NK4< II2>T71II.III77:2 PN 4&ft4.A. TftWQLAR. . '''...' ...... p Hill A~~~~t-rf~~~~~~~s TIA~~~..''~"I' ~TES C'ER iT ~~~~u O f f IITA; i~~~~~~~~~~~Hl THE MODERN GEOMETRY OF THE TRIANGLE, BY WILLIAM GALLATLY, M.A., LATE ASSISTANT EXA.INER, UNIVERSITIES OF LONDON AND WALES. LONDON: FRANCIS HODGSON, 89 FARRINGDON STREET, E.C. CHAPTER I. THE LEMOINE AND BROCARD POINTS. 1. The Lemoine Point.-On the sides of the triangle ABC construct squares externally, and complete the diagram as given, the three outer sides of the squares meeting in Al, B1, Cl. A, FIG. 1. The perpendiculars from A, on AC, AB being b and c, the equation to AA, is /3/b = y/c. Hence AA1, BB1, CC, meet at the point whose normal coordinates are (a, b, c), and whose barycentric coordinates are therefore (a2, b2, c2). This point is called the Lemoine or Grebe Point and will be denoted by K. B 2 MODERN GEOMETRY. AK, BK, OK are called the Synzmedians for A, B, C. The absolute values of the normal coordinates are given by a - ka, 2A &c., where k = a2 + b2 + c2 Produce AK to meet BC in Ki; then, BK1: CK1 = AAKB: AAKC = ratio of barycentric coordinates z and y = 2: b2; so that the segments BK', COKl are as the squares of adjacent sides. 2. K is the centroid of its pedal Adef. For A eKfif =. Ke. Kf si A oc bc sin A;. A. eKf = AfKi- = A dKe. So that K is the centroid of def. 3. If a, /3, y are the normal coordinates of any point, then a2 + /f2 + 72 is a minimum at K.[ Since (ca + +b2 ') (a2 + 32 + y2) = (aa+ b3 + cy)2+ (by - c)2 + (ca-ay) + (a-ba)2 = 4A~+...; therefore (a2+ b2+ c2) (a24 /32+ y2) is always greater than 4A' except when a b c7 or when the point coincides with K. In this case, therefore, a2+- 2+y 2 has its minimum value, 4A2 which is 4a2 a2b +b2+c2 4. The equation to OK, joining (cos A, cos B, cos C) with (sin A, sin B, sin C) is sin (B-() a + sin (C-A) /+ sil (A-B) y = 0, b2 C,2 or 2 caa+... = 0. a The normal coordinates of any point on OK are given by a=m cos(A —), &c. THE LEMOINE AND BROCARD POINTS. 3 For: sin (B — ) cos (A-O) = cos. S cos A sin (B- C) + sin 0. E sin A sin (B - C) = 0. 5. Let figures Y and Z, directly similar, be described externally on AC, AB, so that A in Y is homologous to B in Z, and C in Yto A in Z. FIG. 2. Then, if S be the double point of Y and Z, the triangles SAC, SBA are similar, having angle SBA = SAC; SAB = SCA; ASB = ASC, so that S is the focus of a parabola touching AB, AC at B, 0. Let P2, p? be the perpendiculars from S on AC, AB. Then, from similar triangles SBA, SAC P2: p = AC: AB, so that S lies on the A-symmedian. Again, angle BSK, = SAB + SBA = SAB+SAC = A;.'. SC = 2BSK1 = 2A = BOC; so that S lies on the circle BOC. 4 MODERN GEOMETRY. Also, OSKR = OSB+A = OCB+A (in circle BOSC) = 90~;.. SA = SA1. Hence the double point S of the two similar figures on AB, AC may be found by drawing the chord AA1 through K and bisecting it at S. 6. Let T1T2T3 be the Tangent Triangle, formed by drawing tangents to the circumcircle at A, B, C. FIG. 3. Then AT1, BT2, CT3 are concurrent at K. For, if q, r be the perpendiculars from T1 on AC, AB, q _ T C sin A (T2 _ sin B _ b ~' T1B sinABT3 sin C C Hence AT1 passes through K; so also do BT2, CT3. Since ABC is the inscribed circle of TIT2T3, the point K is the Gergonne Point of this triangle. 7. To prove that AK bisects all chords of the A\ABC, which are parallel to the tangent at A, or perpendicular to OA, or parallel to the side H2H3 of the orthic A Hji2H,. THE LEMOINE AND BROCARD POINTS. 5 Let BAK = 0, CAK =. Then si- = - =b- a most important relation; sin< q/3 b sin C sin BAT, sin B sin CAT2 Therefore AK and T2AT3 are harmonic conjugates with respect to AB and AC. It follows that AK bisects all chords which are parallel to T2T, or H2H3, or are perpendicular to OA. 8. The Apollonian Circles.-Let the several pairs of bisectors of the angles A, B, C meet BC in a, a'; CA in f/, if'; AB in y, y'. (See Fig. 3.) The circles described on aa', 3/3', yy' as diameters are called the Apollonian Circles. Let the tangent at A to the circle ABC meet BC in L,. Then angle L1aA = aBA+ BAa = BR+A, L1Aa = L1AC+ CAa = B+A; so that Lla = L1A. And, since aAa' = 90~, L1A= LIa'; so that L1 is the centre of the Apollonian circle (aa'), passing through A and orthogonal to the circle ABC. Since AL1 is a tangent, the polar of L, passes through A; and, since (BKLCL1) is harmonic (7) I, the polar of L, passes through K1. It follows that AIKKIA1T is the polar of L1; so that LlA1 is the other tangent from L,. Hence the common chords AA1, BB1, CCO of the circle ABC and the Apollonian Circles intersect at K, which is therefore equipotential for the four circles. Note that OL1 bisects AKKIA, at right angles, and therefore passes through the point S of Section (5). The Lemoine Line.-Since the polars of L1, L2, L3 pass through K, the points L1, L2, L3 are collinear, lying on the polar of K. This polar is found by the ordinary method to be a/a + 3/b + y/c = 0, and is called the Lemoine Line or Axis. * The bracketed numbers refer to sections. 6 MODERN GEOMETRY. 9. The Lemoine Point is the point of intersection of lines joining the mid-points of the sides of ABC to the mid-points of corresponding perpendiculars. FIG. 4. Let A', B', C' be the mid-points of BC, CA, AB, and let A'K meet AD in 1. Then, since BK and the tangent BT1 at B form a harmonic pencil with BA, BC, therefore the range (AKKT,)) is harmonic; therefore the pencil A'(AID co) is harmonic, so that AD is bisected at 1. 10. A rectangle XYVU being inscribed in ABC with the side XY on BC, to find the locus of its centre P. The diagram shows that the mid-point of UVlies on A'A, and that P lies on A'l. The Lemoine Point K, common to A'l, B'm, C'n is the common centre of three inscribed rectangles, standing on BC, CA, AB respectively. 11. To determine the "direction angles," 60, 6 063, which OK makes with the sides of ABC. From the diagram, OKcos 0 = A'd = A'D K ID" THE LEMOINE AND BROCARD POINTS. 7 b2-c2 2A Now A'D =- K=2 a, 2a Kd a2.+b+L +c2a ID -= 1h, = A; a so that cos 0h1 o a (b2_-c2);.. cos0 cos 0: cos6 = a (b3-c2): b (c2-c): (a — b2). 12. The Harmonic Quadrilateral.-The angles of the harmonic pencil at A are seen to be B40 or CO6. (See Fig. 3.) The same angles are found, in the same order, at B and C and A1. Hence the pencils at B, C, A, and A1 are harmonic; ABA1C being called, on this account, a Harmonic Quadrilateral. In the triangle ABA1 the tangent at B is harmonically conjugate to BK1, so that BKI is the B-symmedian for this triangle. Similarly, AIK1 is the Al-symmedian in BA1C1; and CK1 the C-symmedian in ACA1. To prove that rectangle AB.A1C = rectangle AC.A1B, AB sin 0 (7) AC sin (b sin BCA1 _ A1B sin CBA1 A1C. AB.AC = AC.A1B. If x, y, z, u are the perpendiculars from K1 on AB, AC, A1B, A,C, then x _ _ z _ U AB AC A1B A1C For, since AK1 is the A-symmedian of ABC, x _y AB AC And, since BK1 is the B-symmedian of AABA1, x z AB- AB And, since AIK1 is the Al-symmedian of A1BC, z _ u A1B AIC 8 MODERN GEOMETRY. 13. The Brocard Points.-On the sides BC, CA, AB let triads of circles be described whose external segments contain the angles (a) A, B, C; (b) C, A, B; (c) B, C, A; the cyclic order (ABC) being preserved. The first triad of circles intersect at the orthocentre H. Let the second triad intersect at Q, and the third at 0'. 0 and 0' are the Brocard Points of A.BC. A B FIG. 5. Since the external segment of B2O contains the angle C, this circle touches CA at C. Similarly, C2DA touches AB at A, and AdB touches BC at B. (Memorize the order of angles for Q by the word " CAB.") In like manner, Bf2'C touches AB at B, ClQ'A touches BC at C, AT'B touches CA at A. Again, since A2qB touches BC at B, the angle BC- = DAB. So UGCA- = BC. Similarly, Q'AC = C2'CB = Q'BA. Denote each of the equal angles 92BC, QCA, OQAB by w, and each of the equal angles Y'CB, &c., by o'. 14. To determine o and o'. In A AMB, B = c Sin = 2B sin. sinB b THE LEMOINE AND BROCARD POINTS. So DA = 2R sin w; a (C = 2R sin o a-. c sin (A - ) 0_ a a2 Hence v ---. = == sill ) - A bc acot - b2+ c2.'. cot to — _ - 4A = cot A +cot B + cot C 1 + cos A cos B cos C sin A sin B sin C The same expressions are found for cot o';.. WI= 0). The angle o, which is equal to each of the six angles QAB, 2B(C, fiCA, O'BA, Q'CB, n'AC, is called the Brocard Angle' of ABC. 1.5. To determine the normal and barycentric coordinates of f~ and O'. From the diagram, a = QB sin o = 2R sin2 o -. b 2a b So / = 2R sin2; v =2R sin2o -. C, a And for 0', R b J. b. c 'a a' = 2R sin2 -; / -2R sin2' oc; y 2R sin -. c a b The barycentric coordinates of Q are given by I 1 1 x' y'=z-'b" c2 '; and for Q' x: y': -: a2 -b2 16. The Brocard angle is never greater than 30~. For cot w = cot A + cot B + cot C, and cot B cot C+... = 1;. ot2 = cot2 A +... +2; 10 MODERN GEOMETRY..(cot B-cot C)2+... =2 (cot2 A +...)-2 = 2 (cot2 o0-3). Hence cot o is never less than,/3, and therefore w is never greater than 30~. 17. Some useful formulae:(i) cosec2 Co = 1 +cot2 = 1+ (a + -2 + c2) bC2 + c2a2 +a a2b2 4A2 and 14* 2 a4 + b+ C4 — b2c2 -_2a2 a2b2 and 1 —4 sin'D += -- -—. band2 + c2a2 + a2b2 sin (A —o) ac2' (ii) Since sin A- =- ' sin oC bc.sin (A-o): sin(B-o): sin(C-w) = a3:b3: c3. (iii) sin (A + ) = sin A (cot o+ cot A) sin o b2 + C2 bc Note that, when b = c, sin (A+ w) = 2 sinw. (iv) cos (A +) = cot A cot o- 1 sin A sin o - (-a2+ b +c2) (a2+b2 +c2) 16A2 A sin A sin b4 C4 a2c22 cos (A+o ( B) 8 in B e+ A oo sin (A -B) sin B + sin (A - C) sin C. 11 CHAPTER II. ANGULAR AND TRIPOLAR COORDINATES. 18. Take any point P within the triangle ABC. Let angle BPC = X, CPA - Y, APB = Z. The angles X, Y, Z are called the Angular Coordinates of P. To determine the normal and barycentric coordinates of P. Let AP, BP, OP intersect the circle ABC in L, M, N. Since BPC = X, we have MPC = 180 —X, P1-C =. PCM= X-A. sin A.'. PC= PH sin (X-A);.. aa =2ABPC =BP.PC.sinX = P PM sin X sin A sin (X-A) sin X sin A sin (X —A)' where II = power of P for circle ABC; sin X or 2B. a= sin X sin (X-A) 19. Let X, /,,, be the angles of LMAN. It is required to determine the power and the coordinates of P in terms of X, p, v. We have X - BPC = BMC + M3ICN = A +; sin (A-4K\) sin A and now aa = II s (A+ ) sin A sin X and 2B a s i sin (A + X) sin X.. 2A/I = S sin2 A cot X+ sin 2A;.8. R2s/ or 2R abc/H = a2 cot X + b2 cot / + 3 cot y + 4A, 12 MODERN GEOMETRY. an expression which will frequently present itself in this work. Denote it by Q.* 2R abc nI 8A *** rI = or - = - Q 1?2 Q' -~and a = abc sin (A+ X) and a =- Q sin X 20. Inversion.-To invert a triangle ABC into a triangle with given angles X, i, v. Determine a point P lying within the circle ABC whose angular coordinates are X=A+A, Y=B+P, Z= C+v. Produce AP, BP, CP to meet the circle again in L, M, N. Then L = X, iM =,.=. (19) Also PA.PL = PB.PMll = PC.PN = I. Hence ABC is inverted into the /ALMN, with given angles A,,A, v; the pole being P, and V/II being the radius of inversion. FIG. 6. 21. There is also a second pole of inversion. The point P', inverse to P, lies on OP, and is such that OP.OP' 2. OP. OP' - V. * The expression a2 cot A +... was first used, I believe, by Dr. J. Schick, Professor in the University of Munich. The relations he deals with are different from those given in this work. ANGULAR AND TRIPOLAR COORDINATES. 13 Let P'A, P'B, P'C meet the circle in L', M', N'. Then, since OL2 = OP.OP', the angle OP'L = OLP or OLA = OAP = OP!A or OP'L' (OA2 = OP. OP'), so that L' is the image of L in OPP'. Similarly, 1', N' are the images of M and N, and thus the triangles LMIN, L'M'N' are in all respects equal, though inverse in position. We then have L' = L = X, M1' = M - =, N = N = v; and, since P'A. P'L' = P'B. P'M' = P' C. P'N' = I', the triangle ABC is inverted into L'M'N', with angles X, /,, v; the pole being P', and the radius of inversion,II'. 22. To determine the angular coordinates X', Y', Z' and the power II' of P' in terms of X)tv. In all cases X, Y, Z are to be measured round in the same direction, so that X + Y+ Z = 360~ always. From the diagram, Y'= AP'C = CL'A-L'CP' = CL'A-L'AN' (cyclic quadrilateral OL'AN') = B —. So Z'= C-v;. X'= 3600-Y'-Z' = 360~-(k-A). And aa' = ' sin X' sin A And aa' = n' sin (X'-A) = ', sin (A-X) sin A sin X Hence, as in (19), II' = 2R b, where Q' = a2 cot X + b2 cot + c2 cot v-4A and,a = abc sin (A-X) Q' sin A 23. Orthologic Triangles.-Take any point P, and draw any straight lines YZ, ZX, XY perpendicular to PA, PB, P; so that the perpendiculars from A, B, C on the sides of XYZ are concurrent (at P). Then shall the perpendiculars from X, Y, Z be concurrent. 14 1MODERN GEOMETRY. We have PY2 —PZ2 = y2_-AZ2; PZ2-PX2 = BZ2-BX2; pyX_ py2 CXY2-_ Cy (BX2-CX) +(CY —AY) + (AZ2 —BZ2) = O. Hence the perpendiculars from X, Y, Z on BC, CA, AB are concurrent, say at Q. Triangles ABC, XYZ, which are thus related, are said to be mutually Orthologic. To determine the trigonometrical relation between P and Q. Since PB, PC are perpendicular to XZ, XY,. angle BPC =rr-X; so that, if x, y, z are the barycentric coordinates of P in ABC, sin (7r — X) sin A sin X sin A x co- c sin (7r- - -A) sin(.X+A) Again for the triangle XYZ, YQZ = r —A; so that, if x', y', z' are the barycentric coordinates of Q in XYZ, sin (r - A) sin X sin X sin A 931 O0 --. sin (7r —A —X) sin(X + A) Hence the barycentric coordinates of P in ABC are proportional to the barycentric coordinates of Q in XYZ; the ratio evidently being that of the areas of the triangles. 24. A Harmonic Quadrilateral such as ABA1C can be inverted into a square. Let the circle described with centre T1 and radius T1B or T1C cut the Apollonian Circle L1 in E and E'. Then, since the tangents OA, OB, OC, OA1 to the two circles are equal, 0 lies on their common chord EE', and OE. O' = R2; so that E, E' are inverse points for the circle ABC. Now the angle BT1C = 7r-2A;.BEC=7r —( 7r-2A) =A+- 7r. Hence, if ABC be inverted from E as pole into LMN, the angle L is a right angle (X = 7r). (19) Again, in the triangle ABA1, the circle AEA1 having centre L1 and radii LA and LA1 touching circumcircle ABA1, corresponds to the circle BEC with centre T1 in the triangle ABC. ANGULAR AND TRIPOLAR COORDINATES. 15 And, since BK1C is the B-symmedian of the triangle ABA1, it follows that the circle BEC with centre T1 is the Apollonian circle of ABA1 for the side AA1. FIG. 7. Hence E, where these two circles intersect, holds the same position in ABA1 as it holds in ABC; so that, when ABAI is projected into LML', with E as pole of inversion, the angle M is a right angle, and so on. Hence, LML'N is a rectangle. And since, in the inverted figure, the pencils at the angular points remain harmonic, the rectangle is a square. Another square is obtained by inverting from E'. 25. Tripolacr Coordinates. - The tripolar coordinates of a point are the distances, or the ratios of the distances, of the point from A, B, 7. To determine the point whose tripolar coordinates are 1, rn, n. Divide the sides BC, CA, AB at P, Q, R, so that BP: PC = i: n, CQ: QA = n:, AR: RB =: m. 16 MODERN GEOMETRY. Then AP, BQ, CR are concurrent at a point L whose barycentric coordinates are 1/1, 1/m, 1/n. Produce QR to meet BC in P'. FIG. 8. Then, since AP is my-nz = 0 in barycentric coordinates, and the range CPBP' is harmonic, it follows that AP' is my + nz = 0; so that P', Q', B' lie on the line lx+my+nz = 0. Bisect PP' in 01; then OiP2 = WI p' = 0lB. (0cC, since (CPBP') is harmonic. Hence the circle ABC cuts the circles (PP'), (QQ'), (BR') orthogonally. m2 BP2 BP'2 B __ -BP Again Again = C - CP' CP'2 — P_ 2 BoU1 so that 01, C02, w3 lie on the line 12x + mIy + n2z = 0. Since (CPBP') is harmonic, for every point T on the circle (PP') we have BT: CT: = BP: P =CP: n. So, for every point 'T on (QQ'), CT: AT-= n: l. ANGULAR AND TRIPOLAR COORDINATES. 17 Therefore at the points T, T', where these circles intersect, AT: BT: CT =: m: n = AT': BT': CT'. The symmetry of the result shows that the third circle (RR') also passes through T and T'. Since these circles (PP'), (QQ'), (RR') cut the circle ABC orthogonally, and their centres wo, %2, %3 are collinear, it follows that their common Radical Axis TT' passes through 0; and therefore OTT' is perpendicular to w1i2w3, OT. OT'= -2, and every circle through T and T'cuts the circle ABC orthogonally. Let the circle (TT'), whose centre is w, cut the circle ABC in M, M', so that the angle wMO = 90~, and let MM' cut Ow in J. Then OJ. Oo =- OM2 = R2, so that J is the pole of wlw2(3 for the circle ABC. Also 0)M2 = (J. W0, so that to%2fs is the Radical Axis of circles ABC and (OJ). Note also that OTJT' is harmonic. 26. As an illustration, consider the Isodynamirc points U and U'. These are the positions of T, T' when L coincides with the in-centre I. Then BP: CP = c: b, and 1:: n = 1/a: 1/b: 1/c; so that (UA, UB, UC) and (U'A, U'B, U'G) are inversely as (a, b, c). The line P'Q'R' is then -+... =0 or a+/+y=0, a the Radical Axis of ABC and I,1I3. The line of centres W1203 is then +... =0 or -+ + y =0, Ca.( b c which is the Lemoine Line; and J, the pole of this line, coincides with K. The points U, U' therefore lie on OK, and are harmonic conj agates with 0 and K. C 18 MODERN GEOMETRY. 27. To determine the tripolar equation to a circumdiameter, which makes angles 0,, 02, 03 with the sides. Let P be any point on the line, xyz the projections of OP on the sides. Then if rl, '2, r be the tripolar coordinates of P, r2-r32 =- a.2x = a.2. OP cos 0. And (r2 —,2_) r + (32 -r12) r2 + (r -r,2) 3 = 0;. a cos 01. rl2 + b cos 02. r + COS 03. r2 = 0, and a cos O1+... = 0. So that an equation of the form Ir-2 + nr2 -nr3 = — 0, where l+m+ n = 0, represents a circumdiameter. Note that, for every point P or (r1, r, r,) on the line, the ratios r2 2-r: r,' -rl: -r' 2 are constant; being, in fact, equal to the ratios a cos 01: b cos 0: c cos 03. 28. The angles 01, 02, 0, made by a straight line OT with the sides of ABC, will henceforth be called the Direction Angles of the line. A B 0,c // TV \ FIG. 9. It is very important to note that these direction angles must be measured in the same sense, from the given line as axis. Draw Oa, 0/,, Oy parallel to, and in the same sense with, BC, CA, AB or CB, BA, AC. ANGULAR AND TRIPOLAR COORDINATES. 19 Then, if TOa is taken as 0, (or the supplement of TOa may be taken as 01) TOy = TOa4-aOy = 03, TO# = TOy+yO3 =, aOy = 7r-B, yO/3 = 7r-A. Then, by projecting the sides on to the line and perpendicular to it, we obtain a cos 1 + b cos 0 + c cos 03 = 0, a sin 01 + b sin 02 + c sin 03 = 0. 29. The tripolar equations to 01, OGH, OK, O( should be noted. (i) The projection of OI on BC = ~(b-c);. cos 0 o (b-c); and the equation to OI is a (b-c) r2... = 0. (ii) The projection of OH on BC = - 2; 2a and the equation to OH is (b2_-2) r2 +..= 0. (iii) The projection of OK on BC o a(b2_-c2); (11) and the equation to OK is a2 (b2- 2) r+... = 0. = a (iv) Noting that BM oo c-, 0 ooa-; b c and that r22~ —2 co a cos 0~, we find for the equation of OD a2 (- a2b2) r2 +2 (4- b2) r- + cM 2 (b4-c2a2) r2 = 0. ObserveThat the point whose tripolar coordinates are (s-a, s-b, s-c) lies on OI; That (a, b, c,) and (cos A, cos B, cos C) lie on OH; That (l/a, l/b, 1/c) and (cot A, cot B, cot C) lie on OK. 30. To find the equation to a straight line with direction angles 0, 02 03, and at a distance d from 0. Transferring to Cartesian coordinates, we see that the equation differs only by a constant-from that of the parallel diameter. 20 MODERN GEOMETRY. It must therefore be of the form a cos 01. r2 +... = k. Let A', B', C' be the mid-points of the sides, and let A'O meet the line in A". B \ \ C MD9 A' FIG. 10. Then if (pl, P2, P3) be the coordinates of A", k = a cos 01. p2 + b ccs 0. p2 + c COS 03. p [P2 = P3] = a cos 01 (p2 —P22) = a cos 0, (AM2 —MB2) = a cos 01. 2c. C'M OD =a O,2c. OA! sin 1B OA" = d. 2ac sin B = 4dA. Hence the required equation is a cos 01. r12 +... = 4dA. 31. When (1+ +n) is not zero. To prove that, if Q be the mean centre of masses 1, m, n at A, B, C, or if (1; m, n) are the barycentric coordinates of Q, then, for any point P whose tripolar coordinates are (r1, r2, r2), lrl,+ nr2 + nr 2 =1. A Q2 +. A BQ2 + Q n. CQ2 + (m + m + n) pQ2. Take any rectangular axes at Q, and let (aa2), (blb2), (clc2), (xy) be the Cartesian coordinates of A, B, C, P. Then Ir,2 = I (a -) + I (a - y)2 1=.AQ2 + 1. PQ2 -2x. 1al-2y 1a2. ANGULAR AND TRIPOLAR COORDINATES. 21 But, since Q is the mean centre for masses 1, m, n,.la, +mb +-nlnc = O, 1a2+qmb2+nc2 = 0; r. 12+ mr22+ nr2 = l. AQ2 +m.BQ 2+n. CQ2 + ( +m+ n) PQ2. This, being true for any point P, is true for 0... +.RB2M.R n.R = I..AQ+(t+m+n) OQ2;. r2 + mr-2 - n12 = (+m n) (Q2- Q0 +R2). 32. Particular Cases.(i) If P so move that Z 1. r12 = constant = k2, then QP is constant; so that P describes a circle with centre Q and radius p, given by p2 = QP2 = k2/(l m+n ) + QO2 — R2. (ii) If lrl2+mr+22+ nr2 = 0; then, as Mr. R. F. Davis has pointed out, p = QP2 = Q02 —2; so that the circle cuts the circle ABC orthogonally. (iii) If the circle passes through 0, then QP = QO; and the equation becomes lr12 + mq.22 + nr32 = (1 + mr + n) -.2. (iv) When P coincides with Q, QP = 0, and 1. QA2 +. QB2+n. QC2 = ( +m+?n) x power of Q, 1.,,A2 + lA. ~ QB2 n + q. QC2 so that the power of Q = 1 QA2+m. QB2 —. QC2 I + m +n Examples.(A) The Circumcircle: Here QP = R, QO = 0, 1 osin 2A;. sin 2A. r.2 sinin 2B. r2 + sin 2C. r.3 = 4A. (B) The inscribed circle: QP = r, QO2 = IO2 = R-2Br, I co a;.'. ar + br22 + cr32 = 2A (r + 2R). (C) The Nine-point circle: QP = R; Q02 = OH2 = -f 2 -2R cos A cos B cos C. Since the normal coordinates of the nine-point centre are cos (B- C)..., the barycentric coordinates are sinA cos (B- C)..., so that I co sin 2B+sin 20... (sin 2B+sin 2C) r2 = 4A ( + 2 cos A cos B cos C). 22 CHAPTER III. PEDAL AND ANTIPEDAL TRIANGLES. 33. From the point P within the circle ABC draw perpendiculars Pd, Pe, Pf to the sides of ABC, and produce AP, BP, CP to meet the circle again in L, M, N, denoting the angles of LMN, as before, by X, /, v. In the cyclic quadrilateral PdCe, Pde = PCA or NGA = NLA, Pdf = PBA or MBA = MLA; d = L = AX; so e =, f=,, and the pedal triangle of P is directly similar to LMN. FIG. 11. 34. If P' be the inverse point of P, produce P'A, P'B, P'C to meet the circle in L', M', N'. Then L' = L =,.... (21) In the cyclic quadrilateral P'd'Bf', angle P'df' = P'Bf' or MBA = M'L'A. PEDAL AND ANTIPEDAL TRIANGLES. 23 So P'd'e' = Pe' = N'L'A (cyclic quadrilateral AL'N'C);.. d' = P'd'e' —P'd'f' = = X,.... Hence the pedal triangle of P' is directly similar to L'M'N', which is inversely similar to LMN and def. It follows that the pedal triangles def, d'e'f of two inverse points P, P' are inversely similar. 35. The sides ef, fd, de of the pedal triangle def, and therefore the sides of LMN, are as ar: br2: cre (AP = r1). For, since AP is a diameter of the circle PeAf, ef = AP sin A. sinX: sin: sinv = ef:fd: de = MN: NL: LM = al.: br,: cr3 = arl: br2': cr3. To prove that the area Al of def HA 2A2 -4-2 Q We have 2-1 = de. df.sin X = rr, sin B sin C sin X and 2A = 4R2 sin A sin B sin 0; A* _ 1 r~r3 sin X A 4R- sinA' But r3 = PM sin A/sin X; (18). = HII/R2 = 2A/Q, since II 2 ab (19) Q 36. All the elements of the pedal triangle def may now be obtained in terms of X, P, v. For example, to find the circumradius p, 2A2 2pi2 sin X sin p sin v = A = 22 Similarly, if A1' be the area of d'e'f', _ H'A e' 2A2 ^A1 = _ 2 (22) He-nce 1 1 Q -Q' 8A _ 4 Hence 2A2 2A2A A~r A/' 2'_X 2A~ 24 IMODERN GEOMETRY. Since II = - O-OP2, it follows that the pedal triangles of all points on a circle whose centre is 0 have the same area. 37. In section (25) the axis W1(0W3 is l2 aa+... = 0 or sin2 a + 0. a The pole J of this line is given by a' o a (sin2 X-sin2 — sin2 v),.... (Salmon, p. 268) But, since sin X, sin p, sin v are proportional to the sides of def, cos k = sins 2 + sin2 v-sin2 X 2 sin p sin v Hence, for J, a' co a cot X or J is (a cot X, b cot /, c cot V'). (Dr. J. Schick obtains this result by a different method. See his Isomnorphopolzentrik, p. 88.) 38. For an example of these formulae, take the Isodynamic Points U and U'. (26) Since 1 a' C_ -, a. ar = br2 =- C3g; X = = v = 60~; so that the pedal triangle def is equilateral, and from either U or U' the A/ABC is inverted into an equilateral triangle, LMN being similar to def. Q = (a2+b 2+ c2) cot 60 + 4A 4A (cot o cot 60~ + 1), (14) and hence the absolute normal coordinates of U or U' are Ca = cotwcot60~1 R sin (A 600) (19) cot o cot 60~ 4-1 sin 60~ ' 2R2 The powers are 2 cot (o cot 600~~ -1 and the areas of the pedal triangles _ 1 A - cot w cot 60~ ~ 1 PEDAL AND ANTIPEDAL TRIANGLES. 39. To determine the power II of a point P in terms of its barycentric coordinates x, y, z. Since a bf 2a x y z x+y+z' 4A2 y z a 2 (X +y + z)2 b c 2R. _-~+.J+0~ {c~yz+...a/abc.IR; = ) — L ayz +... B; I 4 — A _ + Z) 2' (x+y+- Z)2 Examples.(i) For I, xcc a; nI = c — = 211r. a+ b+c (ii) For H, x co tan A; a 2 tan B tan C +... II = =tntanC+... 8R2 cos A cos B cos C. (tan A + tan B- +tan 0)2 (iii) For IK, x co a2; 2 2C2 3a2b2c2 (a 2+ b 2+c2)2' (iv) For G, x = y =z; I = (a2 +b2 + c2) -= cot w. Let A G = m = V /2b2 - 2c2-a... and let M111i21MM be the Median Triangle, whose sides are n1L, 1?)2, 9n3. Producing AG to A1 and making GA1 = GA, we have the AA1BG equal in all respects to MI11M3E2M, so that A' the area of M 2MM, = - A. To determine M11, cot Ml - `n22 + -32-_MI2 _ 5-a2-b 2-c2 c 4A' 12A 2 i ^2 __ a2 and cot A - bc ---- 4A. cot Mi +cot A = 2 cot w. Note that cot M1 + cot 3M2 + cot M; = cot o, so that ABC and M1M2M3 have the same Brocard Angle. The angular coordinates of G are -- Mi,..., so that the angle d of the pedal triangle of G = X-A r- A,.... 26 MODERN GEOMETRY. The side ef of the pedal triangle = ml sin A. A 1 A2 Area of pedal triangle = ELI = - Ri cot o. 40. The Brocard Point Q or Q' supplies another interesting illustration of the properties of the pedal triangle. For Q, A+d = BGC= 180~-C = A+B; (13).. d = B; so e = C, f= A. And if d'e'f be the pedal triangle of 0', d' = C, e' = A, f = B. So that the pedal triangles of the Brocard Points are sinmilar to ABC. To determine p, the radius of the pedal circle of Q, ef = 2p sin d = 2p sin B. But ef = r1 sin A 2R sin o - sin A; (14) a p = sin o. Hence the linear ratio of the similar triangles def, ABC is sin o: 1. To determine II for 2. We have = sin2 0. A R2 But II=4 -A; (35) R2_- 02 _ HI = 4R2 sin2 o;.. 02 = /2 (1-4 sin2 ao). Or HI may be found from the barycentric formula. 1 1 1 For I oo - m oo- n co, for c; I' ao I2 ' CX-l C n'co, for 0'. C2 a2 b2 Hence, for 2 a2b2c2 1 b'2c + c2a2 t a2b I/a2 + / + /c2.-. OQ= O'. 41. Equzation to a Circle.-The powers t2, t22, t32 of A, B, C with respect to a circle U being known, to determine the equation to the circle. PEDAL AND ANTIPEDAL TRIANGLES. 27 Let p, q, r be the perpendiculars from ABC on the Radical Axis of the circles U and ABC, so that px+ qy + rz = 0 is the barycentric equation to the Radical Axis. Let k be thedistance between the centres. Then, by a well known theorem, Power of A for U = t2 = 2kp. Hence the Radical Axis is t12x + t22y + t32z = 0. Let xyz be any point P on U. Then II = power of P for ABC = 2k x perpendicular from P on Radical Axis = 2kPx+qy+rz 2A - t-2x +t2'y + t,2z (x - aa) x+y+z But it has been proved that II = a2!/z. (39) (x+ y+ -z)' a2yz + b2zz + c2=y (x + + y ) (tl2x+ t2 2y + t32z), which is the equation to the circle U. The former of these two expressions for II may sometimes be used when the latter is unworkable. For example, to find II for the Feuerbach Point, whose barycentric coordinates are as (b-c)2(s-a).... This point lies on the Nine-point Circle for which t12 = 4(-_C2 +2 ). It also lies on the In-Circle, for which t2 = (s-a)2. Substitute either set of values of t12, t22, t2 in the former expression for II, and we get IT -- a3(a-b)(a-c). 4 a (a-b)(a-c) ' whereas the form of section (39) would give a numerator of the 8th and a denominator of the 6th degree, practically impossible to reduce. 42. Antipedal Triangles.-Let S be the point whose angular coordinates are 7r-X, 7r —/, 7r-v. Draw QAR, RBP, PCQ perpendicular to SA, SB, SC. Then PQR is the Antipedal Triangle of S, and its angles are A, A, v. 28 MODERN GEOMETRY. On SP, SQ, SR as diameters describe the circles SBPC,... Let AS meet the circle (SP) in A1. Then in the triangle BAIC, angle BAC = BPC = X, ABC = A1SC = AQC = /A, A1CB = ASB = ARB = v. FIG. 12. Similarly, in the triangle CB1A, BAC = X, CB1A = /A, ACB1 = v; and in the triangle AC1B, C1AB = X, C1BA = I, ACIB =v; two angles X being at A, two /t's at B, two v's at C. PEDAL AND ANTIPEDAL TRIANGLES. 29 Since SP is a diameter of the circle SBPC, the angle SA1P is right; so that ASA1, being perpendicular both to A1P and QABR, is equal to the perpendicular from P on QB = hl. By Ptolemy's theorem SA. BC = SB.CA1+ SC..AB. But BC: CA: A1B = sin X: sin /: sin v.. SA, sin = SB sin p + SC sin v. h. i sin X = AA1 sin X = SA sin X+ SB sin/ + SC sin v. 43. If X be any point within ABC, the minimum value of XA sin X +... is SA sin X +... Draw XX1, XX2, XX3 perpendiculars to the sides of PQR. Then SA. QR +... = 2APQR = XX,. QB +... < XA. QR +...;.. SA sinX+... < XA sin X+.... 44. The Isogonic Points. - The point, within ABC, whose angular coordinates are 120~, 120~, 120~, is called the Inner Isogonic Point, and is denoted by V. In this case, V is the point for which r1 + r2 + r3 is a minimum. From the preceding article 91l +-2+3r = AA1. And from the triangle ABA,, AA12 = c +a2 - 2ac cos (B + 60~) = 2A (cot o + /3) (the triangles AiBC, B1CA, C1AB being equilateral)... (VA + VB+ VC)2 2A (cot o+ /3) = i (2 + b2 + c2) + 2A /3. 45. To determine the area A2 of PQR, the Antipedal Triangle of S, in terms of X, /, v. Let 01, 02, 03 be the circumcentres of PBC, QCA, BAB, and A'. B', C' the mid-points of BC, CA, AB. Then 01A' = la cot X;.~. AO1BC = laI2 cot A. Now, 01 is the mid-point of SP... APO1B = ASO1B, APO1C = SO1C;.-. area PBSC = 2 { OBC+SBC} =2 { a' cot + SBC}; 30 MODERN GEOMETRY. A2 = area of PQR = 2 {a2 cot X + b cot +c2 cot v+4A } = Q. FIG. 13. If P2 be the circumradius of PQR, 4p22sin X sin i sin v = 2A2 = Q, and now all the elements of PQR may be determined in terms. of A, u, v. Since A 2A2 1- Q (35) 46. To determine (a, /3, y) the normal coordinates of S. In the circle SBPC, SP = a cosec X; and, since ABC is the pedal triangle of PQR, angle QSR = P+A = A+X; also aa = 2 ABSC = SB. SC sin X. Now 4 APSRx APPSQ = SB.SC.PR.PQ aa 2 AQPR sin X sin A aa - Q. sin2 A PEDAL AND ANTIPEDAL TRIANGLES. 31 Also 4 APSRx APSQ = SP.SR sin(B-+ u) SP.SQ sin(O+v), a2 be sin(B +,) sin(G + v) sin AX sin / sin v Hence a = xabc sin(B + pA) sin (C + v) Q sin,u sin v 47. Twin Points. -Just as two points P and P' can be found whose pedal triangles are inversely similar, so a companion point S' may be found for S such that the antipedal triangles of S, S' are inversely similar. FIG. 14. 32 2MIODERN GEOMETRY. Let I, m, n be the images of S in BC, CA, AB, so that the circles BIC, CmA, AnB are the images of the circles BSC, GSA, ASB; and BIG = BSC-= 180~-X.... Let the circles BIC, CvzA cut in S'. Then CS'A = CinA = 180~ —/ = v+ X. And BS'C = 180~ —1C = A. Hence AS'B = v, so that the third image circle AnB passes through S'. Draw AQ'R' perpendicular to S'A, BR'P' perpendicular to S'B, and CP'Q' perpendicular to S'C, so that P'Q'R' is the antipedal triangle of S'. Then S'CBP' is cyclic, since S'CP' = 90~ = S'BP'; so that P' lies on the circle BIG. So Q' lies on CGA, and R' on AnB. Also the angle Q'P'R' or CP'B = BS'C = X.... Hence S, 8' have inversely similar antipedal triangles with angles X,, v. The name " Twin Points " is usually given to S, S'. Let S'A cut the circle B1C in A2. Then ABC 180~-CS'A, = /, ACB= AS'B = v;.BA2C = X. So that the triangle ABC is the image of ABBC in BC. This result gives another method of determining 8'. Take the images A, and B2 of two of the points A,, B,, C..' is the point of intersection of AA2, BB2. Note that BSC+BS'C = B1C+BS'C = 180~; ASB + AS'B - AnB+AS'B = 180~, while CSA = CimA = CS'A. Hence, if a rectangular hyperbola be drawn through ABCS, the point S' is the opposite end of the diameter through S. The mid-point of SS' is the centre of the rectangular hyperbola, and therefore lies on the Nine-point or Medial Circle. 48. To determine the area (A,') of P'Q'R' in terms of X, /, v. Let 0,', 02', O,' be the centres of the circles P'BC, Q'CA, R'AB; a', 3', y' the normal coordinates of S'; hl, h2, ha the perpendiculars from P' on BC,.... PEDAL AND ANTIPEDAL TRIANGLES. 33 Then, since 0O' is the mid-point of S'P', and lies on OA', hi+ a' = 2A' O'; h.. -a' = 2 (A'Ol-a').. P'BC —S'BC = 2 { A 01'B0-S'BC} = 2 {a cot -S'BC}. So Q'CA + S'A = 2 { b cot o + S'CA}; R'AB-S'AB = 2 { c2 cot v- S'AB}. Adding, we have, on the left side, P'B + R'AB- (ABC- Q'AO) = P'B +R'AB-(Q'BC + Q'AB) = (P'BC- Q'BC) + (R'AB- Q'AB) = P'Q'R'. Hence A2 =- {a2 cotX+ b2 cot p +c2 cot v-4A}, or a, = Q' And A I 2a 2 (36) Hence AI' A2 = A2; and A,-A, = {Q-Q'} = 4A. The normal coordinates of S' are found to be a, = abc sin(B —L) sin(C-v) Q' sin pu sin v I) 34 CHAPTER IV. THE MEDIAL TRIANGLE. 49. It is often preferable to take the Mid-Point or Medial Triangle A'B'C', in place of ABC, as triangle of reference. Let af3y and a'fiy' be the normal coordinates of a point P referred to the triangles ABC, A'B'C'. Then a + a'= - i1; aa + aa' = ah1z = A = 4' (A' = area of A'B'C') =2 (a''+...) = aa' b + +cy';.. aa b/3' +cy'. Hence 2 aa' = -aa + b + cy; or, in barycentric coordinates, x = y' -z', 2x' = -x+y+z. A line whose ABC equation is la + mp3 +ny = 0 or px+q-qy+ ' = 0 has for its Medial equation (- + ca'+... = or (q+ r)x'+(r+p)y'+((p+q)z' =. And, conversely, if the Miedial equation is la' + 02' + ipy'+' = 0 or p'+q'y+ ' 0, then the ABC equation is +( — - ) aa+... = 0 or (-p+q+r)x+... =O. \ a b c0 THE MEDIAL TRIANGLE. 35 Examples.(A) If H1H,2E be the orthic triangle, the ABC equation to H2HI is -a cos A +~ cos B + y cos C 0. The Medial equation is a2 aa+ (a2-C2) bf + (a2-b2) cy = 0. (B) If X, Y, Z are the points of contact of the in-circle, the ABC equation to YZ is -(s-a) aa'+... =0. The Medial equation is a.aa+ (a-c) b/3+ (a-b) cy = 0. 50. The Medial or Nine-point Circle.-Referred to A'B'C', the equation is a2 b2 c2 + + z 0. Referred to ABC, it is - +.. —. =0, — x+y+z reducing to the well known form a cos A a2 +... -af/y-... = 0. Or, by the Power method of (41), t2 = -bc cus A, so that the circle is 2 (a2yz+...) = (x+y +z) ( cos A +...), reducing to the above form. The equation of the circumcircle ABC, a2 b 2 C2 + b + = o, x y z becomes a + c _= 0, y +Z' z'+ x' '+ //' when referred to A'B'C', or a2X2 +... + (a2 + b2 + c2) (y'z' + z'x' + 'y) = 0. By the Power method, tl2 =-A'B.A'C = —la2, A'B, A'C being drawn in opposite directions... (y'' +...) = - ( + y' + ) (' +...), leading to the above equation. 36 MODERN GEOMETRY. 51. To determine the A'B'C' equation to a circumdiameter TOT', whose direction angles are 01,, 03. If p', q', r' are the perpendiculars on TOT' from A', B', C', its A'B'C' equation is p'x+q'y + r'z = 0. But a diagram shows that p' = OA' cos 01 = R cos A cos 0. Hence the equation is cos A cos 0i x'+... =0. Note that, since A' is the mid-point of BC, r= 1 (q++r); and p = -p'+ q' + r' = {-cos A cos 01 +...}. 52. A Note on Feuerbach's Theorern.-Let rr be the Feuerbach Point where the in-circle (I) touches the Medial Circle (0'). A c FIG. 15. Then, since the in-circle also touches A'H1 at X,. rX bisects the angle A'rrH; A''r: 7rrH1 = A'X: XH1 = 01: IR = OA: AR, and A'rlH = B-C = OAR. Hence the figures OARI, Ar'wH1X are similar;.A'r: A'X = AO: OI, and A'X = (b —c);. A'roo (-c); and thus the A'B'C' normal coordinates of 7r are b-1 1 a-b \-c) 'c-a ' -b THE MEDIAL TRIANGLE. 37 53. The following is an analytical proof of Feuerbach's Theorem:It has been shown that if t12, t22, t32 are the powers of A, B, G for a circle U, then the Radical Axis of the circles U and ABC is t2x + t22y + t32z = 0. (41) Take A'B'C' as triangle of reference. Then for the in-circle X YZ t]2A= A'2 = (b-)2. Hence the Radical Axis of the circles XYZ, A'B'C' is (b-c)2 +... 0. But, the barycentric equation to the circle A'B'C' being a2 a +... = 0, the above is the equation to the tangent to A'B'C' X at the point whose barycentric coordinates are l a b c \ b-c' c-a' a-b) that is, at the Feuerbach Point 7r. It follows that the two circles touch at 7r. Similarly it may be proved that the Medial Circle touches the ex-circles 1h, 1, 13 at the several points Tl, Tr2, r3 whose barycentric coordinates are ( —a -c +b ) b-c c+ a' a+bl t a b _c \ a/ Ca b c \ (bc' r c-a +b)' a b+c' c+a' a-b the coordinates being obtained by changing successively the signs of a, b, c in the identity a (b-o) + b (c-a) c (a-b) = 0. The tangents to the Medial Circle at 7r, 7r,... are therefore (b-c)2+ (c-a) y+ (a-b)2z = 0; (b —c)2 +(c+a)2 y +( + b)2 = 0. Hence, by subtraction, we see that the tangents at xr, 7r, intersect on the outer bisector of A', and the tangents at 71r, 7Tr on the inner bisector. 54. The following propositions relating to Feuerbach's Theorem are demonstrated in an article by Prof. M. T. Naraniengar, contributed to the Progress Report of the Indian Mathematical Club, October, 1908.* * Published by S. Murthy & Co., 305 Thumbu Chetty Street, Madras. 38 MODERN GEOMETRY. (A) The straight lines 7rrq and 7r2r, pass through the feet of the internal and external bisectors of A. (B) They divide BC harmonically. (C) L, their point of intersection, has ABC normal coordinates -sin(B -C), sin(C -A), sin(A-B). If HE1H.H3 be the orthic triangle, then H2fH passes through L. (D) L lies on B'C', M on C'A', N on A'B'. (E) AMN, BNL, CLM are straight lines. (F) AL, BM, CN are perpendicular to the Euler Line OO'GH, so that L is the point where the perpendicular from A on OGH cuts B'C'. (G) The triangle LIMN is self-polar for the circle A'B'C'. *(H) The polars of A, B, C with respect to the Medial circle pass through L, M, N. (J) A'7r,, B'r2, Ct'r3 meet two and two on the sides of LMN. 55. A parabola, touching the sides of A'B'C', has TOT' for directrix. To determine the focus, with A'B'C' as triangle of reference. It may easily be shown from Salmon's article (Conics, p. 274) that, if (a',,/', y') be the focus, then the directrix is cos A /+-... = 0. a This coincides with TOT', whose equation is cos A cos 01.x +... = 0. (51) Hence the A'B'C' normal coordinates of the focus are sec 01, sec 02, sec 03; a most important result. 56. Examples of particular diameters:(i) For OI, cos 06 oo (b-c). The Medial equation to 01 is therefore (b- c) cos A. x+... 0; and the focus of the parabola is (b-c c-a a- b ' normal coordinates, or the Feuerbach Point of A'B'C'. * Mr. S. Narayanan has pointed out that these polars also pass through ZL, L2, L3, the centres of the Apollonian Circles. (8) THE MEDIAL TRIANGLE. 39 (ii) For OH, cos o b- C2 a The Medial equation to OH is -~-_c- cos A.x+... '- 0; a and the focus is b — c C -, a2 - 2, or the Euler Point of A'B'C'. (iii) For OK, cos 0oo a(b-c2). (11) The Medial equation to OK is a(b2-c2) cosA. +-... = 0;., ~,,. I 1 1 1 I and the focus is I (b2-c) b(c2 — ) Ic(a-b2).a(&_-c) b(C2-_) c(a 2-_b) or the Steiner Point of A'B'C'. 57. To determine the focus and directrix of a parabola inscribed in A'B'C' and touching the line whose Medial equation a. b/3 cy =0. is + + X JL v Since +... =0 and aa+bf3+cy = O (line at o), are tangents to the parabola VJ(ua) +... = 0,.LX + V+-E - 0v; a b c a. c O; Zt V W.:-: + = -v: v-0: A-; a b c therefore the focus is given by at O c a-; (Salmon, p. 274) U f11-V and the directrix is (0u-v) cot A. +... = 0. For example, putting X oo a, it is at once seen that the Lemoine Line (8) of A'B'C' touches the parabola inscribed in A'B'C', whose focus is the Euler Point and whose directrix is OH. 40 MODERN GEOMETRY. 58. Let (1, m, n) be the tripolar coordinates of a point T and its inverse T', lying on a fixed circumdiameter whose direction angles are 01, 0, 03. (See Fig. 8, p. 16.) C FIG. 16. Let the internal and external bisectors of the angles BTC, CTA, ATB, or BTGC, CT'A, AT'B, meet the corresponding sides BC, CA, AB in P and P', Q and Q', R and R', so that BP: PC = BP': PC ==: n,.... Then AP, BQ, CR are concurrent at a point L, whose barycentric coordinates are 1/1, 1/m, l/n; while AP, BQ', CR' are concurrent at a point Li, whose barycentric coordinates are -1/1, 1/m, 1/n, and so on. And the triads of points P'Q'R', P'QR, Q'RP, R'PQ lie on four straight lines forming the system lx -my4:nz = 0 (barycentric coordinates being used throughout). To prove that these four straight lines touch the parabola which is inscribed in the Medial Triangle and has the given circumdiameter for directrix. THE MEDIAL TRIANGLE. 41 The line lx + my + nz =- 0, when referred to the Medial triangle, becomes (m+n) x+ (n+l) y++ ( +m)z = 0. (49) Now, the barycentric coordinates (x', y', z) of the focus S of the parabola which is inscribed in A'B'C' and touches + Y + -=O a2 are given by x',.... (57) C a, b2 C2 -_n2' n_ —l'2 12-_22 ) are the barycentric coordinates of the focus of the parabola which touches the sides of A'B'C', and the line (rn + n) x +... = 0. This focus remaining unchanged by any change of sign in I, m, n, it follows that each of the four lines of the system touches the parabola just determined. Now draw Tt perpendicular to BC. Then m2 - n2 oc TB2 - TC2 co 2a. A't co 2c.0OT.cos 01; so that the focus is (a sec 60, b sec 2, c sec 03) and is thereforefixed. The directrix is cos A cos 01. x +... = 0, and is therefore the circumdiameter TOT'. (51 and 52) Thus, as T and T' travel along the fixed diameter, the variable system Ixl4-rmynz = 0 envelops the parabola which touches the sides of A'B'C' and has TOT for directrix. Taking the four tangent lines P'Q'R', P'QR, Q'RP, R'PQ in triads, we see that the four circles PQR, PQ'B', QR'P', RP'Q' pass through the focus, which lies on the Medial circle. 59. With the aid of this parabola we can now deduce some interesting properties of tlis four-line system. For these lines, in conjunction with the sides of A'B'C', give a system of seven tangents, forming thirty-five triangles. The circumcircles of all these triangles pass through the focus S, whose barycentric coordinates are 2 i2... or asee,... }. m -9t 42 MODERN GEOMETRY. The thirty-five orthocentres lie on the directrix. (m2 —n2) cot A. +... 0; or cos A cos 0.x+... = 0. The seven feet of perpendiculars from S on the tangents lie on the vertex tangent, whose equation will prove to be cot 01.x+cot 02.y+cot O,.z = 0. (70) The seven tangents also form thirty-five quadrilaterals. The circles described on the diagonals of any one of these quadrilaterals have for their common Radical Axis the four-orthocentre line, which is the fixed directrix TOT'. In each quadrilateral the mid-point line is perpendicular to the directrix. 60. To determine the second point S' in which the circle PQR cuts the Medial Circle. Let p, q, r be the second points in which the circle PQR cuts the sides of ABC. Then BP. Bp = BR. Br; CQ.Cq = CP.. p; AR.Ar = AQ.Aq. And by Ceva's Theorem BP.CQ.AR R = BB,.AQ.CP; therefore Bp. Cq Ar = Br. Aq. Cp, so that Ap, Bq, Cr are concurrent at a point XI. Let Bp: pC = mn: n,..., so that 1/11, 1/m1, 1/n1 are the barycentric coordinates of X1. Now AQ=- b, AR- ==-, n+l +m and AQ.Aq = AR. Ar; a2 b2 2 (m +)(ml + n) (n+-l)(n,+11) (l+m)(ll+-m)' a2 b2 c2 so that -- + +. Hence m2 —n12 = (ml + n1) (m —nl) a2 n b2 c.1 2 mr{-n (n+l l4- m ) TIHE MEDIAL TRIANGLE. 43 Hence the barycentric coordinates of S', which are a2 i 3nl212 5 take the form b (- +I) —c2 (n —... The circle PQ'R' (changing the sign of 1) cuts A'B'C' at S and also at the point SI, whose barycentric coordinates are ( 1 1 b2 (n-1)-c2 (m-1 ' ) ' And so for the circles Qt'P', EP'Q'. 61. 1H2Hf being the orthic triangle, to prove that S'S, and and S2S3 intersect on H2. As these points are on the circle, S'S, and S2,S are of the form a2 -x+... = 0. The chord S'S1 is a2 (b2m- cn)2-(b2-_2)2} x... = 0. The chord S,2S is a2 (b2-2)2 2_-(b2q+ C2,)2} X... = 0. Adding the two equations, we obtain a2x+ (a2 — ) y/ (a2-b2) z = 0, which is H2H3 referred to the Medial triangle. (49) 62. Examples.(A) Let L coincide with the Gergonne Point, whose barycentric coordinates are {1/(s-a)...}. Then P, Q, R are the points of contact of the in-circle, and I co s-a: m2-n2 Co a(b-c). Hence S is {a/(b-c)...} or the Feuerbach Point ir; and the directrix is OI, on which lie the orthocentres of PQR,..., and the points T and T', whose tripolar coordinates are (s-a).... For S', the second point of intersection of the circle PQR with the Medial circle, 1 a b'2 (s-b +-s-a)- c(s-c- + s-a) b-c' Hence S and S' coincide, the circle touching the M3edial circle at the Feuerbach Point. 44 MODERN GEOMETRY. The ABC four-line system (s-a)x:-(s-b)y- (s-c)z = 0 touches the above parabola. (B) Let L coincide with I or (abc). Then AP,..., are the inner bisectors of A, B, C. Since 1 co-..., IT and T' are the Isodynamic Points. a 2 1 Also c '-n2 b2-c2 so that S is the Steiner Point, and therefore the directrix is OK, on which lie the orthocentres of PQ,.... For S', ( 00 — 1a s_+ c I Lb b-c \ 6 a / c a so that S' is the Feuerbach Point vr. The system x/a~y/bhz/c = O, or a:j:y = 0, touches the above parabola. That in this case the circle PQB passes through the Feuerbach Point was first discovered, I believe, by Mr. J. H. Lawlor, of the College, Maritzburg. That the Steiner Point is the other point of intersection has been proved independently by Mr. V. Ramaswami Aiyar. 45 CHAPTER V. THE SIMSON LINE. 63. From any point T on the circumcircle ABC, draw perpendiculars TX, TY, TZ to the sides BC, CA, AB. Produce TX to meet the circle in t. FIG. 17. Then, since BZTX is cyclic, angle BZX = BTX or BTt = BAt;.'. ZX is parallel to At. So XY is parallel to At. Thus XYZ is a straight line, parallel to At. It is called the Simson Line of T, and T is called its Pole. 46 MODERN GEOMETRY. 64. To draw a Simson Line in a given direction At. Draw the chord tT perpendicular to BC, meeting BC in X. A line through X parallel to At is the Simson Line required, T' being its pole. 65.* To prove that XYZ bisects TH (H orthocentre). If Q be the orthocentre of TBC, TQ - 2R cos A = A, and QX =Xt = Ax.. L. H = TX.. HxTX is a parallelogram.. YZ bisects TH, say at h. It follows that h lies on the Nine-point circle. 66, If X, A, v are the direction angles of XYZ, then OAT = OTA = 90~ —AOT = 90~-AtT = 90~- YXT = X. A relation of fundamental imlportance. 67. To determine the absolute normal coordinates of T. TA = 2R. cos OTA = 2 cosX....2R = TB. TO = 4/2 cos / cos v... = 2 cos, cosv; so that T is the point (sec X, sec A, sec ). Another result of great use. 68. To determine the length of the perpendicular from T on its Simson Line. p = TX sin TXZ = 2R cos A cos /I cos v; or, since TA = 2R cos A, TA. TB. TC P 41 ' In (55) the Simson Line of the focus S, being the vertex tangent of the parabola, is parallel to the directrix TOT', so that its direction angles are 0,, 02 63. Hence the latus rectum = 4 x perp. from S on the Simson Line 1 =. SA'. SB'. SC', = R cos 01 cos 0 03. * Proof by Mr. N. McArthur. See Notes of Edinburgh Mathematical Society, July, 1910. THE SIMSON LINE. 47 69. To determine the segments YZ, ZX, XY. In the circle A YTZ, with AT as diameter, YZ = AT sinA - 2R cos X sin A a COS X.... 70. To find the equation to XYZ in terms of X, au, v. Let u, v, w be the perpendiculars from A, B, C on XYZ. Then the equation to XYZ is ux+vIy f w= 0. Now u AY sin AYZ YZ sin AZY sin A YZ sin A = 2R cos A sin A n v, sin Su sin A' = 2R cos X sin,u sin v, cO cot X. The equation to XYZ is therefore cotk. X.- cot /.y+cotv.z = 0. 71. Consider the quadrilateral formed by the sides of ABC and a straight line PQU. FIG. 18. 48 MODERN GEOMETRY. The circles ABC, AQR, BRP, CPQ have the common point M, and their centres 0, 01, 0,, 0 lie on a circle, called the Centre Circle, passing through M. Let X, 1A, v be the direction angles of PQR. If p be the circumradius of A(QR, the perpendiculars from 01 on AQ, AR are p cos R, p cos Q. So that, if a, /3, y are the normal coordinates of 01, f3: y = cos R: cos Q = sec: secv. Hence AO1, BO,, CO, meet at a point N whose coordinates are (sec X, sec M, sec v). a b c And, since + = acos X+... = 0; (28) secX sec A sec v N lies on the circle ABC, and is the pole of the Simson Line parallel to PQR. Again the angle OlNO2 = -— ANB (BNO2 being a straight line) = 7- C. and since 00, 002 are perpendicular to MA, MB,.'. 000= AMB = C; so that N lies on the Centre Circle, and is therefore the second point in which the Centre Circle intersects the circle ABC. 72. T and T' being any points on the circle ABC, it is required to determine S, the point of intersection of their Simson ~\ ~ I x x' FIG. 19. THE SIMSON LINE. 49 Lines. Let (XMv) (X','v') (060203) be the direction angles of the Simson Lines of T, T' and of the chord T'". Let (a3y) be the normal coordinates of S. Then a = SX sin A - X, sin X sin A' sin (X-A') = /T'" cos 0 sin X sin X' sin (X-X') But OAT = A, OAT' = X', (66) so that TAT' = X-X'. Therefore TT' = 2R sin (X-X'), and a = 2 R cos 0 sin /\ sin X'. 73. To prove that the Simson Lines of T and T', the extremities of a diameter, intersect at right angles on the Medial Circle. A FIG. 20. E 50 MODERN GEOMETRY. Let these Simson Lines X7i, X'h' intersect at o. Produce TX, T'X' to meet the circle in t, t', so that Tt''t' is a rectangle, and tt' a diameter. The Simson Lines of T, T' are parallel to At, At', and are therefore at right angles. Again, since h, h' are mid-points of HT, IT', (65). h' = ~TT' = R. But h, h' lie on the Medial Circle; therefore they are the ends of a MIedial diameter. Also hwh' = 90~, as shown above. Therefore o lies on the Medial Circle. 74. If the Simson Lines of any two poles T and T' have. direction angles (XwAv) and (AX''v'), the points.T, T' are (sec A...) and (see X'...). Now, the chord through two points (a/ly1) (a..,y.) on the circle is a a <la2 Hence, TT' is cos X cos X'. aa +... = 0, and the tangent at T is cos2 A. aa +... = 0. If 'T' be a circumdiameter, then the Simson Line of T' has direction angles X=...; so that the absolute normal co2 ordinates of T' are a' = 2R sin sin v,... (67) and T' is the point (cosec X...): the equation to the diameter TOT'being cos X sin X. aa... =0. 75. Let 0, 06, 03 be the direction angles of a diameter TOT'. Since XwX' = 90~ and A'X = A'X'; o lies on the circle (XX'), so that - 'o = A'X =? cos 0,. At the second point (a'), where the circle (XX') cuts the Medial Circle, the Simson Lines of t, t' cut at right angles. The circle (XX') passes through q and r, the feet of the perpendiculars from B, C on TOT'. For, draw A'p' perpendicular to TOT'; then p'r = p'q = ~qr = -a cos 0= i- l sin A cos 01 and Aop' = cos A cos 0; (51).. A'= A'q =.- cos 01. THE SIMSON LINE. 51 Note also that cot qwr = cot qA'p' - p'A'/p'q = cot A, so that q'r, rp, pq subtend angles A, B180~- C at w. Consider the figures ATOT', oXA'X'. Since TAT' and XwX' are right angles, while 0, A' are the mid-points of TOT', XA'X', and the angles OTA or OAT = A'Xw or A'wX; (66) it follows that The figures ATOT', wXA'X are, similar, a fact to be very carefully noted. L FIG. 21. Hence angle wA'X = AOT, so that o may at once be found by drawing angle XA'w = AOT. To determine (a', f', y') the Medial or A'B'C' coordinates of w. 52 MODERN GEOMETRY. Since o'A - R cos 01..., and a'.= oB'.oC';.. a' R cos 0 cos 03. Hence o is the point (sec 0, sec02, sec03), and its Simson Line (in A'B'C') is parallel to TOT'. If therefore the chord oP be drawn perpendicular to BC, then A'P, parallel to the Simson Line of uo in A'B'C' (by 63), is also parallel to TOT'. It will be seen that w is the focus of the parabola of (52). 76. From the similar figures AT~pO, wXp'A', A'': A'X = Op: OT. Therefore p'wP passes through p. Hence the important propositionIf Ap, Bq, Cr be drawn perpendicular to a diameter TOT' and pp', qq', rr' be drawn perpendicular to BC, CA, AB, thenpp', qq', rr' meet at the point ( on the Medial Circle, which is the point of intersection of the Simson S Lines of TOT'; also the (Medial) Simson Line of o is parallel to TOT'. 77. The envelope of the Simson Line is a Tricusp Hypocycloid. Since angle wA'H1 = AOT,.. 0' = 2. AOT. (0' is Medial Centre) But Medial Radius = 112;. arc How - are AT = 2 arc dh, by similar figures Hdh, HAT. Now take arc A'L = -- are A'H1, and draw Medial diameter LO'l. Then, since A'd is also a Medial diameter, arc A'L = arc Id;.. arc Hi, = 2 arc A'L - 2 arc Id. Also arc H1ow = 2 arc hd, as just proved;.'. arc Lw = 2 arc lh. Therefore Xw! touches a Tricusp Hypocycloid, having 0' for centre, the Medial Circle for inscribed circle, and LO'l for one axis. THE SIMSON LINE. 53 78. If P, P' be any pair of inverse points on the given diameter TOT', then the Simson Lines of T, T' are the axes of similitude, and therefore w is the double point, for the pair of inversely similar pedal triangles of P, P'. (Neuberg.) FIG. 22. In the similar figures ATOT', oXA'X' the points P, P' correspond to x, x'. But AT is known to bisect the angle PAP'; therefore wX bisects the angle axwx'. Also, in the similar figures, Wx:' x = AP: AP' = A OA ' =.: OP' = (, y, = wz: w. = wZ: WZ. Hence o is the double point, and wX, wX' are the axes of similitude for the inversely similar pedal triangles xyz, x'y'z'. 54 CHAPTER VI. THE ORTHOPOLE. 79. Let p, q, r be the lengths of the perpendiculars Ap, Bq, Cr from A, B, C on a straight line TT', whose barycentric equation is therefore x + qy +-rz = 0. Draw pp', qq', rr' perpendicular to BC, CA, AB. Then shall pp', qq', rr' meet in a point. A diagram shows that, if 01, 02, 03 are the direction angles of TT', the perpendiculars from A, B, C on pp' are p sin 0,, p sin 01 + c cos B, p sin 01-b cos C, so that pp' is p sin 0 x + (p sin 01 - c cos B) y+- (p sin 0- -b cos C) z = 0; or, noting that q-r = a sin 0,, p (q-r?)( + y+ z) + ac cos B y-ab cos C z = 0, with similar forms for qq', rr'. By addition, all the terms vanish, so that the three lines meet in a point (call it S), which is named by Prof. J. Neuberg the Orthopole of TT'. Solving, we have x cO q (r-p) ca cos B-r (p-q) ab cos C + a2bc cos B cos C. 80. Let T1T2 be any chord of the circle ABC, 1X,1, T2X2 perpendiculars to BC. Draw the chord.Apt, perpendicular to T1L', pS and tX3T, perpendicular to BC, X3S parallel to At,, t3x parallel to BC. Obviously, SXK is the Simson Line of 13. We shall prove that S'X1, SX are the Simson Lines of T1, T2 respectively. Since St3 is a parallelogram,. 13S = t3X3 = xX1. THE ORTHOPOLE, 55 Hence S'XN is equal and parallel to px. Again, the line tl3T subtends right angles at p, and x, so that p3t3Tlx is cyclic..t, FIG. 23..'. angle p3xtl = p3t3T'l OL At3r'l' = supplement of Atjlx; thus p3x, and therefore,S'XI, is parallel to Atl. Hence SX1 is the Simson Line of T1. So SX., is the Simson Line for T2, and thus the Simson Lines of T,T2T3 are concurrent at S, which lies on the perpendicular from p3 on BC. By symmetry, S lies also on the perpendicular from q on CA, and on that from r on AB. Therefore S is the Orthopole of T1T2, so that the Orthopole of a chord TilT is the point of intersection of the Simson Lines of rpT It follows at once that S is also the Orthopole of T1T3 and T'T.:. 81. It is clear that each of the points T1, T., '1T has similar relations to the other two points. 56 M ODERN GEOMETRY. For example.(i) SX3 parallel to At3 is perpendicular to TTT; so SX1 is perpendicular to T2Ta, and SX2 to T3T1. (ii) The point p,, the foot of the perpendicular from A on TT2, lies on the perpendicular from S to BC. Therefore also pl and 2,, the feet of the perpendiculars from A on T2T3 and T,31, lie on this line. It follows that pP2p, is the Simson Line of A for the triangle T1ToT. Hence the Simson Lines of the poles A, B, C for the triangle TTTa pass through 8'. This theorem (ii) is due to Mr. S. Narayanan, B.A. 82. Let (XAyl), (A2P2v,), (A38,,3) be the angles which the Simson Lines of TI, T2, T, make with the sides of ABC. The normal coordinates (afy) of X, as the point of intersection of the Simson Lines of T, and T2, are given by x = 2R cos 01 sin A, sin X2, (72) where 01, 0,, 03 are the angles which T1T2 makes with the sides of ABC. But 01 = - X3, the Simson Line of T, being perpendicular to T'1'T. a = 2R sin Xi sin X2 sin X3; /3 = 2R sin JU sin m2 sin /3; y = 2R sin v, sin v2 sin v,. 83. To determine the normal coordinates of A with respect to the triangle T, T3. These are inversely proportionate to AT,, AT2, ATs. But ATl = 2Rcos XA; AT = 2R cos A; AT, = 2R cos 3. Hence the coordinates are as sec X,, sec A2, sec A%; their absolute values being 2H cos A2 cos A,.... 84. To prove that S bisects HH'. (H' orthocentre of TT.,T.) Since pP2P3 is the Simson Line of A for T1T2T,, therefore it bisects AH'. Hence the projection of S on BC is equidistant from the projections of H and H' on BC. So for CA, AB. Hence S is the middle point of HH'. THE ORTHOPOLE. 57 85. To determine Sp,. Since f3A is parallel to SX, (Fig. 23), the angle t3AB = angle between SX, and AB = V3 -And the angle tBC = t3AC = angle between SX3 and A C = IL3 ~'. Sp t3X3- t3B t3C = 2R sin tAB sin tBC = '2R sin,At sin v,. 86. If T,T2 meet the sides of ABC in a, /3, y, to determine the power of S for the circles on Aa, B/3, Cy as diameters. A X3 FIG. 24. The circle (Aa) obviously passes through p. and H,, and also through the point T3r, such that Ap37r3H1 is a symmetrical trapezium. Then AH —2Am- = p3r3 = Sr7r +Sp = S7r3 + t3X= 87r3 + Hn.~. S7r3 = AH1-Hnz-2Am = An-2Am = (At3 —2Ap) sin XA (angle SX3H1 = p3t3n) = p1a sin X3, where tl3 = p3A. Let d3, a, be the perpendiculars from 0 and S on T1T2. Then 2d3 = p31. 58 MODERN GEOMETRY. And a3 = Sp3 sin A3 = 2 R sin A, sin I.s sin v3.. Power of S for circle (Aa) =,Sgo3:-,= 4Rd3.sin XA sin /3 sin v, = 2cf3:,.. (85) The symmetry of the result shows that S has the same power for the three circles (Aa), (B/3), and (Cy); therefore it lies on the common Radical Axis of these circles, which is the 4-orthocentre line of the four triangles formed by T1T2 and the sides of ABC. 87. Lenoyne's Theorem.-TT' is any chord of the circle ABC, the Simson Lines of T, T' intersecting at S. XYZ is the pedal triangle of any point P on TT'. To prove that the power of S with regard to the pedal circle XYZ is constant. Draw Ap, Bq, Cr perpendicular to iT'. Let TT' cut AO in o. Draw ),/, wy perpendicular to AB, AC, so that fly is parallel to BC. THE ORTHOPOLE. 59 Draw AX' parallel to BC. Then, since the circle Af/y with diameter Ao, and the circle AYZ with diameter AP, pass through p, therefore a parabola can be drawn with p for focus, touching AB, AC, fly, YZ. Hence the angle pYC = p.ry, since the tangents AC, fy cut the tangent YZ in Y and z. But, from the cyclic quadrilateral A YpX' (diameter AP), p YC = pX'A; pxy = pX'A, so that pxX' is a straight line. Hence, as XPX' takes up successive positions XIP1X1', X9PX2', &c., we shall have ax.l. X'X' = px: pX' = nm, a constant;. X, =M m. X1'X2' = 1a. X1X.2 It follows that X1xI, X4x2, &c. are concurrent at some point S'. But at T and T' the points X, Y, x, Z all lie on the Simson Line of the point. Therefore S' coincides with S, the Orthopole, or point of intersection of the Simson Lines of T and T'. Let XS meet the pedal circle XYZ again in ar. In this circle Xx.xo- = Yx.xZ. And in the circle A YpZX' (diameter AP) X'r.xp = Yx.xZ = Xx.xo-. Therefore X'Xpo- is cyclic, the centre of the circle lying on the line which bisects XX' at right angles; that is, on the side B'C' of the Medial Triangle A'B'C'. Let 7r be the image of p in B'C', so that the trapezium AHLpTr is symmetrical. Then, clearly, the circle X'Xpo- passes through 7r; therefore in this circle XS. So- = Sp. S'r. But X and a- are points on the pedal circle XYZ;. Power of S' for X YZ = XS. So = Sp.Sr - 2Sd, (86) where (, d are the perpendiculars from S and 0 on TT'. 88. For a circumdiameter TT2 the orthopole S, as the point of interection of the Simson Lines of T1T2, coincides with the point o (73). 60 MODERN GEOMETRY. 89. To prove that the pedal circles of all points on the diameter T1OT2 pass through the Orthopole o. This fact is at once deducible from Lemoyne's Theorem, for now d = 0, so that the power of the Orthopole for each pedal circle is zero. FIG. 26. For an independent proof, note that angle wA'X = A OP, (75) and A'X: OP = A'o: OA; therefore the triangles wA'X, AOP are similar..angle oXC = APT,. So wYA = BPTo, and oZA = CPT. But oYA + wZA + YwZ+ A = 4 right angles = BPT, + CPT + BPC; YwZ+A = BPC =A+YXZ; (35).Y. W = YXZ, so that the circle XYZ passes through o. THE ORTHOPOLE. 61 Illustrations. (a) The pedal circles of all points on 01 pass through the Feuerbach Point (- -- ). b-c c-a a —b (b) So for OK and the Steiner Point t aT a.... (c) Also for OH and the Euler Point (b2 a.). Isogonal Coijugate Points.-It is at once seen that the common pedal circle of two isogonal conjugates P and P' cuts the Medial Circle in w and o', the Orthopoles of OP, OP'. When w, o/ coincide, or the circles touch, then OPP' is a straight line. 90. To prove that Xo) and YZ intersect on B'C'. This follows at once from Lemoyne's Theorem, for there YZ and XS intersect on fy, which coincides with B'C' when TT' is a diameter. For an independent proof, draw AX' parallel to BC (Fig. 26). Then, since AI)wp is a symmetrical trapezium, and DX, AX' are equal and parallel; therefore oXpX' is a symmetrical trapezium, and therefore cyclic. Again, a circle with AP as diameter goes round PYpAX'Z, and another circle passes through XwYZ. Now let the circles, taken in order, be denoted by U, V, W. The common chord or Radical Axis of U and V is pX', that of U and TV is woX, that of V and W is YZ. Therefore these chords are concurrent. But, since the trapezium Xop'X' is symmetrical, therefore the chords XTp, Xw intersect on B'C'. X. o and YZ intersect on B'C'. NOTE.-Sections 79-80 are based on two articles contributed to Matlhesis by Prof. J. Neuberg; sections 82-86 and 89 are by the present writer. 62 C HAPTE R VII. ORTHOGONAL PROJECTION. *91. AL being a given axis in the plane of the triangle ABC, it is required to determine the shape and size of the orthogonal projection of ABC on a plane X, passing through AL and inclined at an angle 0 to the plane ABC. Let A, be the image of A in BC; draw the circle ALA, cutting BC again in lM, so that LAML = 90~. In AM take a point T, such that AT AM- - = cos. Draw BUb, CVc perpendicular to AL, cutting LT in U and V. Then, since Ub/Bb = Vc/Cc = AT/AIM = cos 0, AUV represents, in shape and size, the orthogonal projection of ABCO on the plane X. 92. On either side of the common base BO, a series of triangles is described similar to the orthogonal projections of ABC on a series of planes X passing through the common axis AL. It is required to determine the locus of the vertices of these triangles. Draw AtS perpendicular to LT, cutting the circle ALA, in S'. The triangle SBC will be similar to A UV, and therefore to the projection of ABC on the plane X. Draw SNS' perpendicular to BC. Then ALT = tAT or SAM = SLM. And A, t, S, N are right angles. Therefore the figures ALtT, SLNMVi are similar. And LU: LB = LV: LC = LT: LM. Therefore the figures ALUVtT and SLBCANM are similar. Therefore the triangle SB is similar to A UV. * In writing Sections 91-96 I have drawn freely on Professor J. Neuberg's " Projections et Contre-projections d'un Triangle fixe," with the author's permission. ORTHOGONAL PROJECTION. bj5:Hence the vertices of all triangles SBC, described on BC, and having SBC- = AUV, SC.B = ATVU, BSC = UTAV, lie on the circle ALA,. A, FIc. 27. The angles of projection range from = 0 to 0 = -r. When 0 = 0, cos 0 = 1, so that T coincides with M, S with A1, and S' with A. As 0 increases from 0 to 7r, T' travels from Mi to A, S from Al to 1M, and S' from A to Mi. Hence the locus of S is the arc AMA, on the side of AA' remote from the axis AL. 93. When a series of variable axes AL,, AL2,... are taken, the point Al, being the image of A in BC, remains unchanged. Hence each position of the axis AL gives rise to a circle ALA,1M passing through the two fixed points A and A,, so that this family of circles is coaxal. Let DA = DA1 = Ik, and let DR = h, where 1 is the centre of the circle ALA,. 64 MODERN GEOMETRY. Then, with D as origin and DA as y-axis, the equation of the circle is 2+y2_ k2 = 2hx, h being the variable of the coaxal system. 94. We may now deal with the problem: To determine the plane X on which the orthogonal projection of ABC has given angles X, x, v. Construct the A SBC, so that SBC = p, SCB = v, and hence BSC - X. Let the circle A SA1 cut BC in L and Ml. Then, if AL and S' are on opposite sides of AA1, AL is the required axis. Draw LtT perpendicular to SA. Then the required inclination 0 of the plane X to the plane ABC is given by A7 cos 0 = AM 95. Counter-Projection,-To determine the counter-projection of ABC on a plane through AL, making an angle 0 with the plane ABC. L / T FIG. 28. Draw M1T' parallel to LT; and Bb', Cc' perpendicular equal to AM1, meeting 1MT in U', V'. Then U'b': Bb' = T'A: LA = MlA: TA = Bb: Ub;. U'b': Bb = Bb': Ub. ORTHOGONAL PROJECTION. 65 Hence the perpendiculars from U, U' on AL, AM are proportional, and therefore A UU' (and similarly 4 VV') is a straight line. If therefore AM be taken as axis, the triangle A U'V', which is the counter-projection of ABC, is similar to AUV. Hence, the variations of the triangle SBC give the shapes of the counter-projections with AM as axis, as well as those of the projections with AL as axis. It follows that points S on the arc AMA1 determine the shapes SBC of projections with axis AL, and of counter-projections with axis AM: while points S on the arc ALA1 determine the shapes of projections with axis AM- and counter-projections with axis AL. 96. The triangle ABC being projected on a series of planes making a constant angle a with the plane of ABC, and the triangles SBC being drawn, as before, similar to the successive projections, it is required to determine the locus of S. Draw AL parallel to the line of intersection of the planes, then the original projection is equal and similar to the projection on a parallel plane through AL. Determine S as before, taking T such that AT/AM = cos a. FIG. 29. F 66 MODERN GEOMETRY. Join SL, SiM, cutting AA1 in H and H'. Then H is the orthocentre of LH'M, so that DH. DH' = DL. DM = k2. AT tan ALT Again, cosa= AT tanALT A M/ tan ALM tan SLM HD tan AL - AI D H. D = k cos a = constant, H'D = k sec a = constant, so that H and H' are fixed points. Hence, since HSH' is a right angle, the point S describes a circle on HH' as diameter. The equation to the circle HSH' is 2 + (y-k cos a) (y- k sec a) = 0. For another series of planes inclined at a constant angle a', the points H and H' would be changed to HI and H1', where DH1 = k cos a', DH1' = k sec a', and DHB. DH1' = k2. The series of circles are therefore coaxal, having A and Al for limiting points, and therefore cutting orthogonally the former series of circles, which pass through A and Al. 97. A triangle ABC, with sides a, b, c and area A, is projected orthogonally into a triangle LMN, with sides x, y, z and angles X,,A, v, its area being A', and 0 the angle of projection. It is required to determine the trigonometrical relation between the two triangles. Let hi, h2, h be the heights of A, B. C above the plane LMN. Then h2-ha = v/a2 —x;...2_X v/r,/^cY2 zt 2 r 0,. a. /a — b'b- y2~ vc2-z2= O 0 the signs of the surds depending on the relative heights of A, B, C. This leads to 4y 2_(y2+ 2_2) + b2_(b2 + c2 _ a2)2 = 2 2 (b2+ C2 2). b2 + C2 -a2 a2 Now cot A =, cot B + cot C -. 4A 2. 16A + 16A'2 = 16 E (cot t +cot v) cot A. AA';.. (cotu+cot v) cotA = (cotB+cot C) cotX = A'/A + A/A' = cos 0 + sec 0; ORTHOGONAL PROJECTION. 67.'. 2 cotA = 2A' (cos + sec ), and, a2 cot X = 2A (cos 0 + sec 0) =sum of areas of projection and counter-projection. 98. The Equilateral Triangle and the Brocard Angle.-When ABC is equilateral, we have (x2 + y2 + 2)1 = 2A'(cos 0 +sec 0). But, if o be the Brocard Angle of LMN, cot w o +y2-z2 4A' /3 cot w = (cos 0 + sec ). The Brocard Angle therefore depends solely on the angle of projection. It follows that all coplanar equilateral triangles project into triangles having the same Brocard Angle. 99. Antipedal Triangles and Projection. —Let Imn, I'mn' be the sides, and A2, A2' the areas of the two antipedal triangles, whose angles are X, u, v. From (45), 2A2 = a2 cot X +4A, and from (48) 2A2' = S a2 cot X-4A;.- 2 A2+A = '-a2cotX = areas of projection and counter-projection. Again, 2A2 = a a2 cot X + 4A = 2A (cos 0-sec 0) +4A; '. A2/a = ( /sec 0 + V/cos 0)2. So a2/A = ( /ssec 0 - Vcos 0)2; VA2 - / v/A' = 2 /A cos 0 /A2 + '/A2/ = 2,/A sec 0. Now, the triangles.2, A2', with the projection and counterprojection, are all similar, each having angles A, Iu, v, so that their sides are as the square roots of their areas; I-. ' = -2x,...; 1+1' = 2x,.... The theorem, of which the above is a new proof, is due to Lhuillier and Neuberg. 68 MODERN GEOMETRY. 100. The triangle ABC is projected orthogonally on to a plane inclined at an angle 0 to the plane of ABC. If S, the line of intersection of the planes, make angles A, /u, v with the sides of ABC, it is required to determine the lengths of the sides of the projection in terms of X, /, v, and 0. Let cos 0 = k. Draw perpendiculars Bb, Cc to the line S, and take b' = k.bB, cC' = k.cC, so that B'C' is equal to the projection of BC on the plane. Now, the projection of B'C' on Bb = B'b - C'c = k (Bb- Cc) — =. a sin X. And the projection of B'C' on S = a cos A. Hence B'C'2 = 2 = a2 cos2 + k sin2 X = a2 1- (1-k2) in2 } = a2 { 1-sin20 sin2 }. The dimensions of the projection depending, of course, only on the direction, not on the position, of S. 101. To prove that, as the plane of projection revolves round S, the projections of ABC are similar to the pedal triangles of points lying on the circumdiameter TOT', where T is the pole of the Simson Line parallel to S. Let XYZ and X'Y'Z' be the Simson Lines of T, T'. Then XYZ, being parallel to S, makes angles A, /t, s with the sides of ABC; and therefore angle OTA = OAT = X. In TOT' take any point P, and let OP = k'. R. (66) Let u, v, w be the sides of the pedal triangle of P. Then, in the OA P, AP2 = R2 + k'2R2 —2k'R2 cos (7r-2X) =R2 (l+ k')2 1- 4k sin2x So that 42 = 4. P2 sin2 A 2 2 4k' sin2x a (1+k')2 1 - (1+ k- sin) But X2 = a2 1 —(1-k2) sin2 XA. ORTHOGONAL PROJECTION. 69 FiG. 30. 2 1 Hence =1 -4u2 (l+ k')2 and x': y: z =: v: w, (so that the projection on the plane 0 is similar to the pedal triangle of P,) provided that 4k' (1+k')2 1-k' - 1-k or k +k; so that P, and its inverse point P", are given by TP/T'P = TP'IT'P' = cos 0. The inversely similar pedal triangles of P and P' correspond to the projections of ABC on a pair of planes equally inclined to the plane ABC. 102. If 0i, 02, 03 are the angles which TOT' makes with the sides of ABC, the tripolar equation to TOT' is a cos 0.rl2+... = 0. (27) For P, r1 sin A = t; ~o '. 2+..= o; Ct 70 MODERN GEOMETRY. COS61,..., 1 +...-=O, a where a cos 0o +... - 0. Hence the sides of the projections are connected by the relation 1x2 + my2 + nz2 = 0, with the condition la + mb2 + nct = 0. 103. Next, let the plane of projection be inclined at a constant angle a to the plane of ABC. -r,1 * 7; 1-k 1-cos a In this case k = 1- cos 1 +k 1 +cos a = tan2 aa;. OP = R tanl2 a, OP' = R cot2 la. If, therefore, circles be drawn with centre 0 and radii tan2 ~a. 1 and cot2 la. R, any point P on one circle, and its inverse point P' on the other circle, will have their pedal triangles similar to the projections of ABC on a pair of planes drawn through an axis in ABC, and making equal angles a with the plane of ABC. 104. Counter-Projections.-To determine the length of the counter-projection of BC on S at the angle 0. Take bB" = sec 0. bB, cC" = sec 0. cC, so that B"C" = xa is the required counter-projection of BC. Now the projection of B"C" on bB = sec 0. a sinA; ="Ca2 = x2 a2 [ cos2 X + sec2 0 sin2 X = a2 sec 20 1- (1- k2) cos2, showing, as already proved, that the counter-projection of ABC for an axis perpendicular to S is similar to the projection of ABC for the axis 8, the angle of projection being 0 in each case. Between 0 and T' take a point Q, such that OQ = k'R. Then OQ2 = R2 + k'3 R2 -2k' -2 cos 2X = ( + k')2 R2 1-( 4 k)2 os2. Hence the counter-projections for the axis S are similar to, the pedal triangles of points ranging from 0 to T'. LONDON: PRINTED BY C. F. HODGSON & SON, 2 NEWTON STREET, KINGSWAY, W.C.