX. 1. Given four points and the direction of an asymptote, construct the conic. Ex. 2. Given three points on a conic and a tangent at one of them, and the direction of one asymptote y constrttct the conic. Ex. 3. Given three points and the directions of both asymptotes, constmct the conic. Ex. 4. Given one point and both asymptotes, construct the conic. Ex. 5. Given four points on a conic and the direction of one asymptote ; construct the meet of the conic with a given line draum parallel to the asymptote. Ex. 6. Given three points on a conic and the directions of both asymptotes; find the meet o/the conic loiSi a given line parallel to one of the asymptotes. Ex. 7. Given four points on a conic and the direction of one asymptote; find the direction of the otiier. 5. As an example of drawing a parabola to satisfy given conditions, let us construct a parabola, given three points and the direction of the axis. Let ABC be the given points, and I any line in the direc- tion of the axis. We may consider D and E to coincide at the point at infinity upon I, so that the line DE is the line at infinity. Then M is the point at infinity on AB. Hence the construction is — To find where any line AG through A cuts the conic again, let AG cut a parallel through C to lia L; let a parallel through L to AB cut BO in N; then a parallel through N to I cuts AG ia the re- quired point F. 4 2 26 Constructions of the First Degree. [ch. JEx. Canstnuit a parabda, given two points and the tangent at one of them, and the direction of the axis. 6. Given ficK points on a conk, to construct tlie tangent at one of them. Let A, B, C, D, E be the five given points, and suppose F to coincide with A ; then AF is the tangent at A. Hence the construction— Let AB and DE meet in M, and BG and AE in N, and let MN cut CD in L ; then LA is the tangent bXA. Ex. 1. Given four points on a conic, and the tangent at one of them ; construct the tangent at another of them. Ex. 2. Given three points on a conic, and the tangents at two of them ; con- struct the tangent at the third. Ex. 3. Given both asymptotes of a hyperbola, and one point; cons&uct the tangent at this point. Ex. 4. Giren three poirUs on a parabola, and the direcHon of the axis ; con- struct the tangent at one of the given points. Ex. 6. Given two points on a parabola, the direction of Oie axil, and the tangent at one of the points ; construct the tangent at the other point. Ex. 6. Given four points on a conic, OTid the direction of one asymptote; con- struct that asymptote. Ex. 7. Given three points on a conic, and the directions of both asymptotes ; construct the asymptotes. Ex. 8. Given two points on a conic, and one asymptote, and the direction of the other ; construct the other asymptote. 7. Given five tangents of a conic, to construct the pointu of contact. Let AB, BC, GE, EF, FA be the five given tangents. Then in the figure of XV. 4, if D is the point of contact of CE, we may consider GD, BE to be consecutive tangents of the conic. Hence the construction — Let BE and GF meet in 0; then AO cuts CE in its point of contact. So the other points of contact can be constructed. Hence given five tangents, we can at once construct five points ; so that every construction which requires five points to be given, is available if we are given five tangents. Ex. 1. Given four tangents and the point of contact of one of them, construct the points of contact of the others. Ex. 2. Given three tangents of o conic, and the points of contact of two of them ; construct the point of contact of the third. XXIV.] Constructions of the First Dep'ee. 227 £x. 3. Given both asymptotes of a hyperbola, and one tangent, construct the point of contact of the tatigefit. Ex. 4. Given f ; let LB meet CA in B', and linE;MLG meet AB in C, and I in -F. Now A and A' are conjugate points for the required conic, and so are L and D. Hence the re- quired conic must pass through XX', the double points of the involution (AA'. LB). So the conic must pass through the double points YY' of the involution (BB', LE), and through the double points ZZ' of the invo- lution (CC, LF). AJso the six points XX'YTZZ' lie on a conic. For draw a conic through XX'YY'Z. Then since LB are harmonic with XX', and LE with YT, I is the polar of L ; also (LF, ZZ') = — I, and the conic passes through Z ; hence the conic passes through Z'. Again, the conic through XX'YY'ZZ' satisfies the required conditions. We have proved that Z is the polar of L. Let BC and B'C meet in H. Then the opposite vertices of the quadrilateral BC, CW, B'C, C'B are BB', CC, and AH. Now BB' are conjugate for the conic, and so are CC ; hence so are AH. Hence the polar of A passes through H ; and also through A'. Hence BC is the polar of ^ ; so CA is the XXV.] Constructions of the Second Degree. 235 polar of B, and AB is the polar of C. Hence ABC is a self- conjugate triangle for the conic. This completes the theoretical solution of the problem ; and we have shown that one, and only one, conic can be drawn satisfying the given conditions. Practically the above solution is worthless ; for any pair of the points XX', YY', ZZ' may be imaginary. The following is the practical con- struction when the conic is real. "We have already found two points upon CL. To find two points on CA. Let AG cut I in Q,. Then AG are conju- gate points ; and so are 5'Q, for Q is the pole of LB. Hence the two points upon GA are the double points of the involu- tion '.4 C, ^Q). So two points can be found on CB. Hence sis points on the conic are known, viz. those on GA, GB, and GL. Now if the conic is real, one of the points ABG (say G) is inside the conic, and hence GA, GB, GL all cut the conic in real points. Hence, by trying ABC in succession, we get six real points on the conic. If on trial we find that neither A nor B nor C gives six real points, we conclude that the conic is imaginary. We see again that two conies cannot have two common self- conjugate triangles ; for since two such triangles more than determine a conic, the two conies would be coincident. Ex. L Given a pentagon ABODE, construct a conic for which each vertex is the pole of the opposite side. Let AB and CD meet in F. The required conic is the one for which ASF is self-conjugate, and E is the pole of BC. Ex. 2. For Otis conic, the iTiscribed conic and the circumscribed conic are reciprocal. Ex. 3. Oiren the centre of a conic and u self-conjugate triangle, construct the asymptotes. Draw QX, OT, OZ parallel to BC, CA, AB ; then the asymptotes are the double lines of (.^-T, BT, CZ). Ex. 4. Given « pde and polar and a self-amjugate triangle, construct the tangents from the pole. Ex.6. GivenfourpointsA, B,C,Dandaline I. With A as pde qf I and urith BCD as a self-conjugate triangle, a conic is drawn; similarly the conies {B, CDA), (C, DAB), {D, ABG) are drawn. Show that these four conies meet I in the same two points. 13. Given five points on each of two conies, to construct the 236 Constructions of the Second Degree. [ch. conic which passes through the meets of these conies and also through a given point. Through the given point L draw any line I ; and construct the points ^j)', qq" in which I cuts the two conies. Then if M be the other point in which the required conic cuts I, we know that pp', q^, LM are pairs in an involution. Hence M is known, i.e. a point on the conic is known on every line through L. Criven five points on each of two conies, to construct the conic which passes through the four meets of these conies and also touches a given line. Construct the points in which the given line cuts the given conies, viz. pp', gg'. Then the points of contact of the required conies are the double points e, / of the involution determined hypp', qq'. Then, taking either e or / we con- tinue as above. Ex. Oive the reciprocal conslnKtians. 14. Given three points on a conic and an involution of eon- jugate points on a line, to construct the conic. If the given involution has real double points, draw a conic through the three given points and the two double points. This conic clearly satisfies the required conditions. If the given involution is overlapping, proceed thus — Let A, B, C be the given points, and I the line on which the involution of conjugate points hes. Let BC cut lin P, and take P', the mate of P, in the involution. Also take P" such that (BC, PP") = - 1. Let P^ cut P'P" in a, and take A' such that {AA', Pa) = —i. So, using CA and QQ', B' can be constructed. Then the conic ABCA'B' is the required conic. For since {BC, PP") = - 1 = (AA', Pa), P"a is the polar of P. Hence PP' are conjugate points. So QQ" are conjugate points. Hence the involution (PP', QQ^ (which is the given involution) is an involution of conjugate points for this conic. If the given involution is overlapping, we have solved XXV.] Constructions of the Second Degree. 237 the problem — To draw a conk through five given points, two of which are imaginary, 15. Construct a conic to pass through four given points and to divide a given segment harmonically. Let LM be the given segment. Let E, F be the double points of the involution determined by the given quadrangle ABCD on LM. Let the double points P, Q of the involu- tion {LM, EF) be constructed. Then the conic through ABCDP is the required conic. For let LM cut this conic again in Q'. Then, PQ" belong to the involution of the quadrangle on LM. Hence {PQ', EF)= — i. Hence ^ coincides with Q. And {LM, PQ) = — i. Hence the conic cuts LM harmonically. If the double points E, F are imaginary, constinict the involution of which L, M are the double points, and let P, Q be the common points of this involution and that of the quadrangle on LM. Then the required conic is ABCDP. For, as before, LM cuts the conic again in Q, and {LM,PQ) = -i. Also, since E, F are imaginary, this construction is real. Ex. Construct a conic which shcM pass ihrtmghfour given points and through a pair {not given) of points of a given invdutim on a line. 16. The following proposition will be used in the suc- ceeding constructions — If a variable conic through four fixed points A, B, C, B meet fixed lines through A and B in P and Q, then PQ passes through a fixed point upon CB. For consider the in- volution in which CD cuts the conic and the four sides AP, BQ, AB, PQ of the quadrangle ABPQ. Five of these points are fixed, viz. the meets with the fixed lines AB, AP, BQ, and the meets C, L with the conic 238 Constructions of the Second Degree. [ch. Hence the sixth meet is fixed, i.e. PQ passes through a fixed point on CD. The theorem may also be stated thus — A system of conies pass throitgh ABCD. A fixed line thrmigh A cuts these conies in PP'. . . , and a fixed line through B cuts them in QQ'.... Then all the lines PQ, P'Qf,--- are concurrent in a point on CD. If A and B coincide, the theorem is — A system of conies touch at A and pass through CD. A fixed line through A cuts tJiese conies in P, P', ..., and another fixed line through A cuts them in Q, Q^.--. Then all the lines PQ, P'(^, ■ ■ ■ are concurrent in a point on CD. \i A, B and C coincide, the theorem is — A system of conies have three-point contact at A and pass through D. A fixed line through A cuts these conies inP, P',..., ami another fixed line through A cuts them in Q, Q',.... Then all the lines PQ, P'Q',... are concurrent in a point on AD. Uz. 1. Beciprocate all these theorems. £x. 2. Given three meets ABC of two five-point conies, prove the following amstruction for the fourth meet D — 2'ake any two points L, M on either conic, and construct the points JJ , M' in which AL, BM cut the other conic. Join the meet of LM, L'M' to C. Then D is the meet of this line with eitlier conic. £x. 3. Given two meets A, B of two five-point conies, prove the following con- struction for the other meets C and B — Take any two points L, M on either conic, and construct the points V, M' in which AL, BM cut the other conic. LM, L'M' meet in one point on CD. Similarly construct another point on CD. Now con- struct the points in which the joining line cuts eitlier conic. Ex. 4. Reciprocate the two preceding constructions. Ex. 5. Prove the following construction for the directions of the axes of a conic given by five points — Draw a circle through three A, B, C of the given points; now construct the fourth meet P of the conic and the circle ; then the directions qf the axes bisect tlie angles between AB and CP. 17. Gtiven five points on a conic, three of which are coincident, to construct the conic. Let ABC be the three given coincident points, and DE the other given points. Then to be given ABC is equivalent to being given the point A, the tangent at A, and the circle of curvature at A. Let AD, AE cut this given circle in D', E'. Then DE, D'E' meet on the common chord of the XXV.] Constructions of the Second Degree. 239 circle and the conic. Hence the point P where this chord cuts the circle can be constructed. Now P is on the conic. Hence we know four points A, B, E, P on the conic and the tangent at one of them. Hence the conic can be con- structed. Ex. Obtain, by using the reciproccd theorem, u soliUion of the probleiii — Given five tangents of a conic, three of which are coincident, constriKt the conic. Notice that the circle of curvature has three-tangent contact witu the conic as well as three-point contact. CHAPTEE XXVI. METHOD OF TRIAL AND EEEOE. 1. Given two homographic ranges {ABC.) and {dbc ...) cm different lines, and given two points V and u, find two points XY of the first range, such that the angles XVT and xm/ may have given values, x and y being the points corresponding to X and T in the homographic ranges. Try any point P on AB as a position of X. To do this, take Q on AB, so that the angle PVQ is equal to the given value of XVY. Take p and q, the points corresponding to P and Q, in the homographic ranges. Also take r on 06, so that the angle pvr may be equal to the given angle xvff. Then if r coincides with q, the problem is solved. If not, try several points P„ P^... . Then (r,r^...) = v{r,r,...) — '"(PiP-z---) since the pencils are superposable = (pjjjj ...) = (PiPj ...) since the ranges are homographic = V{P,P, ...) = F(<2. Q,...) = (Q,Q, ...) = (q,q, ...). Hence the ranges (^iSj...) and {r,rj...) are homographic. Now if q and r coincide, q will be a position of y. Hence y is either of the common points of the homographic ranges {q, fj ...) and {r^r^ ...). Hence Y and X are known. The problem has four solutions. Two are obtained above, and two more are obtained by taking the angles PVQ and pvq in relatively opposite directions. Notice that we need only make three attempts ; for the common points of two homographic ranges can be deter- mined if three pairs of corresponding points are known. Method of Trial and Error. 241 The above process may be abbreviated by writing (r) for the range (r, r^ ...), and so on. The method is called by some writers the method of False Positions. Ex. 1. Find aynespcnding segments XY, X'Y' of two given homographic ranges which shall be of given lengths. Ex. 2. 6iven two homographic ranges on the same line, find a segment XX' bounded by corre^onding points, (i) which is bisected by u, given point, or (ii) which divides a given segment harmonically. If XX' satisfies either condition, X and X' generate ranges which are in involution and therefore homographic. Ex. 3. Find also XX', given that {Vi AX : BX' is a given ratio, or (ii) that XX' is of given length, or (iii) that XX' divides a given segment in a given cross ratio. Ex. 4. If A, A' generate homographic ranges on two littes, show that throtigh any given point two of the lines AA' pass. Ex. 5. Find corresponding points X, X' of two homographic ranges on different lines, such that XO and X'C/ meet at a given angle, and 0/ being given points. The pencils at and 0' are homographic. Sx. 6 . Given on the saTne line the homographic ranges (ABC.) ^ (A'B'C' . . . \ and Oie homographic ranges (LMN...) = (L"M"N"...') ; find a point X which has the same mate in both ranges. Ex. l.IfA and A' generate homographic ranges on two lines, and B and Bf generate homograpkic ranges on tu}0 other lines, find the positions of A, B, A', Bf that both AS and A'Bf may pass through a given point. 2. Between two given lines place a segment whose projections on two given lines shall be of given lengths- Let the projections lie on the lines AB and CD. On AB take a length LM equal to the given projection on AB ; through L and M erect perpendiculars to AB to meet the given lines in X and Y. Let the projection of ZF on CD be PQ. If PQ is of the required length, then the problem is solved. If not, make PQ' of the required length. Then the ranges generated by Qf and P are homographic, being superposable. Again, the ranges P and X are homographic, by considering a vertex at infinity. Similarly range X = range L = range M = range Y = range Q. Hence the ranges Q' and Q are homographic. Either of 242 Method of Trial and Error. [ch. the common points of these ranges gives a true position of q. Ex. 1. Cm, two given lines find points A and B, such that AB subtends giren angles at two given points. Ex. 2. Through u. given point draw two lines, to cut off segments 0/ giren lengths from two given lines. Ex. 3. Given two fixed points and tf on two fixed lines, through a fixed point V draw a line cutting the fixed lines in points A, A', such that (i) OA . (/A' is constant, or (ii) OA : Of A' is constant. Ex. 4. Through a given point draw a line to include with tico given lines a given area. Ex. 5. Two sides of a triangle are given in position and the area is given ; show that the base in two positions subtends a given angle at a given point. Ex. 6. Given four points A, B, C, D on the same line, find two points JC, Y on this line, such that {AB, XT) and (CD, XF) may have given values. Ex. 7. Given two fixed points A and B, find two points P, Q on the line AB, such that {AB, PQ) is given and also the length PQ. Ex. 8. Given three rays OA, OB, DC, find three other rays OX, OY, OZ, such that the cross ratios 0{AB, XY), 0(BC, TZ), 0{CA, ZX) may have given values. Ex. 9. Find the lines OX, OX' such that (AA', XX') may be a given cross ratio and XOX' a given angle, OA and OA' being given lines. Ex. 10. Solve the egualion ax' +bx + c = obya geometrical construction. The roots are the values of x at the common points of the homo- graphic ranges determined by axx' + bx + c — o. Ex. 11. Solve geometrically the equations y = Ix + a, s = my + b, x = m + c. Obtain the common points of the homographic ranges (a;, x') deter- mined by !/ = ir + o, a = my + b, 3^ = m + c. Ex. 12. Solve geometrically the equations xy + lx + my + n^o, xy+px + qy + r=-o. 3. Criven two points L, M on a conic, find a point P on the conic, SMcft that PL, PM shall divide a given segment JJY in a given cross ratio. Take any position of P, and let PL, PM meet UV in A, B, and take B' such that {UV, AB') is equal to the given cross ratio. Then (A) = L(A) = L(P) = M(P) = M{B) = (B). Also, since ( VT, AB') is constant, we have (A) = (B'j. Hence (B) = (B'). Hence the required position of B is either of the common points of the homographic ranges generated by B and B'. XXVI.] Method of Trial and Error. 243 Ex. Give (wo ipoints L, M on a conic, find a point P on the conic such that the bisectors of the angle LPM may have given directions. Draw parallels to PL, PM through a fixed point. 4. Inscribe in a given conic a polygon of a given number of sides, so that each side shall pass through a fixed point. Consider for brevity a four-sided figure. It will be found that the same solution applies to any polygon. Suppose we have to inscribe in a conic a four-sided figure ABCD, so that AB passes through the fixed point U, EC through F, CD through W, and DA through X, On the conic take any point A. Let A U cut the conic again in B. Let BY cut the conic again in C. Let CW cut the conic again in 2). Let BX cut the conic again in A'. So take several positions of A. Then the range on the conic generated by .4 is in involu- tion with the range generated by B, since AB passes through a fixed point U. Hence (A) = (B). So {B) = (C) - {B) = {A'). Hence the ranges (.4) and (A') on the conic are homographic. A true position of A is either of the common points of these homographic ranges. Note that in the exceptional case of XXI. 3. Ex. 14, the common points lie on the line ; and the above solution becomes nugatory. £x. 1. Describe abo^U a given conic a polygon such that each vertex shall lie on a given line. Inscribe in the conic a polygon whose sides pass through the poles of the given lines, and draw the tangents at its yertices. £z. 2. Inscribe in a given conic a polygon o/a given number of sides, such that each pair of consecutive vertices determine with two given points on the conic a given cross ratio. Ex. 3. In the given figure ABCD inscribe the figure NPQS, so that BN, PQ meet in the fixed point U, and NP, RQ in the fixed point V. Ex. 4. Construct a polygon, whose sides shall pass through given points and whose vertices shall lie on given Hrus. Ex. 6. Construct a polygon, whose vertices shall lie on given lines and whose sides shaU subtend given angles at given points. Ex. 6. Construct a triangle ABC, such that A and B shall lie on given lines, and that the angle C shaU be egwd to a given angle, whilst the sides AB, BC, CA pass through fixed poirtts. R 2 244 Method of Trial and Error. Ex. 7. A ray of light starts frmn a given point, and is r^ected successivelij from n given lines ; find the initial direction that the final direction may make a given angle with the initial direction. Ex. 8. Given two hamographic ranges {ABC...) = (A'B'C ...)on a conic, find the corresponding points Z, X', such that XX' may pass through a given point, Ex. 9. Oiven two points AA' on a conic, find tioo points XX' also on the conic, such that (AA', XX') has a given value and XX' passes through a given point. Ex. 10. Through a given point A is draum a chord PQofa conic ; BC are fixed points on the conic ; find the position o/PQ when PB and QC meet at a given angle. Ex. 11. Through two given points describe a circle which shall cut a given are of a drde in a given cross ratio. Ex. 12. Through four given points draw a conic which shall cut off from a given lirte a length which is either given or subtends a given angle at a given point. CHAPTEE XXVII. IMAGINARY POINTS AND LINES. 1. The Prmciple of Continuity enables us to combine the elegance of geometrical methods with t^e generality of algebraical methods. For instance, if we wish to determine the points in which a line meets a circle, the neatest method is afforded by Pure Geometry. But in certain relative positions of the line and circle, the line does not cut the circle in visible points. Here Algebraical Geometry comes to our help. For if we solve the same problem by Algebraical Geometry, we shall ultimately have to solve a quadratic equation ; and this quadratic equation will have two solutions, real, coin- cident and imaginary. Hence we conclude that a line always meets a circle in two points, real, coincident or imaginary. Another instance is afforded by XXIII. 5. Here we prove the proposition by using the points and (/ in which the circles on AA' and BB' as diameters meet. But these circles in certain cases do not meet in visible points. But we might have proved the same proposition by Algebraical Geometry, following the same method. Then it would have been immaterial whether the coordinates of the points and 0' had been real or imaginary, and the proof would have held good. Hence we conclude that we may use the imaginary points and (X as if they were real. In all solutions by Algebraical Geometry, points and lines will be determined by algebraical equations. Hence 246 Imaginary Points and Lines. [ch. imaginary points and lines will occur in pairs. Hence we shall expect that in Pure Geometry, imaginary points and lines will occur in pairs. 2. The best way of defining the position of a pair of imaginary points is as the double points of a given over- lapping involution ; and the best way of defining the position of a pair of imaginary lines is as the double lines of a given overlapping involution. Thus the points in which a line cuts a conic are the double point6 of the involution of conjugate points deter- mined by the conic on the line ; and these double points, i.e. the meets of the conic and line, are imaginary if the involu- tion is an overlapping one. So the tangents from any point to a conic are the double lines (real, coincident, or imaginary) of the involution of conjugate lines which the conic determines at the point. Note that a pair of imaginary points is not the same as two imaginary points. For if AA' are a pair of imaginary points and BB^ another pair of imaginary points, then AB are two imaginary points but are not a pair. 3. The middle point of the segment joining a pair of imaginary points is real. For it is the centre of the involution defining the imaginary points. A pair of imaginary points AA' is determined when we hnmc the centre and the square (a negative guanity) OA \ For the involution defining the points is given by OP. OP=OA\ The fourth harmonic of a real point for a pair of imaginary points is real. For it is the corresponding point in the defining in- volution. The product of the distances of a pair of imaginary points frrnn amy real point on the same line is real and positive. Let A A' he the pair, and P any real point on the line XXVII.] Imaginary Points and Lines. 247 AA'. Take the middle point of the segment AA'. Then TA . PA'= (OA - OP) {OA'- OP) = {OA -OP)(-OA- OP) = OP' - 0A\ Now OA' is negative, or the involution would have real double points. Hence PA . PA' is real and positive. 4. Two conies cut in four real points, or in two real and two imaginary points, or in four imaginary points. Since a conic is determined by five points, two conies cannot cut in more than four points, unless they are coincident. Also we can draw two conies cutting in four points, e. g. two equal ellipses laid across one another. Now if we were solving the problem by Algebraical Geometry, and were given that the problem could not have more than four solutions, and that it had four solutions in certain cases, we should be sure th^t the problem had in all cases four solutions, the apparent deficiencies, if any, being accounted for by coincident or imaginary points. Hence it follows by the Principle of Continuity, that two conies always cut in four points, real, coincident, or imaginary. Also imaginary points occur in pairs. Hence two or four of the points may be imaginary. 5. If two conies cut in two real points, the line joining the other common points is real, even if the latter points are imaginary. For, by the principle of continuity, Desargues's theorem holds, even if two or four of the points on the conic are imaginary. Let any line cut the conies in pp' and qq and the given real common chord in a. Then the real point a', taken such that {aa',pp', qq') is an involution, lies on the opposite common chord. Hence the opposite common chord is real, being the locus of the real point a'. If two conies cut in two real and two imaginary points, one pair of common chords is real and two imaginary. 248 Imaginary Points and Lines. [ch. For if a second pair were real, the four common points would be real, being the meets of real lines. 6. One vertex of the common self-conjugate triangle of two conks is always real. Take any line I ; then the locus of the conjugate points of points on I for both conies is a conic. Take any other line m ; the locus of the conjugate points of points on m for both conies is a second conic. These conies have one real point in common, viz. the conjugate point of the meet of I and m. Hence they have another real point in common, say U. Take the conjugate point Q onl oi U for both conies and the conjugate point R on m oi U iat both eonics. Then QR is clearly the polar of JJ for both conies ; for the polar of U for both conies passes through Q and R. Hence U is a, real vertex of the common self-conjugate triangle of the two conies. Similarly, the other two points, real or imaginary, in which the conies cut, are the other two vertices of the common self-conjugate tiiangle. 7. The other two vertices of the common self-conjugate triangle of two conies are real if the eonics cut in four real points or four imaginary points; hit if the conies cut in two real and two imaginary points, the other two vertices are imaginary. If the four intersections are real, the proposition is obviously true. If the four intersections are imaginary, one conic must be entirely inside or entirely outside the other. Hence the polar of the real vertex Z7 cuts the eonics in either two non- overlapping segments AA', BB', or in one real segment and one imaginary, or in two imaginary segments. Now the other two vertices VW are the points on the polar which are con- jugate for both eonics, i.e. are the common pair of the two involutions of conjugate points on the polar. And the double points AA', BB' of these involutions are either real and non- overlapping, or one pair (at least) is imaginary. Hence by XX. 5, FTT are real. XXVII.] Imaginary Points and Lines. 249 If two intersections are real and two imaginary, the meets of the given real common chord and of the opposite common chord (which is known to be real) gives a real position of U. But the opposite chord does not cut either conic ; hence V is outside both conies. Hence the polar of TJ, passing through the fourth harmonic of U for the two real points, cuts the two conies in overlapping real segments. Hence FW, being the double points of the involution determined by these segments, are imaginary. 8. One pair of common chords of two conies is always real. If all four intersections are real, it is clear that the six common chords are all real. If all four intersections are imaginary, then UVW are real. Take any point P and its conjugate point P' for the two conies. Then the common chords through U are the double lines of the involution U(yW, PP') ; for the polar of P for these common chords passes through P', and the polar of V passes through W. Hence the common chords through U are both real or both imaginary. Also the common chords through two of the three points TJVW must be imaginary ; for otherwise the four real com- mon chords would intersect in four real common points of the conies. Let the chords through V and W be imaginary. Then taking P inside the triangle UVW, we see that since V{UW, PP') overlap, P' must lie in the external angle 7; so P' must lie in the external angle W. Hence P' lies in the internal angle U. Hence U{yW,PP') does not overlap; hence the double lines of the involution are real, i.e. the common chords through TJ are real. If two intersections are real and two imaginary, we have already proved that two common chords are real. 9. Two conies have four common tangents, of which either two or four may he imaginary. If two conies have two real common tangents and two imaginary, the intersection of the real and also of the imaginary tangents is real; and the other four common apexes are imaginary. 250 Imaginary Points and Lines. One side of the common self-conjwgate triangle of two conies is always real ; the other two sides are real if the four common tangents are aM real or aU imaginary ; otherwise the other two sides are imaginary. One pair of common apexes of two conies is always real. These propositions can be proved similarly to the corre- sponding propositions respecting common points and common chords (or by Beciprocation). Ex. 1/ two conies have three-point contact at a point, they have a fourth real common point, and a fourth real common tangent. CHAPTER XXVIII. CIECULAE POINTS AND CIKCULAE LINES. 1. The circular lines through any point are the double lines of the orthogonal involution at the point. Every pair of drcular lines cuts the line at infinity in the same two points (called the circular points). Take any two points P and Q. Then to every ray in the orthogonal involution at P there is a parallel ray in the orthogonal involution at Q, or briefly, the involutions are parallel. Hence the double lines are parallel. Hence the circular lines through P and Q meet the line at infinity in the same two points. The notation w , oo ' will be reserved for the circular points. Any two perpendicular lines are harmonic mth the circular Unes through their meet. For by definition the circular lines are the double lines of an involution of which the perpendicular lines are a pair. The points in which any two perpendicular lines meet the line at infinity are harmonic with the circular points. For the circular lines through the meet of the lines are harmonic with the given lines. 2. The triangle whose vertices are any point C and the circular points, is self-conjugate for amy rectangular hyperbola whose centre is at C. For Caa ,C oa' being circular lines are harmonic with every orthogonal pair of lines through C, and are therefore har- monic with the asymptotes, i.e. with the tangents from C to the r. h., and are therefore conjugate lines for the r. h. 252 Circular Points and Circular Lines, [ch. Also G is the pole of 00 00 '. Hence Coo 00 ' is self-conjugate for the r. h. Ex. L AH rectangular hyperbolas have a common pair qf conjugate points. Hz. 2. Every conic for which the circular points are conjugate is ar. h. 3. All circles pass through the circular points. Let C be the centre of any circle. Then Coo , Coo ' are the asymptotes of the circle. For Coo , C 00 ' are the double lines of the orthogonal involution at C, i. e. are the double lines of the involution of conjugate diameters of the circle. Now a conic passes through the points in which the line at infinity meets its asymptotes. Hence the circle passes through 00 and 00 '. Notice that we have proved that Cw , Coo ' touch at 00 , 00 ' any circle whose centre is at C. 4. Every conic which passes through the circular points is a circle. Let C be the centre of a conic through oo , 00 '. Then since the lines joining the centre of a conic to the points where the conic meets the line at infinity are the asymptotes of the conic, we see that Coo , Coo ' are the asymptotes of the conic. Hence the involution of conjugate diameters of which the asymptotes are the double lines must be an orthogonal involution. Hence every pair of conjugate diameters is orthogonal. Hence the conic is a circle. We now see the origin of the names circular points and circular linea The circular points are the points through which all circles pass. A pair of circular lines is the limit of a circle when the radius is zero ; the circle degenerating into a real point through which pass imaginary lines to the circular points. So that a pair of circular lines is both a circle and a pair of lines. 5. Concentric circles have double contact, the lime at infinity being the chord of contact. For all circles which have C as centre, touch Coo at w and Coo ' at 00 '. XXVIII.] Circular Points and Circular Lines. 253 Ejc 1. JEsery semicxrde is divided harmonically by the circular pointa, Ex. 2. The cirde which circumsaribes a triangle which is self-conjngaie far a rectangular hyperbola passes through the centre. For five of the vertices of the two triangles consisting of the given triangle and Coo co ' lie on the circle. Ex. 3. Gaskin's theorem. The circle about a triangle sdf-cmjuga(e for a conic is orthogonal to the director. Let F be a common point of the circle and the director. Let the polar of V for the conic meet the circum-circle in u, o' ; Fco , Foo ' in fi, a' ; and the tangent at F to the circum-circle, and eo oo ' in 7, y. Then Faa' is a self-conjugate triangle. Hence aa' are conjugate points for the conic. Again, Fco , Fco ' are conjugate lines, for the tangents from V are orthogonal ; hence 00' are conjugate points. And (oa', $0', f/) is an involution. Hence y/ are conjugate points. Hence the polar of •/ passes through 7. Now ■/ is at infinity, hence its polar F7 passes through C; i. e. the tangent to the circum-circle at F coincides with the radius of the director circle. Ex. 4. The axes of any one of the system of conies through four given poirUs on a circle are in fixed directions. Take any point F and join F to the points at infinity AA', BB',... on the conies. Then Y(^AA', BB',...) is an involution pencil parallel to the asymptotes. But Foo , Fco ' is one pair, corresponding to the circle. Hence the double lines are at right angles, and therefore bisect the angles AVA', BVB', .... Hence the axes are parallel to these double lines, and therefore are in fixed directions. Ex. 5. Two conies are placed with their axes parallel ; show that their four meets are concydic. Ex. 6. Give a descriptive proof of the property of the director circle of a eonic. Let A and B be any fixed points, and let PA and PB be any two lines through A and B which are conjugate for a conic. Draw the polar 6 of B cutting PA in Q. Then Q is the pole of PB. Hence A{P)=A{Q)^{,<)) = B{P). Hence the locus of P is a conic through A and B. Now let R be any point on the director circle. Then JBco , Eta ' are conjugate for the conic, since the tangents from B, being perpen- dicular, are harmonic with Ba> , iJoo '. Hence the locus of jR is a conic through 00 and 00 ', i. e. is a circle. 6. If the penal V{ABC . ) be turned bodily through any angle about V into the position V{A'B'C' ...), then the common lines of the two hmiographic pencils V(ABC...)and V{A'B'C'...) are the circular lines through V. The pencils, being superposable, are homographic. Hence if they cut any circle through Fin 06c ... and a'h'c' ..., the two ranges (06c...) and {a'b'c' ...) on the circle are homo- 2 54 Circular Points and Circular Lines, [ch. graphic. One point on the homographic axis of these ranges is the meet of db' and a'h. But these lines are parallel. Hence this point is at infinity. So every point on the axis is at infinity. Hence the common points of the ranges (abc ...) and (a'ftV ...) are the meets of the circle with the line at infinity, i. e. are oo oo '. Hence the common Unes of the pencils V{ABC...) and Vi^'E'C ...) are Foo , Foo'. Hence — The legs of a constant angle divide the segment joining the circular points in a constant cross ratio. Let the constant angles be ALA', BMB', CNC, .... Through any point F draw a circle and let parallels through F to LA, MB, NC,..., LA', MB', NC... cut this circle in a, b, €,..., a', b', c' .... Then, as above, oo oo ' are the common points of the homographic ranges (phc ...) and {a'b'd ...') on the circle. Hence (oo 00 ', aa') = (oo 00 ', bb") = (oo oo ', cc) = .... Hence F(oo oo ' aa') is constant. But the parallel lines LA and Ya cut oo oo ' in the same point ; so LA' and Ya' cut 00 00 ' in the same point. Hence i (oo oo ', AA') is constant. Hence LA and LA' divide the segment oo oo ' in a constant cross ratio. 7. Coaxal circles are a system of four-point conies. For two circles meet in two points (real or imaginary) on the radical axis and also in the circular points. The adjoining ideal figure explains the relation of coaxal circles XXVIII.] Circular Points and Circular Lines. 255 to the circular points. A and B are the common finite points on the radical axis, and A is the point at infinity on the radical axis. L and L' are the limiting points. For since LA and LB are circular lines through A and B, Lisa, point-circle of the system. So for L'. Also LL'Q. is the common self-conjugate triangle of the coaxal system. Foci of a Conic. 8. Every conic has four foci, which are inside tJie conic and lie two on each axis, those on either axis being equidistant from the centre. The tangents from a focus of a conic to the conic are the double lines of the involution of conjugate lines at the focus, i. e. are the double lines of an orthogonal involution, i. e. are circular lines, i. e. pass through 00 , 00 '. Hence a focus is an internal point, since the tangents from it are imaginary. Also every intersection S of the four tangents from 00 , 00 ' to the conic is a focus of the conic. For Sx , Sco ' being the tangents from S and also circular lines, the involution of conjugate Unes at S is orthogonal, i. e. S' is a focus. Hence the foci of a conic are the other four meets of tangents to the conic from oq and x '. Consider the adjoin- ^, ing ideal figure. Here SS'FF' are the foci. Also G is the centre ; for the lines SS', FI", 00 00 ' form a self -conju- gate triangle, hence G is the pole of 00 00 . Again, SS' and FF are the axes. For G{rx>cc',SF) is a harmonic pencil (from the quadrangle SFS'F') ; hence SS' and FF are orthogonal, Hence SS' 256 Circular Points and Circular Lines, [ch. and ¥F', being orthogonal conjugate lines at the centre, are the axes. Hence the foci lie two by two on the axes. Again, FT" cuts 00 00 ' in a point fl, such that (C12, FF) is harmonic; hence C bisects FF'. So G bisects SS'. Hence the foci on each axis are equidistant from the centre. It will be instructive to draw an ideal pic- ture showing the rela- tion of a parabola and of a circle to its foci. In the case of a para- bola 00 CO ' touches the conic. Hence F' coin- cides with 00 ' and F with 00 ■ Also C and S' coincide at the point of contact of 00 00 '. In the case of a circle, 00 and 00 ' are on the conic ; and all the foci coincide with the centre C. IjZ. 1. Tb» sides of a triangle ABC touch a conic a and meet a fourth tanged to a in A'BfCf ; show that the double lines of the involution subtended &V {AA', Bff, CCf) ai a focus are perpendicular. Being conjugate lines at a focus. Ex. 2. The circle described about a triangle which dreumscribes a parabola, passes through the focus. For five of the vertices of the two triangles consisting of the given triangle and Sco co ' lie on the circle. Ex. 3. A circle is draum with centre on the directrix of a parabola to pass through the focus. At B, one of the meets of the parabola and the circle, are draum the tangents to the circle and pardbola, meeting the parabola and circle again in P and (). Show that PQ isa common tangent to the turn curves. Let be the centre on the directrix, and let the tangents from to the parabola meet the line at infinity in fl and (}'. Then con- XXVIII.] Circular Points and Circular Lines. 257 sideling the triangles Onil' and Satco', we see that the conies are related as in Ex. 14 of XIV. 2. 9. The foci on one axis (called the focal axis) are real, and the foci on the other axis (called the non-focal axis) are imaginary. Take any point P, and through P draw the orthogonal pair of the involution of conjugate lines at P, cutting one axis in G and H and the other axis in g and h. Then PG and PH are harmonic with Poo and Poo ' since GPH is a right angle, and with the tangents from P since PG and PH are conjugate. Hence PG and PH are the double lines of the involution P(oo 00 ', SS', FF') to which the tangents belong. Hence P (SS', GH) and P{FF', gh) are harmonic. And G bisects SS' and FF'. Hence OS" = CG . CH and CF^=Cg.Ch. But on drawing the figure, we see that if CG and CH are of the same sign, Cg and Ch are of opposite signs. Hence, taking CG . CH positive, CS' is positive and CF^ is nega- tive. Hence S and S' are real and F and F' are imaginary. Ex. 1. Show that gh subtends a right angle at S and at S'. Now Cg.Ch = -CG. CH hy elementary geometry =-CS' = CS . CS'. Hence SS'gh lie on the circle whose diameter is gh. Ex. 2. Any line through G is conjugate to the perpendicular line through H; and the same is true ofg and h. Ex. 3. In a parabola, S bisects GH. 10. Confocal conies are a system of four-tangent conies. For if S and S' be the real foci, the conies all touch the lines Sco , S'os, Sao ', and S'co '. Hence, the tangents from any point to a system of confocals form an involution, to which belong the pairs {PS, PS'), (PF, PF') and (Poo , Poo '), P icing the given point. Through every point can be dravm a pair of lines which are conjugate for every one of a system of confocals. 258 Circular Points and Circular Lines, [ch. Viz. the double lines PG, PH of the above involution. PG and PH are perpendicular. For they are harmonic with Poo , Poo '■ The pairs of tangents from any point to a system of con- focals and the focal radii to the point have a common pair of bisectors. For the double lines PG and PH of the involution are perpendicular. Ex. 1. In a parabola, PG and PB are the bisectors 0/ the angles betieeen PS and a parallel through P to the axis. Ex. 2. From a given point 0, lines are drawn to touch one of a system 0/ confoeal conies in P and Q ; show that PQ and the normals at P and Q touch a fixed parabola which touches the aaes of the canfocdls. Viz. the polar-envelope of the point for the system of four-tangent conies. The normal PG at P touches the polar-envelope, because it is conjugate to OP for every conic of the system. Also 00 co ' and the axes touch, since they are the harmonic lines of the quadrilateral, Hx. 3. The directrix of the parabola is CO, C being the common centre. For the tangents at to the two confocals through are two positions of PQ. Ex. 4. T?ie cirde about OPQ passes through a second fixed point. Let the normals at P and Q meet in E. Then the circle about OPQ is the circle about PQS, which passes through the focus of the para- bola. Ex. 5. The locus qfthe orthocentre ofPQR is a line. Viz. the directrix of the parabola. Ex. 6. The conic through OPQ and the foci passes Oirough a fourth fixed point. Let the perpendiculars at S, S' to OS, OS' meet in V. Then S(£70, PQ) = S' (VO, PQ) = -I. 11. The locus of the poles of a given line for a system of con- focals is the normal at the point of contact oftJie given line with a confoeal. For let the given line I touch a confoeal at P, and let PG be the normal, and PH{= I) the tangent to this confoeal. Then PG and PH are perpendicular. Hence P(GH, 00 00 ') is harmonic. But PH is one of the double lines of the involution of tangents from P to the confocals, being the pair of coincident tangents from P to the confoeal which xxviii.] Circular Points and Circular Lines. 259 PK touches. And Poo , Poo ' is a pair of this involution. Hence 'PG is the other double line. Hence P(? and PH, being harmonic with every pair of tangents, are conjugate for every confocal. Hence the locus of the poles of I is PG. Reciprocation of circular points and lines. 12. Circular lines are the double lines of the orthogonal involution at a point P. Hence, the reciprocal of a pair of circular lines is a pair of points on a line p which are the double points of the involution on the line which subtends an orthogonal involution at the origin of reciprocation, in other words, are the meets of p with the circular lines through the origin of reciprocation. Circular points are the points on the line at infinity which are the double points of the involution on the line at infinity which subtends an orthogonal involution at 0. Hence the reciprocals of the circular points are the double lines of the orthogonal involution at 0, i. e. are the circular lines through the origin of reciprocation. Theredprocdl of a circle for the point is a conic with focus atO. For since the circle passes through the circular points, the reciprocal touches the circular lines through 0, i. e. is a focus of the reciprocal. To reciprocate confocal conies into coaxal circles. Confocal conies are conies inscribed in the quadrilateral Soo , S'oo , Sco ', iS'oo '. Eeciprocate for iS". Then since Sn , Soo ' touch the given conies, the circular points lie on the reciprocal conies, i. e. the reciprocal conies are circles. Also the given conies have two other common tangents ; hence the reciprocal conies have two other common points, i. e. are coaxal circles. To reciprocate coaxal circles into confocal conies. Coaxal circles are conies circumscribed to the quadrangle ABia CO '- (See figure of § 7.) Eeciprocate for L. Then B 2 26o Circular Points and Circular Lines. the given conies pass through four fixed points, two on each circular line through the origin of reciprocation. Hence the reciprocal conies touch four fixed lines, two through each of the circular points ; i. e. the tangents to all the reciprocal conies from 00,00' are the same, i. e. the reciprocal conies are confoeal. CHAPTEE XXIX. PEOJECTION, BEAL AND IMAGINAEY. 1. To project a given conic into a circle and at the same time a given line to injvnity. Take K, the pole of the given line I which is to be projected to infinity. Through K draw two pairs of conjugate lines cutting I in AA', BW. On AA' and BB' as diameters describe circles cutting in V and v. About AA' rotate Fout of the plane of the paper. With Fas vertex project the given figure on to any plane parallel to the plane VAA'. Then KA will be projected into a line parallel to VA, and KA' into a line parallel to VA'. Hence AKA' will be projected into a right angle. So BKB' wiU be projected into a right angle. Again, since KA and KA' are conjugate for the given conic, their projections will be conjugate for the conic which is the projection of the given conic. So KB and JO' will be conjugate in the figure obtained by projection. Again, K is the pole for the given conic of the given line I which is projected to infinity. Hence in the second figure, K is the pole of the line at infinity, i. e. is the centre of the conic. 262 Projection, Real and Imaginary. [ch. Hence in the second figure KA, KA' and KB, KBf are two pairs of orthogonal conjugate lines at the centre, i.e. the second conic has two pairs of orthogonal conjugate diameters. Hence the second conic is a circle. 2. The above construction fails when the line to be projected to infinity is the line at infinity itself. The problem then becomes — To project a given conic into a circle, so that the centre of the conic may be projected into the centre of the circle. If the conic is an ellipse, this can be done at once by Orthogonal Projection. If the conic is a h3rperbola, we must use an imaginary Orthogonal Projection. If the conic is a parabola, the projection is impossible. Ex. Projed a system of iwmothetic amies into cirdes, 3. To project a given conic into a drck and a given point into its centre. Take K to be the given point and I its polar. To project a given conic, so that one given point may he pro- jected into the centre and another given point into a focus. To project L into the centre and K into a focus, take'Z in the above construction to be the polar of L instead of the polar of K, using E and I as before. Then L is projected into the pole of the line at infinity, i. e. into the centre, and K is projected into a point at which two pairs of conjugate lines are orthogonal, i. e. into a focus. To project a given conic, so that two given points may 6e pro- jeded into its foci. To project K, K' into the foci. Take L and L', the double points of the involution (PP", KK'), P and P' being the points in which KK' cuts the conic. Now project K into a focus and L into the centre. Then {KK', LL') is harmonic ; also L' is at infinity, for since (PP', LL') is harmonic, L' is on the polar of L. Hence KK' is bisected at L, i. e. K' is the other focus. Ex. 1. Project a ffiven conic in a given plane into a circle in another given plane. Take the line AA' parallel to the intersection of the two planes, and take V in the plane through AA' parallel to the second plane. XXIX.] Projection, Real and Imaginary. 263 Ex. 2. Project a given conic into a parabola, and a given point into its focus, and a given point on the conic into the vertex of the paratiola. Suppose we want to project S into the focus, and P into the vertex of a parabola. Let SP cut the conic again in P'. Take the tangent at P' as vanishing line. Ex. 3. Project a given conic into a rectangular hyperbola, and a given point into a focus. Let two conjugate lines at S cut the conic in P and P'. Take PP' as vanishing line. 4. In the fundamental construction of § i , if the point K be outside the conic, the pencil of conjugate lines at £' is not overlapping ; hence the segments AA', BB' do not overlap ; hence the points F and V are imaginary. In this case we say that the vertex of projection is imaginary, and that we can by an imaginary projection stUl project the conic into a circle and I to infinity. Also by the Principle of Continuity proofs which require an imaginary projection are valid ; in fact we need not pause to inquire whether the projection is real or whether it is imaginary. Prove Pascal's theorem by projection. See figure of XV. i. Project MN to infinity and the conic into a circle. Then in a circle we have AB parallel to DE, and BC parallel to EF. It follows by elementary geometry that AF is parallel to CD. Hence in the original figure L is on MN. Ex. 1. Prove by Projection that the harmonic triangle (i) of an inscribed quadrangle, (ii) of a circumscribed quadrilateral are self-conjugate for the conic Project in each case into a parallelogram. Notice that a parallelo- gram inscribed in a circle must be a rectangle. Ex. 2. A,B,G,D are four points on a conic. Show that the harmonic triangle qfthe quadrilateral AB, BC, CD, DA is generally not self-conjugate. Ex. 3. Show thai the harmonic triangles of a quadrangle inscribed in a conic and of the quadrilateral of tangents at the vertices of the quadrangle are coincident. Ex. 4. A, B, C, D, A', B', C, 2/ are eight points on a conic. AB, CD, A'B', eif are cmemrent, and so are BC, DA, B'C, D'A' ; show that CA, DB, CA', D/S meet in a point, and that a conic can be draum touching A' A, B'B, ffC, D'D at A, B, C, D. Ex. 5. The chords PP', W HiJ', SS' of a conic meet in 0. Show that the two conies OPQRS and OF't^R'S' touch at 0. Project the conic into a circle and into its centre. Then the two 264 Projection, Real and Imaginary. [ch. conies are the reflexions of one another in 0. Hence the tangents at coincide. Thr. 6. If two homologous triangles be inscribed in {or cirtMmscribed to) a conic, the c. o/h. is the pole of the a. ofh. Project the polar of the c. of h. to infinity and the conic into a circle. Then in the new figure each triangle is the reflexion of the other in the centre. Hence the sides are parallel. Hence the a. of h. is at infinity ; i. e. the a. of h. is the polar of the c. of h. Hence the same is true in the original figure. Ex. 7. Two homologous triangles are inscribed in (or circumscribed to) a conic ; show thai any transversal through the centre of homology cuts the sides in pairs of points in involution. Ex. 8. Reciprocate Ex. 7. Ex. 0. A is a ficed point ; P is any point on its polar for a given conic ; the tangents from P meet a given line in Q, R. Show that the meets of AR, PQ and of AQ, PR lie on a fixed line. Project the conic into a circle and A into its centre. Ex. 10. The lines joining the vertices of a triangle ABC inscribed in a conic to a poirU meet the conic again in a, i, c ; and Ab, Be, Ca meet the polar of in R, P, Q. Show that the lines joining any point on Ote conic to P, Q, R meet BC, CA, AB in coUinear points. Ex. 11. The lines AB and AC touch a conic at B and C. The lines PQ and PR touch tlie conic at Q and R. Show by Projection that the six points A, B, C, P, Q, R lie on a conic. Through A is drawn a line cutting tlie conic in L and M and cutting QB in N, and a point D is taken such that {LM, NTT) = — i. Show that XJ lies on the conic ABCPQR. Ex. 12. If from three coUinear poiiits X, Y, Z pairs of tangents be drawn to a conic, and if ABC be the triangle formed by one tangent from each pair, and DEF the points in which the remaining three tangertts meet any seventh tangent, the lines AB, BE, CF meet at a point on XTZ. Beciprocating, we have to prove the theorem — ' If AOA', BOA', CO(f be chords of a conic, and P any point on the conic, then the meets of AB, PC, otBC, PA', and of CA, PB' lie on a line through 0.' Project to infinity the line joining to the meet of AB, PC, and at the same time the conic into a circle. The theorem becomes — ' If AA', BBf, CC be parallel chords of a circle and P a point on the circle such that PC is parallel to AB, then PB' is parallel to CA and PA' to BC This theorem follows by elementary geometry. Ex. 13. ABC is a triangle inscribed in a conic of which is the centre. OA', OB', OC bisect BC, CA, AB. Through P, any point on the conic, are drawn lines parallel to OA', OB', OC meeting BC, CA, AB in X, Y, Z; shme that X, Y, Z are coHinear. By an Orthogonal Projection, real or imaginaiy, project the given conic into a circle with as centre. Then in the circle, OA' is perpen- dicular to BC, OB' to CA, and OC to AB. Hence the theorem becomes — ' The feet of the perpendiculars drawn from any point situated on a circle upon the sides of a triangle inscribed in the circle are coUinear.' Ex. 14. Reciprocate Ex. 13. Ex. 15. Through afixedpoint is drawn a chord PP' of a conic ; show that the locus of the middle paint of PP" is a homothetic conic through and through the points of contact of tangents from 0. XXIX.] Projection, Real and Imaginary. 265 5. To project any two given imaginary points into the circular Let the two imaginary points E, F be given as the double points of the overlapping involution {AA', BB'). Take any point K in the given plane and proceed as in § i to project the angles AKA' and BKB' into right angles and AA' to infinity. Then KE and KF are the double points of the orthogonal involution K{AA', BB'), and E and F are at infinity ; hence E and F are the circular points. If E and F are real points, we can project them into the circular points by an imaginary projection ; and proofs in which imaginary projection is employed are valid by the Principle of Continuity. To project any two imaginary lines into a pair of circular lines. Let the given lines KE, KF be defined as the double lines of the involution K(AA', BB'). Draw any transversal AA'BB'. Then proceed as in § i to project the angles AKA' and BKB' into right angles. Then KE and KF, being the double lines of an orthogonal involution, are circular lines. To prefect any conic into a rectangular hyperbola. Project any two conjugate points into the circular points. To project a system of angles which cut a given line in two homographic ranges, into equal angles. Project the common points into the circular points. "B!t , 1. Deduce the construction fvr drawing a conic to towih three lines and to pass through two points from the construction far drawing a circle to touch three lines. Sx, 2. Ttie pole-locus of four given points A, B, C, B and a given line I, ImKhes the sixteen conies which can be drawn through the common conjugate points on I to touch the sides of one of the triangles ABC, ACS, ADB, BCD. Project these conjugate points into the circular points ; then I goes to Infinity. Also AD, BC meet the line at infinity in points harmonic with the circular points ; hence AD, BC are perpendicular. Similarly BD, AC are perpendicular, and also CD, AB. Also the pole-locus becomes the nine-point circle of each of the four triangles ; and this is known to touch any circle which touches the sides of any one of the four triangles. 266 Projection, Real and Imaginary. [ch. 6. To project any two conies into circles. Project any two common points into the circular points, or project one conic into a circle and a conmion chord to infinity. There are six solutions, as there are six common chords. But the projection is only real if we take a real common chord which meets the conies in imaginary points, for the line at infinity satisfies these conditions. To project a system of four-point conies into a system of coaxal dreles. Proceed as above. Ex. 1. Points P, Q, R are taken on BC, CA, AB, and conies are described through AQRLM, BRPLM, CPQLM, where L, M are any tm points. Show that these amies meet in a point. Project LU into the circular points. Ex. 2. Given turn tangents and two points on a conic, tlie locus of the meet of the tangents at these points is ttm lines. Ex. 3. Two conies pass through ABCD. AEF, BGH cui the conies in EG, FH ; show that CD, EG, FH are concurrent. Ex. 4. A variable conic passing through four fixed points A, B, C, D meets a fixed conic through AB in PQ ; s?iow that PQ passes through a fixed point. Ex. 5. A, B, C, D are four fixed points on a fixed conic. BC, DA meet in F, and AB, CD meet in G. A variable conic through ACFG cuts the fixed conic again in PQ. Show that PQ passes through the pole of BD for the fixed conic. Ex. 6. If a conic pass through two given points and touch a given conic at a given point, its chord of irUersecUon with the given conic passes through a fixedpoint. Ex. 7. On each side (UW) of the common self-conjugate triangle of two conies He two common apexes {BB') and the two poles {PP' and QQ') of two cmnnum chords (be and ad) of the conies. Also {PP', BB') and {QQf, BB') are harmonic. See figure of XIX. 8. B, B' lie on UW because BB^ and VW are both sides of the self-conjugate triangle. P, f lie on VW because be passes through r ; so (J, Q* lie on UW. Now project be into the circular points. Then P and P' are the centres of the circles, and B and f are the centres of similitude. Hence (PP", BB' ) = — i. So by projecting ad. into the circular points, we prove that (Qtf, BB') = — i. Ex. 8. Reciprocate Ex. 7. Ex. 9. Of two circles, the poles of the radical axis and the centres of similitude form a harmonic range. Ex. 10. // tangents be drawn from any point on any common chord of two conies, touching one conic in AB and the other in CD ; show that the lines AC, AD, BC, BD meet two by two in the common apexes corresponding to tile common chord. XXIX.] Projection, Real and Imaginary. 267 Ex. 11. If through any common apex of tico conies a line be draion cutting the conies in the points AB anil CD, at which the tangents are ai and cd ; show that the points ac, ad, be, bd lie two by two on the corresponding common chords. Ex. 12. If the joins of any point on any common chord of two conies to the poles of this chord cut the comes in AB and CD; show that the lines ACj ADj BCj BD meet two by tivo in the common apexes cotresponding to the common chords. Ex. 13. If three conies ham two paints in common, the opposite common chords qfthe conies taken in pairs, are concurrent. Ex. 14. If three conies have two poirUs in common, the three pairs of common apexes corresponding to the chord lie three by three on four liTies. Ex. 16. Beciprocate Ex. 13 and Ex. 14. Ex. 16. Two conies a and meet at B, C, and touch at A. DEG touches a at E and at G. DFH touches a at F and at H. Show that EF, BC, GH meet {at K say) on the tangent at A, and that the poles of BC for a and lie on DA and divide it hannonically. Show also that A {KD, BC) = D {AK, EF) = K (FH, ^C) = - 1. Ex. 17. The envelope of a line which meets two given conies in pairs (if hamwnic points is a conic which touches the eight tangents to the conies at their meets. Let the conies meet in ABCD. Project AB into the circular points. Then by Ex. a of 111. 6, the envelope of the line is a conic which touches the four tangents at C and D. So by projecting CD into the circular points, we prove that the envelope touches the tangents at A and B. Ex. 18. Proce Ex. it by one projection. Ex. 19. If the given conies be two parabolas with axes parallel, the envelope is a parabola with axis paralld to these axes. Ex. 20. The locus of a point the tangents from which to two given cmks are pairs of a harmonic pencil is a conic on which lie the eight pq^nts in which the given conies touch their common tangents. Ex. 21. Two equal circles touch. Show that the locus of a point, (he pairs of tangerdsfrom which to Die circles are harmonic, is a pair of lines. For if the circles touch at A and the common tangents touch them at BC, DE, the lines BAE, CAD contain the eight points, four being at A. Ex. 22. IfSA, SA', S'A, S'A' be the common tangents of two circles, S and S' being the centres of similitude, and if the angles at A and A' be right, show that the above locus breaks up into a pair of lines. For the four polars of the other tvro common apexes bisect the angles between SS' and AA'. Ex. 23. The tangents to a system of four-point conies at their meets form four homographic pencils. Ex. 24. Sedprocate Ex. 23. Ex. 25. Jf two conies be so situated tluU two of their meets AB subtend at an/other meet C an angle which divides harmonically the tangents at C, the same is true for AB at D, for CD at A, and for CD at B. 268 Projection, Real and Imaginary. [ch. Apply Ex. S3 to the four conies consisting of the two given conies iind the pair of lines AC, BD and the pair AD, BO. Ex. 26. In such conies, the envelope of the lines which divide the two conies harmonicaUy degenerates into two points. Sz. 27. Reciprocate Ex. as and Ex. 26. Ex. 28. Four parabolas are drawn with their axes in the same direction to touch the four triangles formed by four points ; show titat they have a common tangent. A particular case of the more general theorem — ' Four conies are drawn to touch two given lines and to touch, &o.' Reciprocate, and project the given points into the circular points. Ex. 29. A polygon is inscribed in one of » system of four-point conies, and each side but one touches a amic of the system ; show that the remaining side also touches a conic of the system. For the theorem is true for coaxal circles by Poncelet's theorem. Ex. 30. Beciprocate Ex. 29 ; and deduce a property ofcanfoaA arnica. 7. To project any two conks into confocal conies. Let the opposite vertices of the quadrilateral circumscribed to both conies be AA', BB', CC. Project AA' into the circular points ; then the conies have the foci BB", CC in common, i.e. are confocal. To project a system of conies inscribed in the same quadrilateral into confocal conies. Project a pair of opposite vertices of the circumscribing quadrilateral into the circular points. Ex. 1. A variable conic touches four fixed lines ; from the fixed points B, C taken on two of these lines the other tangents are drawn ; find the locus of Iheir meet. Project BC into the circular points. Ex. 2. The line PQ touches a conic. Find the locus of the meet of tangents qf the conic which divide PQ {i) harmonicaUy, (ii) in a constant cross ratio, Ex. S.ffa series of conies be inscribed in the same quadrilateral of which A A' is a pair of opposite vertices, and from a fixed point 0, tangents OP, OQ be draum to one of the conies, the conic drawn through OPQAA' toill pass through a fourth fixed point. Project AA' into the circular points, and see Ex. 4. of XXVIII. 10. Ex. 4. Seciproeate Ex. 3. Ex. 6. ]f two conies be inscribed in the same quadrilateral, the two tangents at any of their meets cut any diagonal of the quadrilateral harmonically. Ex. 6. Given the cross ratio of a pencil, three of whose rays pass through fixed points and whose vertex moves along a fixed line, the envelope of the fourth ray is a conic touching the three sides of the triangle formed by the given poirtts. XXIX.] Projection^ Real and Imaginary. 269 Ex. 7. Thx locus of the point where the intercept of a variable tangent of a central conic between two fixed tangents is divided in a given ratio is a hyperbola whose asymptotes are parallel to the fixed tangents. This is a particular case of the theorem^' If a tangent of a conic meet two fixed tangents AS, AC in P, Q and a fixed line I in U, and if B be taken such that (PQ, RU) is constant ; then the locus of iJ is a conic through the meets B, C of I with the fixed tangents.' To prove this project BC into oo oo '. Then we have to prove that — ' If through the focus S of a conic, a line SB be drawn making a given angle with a variable tangent QB, then the locus of B is a, circle.' This can be proved by Geometrical Conies. 8. To project any two conies into homothetie conies. Project any common chord to infinity. The new conies will pass through the same two points at infinity, and hence are homothetie. (See XIX. 11, end.) To project any two conies which have double contact into homo- thetie and concentric conies. Project the chord of contact to infinity. The pole of the chord of contact projects into the common centre. Ex. The point V on a conic is connected with two fixed points L and M. Show that chords of the conic which are divided harmonically by VL and VM pass through a fixed point 0. Also as V varieSj the locus of is a conic touching the given conic at two points on the join of the fixed points L and M. 9. To project any two conies having double contact into con- centric circles. Project the two points of contact into the circular points. Then the conies will both pass through the circular points, i.e. will both be circles. Also they will both have the same pole of the line at infinity, i.e. they will be concentric. Ex. 1. Conies having the same focus and corresponding directrix can be pro- jected into concentric circles. For the focus S has the same polar, and the tangents from S are the same. Hence the conies have double contact. Ex. 2. Through the fixed point is drawn a chord OAB of a conic, and an OAB is taken the point P such that (OABP) is constant. Find the locus of P. 10. The lines which jmn pairs of corre^onding points of two homographic ra/nges on a conic, touch a conic having double contact with the given conic at the common points of the ranges. Let {ABC.) and {A'B'C'...) be the two homographic ranges, and E, F their common points. Project the conic into a circle and the homographic axis EF to infinity. Then E, F are projected into the circular points. 270 Projection, Real and Imaginary. [ch. Now in the second figure, AB' and A'B meet on the homographic axis. Hence A'B! and A'B are parallel. So AC and A'C are parallel, and so on. Hence the arcs AA', BB', CC, ... are all equal. Hence the envelope of AA' is a concentric circle, i.e. a circle having double contact with the circle which is the projection of the given conic at the circular points E, F. Hence in the original figure the envelope of A A' is a conic having double contact with the given conic at the double points of the two homographic ranges. Ex. 1. Two conies have double contact, and a tangent to one conic meets the other conic in A and A'. Show that A and A' generate homographic ranges, and find the common points of these ranges. Ex. 2. If {ABC ..) and (A'B'C'...) be two homographic ranges on a conic, shcm that the locm of the poles of A A', BBf, ... is a conic having doiMe contact with the given conic. Ex. 3. Tiie points of contact of the tangents AA', BB'. CC', ... form a range on. the envelope homographic xoith the ranges ABC... and A'B'C... Ex. 4. Show that the tangents at ABC... and A'B'C... cut the homographic axis in homographic ranges. For equal angles cut the line at infinity in homographic ranges. Ex. 6. IfObe the pole of the homographic axis of the ttoo homographic ranges on a conic, then 0{ABC...) = 0(A'ffC'...). Ex. 6. IfaUbut one of the sides of a polygon pass through fixed points and ail the vertices lie on a conic, then the envelope of the remaining side is a conic having double contact with the given conic. For the last side determines homographic ranges on the conic. Ex. 7. If all but one of the vertices of a polygon move on fixed lines and aU the sides touch a conic, the locus of the remaining vertex is a conic having doubU contact with the given conic. Ex. 8. Turn sides of a triangle inscribed in a conic pass through fixed points ; show that the envelope of the third is a conic touching the given conic at the meets of the given conic with the join of the given points. Ex. 9. A triangle PQB, is inscribed in a conic; PQ, PR are in given direc- tion ; show that QB envelopes a conic. Ex. 10. The envelope of chords of a conic which subtend a given angle at a given point on the conic is a conic having double contact with, tfie given conic. Ex. U. A, B are ttoo fixed points on a conic, and P, Q two variable points on the conic such that (AB, PQ) is constant; sliow thatPQ envelopes a conic which touches the given conic at A and B. Ex. 12. Show also that the locus of t?ie meet ofAQ and BP, and the locus of the meet of AP and BQ, are both conies having double contact wiOi the given conic at A and B. For A {ABQ...) =B{ABP...) and A {ABP...) = B{ABQ...). XXIX.] Projection, Real and Imaginary. 271 Ex. 13. Inserihe in a given conic a polygon of any given number of sides, each side of which sliaU Umck some fixed conic having doubk contact with the given conic. Ex. 14. If tangents be drawn from points on a conic to a conic having double contact with it, the points of contact generate homographic ranges on the cordc. Ex. 15. A conic is drawn through the common points E, F of two homographic ranges A, B, C, ... and A', B', C, ... on the same line. A pair of tangents nunes so as to pass through a pair of points of these ranges. Show that the points of contact generate homographic ranges on the conic, whose common points are E and F. Ex. 16. Also if P be any point, and PA cut the conic in aa, and A' a cut the conic in a'; show that aa' generate homographic ranges on the conic. Ex. 17. Through a point P is drawn a chord cutting a conic in a a, and a point a' is talon on the conic such that the angle aaa' is constant; show that aa' generate homographic ranges. Here AB... A'B' ... is at infinity. Ex. 18. Reciprocate examples 15, 16 and 17. Ex. 19. If two conies a and 3 have double contact at the points L and M ; and through LM be described any conic y, then the opposite two common chords of ay and 0y meet on LM. Ex. 20. Any angle whose legs pass through L and M res/pecUvely, intercepts chords on a and which meet on LM. A particular case of Ex. 19. Ex. 21. If two hyperbolas have the same asymptotes, any two lines parallel to the asymptotes intercept parallel chords of the hyperbola. Ex. 22. Any two lines parallel to the asymptotes of a hyperbola intercept parallel chords on the hyperbola and its asymptotes. Ex. 23. Sedprocaie Ex. 23. Ex. 24. If tangents at the two points P, Q on one of two conies having double contact at L and M meet the other in AB and CD, show that two of the chords AC, AD, BO, BD meet PQ on LM, and the other two meet PQ in points UV sudi that a conic can be drawn touching these chords at U and V and touching the conies at L arid M. Ex. 25. Reciprocate Ex. 34. Ex. 26. If a tangent to a conic meet a homothetic and concentric conic in P and P', show that CP and CP' generate homographic pencils whose common lines are the common asymptotes, C being the common centre. CHAPTEE XXX. GENERALISATION BY PROJECTION. 1. In the previous chapter we have investigated theorems by projecting the given figure into the simplest possible figure. In this chapter we shall deal with the converse process, viz. of deriving from a given theorem the most general theorem which can be deduced by a projection, real and imaginary. This process is called Generalising iy Pro- jection. In our present advanced state of knowledge of Pure Geometry, Generalisation by Projection is not a very valuable instrument of research. In fact the student Trill often find that it is more easy to prove the generalised theorem than the given theorem. Many things are as general already as they can be. For instance, if we generalise by projection a point, a line, a conic, a harmonic range, a range having a given cross ratio, two conies having double contact, and so on, we obtain the same thing. 2. The properties of any figure have an intimate relation with the circular points 00,00'. Hence the generalised figure will have an intimate relation with the projections of the circular points. But in the second figure there will also be a pair of circular points. Hence, to avoid confusion, we shall call the projections of the circular points ■sr and or'. 3. Since any two points can be projected into the circular points, the circular points generalise into any two points w and ■a', real or imaginary. Generalisation by Projection. 273 Since a pair of circular lines pass through the circular points, a pair of circular lines generalise into a pair of lines, one through w and one through ■as'. Since all circles pass through the circular points, a circle generalises into a conic which passes through ts and nr', where ■ST and is' are any two points. Since concentric circles touch one another at the circular points, concentric circles generalise into conies touching one another at m and at m'. Since the line at infinity touches a parabola, a parabola generalises into a conic touching the line ra-sr'. Notice that we cannot generalise the distinction between a hyperbola and an ellipse ; for by an imaginary projection a pair of real points may be projected into a pair of imaginaiy points and vice versa. Since a rectangular hyperbola is a conic for which the circular points are conjugate, a rectangular hyperbola gene- ralises into a conic for which -or, ■a' are a pair of conjugate points. Since the centre of a conic is the pole of the line at infinity, the centre of a conic generalises into the pole of the line war'. Hence a circle on AB as diameter generalises into a conic passing through AB-ara', and such that the pole of the line ■mxn' is on AB. Since parallel lines meet on the line at infinity, parallel lines generalise into lines which meet at a point on the line 'GTZT . Note that throughout this chapter, sr and m' are any two points, real or imaginary. 4. If £ bisects the segment AC, then the range {AC, BQ) is harmonic ; hence 'B bisects AC generalises into 'J£ AC meet crts'' in I, then B is such that {AC, BI) is harmonic, -sr and «/ being any two points.' Generalise by Projection the theorem — ' CHven two concentric circles, (my chord of one which touches the other is bisected at the point of contact' 2 74 Generalisation by Projection. [ch. The result is — ' Given two conies touching one another at any two points or and or', if any chord 'P'P' of one, touch the other at Q and meet oror' in J, then (PP', Qi") is harmonic' Or, without mentioning is and is/, — ' Given two conies having double contact, if any chord PP' of one, touch the other at Q and meet the chord of contact in I, then (PP', Q/) is harmonic' The student should convince himself by trial that the second theorem can be projected into the first, and that the second theorem is the most general theorem which can be projected into the first. Generalise by Projection the following theorems — Bz. 1. Qiven three concentric circles, any tangent to one is cut by Oie other turn in /our points whose cross ratio is constant. Sx. 2. The middle points qfpardOd chords of a circle lie on a line which passes thro^h the centre of the cirde. Ex. S. If the directions of two sides of a triangle inscribed in a cirde are given, then the envelope of the third is a concentric circle. lEx. 4. CHven four points on a conic, the locus of the centre is the conic through the middle points qf the six sides of the quadrangle formed by the four given points. 5. If A VA' is a right angle, then VA and VA' divide the segment joining the circular points harmonically ; hence a right angle A VA' generalises into an angle A VA', such that VA and VA' divide the segment joining any two points or, ■a' harmonically. Generalise by Projectum the theorem — ' The perpendiculars to the sides of a triangle at the middle points of the sides meet at the centre of the drcum-circle.' The result is — 'If the sides BC, CA, AB of a triangle meet the segment joining any two points nr and ■a/ ia L, M, N ; and if X, Y, Z he taken such that (ww', XL), (ctw', TM), (wo/, ZN) are harmonic ; and if D, E, F be taken such that {BC, JDL), {CA, EM), {AB, FN) are harmonic ; then BX, ET, FZ meet at the pole of oror' for the conic which passes through ABCtsris/.' Generalise by Projection the following theorems — Ex. L A tangent of a circle is perpendicular to the radius to the point of coTitact. XX X.J Generalisation by Projection. 275 "Ex.. 2. The feet of the perpendiculars from any point on a circle on the siden of an inscribed triangle are collinear. XjX. 3. The hcus of the meet of perpendicular tangents of a conic is a concentiic circle. XiZ. 4. The circle dboiU any triangle self-conjugate for a conic is orthogonal to its director circle. £x. 6. The chords of a conic which subtend a right angle at a fixed point on the conic pass through a fixed point on the normal at the point. Ex. 6. If a triangle PQB, right-angled at P, be inscribed in a rectangular hyperbola, the tangent at P is the perpendicular from P on QR. 6. Since all circles pass through the circular points, a system of circles generalises into a system of conies passing through the same two points (or and «/). Since coaxal circles pass through the same four points of which two are the circular points, coaxal circles generalise into a system of conies which pass through the same four points (of which two are ■or and cr'). Since the limiting points of a system of coaxal circles are the two vertices of the common self-conjugate triangle which lie on the line joining the poles of 00 00 ', the limiting points generalise into the two vertices of the common self-conjugate triangle of a system of four-point conies which Ue on the line joining the poles of any common chord (■fw'), i.e. they generalise into any two vertices of the common self-conjugate triangle. Since the centres of similitude of two circles are the tw^o intersections of common tangents which lie on the line joining the poles of 00 00 ' for the circles, the centres of simi- litude of two circles generalise into the two intersections of common tangents of two conies (through ■or and isr') which lie on the line joining the poles of any common chord (■atr') for the conies, i.e. they generalise into any pair of opposite common apexes of two conies. 7. Generalise by Projection the theorem — 'Any common tarn- gent of two circles subtends a right angle at either limiting The result is — ' If w and w' be any two common points of two conies, and if L and L' be the two vertices of the T 2 276 Generalisation by Projection. [ch. common self-conjugate triangle which are collinear with the poles of wcr', then any common tangent of the conies sub- tends at L (and at L') an angle whose rays divide the segment 'srsr' harmonically.' In other words, — ' Any common tangent of two conies sub- tends at any vertex of the common self-conjugate triangle an angle which divides harmonically every common chord which does not pass through this vertex.' Generalise by Projection the theorems — Ex. 1. Any Iramxiffrsal meets a system of coaaxil circles in pairs 0/ points in invoiyMcn. £!z. 2. The circle of simiiiiitde of turn circles is coaxal with them. 8. Since a focus of a conic is one of the four meets of the tangents from the circular points to the conic, a focus of a conic generalises into one of the meets of the tangents from any twff points (ra- and ot') to the conic. The two fod of a conic generalise into a pair of opposite vertices of the quadrilateral of tangents from any two points (w and o/). Since the line joining the circular points touches a para- bola, the focus of a parabola generalises into the meet of tan- gents from any two points {isr and w') lying on any tangent of a conic. Since confocal conies touch the same four tangents from the circular points (viz. Soo , S'co , Sx', S'co '), confocal conies generalise into conies inscribed in the same quadri- lateral (of which ■a and ts' are a pair of opposite vertices). Since conies which have the same focus 8 and the same corresponding directrix I touch Soo , 8 ', where 2 meets these lines, conies which have the same focus 8 and the same corresponding directrix I generalise into conies having double contact, the common tangents passing through S (and through isr and v/), and touching the conies at points on I. A conic having 8 as focus generalises into a conic touching any two lines (Sw and Sw') through 8. XXX.] Generalisation by Projection. 277 9. Generalise hy Projection the theorem — 'The circle which circumscribes a triangle whose sides toiKh a parabola passes through the focus of th£ parabola.' The result is — ' The conic which passes through the points A, B, C, ta, ii/, where ■nr and •bt' are any two points, and A, B, C are the vertices of a triangle whose sides touch a conic which touches the line ■bt's/, passes through the meet of tangents to the latter conic from cr and t!',' In other words — 'The conic, which passes through five out of the six vertices of two triangles which circumscribe a given conic, passes through the sixth also'. Sx. 1. Given iteo points on a cofUc, find the locus of the pole of their join, given also either (i) turn tangents, or (ii) a tangent and a point. Generalise by Projection the following theorems — Ex. 2. Any line through a focus of a amic is perpendicular to the line joining its pole to the focus. lix. 3. Oiven a focus and tu>o tangents of a conic, the locus of the other focus is a line. Sx. 4. The locus of the centre of a circle which touches two given cirdes is a conic having the centres of the circles as foci. Six. 5. The locus of the centre of a circle which passes through a fixed point and touches a fixed line is a parabola of which the point is the focus. "Ex. 6. Confocal conies cut at right angles. Ex. 7. The envelope of the polar of a given point for a system of cmfocah is a parabola tomihing the axes of the amfocals and having the given point on its directrix. 10. Since the rays of an angle of given size divide the segment joining the circular points in a given cross ratio, a constant angle generalises into an angle whose rays divide the segment joining any two points {ist and ts/) in a constant cross ratio. Generalise by Projection the theorem — ' The envelope of a chord of a conic which subtends a constant angle at a focus S is another conic having 8 as focus; and the two conies have the same directrix corresponding to S.' The result is — 'The envelope of a chord of a conic which subtends at S, one of the meets of a tangent from any point «r with a tangent from any point w', an angle whose rays divide isia' in a constant cross ratio, is another conic. 278 Generalisation by Projection. [ch. touching Sot and Sor'; and the two conies have the same polar of iS.' In other words — ' If SQ and SU be the tangents from any point S to a conic, the envelope of a chord 'PV of the conic such that S{QEPP') is constant, is a conic having double contact with the given conic at the points of contact of SQ and SB.' Hz. 1. Oeneralise — ' a regular polygon.' A regular polygon may be defined as h polygon which can be inscribed in a circle so that each side subtends the same angle at the centre of the circle. Generalise by Projeetion the follorring theorems — Hz. 2. 7%e moelope of a chord of a circle which svhtends a given angle at any point of the circle is a coruxntric circle. Ex. 3. If from a fixed poM 0, OP be drawn' to a given drcU, and TP be drawn making the angle TPO constant, the envelope ofTPis a conic with as focus. Ex. 4. If from a focus of a conic a line he drawn maJcing a given angle with a tangent, the locus of the point of intersection is a cirde. Ex. 6. The locus of the intersection of tangents to a parabola which meet at a given angle is a hyperbola having the same focus and corre^aonding directrix. 11. Generalise — ' The bisectors of cm angle.' If AB, AE are the bisectors of the angle BAG, then A {BC, BE) is harmonic, and also .4 (00 00,', BE) since EAB is a right angle. Hence the bisectors of the angle BAC generalise into the double lines of the involution ^(J?(7, wo-'), where cr and ■sj' are any two points. Ex. Generalise by Projection — ' The pairs of tangents from any point to a system of amfocals have the same bisectors.' 12. Generalise — ' a segment divided in a given ratio.' Let AB be divided at C in a given xatio. Then AC:CB is constant ; hence {AB, Oil) is constant, where il is the point at infinity upon AB. Hence a segment AB divided in a given ratio at C genei-alises into a segment AB divided at C so that {AB, CI) is constant, I being the meet of AB and the segment joining any two points (w and vr'), , ' XXX.] Generalisation by Projection. 279 Ex. 1. Qeneralm the equation AB + BC+CA = o connecting three colUnear points. The given equation may be written -{AC,Ba) + i-{AB,Cn) =0. This generalises into - {AC, £/) + i - (AB, CT) = o, i. e. into AB.CI+AI.SC + AC.IB = o. Hence the generalised theorem is—' If A, B, C, D be any four collinear points, then AB.CD + AC.BB + AD.BC ^o.' Ex. 2. If ABCB be collinear, show thai the ratio AB-i-CB generalises into -{BC,AE)-~-{BA,GE). Ex. 3. ff AB and CB be parallel and if AC, BB meet in M, show that the ratio AB-i-CB generalises into (AC, ME), E being the meetqfAB and CD. 13. Ttvo fixed points A and B on a conic are joined to a variable point P on the conic, and the intercept QR cut off from a given line I hy PA amd PB is divided at M in a given ratio ; show that the envelope of PM is a conic touching parallels to I throitgh A and B. Let il be the point at infinity on I. Then (QB, MQ) is a given cross ratio. Hence P (AB, MQ,) is given. Project A and B into the circular points and let I be the projection of i2. Then P(oo oo ', MI) is given, i.e. IPM is a given angle. Hence the theorem becomes — 'A fixed point / is joined to a variable point P on a circle, and PM is drawn making a given angle with IP : show that the envelope of PM is a conic touching 7 00 and Joo ', i.e. is a conic having I as focus.' And this is true (see VIII. 1 7). Hence the original theorem is true. Generalisation by Reciprocation. 14. If we first generalise a given theorem by projection and then reciprocate the generalised theorem, we obtain another general theorem. This process is called Generalising by Projection and Beciprocation, or briefly Generalising by Be- ciprocation. Generalise by Beciprocation the theorem — 'All normals to a circle pass through the centre of the drcle.' Generfilising by Projection we get — 'If t be the tangent at any point P of a conic which passes through any two 28o Generalisation by Projection. points v, ■a', and if the line n be taken such that t and n are harmonic with Per and Pw', then w passes through the pole of ■ere/ for the conic' Seciprocating this theorem we get — ' If on the tangent at any point T of a conic, a point N be taken such that the segment TN is divided harmonically by the tangents from the iixed point 0, then N lies on the polar of for the conic' This is the required theorem. XjZ. Oeneralise by Projection and Reciprocation the theorem — ' Tfie envelope of a chard qfa cirde which subtends a constant angle at the centre is a concentric circle.' CHAPTEE XXXI. HOMOLOGY. 1. Two figures in the same plane are said to be in homology which possess the following properties. To every point in one figure corresponds a point in the other figure, and to every line in one figure corresponds a line in the other figure. Every two corresponding points are colUnear with a fixed point called the centre of homology, and every two corresponding lines are concurrent with a fixed line called the axis of homology. The line joining any two points of one figure corresponds to the line joining the two corresponding points of the other figure. The point of intersection of any two lines of one figure corresponds to the point of intersection of the two corresponding lines of the other figure. The two figures are said to be homologous, and each is called the homologue of the other. The figures may be said to be in plane perspective ; and the centre of homology i§ then called the centre of perspective, and the axis of homo- logy is called the axis of perspective. 2. Homologous figures exist for — If we take two figures in different plcmes, each of which is the projection of the other, and if we rotate one of the figures about the meet of the two planes wntil the planes coincide, then the figures will be homologous. For let ABC be the projections of A'B'C from the vertex F. Then AA', BB', GO' meet in V. That is, the triangles ABC, A'B'C (in different planes) are copolar. Hence they 282 Homology. [ch. are coaxal ; i.e. BC, B'C meet in a, and CA, C'A' meet in P, and AB, A'B' meet in y on the' meet of the two planes. Similarly every two lines which are the projections, each of the other, meet on the intersection of the two planes. Now rotate one figure about the line a/3y untU the two figures are in the same plane. Then the two triangles are still coaxal (for BG, B'C still meet at a, and so for the rest). Hence the two triangles are also copolar ; i.e. AA', BB', CC meet in a point. Call this point 0. Then may be defined as the meet of A A' and BB', and we have proved that every other line such as CC passes through 0. Now the two figures are in the same plane. Also to every point in one figure corresponds a point in the other figure, viz. the point which was its projection ; and to every line corresponds a line, viz. its former projection. Also, cor- responding points are concurrent with a fixed point 0, and corresponding lines are coUinear with a fixed line a/3y. Also, the join of two points corresponds to the join of the corresponding points ; for in the former figure the one is the projection of the other. For the same reason, the meet of two lines corresponds to the meet of the corresponding lines. Hence the two figures are homologous. XXXI.] Homology. 283 3. If two figures are homologous, and we turn one of them about the axis of homology, the figures mil be the projections, each of the other. For suppose the three lines BC, CA, AB in one figure to be homologous to B'C, C'A', A'B' in the other figure. Let BG, B'C meet in a, let CA, C'A' meet in /3, and let AB, A'B' meet in y. Rotate one of the figures about the axis of homology afiy, so that the figures may be in different planes. The figures will now be each the projection of the other. For the triangles ABC, A'B!G' (in different planes) are coaxal ; hence they are copolar. Hence AA', BB', CC meet in a point V. This point V may be defined as the meet of AA' and BB'; and we have proved that in the displaced position the join CC of any two homologous points passes through a fixed point Y. Hence the homologous figures in the displaced position are projections, each of the other. A homologue of a conic is a conic. For after rotating one figure about the axis of homology, the figures are each the projection of the other ; and the pro- jection of a conic is a conic. A homologue of a figure has all the properties of a projection of the figure. For it can be placed so as to be a projection of the figure. Hence a range and the homologous range are homographic ; also a pencil and the homologous pencil are homographic. 4. If one of two figures i/n perspective {i.e. either homologous or each the projection of the other), be rotated about the axis of perspective, the figures will be in perfective in every position ; and the locus of the centre of perspective is a circle. For take any two corresponding triangles ABC and A'B'C- Then in every position these triangles will remain coaxal ; hence in any position they wiU be copolar, i.e. CC will pass through the fixed point Y determined as the meet of AA' and jRB'. Hence the figures will be in perspective in any position obtained by rotating one figure about the axis of perspective. To find the locus of Y. Take any position of F, and 284 Homology. [ch. through F draw a plane P'iP at right angles to the planes of the figures, cutting them in L'P' and HP. Let a parallel to iP' through Y cut LP in J, and a parallel to LP through F cut LP' in T. Let the point at infinity on LP be called I, and the point at infinity on LP' be called J'. Then, since /' F and LJ are parallel, we see that I' F passes through 1, i.e. I' is the projection of I for this position of F; and so J is the projection of J'. Now rotate the moving plane about the axis of perspective into any other position. The new position of the centre of perspective (or vertex of projection) is got by joining any two pairs AA', BB' of corresponding points. Hence in the new position II' and J J' will cut in F. Also LJ is still parallel to I'V, for I is at infinity; so JV is parallel to LI'. Also if LJ is the trace on the fixed plane, then LJ is con- stant in magnitude and position. Also LI' is constant in magnitude, although it changes its position by rotation about the axis of perspective. It follows that LJVI' is a parallel- ogram, in which J is fixed, and JV is given in magnitude. Hence the locus of F is a circle in a plane perpendicular to the planes of the figures, with centre J and radius LI'. To form a clear conception of figures in homology, imagine that they are the projections, each of the other, the vertex of projection very nearly coinciding with the centre of homo- logy, and the planes of the figures very nearly coinciding with one another. XXXI.] Homology. 285 5. Coaxal figures are cqpolar, and copolar figures are coaxal ; that is to say, if two figures, {in the same plane or not, ) correspond, point to point, line to line, meet of two lines to meet of correspond- ing lines, amdjoin of two points to join of corresponding points; then, if corresponding lines cut on a fixed line, the joins of cor- responding points tvill pass through a fixed point, and if the joins of corre^onding points pass through a fixed point, corresponding lines will cut on a fixed line. Coaxal figures are copolar. Take two fixed points A, B in one figure, and let A', B' be the corresponding points in the other figure. Take any variable point P in one figure, and let JP' be the corresponding point in the other figure. Then, by definition, AP, A'P' are corresponding lines, for they join corresponding points ; hence AP and A'P' meet on the axis. Similarly BP, B'P' meet on the axis ; and AB, A'B' meet on the axis. Hence the triangles ABP, A'B'P' are coaxal, and therefore copolar. Hence AA', BB', PP' meet in a point, i.e. PP' passes through a fixed point, viz. the meet of A A' and BB'. Copolar figures are coaxal. Take two fixed lines, viz. AP and AQ, and a variable line PQ in one figure, and let A'P', A'(^, P'§' be the corresponding lines in the other figure. Then the points A, P, Q correspond to A!, P', (^. Hence the triangles .4PQ, A'P'Qf are copolar, and therefore coaxal. Hence PQ, P'Q' meet on a fixed Une, viz. the join of the meets of AP, A'P' and of AQ, A'Qf. Hence the figures are coaxal. "Ex.. 1. Xf one of two figures in homology be turned through turn right angles about the ancis of homology, the figures wiU again be in homology. Ex. 2. If one of two figures in homology be turned tlirough two right angles about an axis which passes through t?te centre of homology and is perpendicutar to the plane of the figures, the figures %Dill again be in homology. Ex. 3. Oiven two homologous figures ABC..., A'B'C...; let A"B"Cf'... be a prcgection of ABC... on any plane through the axis of homology ; then uM A"S'0' ... be also a projection of A'B'C... , and the vertices of projection and the centre of homology will be coUinear. For VO is one of the lines A' A", &c. This construction enables us to place any two homologous figures in projection with the same figure. 286 Homology. [ch. Ex. 4. ShmD that tfie two complete guadrangles determined by ABCD and A'B'&I/ wUl be homologous provided the five points of intersection qf AB with A'B'jOfBCwithB'C, qf CA with C A' , of AD with A' 1/ , and of BD with ff 1/ are cMiruar. Ex. 5. Show that the turn complete qaadrUaierals whose vertices are ABCDEF and A'B'CI/E'F' wOl be homologous if AA', BB', CC, BBf, EE' meet in a point. Ex. 6. The sides PQ, QR, RP of a variable triangle pass through fixed points CAB in a line. Q moles on a fixed line. Shou) that P and R describe homologous curves. For PR and P'R' pass through B, and RR', PP' meet on QCf, P'tyR' being a second position otPQR. Ex. 1. If the axis cf homology be at infinity, show (i) that corresponding lines are parallel, (li) that corresponding sides of the figures are proportional, (iii) that corresponding angles of the figures are equal. Such figures are called homothetic figures, and the centre of homology in this case is called the centre of similitude, and the constant ratio of corresponding sides is called the ratio qf similitude. Ex. 8. If, with any vertex of projecUvn, we project homologous fiffures on, to any plane, we obtain homologous figures ; and if the plane of projection be taken parallel to the plane containing the vertex of projection and the axis of homology, we obtain homothetic figures. Hence homologous figures might have been defined as the projections of homothetic figures. Ex. 9. If the centre of homology be at infinity, show that the joins of corre- sponding poiTits are all parallel ; and that if tme figure be rotated about the axis of homology, the vertex of projection wiU always be at infinity. This may be called parallel homology. Ex. 10. In parallel homology, s?u)w thai to a point at infinity corresponds a point at infinity, and that the line at infinity corresponds to itself. Ex. 11. In parallel homology, show that a parallelogram corresponds to a parallelogram. Ex. 12. In parallel homology, show tliat, when rotated about the axis of homology into different planes, the figures have the same orthogonal projection ; and that the ratios of two areas is the same as that of the corresponding areas. 6. The abbreviation c. of h. will be used for centre of homology, and a. of h. for axis of homology. Given the.c. ofh. and the a. ofh. and a pair of corresponding points, construct the homologue of a given point. Let be the c. of h., and let A' be the given homologue of A. To find the homologue of X ; let AX cut the a. of h, in L, then LA' cuts OX in the required point X^. With the same data, construct the honwhgue of a given line. Draw through any transversal cutting the given line in X ; construct the homologue X' of X, then the join of X' to XXXI.] Homology. 287 the point M, where the given line cuts the a. of h., is the homologue of the given line. Gimn the c. ofh. and the a. ofh. and a pair of coiresponding lines, construct the homologue (i) of a given point, (ii) of a given line. Let any transversal through cut the given Unes in A and A'. Then A, A' are corresponding points, and we may proceed as above. Criven the c. ofh. and the a. ofh. and a pair of corresponding points, one of which is at infinity, construct the homologite of a LX' is parallel to AA', if A' is at infinity. 7. The hymohgue of the c. ofh. is the c. ofh. ; the homologue ofamypoinit on the a. ofh. is the point itself; if the homologue of any other point be itself, then the homologue of every point is itself. For let us construct the homologue of 0. We draw AO cutting the a. of h. in ^; we draw NA' cutting 00 in the required point. Now 00 is indeterminate, but NA' cuts every line through in 0, and hence cuts 00 in 0. Hence the homologue of is 0. Next, let us construct the homologue of any point L on the a. of h. We draw AL cutting the a. of h. in Z ; we draw A'L cutting OL in the required point. Hence the homologue of L is L. Lastly, suppose a point (which is not at the c. of h. nor on the a. of h.) to coincide with its homologue. Take these as 288 Homology. [ck. the points A, A' in the above construction. To construct the homologue of X, we draw A'X. cutting the a. of h. in X ; then A'li cuts OX. in the required point X'. Hence Z' coincides with X, for A'L coincides with AL, i.e. with AX. "Ex.. 1. Show that the only lines which coincide with their homologues are the u. o/h, and lines through the c. ofh. £iX. 2. Qiven the homologues A', ff, C of three points A, S, C; construct Oie homologue of a given point D. The triangles give the centre and axis of homology. 8. The homologue of a point at infinity of one figure is called a vanishing point of the homologous figure. The homologue of the line at infinity considered as belong- ing to one of the figures is called the vanishing line of the homologous figure. All the vanishing points of either figure Ik on the vanishing line of that figure. For a vanishing point is the homologue of a point on the line at infinity of the other figure, and hence lies on the homologue of the line at infinity. Each vanishing line is parallel to the a. o/h. For corresponding lines meet on the a. of h. Hence a vanishing line meets the a. of h. at a point on the line at infinity, Le. a vanishing line is parallel to the a. of h. Six. 1, ff any transversal through cut the axis in N, and the vanishing lines in I and J' , then 01 = ^N, and 01 . OJ' = J'N . IN. For {NI, on) = (JVfl', OJf). Ex. 2. The product of the perpendiculars from any tioo homologous points, each on the vanishing line of Us figure, is constant. For (PQ, la) = {P'tf, n'J'). Ex. 8. Oiven a parallelogram ABCD, prove the following construction for drawing through a given point E a paraM to a given line I — Let AB, CD, AC, BC, AD cut I in K, L, U, N, B. Through M draw any Une cutting EK, EL in A', C. Let BA' and N'& cut in F. Then EF isparallel to I. For EF is the vanishing line. 9. Given the c. ofh., the a, ofh., and a pair of cmrrespondvng points ; construct the vanishing lines. Let AA' be the pair of corresponding points. Let us first construct the homologue of the line at infinity, considered to belong to the same figure as il. In the construction of XXXI.] Homology. 289 § 6, X and M. are both at infinity. Hence the construction is— Through the c. of h. 0, draw any line OX (X being at infinity). Through A draw AL parallel to OX, cutting the a of h. in i ; then LA' cuts OX in X' ; and the required line is X'M, i.e. a parallel through X' to the a. of h. Similarly we construct the vanishing line of the other figure. Given the c. ofh., the a. ofh., cmd one vanishing line, to con- struct the homologue of a given point. Let any transversal through the c. of h. cut the vanishing line in A. Then the homologue of the point A is the point at infinity A' on OA. Two cases arise, (i) The given point X belongs to the same figure as the finite point A. Let AX cut the a. of h. in L ; draw through L a parallel to OA to cut OX in X'. Then X' is the homologue of X. (ii) The given point X' belongs to the same figure as the point at infinity A'. Through X' draw a parallel to OA cutting the a. of h. in L. Then AL cuts OX' in the required point X. ISx. GUven the c. ofh. and the a. ofh. and one vanishing Une, construct the other vanishing line. 10. The angle letween two lines in one figure is equal to the angle subtended at the c. of h, by the vanishing points of the homologous lines. Let AP and AQ he the given lines, P and Q being at infinity. Then P' and Q' are the vanishing points of the homologous lines A'P' and A'Q'. Also OP' is parallel to AP, and OQ' to AQ. Hence the angles P'OQ' and PAQ are equaL 11. Construct the homohgue of a given conic, so that the homologue of a given point S shall be a focus. Take any line as a. of h., and any parallel line as vanish- ing line ; and let two conjugate lines at iSmeet the vanishing line in P and Q, and let two other conjugate lines at S meet it in U and T. On PQ and UV as diameters describe V 290 Homology. [ch. circles, and take either of the intersections of these circles as c. of h. Then since the vanishing points P and Q of the lines SP and SQ, subtend a right angle at the c. of h., the homologues S'P', S'q will be at right angles. So S'V, S'Y' will be at right angles. Hence at S' we shall have two pairs of con- jugate lines at right angles. Hence jS' is a focus of the homologous conic. 12. The homologue of a conic, toMng a focus as c. of h. and the corresponding directrix as vanishing line and amy jpardUel as a. ofh., is a circle, of which the focus is the centre. Let S be the given focus and XM the corresponding directrix. With S as c, of h. and XM as vanishing line, and any parallel line as a. of h., describe a homologue of the given conic. The homologue of S is S, and of XM is the line at infinity ; hence in the homologous conic, 8 is the pole of the line at infinity, i.e. 8 is the centre of the homologous conic. Let SP, SP' be a pair of conjugate diameters of the homo- logous conic. The homologue of SP is SP, the homologue of SP' is SP' ; and the homologues of conjugate lines are conjugate lines. Hence in the given figure, SP and SP' are conjugate lines ; and 8 is the focus, hence 8P and SP' are perpendicular. Hence every pair SP, SP' of conjugate diameters of the homologous conic is orthogonal. Hence the homologous conic is a circle. And we have already proved that the focus is the centre of the circle. Note that the homologue of an angle at S is an equal (in fact, coincident) angle at S. This case of homology is the limit of Focal Projection when the two figures are in the same plane. IjX. 1. Any homologue of a conk, taking afocusS as c ofh., is a amicunih S ' that the homology is harmonic, and that the homologue qf the line at infinity is halfway between the c. ofh. and its polar. Also shoto that a conic is an ellipse, parabola, or hyperbola, according as the Hru halfway between any point and its polar cuts ttte conic in imaginary, coincident, or real points. Ex. 6. AA', BBf, CC are the three pairs of opposite vertices of a guadrilateraL Through any point D on CC are draurn DA meeting BA'ff «n E*, and DA' meeting AB'C in E. Show that EE', AA', BB' are concurrerU, and also B'E', BE and CC. By harmonic homology, taking the meet of AA', BB' as c. of h. and CC as a. of h. xxxi.] Homology. 297 Ex. 7. ShavB OuU two figures in homology reciprocate into two figures iri homology, and that the parameters of homology are numerically equal. Ex. 8. The parameter of homology of turn liomothelic figures is the reciprocal of the ratio qf similitude. Ex. 9. The parameter of homology of turn figures in parallel homology is the constant ratio of the ordinates. Ex. 10. Keeping the same c. ofh., show that the tioo parameters of homology of turn circles are equal but of opposite signs. Ex. 11. Keeping the radical axis as a. of h., show that the two parameters of homology of two circles are equal but of opposite signs. Ex. 12. The poles of the radical axis, of two cirdes divide the join of the two centres of similitude harmonically. For the poles X and X' are homologous if the radical axis be the a. of h., whichever centre of similitude we take as the c. of h. Hence {SN, JCX') ^-{S'N, XX'). Ex. 13. If the radical axis of two circles be taken as the a. ofh., and if the vanishing lines and the radical axis cut the line of centres in IJ'N ; show that SI:IN::r:i', and SJ'-.J'N-.-.r'-.r. Ex. 14. OX is the perpendicular to the line I from 0, and A is any point on OX. From a variable point P the perpendicular PM is drawn to I, arul MA, PO meet in P'. Shew that P and P' generate homologous figures. Ex. 15. If A, B,C be fixed points and P, P' variable paints such that B {APP'C) — A {BPP'C) = constant; show that P and P' generate homologous figures, of which C is thee, of h. and AB is the a. of h. 19. If PP' be any homologous points, and PM the perpen- dicular from P on the vanishing line of the figure generated by P, then OP/PM oc OP', being the c ofh. Let OP cut the vanish- ing line in I and the a. of h. in L. Then, since I corresponds to the point Xi' at infinity upon OP, we have (07, PL) = (012', P'L). Hence OP qL_ op;_ PI "^ LI ~ P'il' ' LQ' i.e. OP:OP'::PI:LI::PM:h, where h is the perpendicular distance between the vanishing line and the a. of h. 298 Homology. Hence OT:FM::OP':K Ex. Prmt the SP : Fit property of a focus. Form a homologue of the conic, taking S as the c. of h. and the cor- responding directrix as vanishing line. Then SP -H PM « SP*. But by § la the locus of P' is a circle with centre S. Hence SP -j- PM is constant. 20. In two homologous figures, if (X, p) and (X, q) denote the perpendiculars from the variable point X on two given lines p and q, and if (X', p') and (X', q') denote the perpendiculars from the corresponding point X' in the homologous figure on the corresponding lines p' and ((, then kJ i -^ p y,' ,[ is constant. For take another point Y, and let XY cut the lines p and qinA and B. Then X'Y' yniX cut p' and g^ in the homo- logous points A' and ff. Hence, since homologous figures are projective, we have {AB, XY) = {A'B', XT'), i.e. AX/AY-^ XB/YB = A'X'/A'Y'-i- X'B'/Y'B', i.e. (X, p)/{Y, p) ^ {X, q)/{Y, q) = (X', p')/(Y', p') ^ (X', g')/(r', «'). Hence {X, p)/{X, q) -r- {X', p')/(X', q') is constant. XiX. l.If^ and T he fixed and p vary, then {X,p)/{T,p)^{J[',p>)/{J>,pr) is constant S!x. 2. If i be the vanishing line qf the unaccented figure, then {X, P)/{JC, t) -r (X', p') is constant. Take 9' at infinity ; then (.Z', 4') = (i", q'). XiZ. 3. If i and f be the vanishing lines, then (X, i) . {Z',jf) is constant. Take p and 3' at infinity. Ex. 4. 0J7(X, p) -i- OX'l{X', pT) is constant. Take q and g' as the axis of homology a, and notice that OXl(X, a) -7- OZ'II^X', a) is constant, since (0<7, XX') is constant. Ex. 5. OX/{X, i)-i-OX' is constant. MISCELLANEOUS EXAMPLES. I . Gekesalise by projection and reciprocation the theorems — ( i) ' The director circles of all conies inscribed in the same quadrilateral are coaxal,' (a) ' The locus of the centre of an equilateral hyperbola which passes through three given points is a circle.' a. The portion of a common tangent to two circles a and fi between the points of contact is the diameter of the circle 7. If the common chord of y and a meets that of 7 and in B, show that S is the pole for 7 of the line of centres of a and fi. 3. Generalise by projection the theorem — ' The straight lines which connect either directly or transversely the extremities of parallel diameters of two circles intersect on their line of centres.' 4. A pair of right lines through a fixed point meet a conic in PQ, P'tf ; show that if PI* passes through a fixed point, then <)e of a straight line which moves so that two fixed circles intercept on it chords of equal length is a parabola. 64. Given a conic and a pair of straight lines conjugate with regard to it, project the conic into a parabola of which the projections of the given lines shall be latus rectum and directrix. Miscellaneous Examples. 305 65. An ellipse has the focus of a parabola for centre and has with it contact of the third order at its vertex. Tangents are drawn to the two conies from any point on their common tangent, and the harmonic conjugate of this latter with regard to them is taken. Prove that its envelope is the common circle of curvature of the two conies at the common vertex. 66. ABC^ DEF are two triangles inscribed in a conic. BC, CA, AB are parallel respectively to EF, FS, DE. Prove that AD, BE, CF are diameters of the conic. 67. Find the double rays of the pencils 0{ABC...) and 0{A'ffff...), each of which is in perspective with the pencil Y(A"ff'C"...). 68. DI/ is a fixed diameter of a conic and PF" is a double ordinate of this diameter. A parallel through I/to DP meets DP' in X Find the locus of X. 69. Through a point is drawn a straight line cutting a conic in AB, and on AB are taken points CD, such that (i -j- OCf) + (i -v OD) = (i -!- OA) + (i-i- OB). Then if MN be the points of contact of tangents from D, and LB those of tangents from C, show that either LM and BN, or LN and BM, meet in 0. 70. Construct the conic which passes through the four points ABCD and is such that AS and CD are conjugate lines with regard to it. 71. AOB and COD are two diameters of a circle and QB is a chord parallel to AB ; if P be the intersection of CQ and DB, or of DQ and CB, and if irom P be drawn PM parallel to AB to meet CD in M, then 01i' = 0Dl' + PM\ 72. AB, AC are two chords of an ellipse equally inclined to the tangent at A ; show that the ratio of the chords is the duplicate of the ratio of the diameters parallel to them. 73. Construct, by means of the ruler only, a conic which shall pass through two given points and have a given self-conjugate triangle. Also construct the pole of the connector of the given points with respect to the conic. 74. Through a fixed point A any two straight lines are drawn meet- ing a conic in B, B' and C, ff respectively ; parallels through A to BC, ffC meet B'C, BC' respectively in D, E ; find the locus of D and of E. 75. Two equal tangents TP and TQ of a parabola are cut in U and N by a third tangent ; show that Tlf=QN. 76. The tangents at two points of an ellipse, whose foci are S, H, meet in 7, and the normals at the same points meet in ; prove that the perpendiculars through S, H to ST, RT respectively divide OT har- monically. Deduce a construction for the centre of curvature at any point of the ellipse. X 3o6 Miscellaneous Examples. 77. An ellipse may be regarded as the polar reciprocal of the auxiliary circle with respect to an imaginary circle of which a focus is centre. Prove this, and find the lines which correspond to the centre and the other focus of the ellipse. 78. Two conies «, v intersect in A, B, C, D ; E, F are the poles of CD with regard to the conies «, v respectively, and AE, AF meet CD in 0, H respectively ; a straight line is drawn through A meeting u, v in P, Q respectively ; show that the locus of the intersection otPH, QBia a straight line passing through B and through the intersection of EF, CD. 79. Two triangles, one inscribed in and the other escribed to a given triangle, and both in perspective with it, are in perspective. Each of the triangles determined by the common tangents of two conies is in perspective with each of the triangles determined by the common points of the conies. 80. Two circles cut each other orthogonally; show that the distances of any point from their centres are in the same ratio as the distances of the centre of each circle from the polar of the point with respect to the other. The directrix of a fixed conic is the polar of the corresponding focns with respect to a fixed circle ; with any point on the conic as centre a variable circle is described cutting the fixed circle orthogonally ; find the envelope of the polar of the focus with respect to the variable circle. 81. Obtain a construction for projecting a conic and a point within it into a parabola and its focus. 8a. A conic circumscribes a triangle ABC, the tangents at the angular points meeting the opposite sides on the straight line DEF. The lines joining any point P on DEF to A, B,C meet the conic again in A', B', Cf. Show that the triangle A'B'Cf envelopes a fixed conic inscribed in ABC, and having double contact with the given conic at the points where it is met by DEF. Show also that the tangents at A', B', to the original conic meet B'0, CfA', A'B" in points lying on DEF. 83. ABCD is a quadrilateral whose sides AB, CD meet in E, and AD. BCinF; A is a. fixed point, EF a fixed straight line, and B, C lie each upon one of two fixed straight lines concurrent with EF; find the locus of D. 84. All the tangents of a conic are inverted from any point. Show that the locus of the centres of all the circles into which they invert is a conic. 85. If A, B, C, D be four collinear points, and any point whatever, prove that , ^ {OA^ -i- (AB . AC . AD)} = o. Also show that if A', B', C, D' be four concyclic points, then 2 {i -H {A'B' . A'C . A'D')] = o, the sign of any rectilinear segment being the same as in the preceding identity. Miscellaneous Examples. 307 86. If be the intersection of the common tangents to two conies having double contact, and if a straight line through meet the two conies in P, P' and Q, er- pendicular from M on its vanishing line is to the perpendicular from M on the axis of perspective. 93. Given two points .^, £ on a rectangular hyperbola and the polar of a given point in the line AB ; determine the points of intersection of the curve with the straight line drawn through perpendicular to AB. * 94. Show how to project a given quadrilateral into a quadrilateral ABCD such that AB is equal to AC, and that D is the centre of gravity of the triangle ABC. 95. A circle has double contact with an ellipse, and lies within it. A chord of the ellipse is drawn touching the circle, and through its middle point is drawn a chord of the ellipse parallel to the minor axis. Show that the rectangle contained by the segments of this chord ia X 2 3o8 Miscellaneous Examples. equal to the rectangle contained by the segments into which the first is divided by the point of contact. 96. ABCDEF is a hexagon inscribed in one conic and circumscribing another. The connectors of its vertices with any point in its plane meet the former conic again in the vertices of a second hexagon A'ffffl/E'F'. Prove that it is possible in this to inscribe another conic. 97. ABCD, ASKJI/ are two parallelograms having a common vertex A and the sides AB, AT) of the one along the same straight lines as the sides Aff, Alf respectively of the other. Show that the lines Bl/, BfD, Cff are concurrent. 98. Three conies u, p, 7 are inscribed in the same quadrilateral. From any point, tangents a, h are drawn to a, and tangents a', 1/ to P. Show that if a, a' are conjugate lines with respect to 7, so are 6, V. 99. If three tangents to a conic can be found such that the circle circumscribing the triangle formed by them passes through a focua, the conic must be a parabola. 100. From each point on a straight line parallel to an axis of a conic is drawn a straight line perpendicular to the polar of the point ; show that the locus of the foot of the perpendicular is a circle. loi. AB is a diameter of a circle, and C and D are points on the circle. AC, BD meet in B. Show that the circle about CDE is ortho- gonal to the given circle. IDS. Find the locus of the centre of a circle which divides two given segments of lines harmonically. 103. The sides AB, AD of a parallelogram ABCD are fixed in posi- tion, and C moves on a fixed line ; show that the diagonal BD envelopes a parabola. 104. A tangent of a hyperbola whose' centre is C meets the asymp- totes in F and Q ; show that the locus of the orthocentre of the triangle CPQ is another hyperbola. 105. Through fixed points A and B are drawn conjugate lines for a given conic. Show that the locus of their meet is the conic through A, B and the points of contact of tangents from A and B. 106. A, B, C, D are four points on a conic, and is the pole of AB. Show that (^AB, CD) is the square of (AB, CD). 107. A, B, C, D are four points on a conic. The tangent at A meets BC, CS in a,, a, ; the tangent at B meets CD, DA in hu h, ; and so on. Show that the eight points 01,%, hi, b^, Cj, Ca, d^, di lie on a conic. 108. The centre of a conic lies on the directrix of a parabola, and a triangle can be drawn circumscribed to the parabola and self-conju- gate for the conic. Show that the tangents from to the parabola are the axes of the conic. Miscellaneous Examples. 309 109. Two sides A% AR of a triangle AQfi circumscribed to a given circle are given in position ; the circles escribed to AQ and AH touch A(i and AB. in Y and V ; show that the locus of the meet of Ql/ and B.V is a hyperbola with A(i and AU as asymptotes. 1 10. If the chords OP, OQ of a conic are equally inclined to a fixed line ; then, if be a fixed point, PQ passes through a fixed point. 111. A fixed line 2 meets one of the system of conies through the four points A, B, C, B in P and Q ; show that the conic touching AB, CD, PQ and the tangents at P and Q touches a fourth fixed line. IIS. Triangles can be inscribed in a which are self-conjugate for $ ; ABC is a triangle inscribed in a and A'B'C is its reciprocal for ; show that the centre of homology of ABC and A'B'C is on a. 113. Six circles of a coaxal system touch the sides of a triangle ABC inscribed in any coaxal in the points aa', bV, of ; show that these points are the opposite vertices of a quadrilateral. ' 114. A, B, C, D are four points on a circle, and A', Bf , C, Bf are the orthocentres of the triangles BCD, CDA, DAB, ABC. Show that the figures ABCD, A'B!C'I/ are superposable. 115. Any conic a which divides harmonically two of the diagonals of a quadrilateral is related to any conic & inscribed in the quadri- lateral in such a way that triangles can be inscribed in a which are self-conjugate for 0. 116. The envelope of the axes of all conies touching four tangents of a circle is a parabola. 117. If {AA>, BBf) = - I = {AA', PQ) = (SB', iV) ; show that {AA', BBf, QQf) is an involution. 118. If two conies can be drawn to divide four given segments har- monically, then an infinite number of such conies can be drawn. 119. If {AA', BB', CC) be an involution, show that {A' A, BC) + {B'B, CA') + (C'C, AB') - i. lao. T is a point on the directors of the conies a and $. The reci- procal of a for meets the polar of T for inQ, B. Show that the angle QTR is right. 131. Through the centre of a circle is drawn a conic, and A and A' are a pair of opposite meets of common tangents of the circle and conic; show that the bisectors of the angle AOA' are the tangent and the normal at 0. 122. A given line meets one of a series of coaxal circles in P, Q. The parabola which touches the line, the tangents at P, Q, and the radical axis has a third fixed tangent. 123. If a series of conies be inscribed in the same quadrilateral of which A, A' is a pair of opposite vertices, and if from a fixed point tangents OP, OQ be drawn to one of the conies, the conic through OPQAA' will pass through a fourth fixed point. 310 Miscellaneous Examples. 124. On a tangent to a circle inscribed in a triangle ABC are taken points a, &, 1^ such that the angles subtended by Aa, Bb, Cc at the centre are equal ; show that Aa, Bb, Cc are concurrent. 135, Through two given points, four conies can be drawn for which three given pairs of lines are conjugate ; and the common chord is divided harmonically by every conic through its four poles for the conies. 126. The locus of the pole of a common chord of two conies for a variable conic having double contact with the two given conies consists of a conic through the two common points on the given chord together with the join of the poles of the chord for the two conies. 127. Find the locus of the centre of a conic which passes through two given points and divides two given segments harmonically. 128. A variable conic passes through three fixed points and is such that triangles can be inscribed in it which are self-polar for a given conic. Show that it passes through a fourth fixed point. 139. If a variable conic touch three fixed lines, and be such that triangles can be drawn circumscribing it which are self-polar for a given conic, then the variable conic will have a fourth fixed tangent, and the chords of contact of the variable conic with the fixed lines pass through fixed points. 130. The directrix of a parabola which has a fixed focus and is such that triangles can be described about it which are self-polar for a given conic, passes through a fixed point. 131. A conic U passes through two given points and is such that two sets of triangles can be inscribed in it, one self-polar for a fixed conic V and the other self-polar for a fixed conic W. Show that U has a fixed self-polar triangle. 133. A variable conic U cuts a given conic V in two given points and also touches it and is such that triangles can be inscribed in it self-polar for a given conic W. Show that U touches another fixed conic. 133. Three parabolas are drawn, two of which pass through the four points common to two conies and the third touches their common tangents. Show that their directrices are concurrent. 134. If a system of rectangular hyperbolas have two points common, any line perpendicular to the common chord meets them in an invo- lution. 135. The reciprocal of a circle through the centre of a rectangular hyperbola, taking the r. h. itself as base conic, is a parabola whose focus is at the centre of the r. h. 136. The reciprocal of any circle, taking any r.h. as base conic, is a conic, one of whose foci is at the centre of the r.h. ; and the centre of the circle reciprocates into the corresponding directrix. Miscellaneous Examples. 311 137. The chords AB and A!B! of a conic a meet in V. & is the conic touching AB, A'S and the tangents at A, B, A', B'. YL and YL' divide AYA' harmonically and cut the conic a in LM and L'M'. Show that the other joins of the points L, M, L', M' touch 0. Also any tangent of P meets AB and A'B^ in points which are conjugate for a. 138. The director circle of a conic is the conic through the circular points and the points of contact of tangents from these points to the conic. 139. Tangents to a circle at P and Q meet another circle in AB and CD ; show that a conic can be drawn with a focus at either limiting point of the two circles and with PQ as corresponding directrix which shall pass through ABCD. 140. Tangents to a conic from two points PP' on a confocal meet again in the opposite points QQ' and RR. Show that QQf lie on one confocal and RR' on another ; and that the tangents to the confocals at PP'QQ^BRf are concurrent. 141. The centroid of the meets of a parabola and a circle is on the axis of the parabola. 142. A variable tangent of a circle meets two fixed parallel tangents in P and Q, and u fixed line through the centre in R. JT is taken so that (Pft JIZ) = — I. Show that the locus of X is a concentric circle. 143. A triangle is reciprocated for its polar circle. Show that the reciprocal of the centroid is the radical axis of the circum-circle and the nine-point circle. 144. The reciprocal of a triangle for its centroid is a triangle having the same centroid. 145. Triangles can be circumscribed to a which are self-conjugate for 0. A tangent of o cuts ^ in P and Q ; and u conic 7 is drawn touching ^ at P and at Q. Show that triangles can be circumscribed to a which are self-conjugate for 7. 146. PP' is a chord of a parabola. Any tangent of the parabola cuts the tangent parallel to PP' in X and the tangents at P and P' in R and Sr ; show that iJX = Xi?'. 147. If the conic o be its own reciprocal for the conic 0, then is its own reciprocal for a. 148. Given a conic o and a chord BC of a, & conic can be found having double contact vrith a at £ and C, such that a is its own reci- procal for 0. 149. A conic cannot be its own reciprocal for a conic having four- point contact with it. 150. If the conic a be its own reciprocal for the conic 0, then a and can be projected into concentric circles, the squares of whose radii are numerically equal. 312 Miscellaneous Examples. 151, Any point P on a conic and the pole of the nonnal at P are coigugate points for the director circle, 15a. The pole of the normal at any point P of a conic is the centre of currature of P for the confocal through P. 153. ABC is a triangle, and AL, BM, CN meet ia a point, LMK being points on BC, CA, AB. Three conies are described, one touching BM, CN a.t M, N and passing through A ; so the others. Prove that at A, B, C respectively they are touched by the same conic. 154. The lines joining four fixed points in a plane intersect in pairs in points OfiiOs, and P is a variable point. Prove that the harmonic conjugates of O^P, 0^, O3P for the pairs of lines meeting in Ofi20, respectively, intersect in a point. 155. If a parabola touch the sides of a fixed triangle, the chords of contact vrill each pass through a fixed point. 156. The six intersections of the sides of two similar and similarly situated triangles lie on a conic, which is a circle if the perpendicular distances between the pairs of parallel sides are proportional to the sides of the triangle. 157. Two conies have double contact, being the intersection of the common tangents. From P and Q on the outer conic pairs of tangents are drawn to the inner, forming a quadrilateral, and B is the pole of PQ with respect to the inner conic. Prove that two diagonals of the quadrilateral pass through B, and that one of these diagonals passes through 0. 158. A conic is drawn through the middle points of the lines joining the vertices of a fixed triangle to a variable point in its plane, and through the points in which these joining lines cut the sides of the triangle. Determine the locus of the variable point when the conic is a rectangular hyperbola ; and prove that the locus of the centre of the rectangular hyperbola is a circle. 159. The feet of the normals from any point to a rectangular hyper- bola form a triangle and its orthocentre. 160. ABC is a triangle and A'SCf are the middle points of its sides. is the orthocentre. AO, BO, CO meet the opposite sides in DBF. BF, FD, DE meet the sides in LUN. Prove that OX is perpendicular to AA', OM to BB', and ON to CC. 161. A variable conic touches the sides AB, AC of a triangle ABC at B and C. Prove that the points of contact of tangents from a fixed point P to the conic lie on a fixed conic though PABC. 163. Given two tangents to a parabola and a fixed point on the chord of contact, show that a third tangent is known, 163. Tangents to a conic from two points on a confocal form a quad- rilateral in which a circle can be inscribed. Miscellaneous Examples. 313 164. AA', BS, CC are opposite vertices of a quadrilateral formed by four tangents to a conic. Three conies pass respectively through AA', BS!, CCf and have three-point contact with the given conic at the same point P. Show that the poles of A A', BS, CC with respect to the conies through AA', BB", CC respectively coincide, and the four conies have another common tangent. 165. If two conies, one inscribed in and the other circumscribed to a triangle, have the orthocentre as their common centre, they are similar, and their corresponding axes are at right angles. 166. A fixed tangent is drawn to an ellipse meeting the major axis in T. Q(jf are two points on the tangent equidistant from T. Show that the other tangents from Q and tf to the ellipse meet on a fixed straight line parallel to the major axis. 167. With a fixed point P as focus a parabola is drawn touching a variable pair of conjugate diameters of a fixed eonie. Prove that it has a fixed tangent parallel to the polar of P. 168. A conic is described having one side of a triangle for directrix, the opposite vertex for centre, and the orthocentre for focus ; prove that the sides of the triangle which meet in the centre are conjugate diameters. 169. The radius of curvature in a rectangular hyperbola is equal to half the normal chord. 170. The radius of curvature in a parabola is equal to twice the in- tercept on the normal between the directrix and the point of inter- section of the normal and the parabola. 171. Two ellipses touch at A and cut at B and C. Their common tangents, not at A, meet that at .^ in Q and S and intersect in P. Prove that BQ and CR meet on AP, and so do BR and CQ. 17a. A transversal is drawn across a quadrangle so that the locus of one double point of the involution determined on it is a straight line. Show that the locus of the other is a conic circumscribing the harmonic triangle of the quadrangle. 173. PQ is a chord of one conic o and touches another conic P. Prove that P, Q are conjugate for a third conic y. 174. JTTZ is a triangle self -conjugate for a circle. The lines joining XrZ to a point D on the circle meet the circle again in A, B, C respectively. Show that as Z) varies, the centre of mean position of ABCD describes the nine-jmint circle of XYZ. 175. Two conies are described touching a pair of opposite sides of a quadrilateral, having the remaining sides as chords of contact and passing through the intersection of its diagonals ; show that they touch at this point. 314 Miscellaneous Examples. 176. With a given point as focus, four conies can be drawn having three given pairs of points conjugate ; and the directrices of these conies form a quadrilateral such that the director circles of all the inscribed conies pass through 0. 177. The line joining two points A and B meets two lines 0(ii OP in Q and P. A conic is described so that OP and OQ are the polars of A and B with regard to it. Show that the locus of its centre is the line OR where R divides AB so that AR : RB : : QR t RP. 178. A chord of a conic passes through a fixed point. Prove that the other chord of intersection of the eonic and the circle on this chord as diameter passes through a fixed point. 179. One of the chords of intersection of a circle and a r. h. is a diameter of the circle. Prove that the opposite chord is a diameter of the r. h. 180. Tangents are drawn to a conie o parallel to conjugate diameters of a conic 0. Prove that they will cut on a conic 7, con- centric with a and homothetic with P. Also y will meet a in points at which the tangents to a are parallel to the asymptotes of P. 181. Four concyclic points are taken on a parabola. Prove that its axis bisects the diagonals of the quadrilateral formed by the tangents to the parabola at these points. 183. If four points be taken on a circle, the axes of the two parabolas through them are the asymptotes of the centre-locus of conies through them. 183. The locus of the middle point of the intercept on a variable tangent to a conie by two fixed tangents is a conic having double con- tact with the given one where it is met by the diameter through the intersection of the fixed tangents. 184. On two parallel straight lines fixed points A, B are taken and lengths AP, BQ are measured along the lines, such that AP + BQ is constant. Show that AQ and BP cut on a fixed parabola. 185. Chords AP, AQ of a conic are drawn through the fixed point A on the conic, such that their intercept on a fixed line is bisected by a fixed point. Show that PQ passes through a fixed point. 186. Three tangents are drawn to a fixed conic, so that the ortho- centre of the triangle formed by them is at one of the foci ; prove that the polar circle and circum-circle arc fixed. 187. Given four straight lines, show that two conies can be con- structed such that an assigned straight line of the four is directrix and the other three form a self-polar triangle ; and that, whichever straight line be taken as directrix, the corresponding focus is one of two fixed points. 188. Parallel tangents are drawn to a given conic, and the point where one meets a given tangent is joined to the point where the Miscellaneous Examples. 315 other meets another given tangent. Prove that the envelope of the joining line is a conic to which the two tangents are asymptotes. 189. With a point on the circum-circle of a triangle as focus, four conies are described circumscribing the triangle : prove that the corre- sponding directrices will pass each through a centre of one of the four circles touching the sides. 190. Three conies are drawn touching each pair of the sides of a triangle at the angular points where they meet the third side and passing through a common point. Show that the tangents at this common point meet the corresponding sides in three points on a straight line, and the other common tangents to each pair of conies pass respectively through these three points. 191. ABCI)\s a quadrilateral circimi scribing a conic, and through the pole of ^C a line is drawn meeting CD, DA, DB, BC, and CA in PQRST respectively. Show that PQ, BS subtend equal angles at any point on the circle whose diameter is OT. 19a. The normal at a fixed point P of an ellipse meets the curve again in Q, and any other chord PP' is drawn ; QP' and the straight line through P perpendicular to PP' meet in B ; prove that the locus of ii is a straight line parallel to the chord of curvature of P and passing through the pole of the normal at P. 193. Two tangents of a hyperbola a are asymptotes of another conic 0. Prove that if $ touch one asymptote of a, it touches both. 194. A conic is drawn through four fixed points ABCD. BC, AD meet in A' ; CA, BD in B' ; AB, CD in C; and is the centre of the conic. Prove that [ABCD] on the conic = {A'B'CO} on the conic which is the locus of 0. 195. Tangents drawn to a conic at the four points ABCD form a quadrilateral whose diagonals are aa', W, of (the tangents at ABC forming the triangle aha and being met by the tangent at D in a'Vd). The middle points of the diagonals are A'PIO and the centre is 0. Prove that {A'PIOO] = {ABCD} at any point of the conic. 196. If a right line move in a plane in any manner, the centres of curvature at any instant of the paths of all the points on it lie on u conic. 197. Defining a bicircular quartic as the envelope of a circle which moves with its centre on a fixed conic so as to cut orthogonally a fixed circle, show that it is its own inverse with respect to any one of the vertices of the common self-conjugate triangle of the fixed circle and conic, if the radius of inversion be so chosen that the fixed circle inverts into itself. 198. A quadrilateral is formed by the tangents drawn from two fixed points on the radical axis of a system of coaxal circles to any 3i6 Miscellaneous Examples. circle of the system. Prove that the locus of one pair of opposite vertices is one conic, and of the remaining pair is another conic, and the two fixed points are the foci of both these conies. 199. Two fixed straight lines through one of the foci of a system of confocal conies meet any one of the conies in PF', (jQ'. Prove that the envelope of PQ and P'tf is one parabola, and of PQ', P'Q is another parabola. Also the points of contact of PQ, P'Q', P'Q, PQf with their respective envelopes lie on a straight line parallel to the conjugate axis of the system, which axis touches both parabolas. aoo. A parallelogram with its sides in fixed directions circumscribes a circle of a coaxal system. Prove that the locus of one pair of opposite vertices is one conic and of the remaining pair is another conic, and the common tangents of these two conies are the parallels through the common points of the system to the sides of the parallelo- gram. Prove also that the tangents at the vertices of any such parallelogram to their respective loci meet in a point on the line of centres of the system. aoi. is the centre of a conic circumscribing a triangle, and 0' is the pole of the triangle for this conic. Show that is the pole of the triangle for that conic which ciroumscribes the triangle and has its centre at (/. ao9. AA', BPl, Cer- bola circumscribes triangles self-conjugate for the r. h. Prove that the corresponding directrix touches the r. h. 370. Triangles can be inscribed in each of two conies a and /3, self- conjugate for the other. Prove that the reciprocal of a for and of fi for a is the same conic 7 ; and a, 0, 7 are so related that each is the envelope of lines divided harmonically by the other two and also the locus of points from which tangents to the other two form a harmonic pencil. Also any two of these conies are reciprocals for the third. 971. Two hyperbolas pass each through the centre of the other and determine a harmonic range on the line at infinity. Prove that the reciprocal of either for the other is the parabola inscribed in the quadrilateral formed by parallels through each centre to the asymp- totes of the hyperbola passing through it. 279. A conic is inscribed in a given triangle and passes through its circum-centre. Show that its director circle touches the circum-circle and the nine-point circle of the triangle. 273. Find the locus of the centre of the conic in the last example. 374. The locus of the centre of a conic touching three given straight lines and passing through a given point is the conic touching the triangle formed by the middle points of the sides of the fixed triangle and such that if I) be the fixed point, 6 the centroid of the triangle and the centre of the locus, then ODG are coUinear, and DO = | DG. 275. If the fixed point be the centroid of the triangle, the locus is the maximum ellipse inscribed in the triangle formed by joining the middle points of the sides. 376. A circle inscribed in a triangle self-conjugate for a hyperbola cuts the hyperbola orthogonally at a point P. Show that the normal at P is parallel to an asymptote. 277. A circle is inscribed in a triangle self-conjugate for a conic and has its centre on its director circle. Prove that it touches the reciprocal of the director circle for the conic. Y 322 Miscellaneous Examples. 378. A circle a with centre is inscribed in a triangle self-conjugate for a conic 0. H P and Q be the points of contact of tangents to fi-om 0, then the tangents from P and Q to the conic which is the reciprocal for of its director, are also tangents to the circle a. 279. The six tangents to a conic from the vertices of a triangle cut again in twelve points which lie by sizes on four conies. 280. The six points in which a conic cuts the sides of a triangle can be joined in pairs by twelve other lines which are tangents by sixes to four conies. 281. If tangents are drawn to a parabola from two points A and B, the asymptotes of the conic through AB and the points of contact of the tangents from A and B, are parallel to the tangents to the para- bola from the middle point of AB. 282. If tangents are drawn to a parabola from A and B, the conic through AB and the points of contact will be a circle, rectangular hyperbola or parabola as AB is bisected by the focus, directrix, or parabola respectively. 283. Tangents are drawn to a circle from two points on a diameter. Show that the foci of the conic touching the tangents and their chords of contact lie on the circle. 284. If tangents are drawn to a central conic from P and Q, and C be the centre and S a focus, then the conic through P, Q, and the points of contact of tangents from P, Q will be a circle if the angle PCQ is bisected internally by CS, and if CP . CQ = CS'. 285. The conic in the previous example will be a rectangular hyper- bola if P and Q are conjugate for the director circle. 286. A point and the orthocentre of the triangle formed by tangents from it to a conic and their chord of contact are conjugate points for the director circle of the conic. 287. If a conic a pass through two points A, B and the points of contact of tangents from them to a given conic, and if /3 be the similarly constructed conic for two points A', B'; then it AB are con- jugate for 0, A'B' are conjugate for o. 288. The reciprocal of the director circle of a conic u for a is confocal with o. 289. Along the normal to a conic at a point are taken pairs of points PQ such that OP. OQ is equal to the square of the semi-diameter parallel to the tangent at 0. Show that tangents to the conic from P and Q intersect on the circle of which a diameter is the intercept on the tangent at by the director circle. 290. The orthocentre of a triangle formed by two tangents to a conic and their chord of contact lies on the conic. Prove that the locus of the vertex of the triangle is the reciprocal of the conic for its director circle or the reciprocal for the conic of its evolute. Miscellaneous Examples. 323 291. The centre of the circle inscribed in a triangle formed by two tangents to an ellipse and their chord of contact lies on the conic. Prove that the locus of the vertex of the triangle is a hyperbola con- focal with the ellipse, and having the equi-conjugate diameters of the ellipse for its asymptotes. 29a. The centre of gravity of a triangle formed by two tangents to a conic and their chord of contact lies on the conic. Prove that the locus of the vertex of the triangle is a concentric homothetic conic. 293. From two points BC, tangents are drawn to a fixed conic, and the sides of the two triangles formed by these two pairs of tangents and their chords of contact touch the conic a. Similarly the pairs of points CA, AB determine the conies P and 7 respectively. Prove that if A lies on a, then B lies on P, and C on 7. 394. A'B'ff are the middle points of the sides of a triangle ABC. Prove that the conic which is concentric with the nine-point circle of A'B'C smdi, inscribed in A'ffff has double contact with the polar circle of ABC at the points where the circum-circle of ABC meets the polar circle, and also has double contact with the nine-point circle of A'BfCf- 295. A triangle is self-conjugate for a conic. Prove that the sides of the pedal triangle touch a confocal. 296. A triangle is self-polar for a conic ; show that an infinite number of triangles can be at once inscribed in the conic and circum- scribed to the triangle, and vice vers4. 297. If two conies a and /9 are related so that the poles for a of two opposite common chords lie on J3, then the polars for /3 of two opposite common apexes touch a. 298. Of all conies inscribed in a given triangle, that for which the sum of the squares of the axes is least has its centre at the orthocentre of the triangle. 299. B, F are a pair of inverse points with respect to a circle whose centre is A ; B is the harmonic conjugate of A with respect to E, F; AP, BP and the tangent at P, any point on the circle, meet the polar of E in L, M, T respectively ; show that LT, TM subtend equal angles at A. 300. The connector of a pair of conjugate points with respect to a given conic passes through a fixed point and one of the pair lies on a given straight line ; show that the locus of the other is a conic, and determine six points upon the locus. Mfiitiii ttitivBrsiiy uoranes OCT 17 1991 MATHEMATICS UBRARM