SOURCEBOOK :v.;a.-:oe-v, mmms mmm SYKES Property of CORNELL UlIIVERSITY Price $2.50 Copied supplied to teachers at a special discount. Correspondence invited. ALLYN and BACON The date shows when this volume ok c(»>y the ca the Ubrarian. To renew tMs tx>ok copy the call No. and give to HOME USE RULES All books subject to recall ' All borrowers must regis- .p..^j.y, ter in the library to borrow ^ ^HlJ S i ^ books for home use. *** All books must be re- turned at end of college year for inspection and repairs. Limited books must be ^ returned within the four week Uroit and not renewed. Students must return all books before leaving town. Of&cers should arrange for the return of books wanted during their absence from town. Volumes of periodicals and of pamphlets are held in the library as much as possible. For special pur- poses they are given out for a liriiited time, Borrowers should not use their library privileges for — - the benefit of other persons. Books of special value and gift books, when the , giver wishes it, are not allowed to circulate. Readers are asked to re- port all cases of books marked or mutilated. Db not deface books by laarks and writing. Cornell University Library arV17041 A source book of JP'iSJiJlf'S.^iJSIiffl 3 1924 031 218 187 olin,anx Cornell University Library The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031218187 A SOURCE BOOK OF PROBLEMS FOR GEOMETRY BASED UPON INDUSTRIAL DESIGN AND ARCHITECTURAL ORNAMENT BY MABEL SYKES INSTRUCTOR IN MATHEMATICS IN THE BOWEN HIGH SCHOOL CHICAGO, ILLINOIS WITH THE COOPERATION OF H. E. SLAUGHT and N. J. LENNES Boston ALLYNK'AND BACON I 91 2 1 ^ — I V^v^ COPYRIGHT, 1912, BY MABEL SYKES. N'otinooti 3^res! J. 8. Cushing Co. — Berwick & Smith Co. \fi .Norwood, Mass., U.S.A. PREFACE The author is fully aware of the danger involved in offering this book to teachers of geometry. At present there is a widespread tendency in education to sub- * stitute amusement for downright work. This temptation pre- sents itself in varied and subtle forms to all teachers. Even the best teacher cannot hope to escape. He must be continu- ally on his guard lest his work call forth mere superficial attention and begin and end in entertainment. One form of this temptation may be found in the copying of designs. Not only may one spend more time than is profitable on this work, but unless care is exercised in the selection of the designs, all of the time spent may be wasted. Many inter- esting figures that might be suggested for this purpose are so complicated that pupils cannot reasonably be expected to determine all of the lines by constructions already mastered. If, attracted by the beauty of the copy, the pupil is allowed to present a result that will not bear detailed geometrical an- alysis, entertainment has been substituted for education. A design ivhich is not an easily recognized application of a known construction is certainly of doubtful value. Many of the designs in this book are very simple. A few are complicated, but even these can be so broken up that only the simple part need be given to the class. While it must remain true that for training in clear thinking and accurate expression nothing in the teaching of geometry can take the place of abstract originals and of formal proof of theorems, nevertheless, exercises of the type here given have a unique and important place. IV PREFACE We no longer teach geometry solely because it develops the reasoning powers or trains in exact thought and expression. Such development and training can be obtained from_ other subjects rightly taught. But geometry gives, as no other sub- ject can give, an appreciation of form as it exists in the material world, and of the dependence of one form upon another. This appreciation is apart from and superior to any particular utili- tarian value that the subject may or may not have. Surely, one of the strongest reasons for teaching geometry in the high school is that it develops in pupils a sense of form. To culti- vate and develop this sense in the highest degree, pupils should learn to recognize applications of geometric form which are omnipresent. The figures in the text-books which may have seemed unattractive and devoid of human interest, will then appear not only useful, but even beautiful. It is for this pur- pose that these exercises are of value. This book is written in the hope that it may meet a real need of both teachers and pupils. Hitherto much material from geometric designs or historic ornament suitable for class use could be obtained only by long and careful search through widely scattered sources. The difficulties of this search and the labor of preparing the results have been such that a valu- able and interesting element in the study of geometry has been for the most part lost until very recently. Much time has been spent in collecting these illustrations. Moreover, care has been taken to select not only many simple designs, but also illustrations of simple geometric figures, inas- much as these are in every way better suited to the purposes mentioned above than are complicated ones. A simple design is not necessarily a poor one. The references inserted amply justify the claim that these illustrations have their origin in what is best in historic ornament. Reasonable limits forbid the insertion of many problems that are special cases under the sanie type. Instead, the PB SPACE V general problem with its answer is given from which, any number of numerical problems may be obtained. , The problems that 'combine both algebra and geometry are sure to be of value in bringing out the essential unity of the two subjects. Wherever data for theorems and problems appear to be in- sufficient, what is lacking may be readUy supplied from the figure. Attention is called to the index of theorems and prob- lems as an aid to selecting material. Without the assistance and cooperation of Dr. H. E. Slaught, of the University of Chicago, and Dr. N. J. Lennes, of Columbia University, this book could never have gone to press. The author wishes to express her deep appreciation of their con- tinual encouragement and of the hours spent by each of them in detailed work on both manuscript and proof. Thanks are due to Mr. Dwight H. Perkins, the Chicago architect, for criticisms and suggestions regarding the notes on trusses and arches, and also to many former pupils ^nd fellow teachers who have assisted in verifying answers. M. S. Chicago, Illinois, November, 1912. TABLE OF CONTENTS Chapter I. Tile Designs PART PAGE 1. Sizes and shapes of standard tiles 1 2. Tiled floor patterns 13 Chapter II. Parquet Floor Designs 1. Designs in rectangles and parallelograms .... 24 2. Designs based on the isosceles right triangle .... 31 3. Designs containing squares within squares . . . .35, 4. Designs containing stars, rosettes, and crosses within squares 54 Chapter III. Miscellaneous Industries 1. Designs based on octagons within squares 2. Designs containing eight-pointed stars . 8. Designs formed from arcs of circles 4. Designs formed from line-segments and circle-i 5. Designs based on regular polygons . 71 81 96 115 126 Chapter IV. Gothic Tracery: Forms in Circles 1. Forms containing circles inscribed in triangles . . . 151 2. Rounded trefoils 155 3. Rounded quadrifoils ........ 161 4. Rounded multifoils 177 5. Other foiled figures 189 Chapter V. Gothic Tracery: Pointed Forms 1. Forms based on the equilateral triangle 206 2. Forms containing tangent circles : geometric constructions . 219 3. Forms containing tangent circles : algebraic analysis . . 248 4. Venetian tracery 293 VlU TABLE OF CONTENTS Chapter VI. Trusses and Arches PART PAGK 1. Raftered roofs and trusses 297 2. Arches 303 Alphabetical Bibliography 327 Classified Bibliography 335 Commercial Catalogues 337 Index to Problems and Theorems 338 Index to Notes axd Illustrations 363 Index to References 371 SOURCE BOOK OF PROBLEMS FOR GEOMETRY CHAPTER I TILE DESIGNS PART 1. SIZES AND SHAPES OF STANDARD TILES 1. Occurrence. — Modern standard tiles are made in cer- tain definite sizes and shapes, all of which are directly derived from the six-inch square. Some of the more common shapes and their relations to the square are shown in this Part. When tiles are intended for floor use, the pattern is worked out with plane unglazed tiles of a large variety of colors and shapes. Tiles intended for the walls of kitchens or bathrooms or for fireplaces are usually rectangular and often highly decorated. Among them are some of the most artistic products of modern industrial art. 2. History. — ^The earliest tiles of which we have any record are from the valleys of the Tigris, Euphrates, and Nile rivers. The most famous examples of tiled pavements are those which occur in the churches and abbeys erected during the middle ages throughout western Europe. The tiles were mostly square, and highly ornamented. Reference will be made to some of the designs later. Tiles for wall decoration were characteristic of the work of the Moslems of the eleventh and twelfth centuries, when they brought the art to great perfection. The tiles used were elaborately decorated and made in a large variety of shapes. 1 TILE DESIGNS SQUARES, RECTANGLES, AND RIGHT TRIANGLES D^ S C 3. In Fig. 1 ABCD represents a standard tile six inches square. EG and FH are the diameters of the square and AC and BD the diagonals. The points are joined as indicated. EXERCISES 1. Prove that EFGH and XYZW are squares. 2. Find the length of one side of EFGH and its area. Ans. 4.242 in., 18 sq. in. 3. ^ind the length of one side of XYZW and its area. Ans. 3 in., 9 sq. in. 4. How many squares of the size of XYZW would be sold for one square yard? ^n«. 144. D C 4. In Fig. 2 ABCD represents a stand- ard tile six inches square. AE = AG and EF and FG are parallel to AG and AE respectively. N r M F 1 U /\ T L / \ iK/i>' K E Fig. 2. EXERCISES 1. Prove iha-t AEFG is a square. 2. Square KLMN is formed by joining middle points of sides of EFGA as in Fig. 1. Find one side of square KLMN and its area if (1) AE^^AB; (2)AE = iAB; (i)AE = \AB. Ans. to (3), 2.12 in., 4 J sq. in. SIZES AND SHAPES OF 8TANVAED TILES 3 3. If AB = I- AB, how many tiles of size of ^XAfiV would be sold for one square yard? Ans. 648. 4. How many whole squares of size of KLMN could be cut from one square yard ii AE = \AB'i Ans. 1152. 6. Squai-e RSTU is formed by joining the middle points of sides of square KLMN, as in Fig. 1. Find one side of square RSTU and its area if (1) AE = \AB; (2) AE = \AB; {^) AE = IAB. Ans. to (3), IJ in?, 2| sq. in. 6. li AE = \ AB, how many tiles of size oi RSTU can be cut from one square yard ? Ans. 2304. E c 5. In Fig. 3 ABCD is a six-inch square. CE = J CD and DG = J DA. EXERCISES X Fig. 3. 1. Show how to construct the isosceles right ti-iangles with CE and DG as hypothenuses. 2. Find CF and DH and the areas of the triangles CEF and DGH. Ans. Sides, ^2,1-^/2; Areas, 1, jV 3. If CF is iV2 in., find the ratio of CE to CD. Ans. i. 4. If CF = a, find the ratio of CE to CD. Ans. ^^. 5. If CE is — , find the length of CF. Ans. - y/2. 6. If CE = J CD and DG = \DA, how many tiles of each size would be sold for one square yard? Ans. 1296 and 2304. TILE DESIGNS 6. InFig. 4ABCDTeiire5entsasix-iiich square. E, F, G, and H are the middle points of the sides. The points are joined as indicated. EXERCISES 1. Prove that (a) KL = LO and PS extended will bisect GN; (b) GKLM is a rectangle ; (c) Square GKON^ PQRS, and in area equals i A DOC; (d) The area of rectangle GKLM is equal to that of A DKG and tojthat of A Z»OC. 2. How many rectangles of the size of GKLM can be cut from one square yard ? Ans. 57S. 3. Find the dimensions of rectangle GKLM. Ans. 2.121, 1.06. 7. In Fig. 5 ABCD represents a six-inch square. E and F are respectively the middle points of the sides AB and BC. EH and FG are drawn from E and F respectively perpendicular to AC. EXERCISES 1. Prove that EFGH is a rectangle. 2. Prove that the area of EFGH is equal to : (a) twice that of AEBF; (b) half that of A ABC; (c) four times that of AAEH; (d) half that of the square formed by joining the middle points of sides of square ABCD. 3. Find the dimensions of rectangle EFGH, and the number that make up one square yard. Ans. (1) 4.242 x 2.121 ; (2) 144. SIZES AND SHAPES OF STANDARD TILES IRREGULAR PENTAGONS, HEXAGONS, AND OCTAGONS Fig. 6. Fig. 60. 8. Irregnilar pentagons. — in Figs. 6 and 6a ABCD represents a six-inch square. E, F, G, and H are the middle points of the sides. In Fig. 6 K, L, R, and S are the middle points of the sides of the square EBFO. In both figures the points are joined as indicated to form the irregular pentagons numbered 1, 2, 3, and 4. EXERCISES 1. In Pig. 6 find the length of HP and PQ, and the area of penta- gon 1. Ans. (1) 2.121 in.; (2) 4.242 in.; (3) ISJ sq. in. 2. In Fig. 6 find the length of KN and MN and the area of pen- tagon 3. 3. In Fig. 6a find PQ, PG, and the area of pentagon 2. Ans. (1) 1.5 in.; (2) 2.121 in.; (3) 6f sq. in. 4. In Fig. 6a prove that RS extended bisects OE and FB. 5. If the points E, B, L, M, and K (Fig. 6a) are joined as indicated, prove that pentagon 2 is congruent to pentagon 4. 6. Prove that pentagons 1 and 3 in Fig. 6 and pentagon 2, Fig. 6a, are similar. Show that their areas obey the law for the ratios of the areas of similar polygons. 7. Show how td construct from the square ABCD a pentagon sim- ilar to pentagon 3, but having one half its area. Show how to construct one similar to pentagon 2, but having one half its area. Suggestion. — See Ex. 6 above. TILE DESIGNS p% t bmiRx Vv — M .- "^ \ ^ ^fW' ^ f^WL y# S^_\^ f r 1 kl/ 5f T^ll'^ ^ i R^ Fig. 7a. Tiled Flooring. Scale } in.=l ft. 9. In Fig. 7 ABCD represents fhe six -inch square, and E, F, G, and H are the middle points of the sides. The points are joined as indicated. EXERCISES 1. Prove that the line joining iV and K bisects AE, HO, and DG. 2. If the three pentagons are formed as indicated, prove them similar. 3. Find (a) the ratio of the sides FM, GP, and KQ, and (6) the ratio of the areas of pentagons 1, 2, and 3. '--Ins. (a) V2:l:-^; (*) 2:1:}. 4. Find the areas of pentagons 1, 2, and 3. Ans. \\\ sq. in., .5| sq. in., 2}| sq. in. 5. Figure 7a contains two sizes of tiles that are irregular pentagons. Show the ratio between each and the iix-iuch square. SIZES AND SHAPES OF STANDARD TILES 10. Irregular hexagons. — in Fig. 8 and 8a ABCD represents the six-inch square. Points E, F, 6, and H are the middle points of the sides. Each side of squares ABCD and EBFO in Fig. 8 and each side of EFGH in Fig. 8a is divided into four equal parts and the points joined to form the hexagons num- bered 1, 2, 3, and 4. EXERCISES 1. Prove that in each hexagon the opposite sides are parallel. 2. Find the length of the sides and the area of each hexagon. Ans. for No. 4, 2.121 in., 1| in., 6f sq. in. 3. How many hexagons of each kind would be sold for one square yard? Ans. 96 of number 1. 4. Show that hexagons 2 and 4 are congruent. 5. Prove that hexagons 1, 2, 3, and 4 are similar, and that their areas obey the law for the ratio of the areas of similar polygons. 6. From Ex. 5, show how to con- struct a hexagon from square ABCD, similar to hexagon 1, but having twice its area. 7. Figure 8J shows two tiles that are irregular hexagons. Show the relation between each and the six- inch square. \.. I N \ I » Fig. 86. Tiled Flooring. Scale f in. = I ft. TILE DESIGNS 11. Irregular octagons. — in Fig. 9 ABCD represents a six-inch square. Each side ^ is divided into four equal parts and the points joined to form the octagon KIM'S, etc. A S E K B Fig. 9. For applications see Fig. S6. EXERCISES 1. Find the ratio of KL to LM and of GE to XY. Ans. — and . 2 4 2. Prove that X F is the perpendicular bisector of MN. 3. Find the area of the octagon. Ans. 311 sq. in. 4. If similar octagons were formed from squares EFGH and BFO, -what would be their areas ? Ans. 15| sq. in., 7J sq. in. 5. Prove that EKLF is a trapezoid and find its area. Ans. 3f sq. in. 6. Join NK. Prove that NK is parallel to BC Complete the rhombus MNUV, and find its area. Ans. 3.181 sq. in. 7. Find the lengths of the diagonals iVF and UM. Ans. |V4 + 2\/2, fV4-2 V2. Suggestion. — Use A VNC and NZM to find lengths of NV and UM respectively. 8. Find the area of square constructed on CY as diagonal. Ans. y\ sq. in. Suggestion. — Use the theorem : " The bisector of an angle of a tri- angle divides the opposite side into segments proportional to the adjacent sides." SIZES AND SHAPES OF STANDARD TILES 9 D_ G a 12. In Fig. 10 ABCD represents the siz-inch square. E, F, G, H are the middle points of the ^p sides. AP = QE = NO, etc. = \ AE. M AP QE Fig. 10. Arts. 8 J sq. in. Ans. 2v'2:l. EXERCISES 1. Find the area of the octagon PQMN etc. 2. Find the ratio of the sides PQ and QM. 13. In Fig. 11 ABCD represents the siz- inch square. E, F, 6, H are the middle points of the sides. L and M are the mid- dle points of KF and FN respectively, and the points are joined to form the figure KLMNO. R and Q are the middle points of HX and XE respectively. RS and QP are parallel to HO and OE respectively, forming figure APQRS. EXERCISES 1. Find the areas of KLMNO and APQRS. Ans. 3}f sq. in. 2. Construct a figure similar to KLMNO from octagon KLMN etc., Fig. 9, and find its area. Ans. 7 J sq. in. 3. Show how to construct from square ABCD a figure similar to APQ.RS, but having twice its area. Suggestion. — It is J the octagon KLMNP etc., Fig. 9. 4. Show how to construct from square ABCD figTires similar to KLMNO and APQRS, but having I their areas. Suggestion. — Use the octagon derived from square AEOH. 5. Show that figures MNOY and RSAX are congruent. Are KLMNO and APQRS congruent? 10 TILE DESIGNS REGULAR HEXAGONS AND RELATED FIGURES 14. In Fig. 12 ABCD represents a six- w^- inch square with its inscribed circle. Sides BC and DA are each divided into four equal parts and the points joined by lines WZ, HF, and XY. Lines WZ and XY cut the circle at points N, K, L, and M. Points X'^ Q and R are the middle points of OM and GL respectively. t A Fig. 12. — For applications see Figs. 17-19. EXERCISES 1. Show how to inscribe a circle in the given square, and prove that the points of tangency are the middle points of the sides. 2. Prove that EKLGMN is a regular hexagon, if E and G are the middle points of AB and DC respectively. 3. Prove KL parallel to BC. 4. Prove that NPQ, is an equilateral triangle, and ORGM is a trapezoid. 5. Find the area of the circle and of the curved figure EKFB. Ans. 28.27 sq. in., 1.93 sq. in. 6. How many circular tiles six inches in diameter would make one square yard ? A ns. 45. 7. Find the areas of (a) hexagon EKLG etc., (V) APQN, Ans. 27v^ 27V3 9V3 2 16 ' 16 -Show that OS is perpendicular to PQ and that (c) A OPQ. Suggestion for (e). OS = J in. 8. What would be the length of one side of the circumscribed square, if the area of the hexagon is (a) J that of hexagon EKLO etc., (b) J that of hexagon EKLG etc. Ans. for (b), 3. SIZES AND SHAPES OF STANDARD TILES 11 9. Construct each of the hexagons mentioned in Ex. 8 from a six-inch square. ' Suggestion. — They must be constructed in squares EFGH and AEOH respectively. 10. Find the area of trapezoid OROM (1) by the formula for the area of the trapezoid, (2) by showing that it is J of the hexagon EKLG etc. Ans.i^Vz. 11. How may a figure similar to ORGM, but having J its area, be constructed from a six-inch square ? Suggestion. — Use square EFGH. 12. How may a figure similar to ORGM, but having \ its area, be constructed from a six-inch square ? CIRCULAR FORMS 15. In Fig. 13 ABCD represents the six- inch square. EG and FH are its diameters. Arc AKLC is drawn with D as center and DA as radius. Lines DK, DL, and AL are drawn. EXERCISES 1. Prove that AADK = /L KDL = Z LDC = 30°. Suggestion. — Prove ALD z,^ equilateral A. 2. Find the length of HL and the area of A HLD. Ans. (1) 3\/3iu. ; (2) fVS sq. in. 3. Find the area of the sector LDC. Ans. 9.4248 sq. in. 4. Find the area of the figure LFC. Ans. 1(12 - 3V3 - 2 tt). Suggestion. — Subtract the sum of the areas of the sector LDC, and the ALDH from the rectangle FCDH. 12 TILE DESIGNS 5. Find the area of EBFLK. Suggestion. — Subtract the sum of the areas of FCL and EKA and the sector CDA from the area of the square. Ans. 6.16+ or 3(3vl5-7r). D^ PG N H A j: Fig. 14. i ^ I S E K Fig. 140. ? ip ^ ^^ v. i 1 ? "^ ^ % 'vf' i4iiTi\ :l ^ A* H ? >*^« w* ? ? i i ^ m^ "i \ 1 ^^ Fig. 146. Tiled Flooring. Scale ^ in. = I ft. 16. In Figs. 14 and 14a ABCD represents the siz-inch square with the di- ameters EG and FH. In Fig. 14 6F and KL are drawn with C as center and CG and CD respectively as radii. In Fig. 14^ AB is the radius for KL, MN, PQ, and R§. EXERCISES 1. If SEOHR in Fig. 14a is congruent to AELKH in Fig. 14, find the center for RS. 2. Show that AELKH in Fig. 14 is congruent to EBFLK in Fig. 13, if the figures are on the same scale. 3. Find the area of EBFGDHKL in Fig. 14 and of KLMNP etc. in Fig. 14a. Ans. (1) 22.77 sq. in. ; (2) 24.64 sq. in. 17. In Fig. 15 ABCD represents the siz-inch square, with EG and FH the diameters. Arc KOL is B drawn with B as center and the half diagonal BO as radius. TILED FLOOR PATTERNS 13 EXERCISES 1. Find the area of the figure FOL. Ans. f(ir - 2). Suggestion. — Show that the sector OBL = J of the circle. 2. Find the length of FL. Ans. 3(V2 - 1). 18. In Fig. 16 ABCD represents the siz-inch square with EG one diameter. EXERCISES 1. Show how to construct the arc KGL tangent to CD at G, if its radius is J the diagonal of the square. 2. Show how to construct the arc LNK so that it shall have the same radius as the arc KGL and shall pass through the points K and L. 3. Prove that /. LMK is 90°. Suggestion. — Since MK = 3 V2 and if = 3, OM = 3, then Z OMK = 45°. 4. Find the area of the oval ^»£7V. ^ns. 9(7r-2). Suggestion. — Find the area of the segment KGL from the areas of the sector MKGL and the A MKL. 5. Find the length of GN. Ans. 6( v^ - 1). 6. If arc KGL is a quadrant and is constructed tangent to CD at G, prove that its radius is \ the diagonal of the square. PART 2. TILED FLOOR PATTERNS 19. Designing in systems of parallel lines. — Many re- peating patterns commonly used in industrial designs are planned on one or more systems of parallel lines. When the design is to be in stripes, one system of parallels is 14 TILE DESIGNS necessary. (See Fig. 112.) When it is based on parallel- ograms, two systems are needed. Rhombuses are often used for wall papers and fabrics, and squares for tiled or parquet floors and linoleums. When the design is based on triangles or hexagons, three systems of parallels are used. Such designs are freely used for tiled floors. They were highly characteristic of Arabic ornament. When octagons are desired, four systems of parallels must be drawn. Such designs are frequently used for linoleums and steel ceilings as well as for tiled floors. 20. Designing tiled floors. — Tiled floor designs are planned on systems of parallel lines. Moreover, modern tiles are made in certain definite sizes and shapes which are related to each other as shown in the previous section. In planning tiled floor designs, therefore, the designer so regulates the distances between the parallels and the angles at which they intersect as to determine the tiles which he wishes to use. This is evident from a study of the designs here shown. 21. History. — The patterns used for tiled floors are among the simplest found in industrial art. These simple designs abound in the geometric mosaics of southern and eastern Europe. ^ Italy and Sicily are full of famous examples, most of which are of Byzantine origin, although those in Sicily strongly suggest Moslem workmanship. There were many different kinds that received special names; among them the Opus Alexandrinwm was an ar- rangement of small cubes of porphyry and serpentine set in grooves of white marble. 1 See Gravina and Ongania for sample pavements. See also J. Ward, (1), Vol. I, pp. 288-2M, Vol. II, p. 344. TILED FLOOR PATTEBNS 15 DESIGNS BASED ON THREE SYSTEMS OF PARALLEL LINES 4 K B rTTTTT fTTTTT TTTTTl Fig. 17. Fig. 17a. —Scale J in. = I ft. 22. Figure 17 shows a pattern inTolving equilateral triangles con- structed by drawing three sets of parallel lines. EXERCISES 1. At what angle must the lines intersect? 2. If Fig. 17a is drawn on a scale of | in. to 1 ft., find the length of each side of one triangle. Ans. 3 in. 3. How may one of the triangles shown in Fig. 17a be derived from the six-inch square ? Suggestion. — See Fig. 12. 4. If a side of each triangle is 3, find the area of each. Ans. fV3. 5. If a side of each triangle is |v'2, find the area of each. Ans. |V3. 6. Construct the pattern shown in Fig. 17a, using if possible draw- ing board, T square, and triangle. Let one side of each triangle rep- resent 3 inches. Scale 3 in. = 1 ft. 7. What will be the actual number of square inches of drawing paper, covered by one triangle, if the drawing is made as suggested in Ex. 6? Ans.-i^V3. 8. Parallel lines may be constructed by using T square and triangle. On what theorems may the construction be based? 16 TILE DESIGNS Fig. 18. — Parquet Flooring. Fig. 18a. — All-over Lace Pattern. EXERCISES 23. 1. If each side of a hexagon shown in Fig. 18 represents 3 inches, measure the accompanying cut and find the scale to which it is drawn. 2. Construct Fig. 18 from three systems of parallel lines. 3. If each side of a hexagon shown in the figure represents 3 inches, what is the distance represented between the oblique paral- lels? 4. Construct the design shown in Fig. 18 from an all-over pattern of equilateral triangles. Let each side of hexagon represent 3 inches. Scale J in. = 1 ft. Fig. 19. —Tiled Flooririg. Scale | in. = I ft EXERCISES 24. 1. Construct the three sets of parallel lines on which Fig. 19 is designed. Is the pattern of equilateral triangles necessary ? TILED FLOOR PATTERNS 17 3. If one side of the hexagon is 3, find the area of the star formed of one hexagon and six triangles. Ans. 27 VS. 3. Prove that if the sides of a regular hexagon are extended until they intersect, six equilateral triangles are formed whose combined area is equal to the area of the given hexagon. 4. Construct the design shown in Fig. 19, letting a side of the hexagon represent 3 inches. Scale 2 in. = 1 ft. 5. If the drawing is made as suggested in Ex. 4, what will be the actual number of square inches of drawing paper included in the star? Ans. |V3. D__ G I X _Z_T T^ Z X X z X X X: .x_x_x_x. tVififi ^T^ T ^T^X^ A E F G -o Fig. 20, — Tiled Flooring Scale g i = lft. EXERCISES 25 ■ 1 . Construct the three systems of parallel lines on which Fig. 20 is based. Suggestion. — The oblique lines form a system of squares. The horizontal lines bisect the sides of the squares. 2. Find the ratio (1) between a side of one of the squares and KL, and (2) between EF and KL. Ans. (1) -^ ; (2) 2. V2 3. If the scale of the figure is f in. = 1 ft., find (1) the true length of EF; (2) the length of one side of one square; and (3) the sides of the triangle and of the irregular hexagons. Ans.il) 6; (2) 3%/2; (3) 3 and |v^. 4. Construct a design like Fig. 20. Let EF represent 3 inches. Scale 4 in. to 1 ft. 18 TILE DESIGNS Q \ K XXH/ /r"^ 1 \ ^ ^-d $3 X 52 Vt] 3. p f b^ O DESIGNS BASED ON FOUR SYSTEMS OF PARALLEL LINES A B 26. Irregular octogons — Figure 21 is constructed on four sets of parallel lines, as shown. Fig. 21. —Tiled Flooring. Scale J in. = I ft. EXERCISES 1. Find AB if one side of the larger dark square is4J inches. Ans. 9.01+. 2. Ji AB is exactly 9 inches, what is the length of one side of the larger square? Ans. 3V2. 3. Construct the design for Fig. 21. Let each side of the large square represent 3 inches. Scale 4 in. to 1 ft. 4. If the drawing is made as in Ex. 3, how much actual paper is contained in the smallest square and in the irregular hexagon ? Ans. (1) I sq. in ; (2)| sq. in. B D F H A C E G Fig. SS. — Tiled Flooring. Moorish. Scale } in. = I ft. Calvert f I), p. 301. 27. Figure 22 is based on four sets of parallel lines. AC = 2 CE = EG, etc. MO = 2 MK. Each side of the larger square is divided into four equal parts and diagonal lines are drawn to form the crosses and octagons. TILED FLOOR PATTERNS 19 EXERCISES 1. Construct a design like that shown in Fig. 22. Let A C repre- sent 6 inches. Scale 2 in. = 1 ft. 2. If ^C = 6, find the area of (1) the octagon ; (2) the irregular hexagon ; (3) one of the small squares. .4ns. (1) 31i;(2) 13^; (3) H- 3. If the drawing is made as suggested in Ex. 1, find the actual number of square inches of paper covered by each of the figures men- tioned in Ex. 2. Ans.(l) J; (2) f; (3) \. Fig. 23. Fig. 23a. — Tiled Flooring, Scale f^ in. = I ft. 28. In Fig. 23 the horizontal and vertical sides of the octagons are equal, and the triangles included are isosceles right triangles. EXERCISES 1. Find the ratio between the sides of the octagon. Ans. 1 : V2. 2. Show how the octagon may be constructed from a square. 3. Show how the design may be constructed from four sets of parallel lines. Suggestion. — How does the distance between the vertical parallels compare with the distance between the horizontal ones ? 4. Make a drawing to scale for the pattern shown in Fig. 23. Let the vertical side of the octagon represent 3 inches. Scale 3 in. = 1 ft. 5. In the drawing made for Ex. 4 how much actual paper is there (1) in one of the octagons and (2) in one of the triangles ? Ans. (1) II sq. in. ; (2) ^^ sq. in. 20 TILE DESIGNS Fig 24. Fig. 24a. —Tiled Flooring. Scale g in. = I ft, 29. Regular octagons. — In Figs. 24 and 24a the octagons shown are regular. EXERCISES 1. Show how to inscrihe a regular octagon in a square. Suggestion. — First Method: AVith the vertices of square ABCD as centers and the half diagonal OA as radius draw arcs cutting the sides of the square at points E, F, G, H, etc. To prove EFGH etc. a regular octagon, show that its sides and angles are equal. Use A EOF, FOG, GOH, etc. Notice that A .4 OF, OBE, etc. are con- gruent and isosceles and that /.FOB is measured by \ arc OF. Second Method: With as center and the half diameter OQ as radius cut the diagonals at points P, R, S, and T. Draw HK, LM, NE, and GF perpendicular to the diagonals through points P, R, S, and T. Now prove EFGHKL etc. a regular octagon. 2. Find the ratios AF: AB, AE:AB, AE : EF. V2 2 - V2 V2 2 ■ ■ Ans. 2 ' 2 3. If AB = a, find (1) the length of AE, (2) the area of AANE, and (3) the area of the octagon EFGH etc. Ans. (1) |(2-V2); (2) |(3-2^); (3) 2 a2(V2 - 1). 4. If the area of the octagon is 100 square inches, find the length oiAB. Ans. 5V'2-v/2 + 2. TILED FLOOR PATTERNS 21 6. If the area of the octagon is S, find the length of AB. Ans. iVs(2V2 + 2). 6. Construct the diagram for Fig. 24a. Let wx represent 6 inches. Scale 4 in. = 1 ft. Suggestion. — See Fig. 24. Construct a network of squares. In- scribe a regular octagon in each square. 7. In Fig. 24 prove that FGG'F' is a square. 8. Prove that UVWXYHOE is a regular octagon. 9. Prove that UFOE and WXYG' are congruent rhombuses. 10. Prove that corresponding diagonals as EK and G'Z are parallel. 11. Prove that points Z', F, and Z are coUinear. 12. Construct Fig. 24a by drawing four sets of parallel lines and erasing those parts which are not necessary for the figure. 13. If in Fig. 24a wx = a, find the length of (1) CB; (2) AB; (8) the perpendicular distances between AD, BE, and CF. Ans. (1) a(2 - V2); (2) a(V2-l); (.3) ? (2 - V2) and a(V2 -1). Fig. 25. Fig. 25a. — Linoleum Pattern. 30. In Fig. 35 the octagons are regular, and the small quadrilaterals are squares. EXERCISES 1. Find the ratio ^ = ?^^ AB AB Ans. 2-^2 22 TILE DESIGNS 2. Construct a diagram for Fig. 25a by drawing four sets of par. allel lines and erasing those parts of each which are not necessary in the figure. Let ^B = IJ inches. Take BP = 2 BF. 3. Prove that the construction suggested in Ex. 2 will make all the octagons regular and all the small quadrilaterals squares. 4. liAB = a, find the length of FR. Ans. a(3 - 2V2). Suggestion. — Piove HS = BP; GS = FR = US - HG ; HG = a(V2-l). 5. If AB = a, find the perpendicular distances between the oblique parallels, ^ns. (1) a(\/2 - 1) ; (2) ^(2 - V2) ; (3) ^(3V2 -4). 6. Can you suggest any other method for developing Fig. 25? Suggestion. — See § 84. 31. In Fig. 26 the octagons are regular. A B X Y Fig. 26. — Linoleum Pattern, EXERCISES 1. Determine the sets of parallel lines on which Fig. 26 may have been constructed. Is there more than one possible method for developing this figure ? Suggestion. — See § 85. 2. Find the relation between the long and the short side of the irregular hexagon. ^„^. 2^/2 : 1. 3. If AB = a, find the area of the cross, and of the small square. Ans. (1) a2(4v^ - 5) ; (2) a2(3 - 2V2). CHAPTER II PARQUET FLOOR DESIGNS 32. Parquet floors ^ are made by glueing or nailing into geometric patterns strips or blocks of various kinds of hard wood. These pieces of wood are cut with a scroll saw at the mills. In preparing the pattern, therefore, the designer is restricted to the use of pieces of such shapes as can be cut out conveniently. This accounts for the predominance of straight lines and for the simple geo- metric figures that enter into the combinations. 33. History. — Like tiled floor designs, most of the par- quetry patterns are extremely simple, so that while sug- gestions for these designs seem to have been taken from every epoch and every country, only the simpler forms of historic ornament have been used. Of the designs here given Figs. 39, 40, 55, 56, 64 are found in Medieval Byzantine Mosaics. Figures 48, 52, 55, 57, 60, 62 are of Arabic origin. Figure 51 is a closely related form from a Roman mosaic. Greek frets are also found in the catalogues. The simpler Mohammedan designs here referred to are from Cairo. Simple geo- metric patterns are also found in Chinese and Japanese art. ^A large proportion of the designs in this chapter are from parquet floor catalogues. See list of references. 23 24 PARQUET FLOOR DESIGNS PART 1. DESIGNS IN RECTANGLES AND PARAL- LELOGRAMS IXKXX S F a H L W¥&&W^ Fig. 27. Fig. 27a. —Tiled Floor Border. Fig. 276. — Parquet Floor Border. EXERCISES 34. 1. Construct a series of squares in a given rectangle, as shown in Fig. 27. Suggestion. — Divide the rectangle into squares by a series of parallels. 2. Prove that squares BF and CG are congruent. 3. How may the points BCD and FGH be laid off on the sides of the rectangle so that the inscribed squares may be constructed by drawing the straight lines EB, FC, AF, BG, etc.? 4. What relation must exist between two adjacent sides AE and AD ot the rectangle in order that an integral number of squares may be inscribed in it? 5. If the border is 12 feet long, what must be its width in order that 16 squares may be exactly inscribed in it? Ans. 9 in. 6. If the border on one side of a floor is 14 feet long, for what widths of border between 8 and 12 inches will it exactly contain an integTal number of squares ? Ans. 8, 12, 10^. If fractions are permissible, there is no limit to the possible number of results. 7. What is the length of a side of one of the inscribed squares if AE='2'>. liAE = a'> Ans. V2, 2^2. 8. Find the length and width of a border which contains 24 in- scribed squares whose sides are each 8 inches. Ans. ?>y/2, 192 V2. DESIGNS IN RECTANGLES AND PABALLELOGRAMS 25 A E F a N D Fig. sea. Fig. 286. — Parquet Floor Bonder. 35. EXERCISES 1. In Pig. 28 how are the points E, F, G, N and H, K, L laid ofE on the sides of the rectangle so that the lines BF, LG, EH, etc., will form two series of squares superimposed with the corner of one at the center of the next ? 2. Prove that the lines BF, LG, etc., are parallel. 3. Prove that the lines BF and EH are perpendicular. 4. Prove that BF bisects EZ. 5. Prove that the vertices X and Y of the square XKYF are the centers of the squares OLZE and ZHWG. 6. Prove that the square OEZL is divided into four equal squares. 7. What fraction of the square XFYK does not form a part of any other square? Ans. \. 8. Make a drawing for the border shown in Fig. 28a. Fig. 29. — Art-glass Design. EXERCISES 36. 1. In Fig. 29 what must be the relation between the length and width of the rectangle ABCD in order that the tangent circles may be inscribed as shown? 26 PARQUET FLOOR DESIGNS 2. If in a rectangle three tangent circles are inscribed, what is the ratio of the area within the circles to the area outside the circles? Ans. 3.66. 3. If in a rectangle three tangent circles are inscribed, what is the ratio of the area within the circles to the area of the whole rec- tangle ? Ans. .7854. 4. Does this ratio depend upon the shape and size of the rectangle? 5. Does the number of inscribed circles depend upon the size or shape of the rectangle or upon both ? Wliy ? ALB N C F Fig. 30. — Art-glass Design. EXERCISES 37. 1. How must the points A, L, B, E, and D, N, C, F be laid off on the sides of the rectangle shown in Fig. 30 so that two series of tangent circles may be inscribed in the rectangle with the center of one circle on the circumference of the next? 2. If KH is the line of centers and the point of tangency, prove that A KOX is equilateral. 3. li AD is 12 inches, find the area of the oval KZOX. Ans. 6(47r-3-\/3). 4. If XO = a, find the area of the oval KZOX. Suggestion. — Draw the chords KX and XO. The sector KOX is one sixth of the circle and its area is — — The segment bounded by the chord KX and the arc KX is the difFerence between the area of an equilateral triangle whose side is a and the area of the sector KOX. The area of the segment is "tj — -j v3. Adding this to the area of the sector, we find the area of the half oval to be -^(4 ir—ZVZ). 5. If AD = 12, find the area of the figure LXOY. Ans. 6(3V3-7r), or 12.32+. DESIGNS IN RECTANGLES AND PABALLELOGSAMS 27 6. If AD = a, find the area of the figure OYLX. ^n..g(3V3-^). 7. If the area of the oval KXOZ is 49 square inches, find AD. Ans. 12.6+. 8. If the area of the figure OYLX is s, find AD. Ans. JZlpZ. 3 VS — TT Fig. 31. — Stair Railing. Fig. 3ia. 38. In Fig. 31 ABCD is a paraUelograin with four tangent circles inscribed. EXERCISES 1. If the lines A C and BD are supposed to be indefinite in extent, show how to construct circle tangent to the lines A C, AB, and BD, and circle X tangent to lines 4 C and BD and to circle 0. 2. If E is the middle point of AB and and X are the centers of the circles, prove that the points E, 0, and X are coUinear. 3. Prove that A AOB is. 90° and that OE = 1 AB. 4. If Z ABL = 60°, prove that ^0 = J AB, and that the radius of circle Oi% — \% 4 28 PARQUET FLOOR DESIGNS 5. In Fig. 31a A C is tangent to circle and DB to the other circle. CA and DB are parallel, and the cii-cles are tangent at point X. Prove that AB is bisected at point X The two circles are equal. Suggestion. — Use two pairs of congruent triangles. Additional construction lines will be necessary. 6. In Fig. 31a iiZCAO is 30°, show that A C passes through the center of the left circle. 7. Construct a figure like Fig. 31a, making Z CAB = 60°, and prove that ^B = | OZ (2 V3 - 3). iJ J Fig. 32a. — Karquet Floor Border, 39. Figure 32a shows a series of regular hexagons inscribed between two parallel lines. Figure 32 is the diagram for the construction of Fig. 32a. EXERCISES 1. In Fig. 32 find the number of degrees in each angle of AFNA. 2. Prove that the line joining F and C is parallel to RO. 3. How must the points E, D, P, etc. and A, B, 0, etc. be laid off from line MW in order that the hexagons may be constructed by drawing lines DO, BP, etc. ? 4. If MN = 4, find the lengths of the sides of A FNA and the length of FC. Find these lengtlis if MN = a. Ans. 2 6 V3, ^V3, ^V3. '3 3 5. If a rectangle is 16 inches wide, how long must it be if 16 hexa- gons are inscribed in it? Ans. 24.63 ft. 6. What must be the relation between the length and width of the rectangle in order that a whole number of regular hexagons may be inscribed in it? 7. What is the ratio of the sum of the areas of the hexagons to the area of the whole rectangle ? Ans. J. 8. Does this ratio depend upon the shape or size of the rectangle ? DESIOlSrS IN BECTANGLE8 AND PARALLELOGRAMS 29 A E F G H C Fig. 33a. — Parquet Floor Border. 40. Figure 33 shows a series of rhombuses KPEO, LQFP, etc. inscribed between two parallel lines AC and BD. The angles of the rhombuses are 60° and 120°. The hexagons shown are formed by joining the middle points of the sides of the rhombuses. EXERCISES 1. How must the points E, F, G, H, etc., and K, L, M, N, etc., be laid o&on AC and BD in order that the rhombuses may be con- structed by drawing the lines EL, FK, LG, etc. ? 2. Prove that the hexagons shown are regular. 3. Prove that XZ YE is a rhombus. 4. ltAB = ?,, find the area of (1) OKPE, (2) one of the hexagons, (3) XZYE, and (4) EYWF. Ans. (1) ^^v^. 5. Find these areas if .4.B = a. n2 Ans. (1) I V3; (2) | V3 ; (3) ^ VH; (4) ^ V3. 6. In Fig. 33a if the width of the border is 12 inches, find the total length of the dark bands in a border of 12 feet. Measure through the center between the bands. 30 PABQUST FLOOR DESIGNS /[\ i\ X-- -'X / X ' ^^ 1 ^\ / \'' \ / / 1 ~\ \ ~^^ x! '' 1/ \ '' "-■ / \ xi' y 1 \.' \/ \ U M Fig. 34. N Fig. 34a. — Ceiling Pattern. 41. Figure 34 shows a series of regular octagons inscribed between two par- allel lines. EXERCISES 1. Show how to consti-uot the series of octagons. Suggestion. — Draw parallels BA, UV, etc., so as to construct a series of adjacent squares. In each square inscribe an octagon. See § 29. 2. If the octagon CELO' etc. is regular, prove that CO' and DO are parallel to .4 fT" and B V, respectively. 3. Find the number of degrees in each angle of the A CEK and DHF. 4. If one side CE of the regular octagon is 1 inch, find the lengths of CK and EK. ■ Ans. J V2. 5. Find lengths of these lines if CE = a. Ans. ?V2. 2 6. If CE = 2, find the area of the trapezoid CEFD, of the rec- tangle ELPF, and of the complete octagon ELO'QPFDC. Ans. Area of octagon 8(1 + v^). 7. Find these areas if CE = a. Ans. 2 a^{y/2 + 1). 8. Prove that the points E, 0, Q and also L, O', R are coUinear. 9. If EL = a, find the distances AE, EL, LM, MN, NT, and so on, and lay off the points E, L, M, N, T and F, P, Q, R, S. How is LAf re- lated to AE^ Construct the octagons. Ans. - V2, a, aV2, a, aV2. DESIGN'S BASED ON ISOSCELES BIGHT TRIANGLE 31 10. What must be the relation between the length and width of a rectangle in order that a whole number of regular octagons may be inscribed in it? 11. 11 AB = i, &nd AE md EL. 12. If AB = a, find these lengths. Ans. AE = ^(2- V2), EL = a(V2 - 1). 13. If EL = b, find AB. Ans. AB = b(V2 + 1). 14. If EL = b, find the area of the octagon by subtracting the areas of the triangles AEC, LUG', OVP, and DFB from the area of the square A UVB. Ans. Area of octagon 2 6^(1 + y/T). 15. If 12 regular octagons whose sides are each 4 inches are inscribed in a rectangle, find the dimensions of the rectangle. Ans. 9.656; 115.8. 16. What is the ratio of the sum .of the areas of the inscribed octa- gons to ihe area of the rectangle? Ans. 2(v^ — 1). PART 2. DESIGNS BASED ON THE ISOSCELES RIGHT TRIANGLE 42. The designs in this section are all based on the unit shown in Fig. 35. It is interesting to note the num- ber of times this unit is used in each design and the effect of the various positions in which the unit is placed. Fiff. 35a. — Parquet Flooring, 43. In Fig. 35 ABC is an isosceles right triangle with a right angle at C. CO is perpendicular from C to AB. 32 PARQUET FLOOR DESIGNS EXERCISES 1. Show how to construct XY parallel to AC so that CY= XY. Suggestion. — Bisect Z OCA . 2. Draw YZ from Y parallel to AB and prove that XY = CY .= YZ = BZ = AX. 3. Show that the lines mentioned in Ex. 2 would have been equal if Z CBA or Z. CAB had been bisected. 4. Prove that BZYX and AXYC are congruent trapezoids. \Y r~ 5. Prove that OY = OX, and hence show that ^—— = v 2. 6. Find the ratio ^j-= and -—■ 7. Prove that XF extended is perpendicular to CB and that the altitude of A CYZ on CZ as base is equal to the altitude of trapezoid AXYC. 8. If ^C = 8, find CY and the area of AXYC. Ans. (1) 8(-\/2 - 1) ; (2) 32(V2 - 1). Suggestion. — For the area of ^^FC show that rO = 4(2 — V'2). Subtract the area of A XOYiroxa the area of A ^ OC. 9. li AC=a, find CY and the area of AXYC. Ans. (1) a(V2-l); (2) |'(V2- 1)- 10. If AB = a, make these same computations. Ans.(l) ^(2-V2); (2) ^(^V2-\). 11. If AB = a, find the area of A CYZ. Ans. — (.3 - 2 v^). 12. If ^£ = a, show that the area of AXYC is about 10 Suggestion. ^ ; '- is a^(.10l+). From the figure, AXYC is less than J ABC, and hence less than ^ a^. DESIGNS BASED ON ISOSCELES BIGHT TBI ANGLE 33 H G W N p-'-'U '' L\ Y E Fig. 36. B J ^ ^ (/. ■^ /^ m^ ^ Fig. 36a. — Parquet Flooring. 44. In Fig. 36 ABCD is a square with the diameters EG and HF and the diagonals AC and DB. OX and OY bisect A FOC and EOA, respectively. The line XY cuts OE and OF at E and M, respectively. KL and ML are drawn from K and M parallel to the sides AB and BC. EXERCISES 1. Prove that Jfy is parallel to ^4 C 2. Prove that OK = OM and that the two A OKL and LMO form a square with the vertex L on the diagonal OB. Construct the figure. 3. ItAB^a, find the area of the square OL. Ans. —(3 - 2 V2). 4. What per cent of the design in Fig. 36a is made of dark wood? Frg. 37a — Parquet Flooring 4S. In Fig. 37 ABCD is a square with the diagonal AC and BD and the diameters EG and HF. 34 PASQUET FLOOB DESIGNS EXERCISES 1. How many times is 'the unit shown in Fig. 35 used in this figure? 2. Show how to construct EL = LP = LK = KB. Show that this may be done by bisecting any one of three different angles. 3. Construct the entire figure. 4. Prove that points K, L, M, N; X, Y, Z, W; and Q, M, P, S, R are collinear. 5. If ^5 = a, find the area of A i:iP. Ans. ^ (S - 2V2) . 8 6. If AB = a, find the area of the wheel-shaped figure. Ans.a\V2-l). Suggestion. — l\-ove that A LKT = A TXY. Then the area of OPLKYXFO is the area of A EFO less the area of A ELP. 7. If AB = a, find the total area of the white triangles about one of the wheels in Fig. 37a. Ans. 0^(3 - 2v^). 8. Find the per cent of each kind of wood in one square in this design. 46. In Fig. 38 ABCD is a square with AC and BD the diagonals and EG and FH the diameters. p G c \ P /i\ / N X 1 w >< H <- M \ 1 / Y > F \ / ^ 1 K 1 X /'- ^'"'^ \i/ \ A E J 3 Fig. 38. — Parquet Flooring. EXERCISES 1. Show how to obtain Fig. 38 from Fig. 37. 2. Show how to construct the lines in AAEH so that LM = LK = LA. DESIGNS CONTAINING SQUABES WITHIN SQUARES 35 3. Construct the entire figure by repeating the construction sug- gested in Ex. 2, and then prove that the points L, M, N, P are coUiuear. 4. Give more than one method for constructing the lines shown in the square EFGH. 5. For each construction given in answer to Ex. 4 prove that LK = KX = XO. 6. Prove that XYZ PF is a square. 7. llAB = a, show that the area of XYZ W is, ^(% - 2 V2). PART 3. DESIGNS CONTAINING SQUARES WITHIN SQUARES FIRST POSITION OF THE SECONDARY SQUARE 47. Figures 39 to 42 show the secondary square with its sides parallel to the diagonals of the given square. The most common form of the design is shown in Fig. 39. The square formed by joining the middle points of the sides of a given square is of universal occurrence. A frequent Roman design is made from Fig. 42. See § 52. D G c H \^ \^/ \ \ / / \ Q /i \ Fig. 39. F >o \ ^^ ^ ^ '^^Sp-Av/ It ^^)^^^K. ^ > ^^5 B -Tile Pattern, Fig. 39a. 48' In Fig. 39 ABCD is a square with AC and BD its diagonals. E, F, G and H are the middle points of the sides and the points are joined as indicated. 36 PABQUET FLOOR DESIGNS EXERCISES 1. Prove that EFGH is a square. 2. If the sides of square EFGH cut A C and BD at points X, Y, Z, and W, prove that XYZ M'' is a square. 3. If AB = 4, find the area of the .squares EFGH and XYZW. Find the area of the trapezoid ABYX and of the AZI'O. Add the areas of the trapezoid and of the triangle. The sum should be \ the area of square ABCD. Why? 4. If AB = a, find the area of (1) (3) ABYX; (4) XYO. Ans. (1) |-; (2) EFGH; (2) XYZW; 3a2 (3) (4) 4 ' ' ' 16 ' '"' 16' 5. How does the area of the AABO compare with the area of the square EFGH'i With the area of the square XYZWt Does this depend upon the length of ABl D L Jl c 49. In Fig. 40 ABCD is a square with diameters and diagonals drawn. The points E, F, G, and H are joined. Various methods may be used to construct the re- mainder of the figure as indicated below. A K E B Fig. 40, — Parquet Floor Design. IVIedi- eval Mosaic, Gayet, Vol, II, Parenzo, PI, 32, EXERCISES 1. Let L and K be the middle points of DG and AE, M and N the middle points of DH and OF, etc. Draw LK, MN, etc. Prove that (1) DO, GH, LK, and MN are concurrent, and (2) MWLD is a square. 2. As an alternative method of construction, form squares in the A A EH, EBF, etc. with AX, BY, etc. as diameters, and prove that . the points K, X, W, and L are coUiuear. DESIGNS CONTAINING SQUARES WITHIN SQUARES 37 3. As a third method, join W and A', X and F, etc., and prove that (1) WX bisects HO and if extended bisects AE and GD, and (2) MWLD is a square. 4. If the diameter GE cuts WZ and J^F at T" and [7, respectively, and if the diameter HF cuts WX and Z^F at iil and V, respectively, prove that HR TG is a trapezoid. 5. Prove that iff/ FT" is a square. 6. JiAB = A, find the length of WS. 7. If AB = a, find the length of WS. Ans. -VJ. 8 8. If ^ B = a, find the area of HRTG by (1) Finding the length of GH, iJT, and WS, then using the for- mula for the area of a trapezoid ; (2) Finding the area of OGDH and then subtracting the areas of 3a2 HGD and OTR. 9. If AB = a, find the area of the square MWLD. 10. If AB = a, find. the area of the square RUVT. D M 50. In Fig. 41 ABCD is a square, DB one diagonal, and E, F, G, H, the H middle points of the sides. Ans. '■ A E B Fig. 41, — From Tiled Vestibule Design. EXERCISES 1. Construct two equal squares between the points D and N so that the sum of their diagonals is DN. 38 PARQUET FLOOR DESIGNS 2. If AB = a, find the axea of the square DKLM. Ans. 3. If the line DL = s, find the length of AB. 64 Ans. 4:sV2. 4. If n equal squares are constructed between D and N so that the sum of their diagonals is DiV, find the' area of each square if AB = a. Ans. ■ 16 n2 B B M C Fig. 42a. — Parquet Flooring. 51 . In Fig. 42 ABCD is a square ; E, F, G, H, the middle points of the sides; AK = BP = BQ = CL = CM, etc. ; and points are joined as shown. EXERCISES 1. Prove that: (1) KL is parallel to AC; (2) KLMN is a rectangle; (3) XYZW is a square; (4) KL is parallel to EF; (5) points E, X, O, Z, and G are collinear. Suggestion for (5). — EG is given a diameter of the squarp ABCD. Prove that KX = PX, and hence that X lies on EG. 2. What must be the relation of BP to 45 in order that EF may be trisected by SP and QR ? Ans. BP = AB 6 ■ Suggestion. — Prove PQ parallel to EF. Then prove ET:TU::EP:PB. DESIGNS CONTAINING SQUARES WITHIN SQUARES 39 3. Draw a figure to illustrate the special case referred to in Ex. 2. 4. If EF is trisected by the lines PS and QR, prove that KL is divided into five equal parts by the lines EH, PS, QR, and FG. 5. If AB = 12 and AK = 3, find the length of KL and KN. Find the area of figure AKLCMN. Ans. Area = 63. 6. liAB = a&aAAK^ ?, find the lengths of (1) KN; (2) KL ; n (3) KP; and (4) PT. Find the area of (5) AKLCMN, and (6) of KLMN. Ans.{\)^V2; (2) °(»-l)V2; (3) ^(n - 2); (4) -A.(„ _ 2)V2; n » n 4 » (5)2!(2„_1); (6) ^-'(n-1). 7. .If ^B = a and if the area of the square ZF.^JF = —, find ^ is: in terms of a. Ans. - ■^. 8 8. If the area of XYZ W is — , find ^iT in terms of a and n. n a Ans. V2n 9. What must be the position of point K if the area of KLMN is J the area of the square ABCDt Ans. AK = J AB. 2 a^ a^ Suggestion. — Solve for n the equation (n — 1) = — . 10. What is the value of n if the area of KLMN is — the area of the square ABCDI Ans. k±Vk'' -2 k. 2 a^ (i^ Suggestion. — Solve for n the equation, — — (n — 1) = — . 11. What must be the value of n if the area of AKLCMN is — of the area of the square ABCD1 Ans. k ± v'A'^ — K. 40 PABQUET FLOOR DESIGNS 52. Figure 43 is closely related to Fig. 42. EXERCISES 1. Show how to construct Fig. 43. Suggestion. — Make DE = DF= etc. Inscribe semicircles in i^ DEF, etc., so that their diameters shall be on the lines EF, etc., and be tangent to the sides of the square. 2. If AB = a and DE = -, find (1) the radius of semicircle LKQ; Ans. -^,5v'2(«-l). 2n n (2) the length of LM. 3. If AB=a and DE =-, find the area of the figure LMNPQK. Ans. -(nv'2-V2 + ?). 4. If AB = 6, DE = 2, and HR = \, find the area included between LMNP, etc. and RSTU, etc. Ans. 6V2 + — . 16 SECOND POSITION OF THE SECONDARY SQUARE 53. In Figs. 44-47 the sides of the secondary square are parallel to the sides of the given square. Of the designs shown Fig. 47 is the most interesting, giving rise to an unusually large variety of patterns. DESIGNS CONTAINING SQUARES WITHIN SQUARES 41 DM L O Fig. 44a. — Linoleum Pattern. 54. The design shown in square ABCD in Fig. 44 may be constructed by different methods, as indicated below. EXERCISES 1. Construct OX=OY=OZ = 0W. Join X, Y, Z, and W. Draw EF, GH, etc., perpendicular to the diagonals at points X, Y, Z, etc. Prove that (1) XYZW is a square; (2) XY is parallel to AB; (3) AE = AF= GB, etc. 2. Let W be any point on OD. From W draw WX parallel to DA, cutting ^O at ^ ; from X draw XY parallel to AB, etc. Prove that OW = OX. 3. As a third method of construction, make AE = AF = GB, etc. Let EF, GH, etc., intersect the diagonals at points X, Y, etc. Prove that (1) XYZW is a square; (2) XY is parallel to AB) (3) EF is perpendicular to A O. 4. Prove that XY = FB. 5. If AB = 16 and AF = 6, find the areas of XYZW, A AFX, and trapezoid FGYX. 6. If AB = a and ^i^ = -, find areas mentioned in Ex. 5. n ^»-'-0)^(»-l)=;(3)^(2«-3). 7. Find the area of the trapezoid FGYX (1) by using the formula for the area of the trapezoid, (2) by subtracting the areas of A AFX, GBY, and XYO from the area of A ABO. 42 PARQUET FLOOR DESIGNS 8. l£AB = a and AF = -, find the area of trapezoid FGYX when (1) n = 2 ; (2) f< n < 2 ; and (3) n < |. Suggestion. — Draw figures to illustrate each of these cases. When n = 2, the trapezoid becomes a triangle. For the other cases the dif- ference between the areas of two triangles is found. 9. For what value of n in Ex. 6 will the area of trapezoid FGYX be zero? Draw a figure to illustrate this case and prove it by geometry. 10. If the area of A AFE is equal to the area of FGYX, find the ratio of .4 F to ^ B. Arts. i. Suggestion. - where AF -. , AB n a* Or ■ Solve for n in the equation (2 n — 3) = - — -" 11. If the area of the A A FE is equal to i the area of the square XYZW, find the ratio of ^ F to AB. Am. J. Suggestion. — Solve for n in the equation — (n — 1)^ = —, where AF=.^. n 12. Construct figures to illustrate the special cases mentioned in Exs. 10 and 11. 13. If the area of A AFE is equal to the area of A XOY, find the ratio of -4F to AB. Ans. —^ . V2 + 1 14. Show that in the case mentioned in Ex. 13, EF = XY, and draw an illustrative figure. Suggestion. — Divide AB in the ratio 1 : y/2\ that is, in the ratio of a side of a square to its diagonal. See Fig. 446, where rz;-;- = . — ■ ^ FB y/2 , AF \ ^^ and .•. — — = . AB v^+i 15. In the case mentioned in Ex. 13 find the area of XYZW, '^AB = a. Fig. 446. Ans. 2 0^(3 -2V2). DESIGNS CONTAINING SQUARES WITHIN SQUARES 43 16. UAB = a, find the value of AF so that the area of FGYX may be ^'j that of the square ABCD. . AB AB Ans. or 6 2 Suggestion. — Solve for n in the equation -^ (2 « — 3) = -^. Then a 17. Construct a figure shovfing the meaning of each answer ob- tained in Ex. 16. 18. If AB = a, find AF so that the area of FGYX is - that of ABCD. Suggestion. — Solve for n the equation ^^ a(k ± y/k^ - 12 k) 3/fc q'(2 n - 3) ^ a^ 19. Show that the area of FG YX cannot be more than j\ that of of ABCD. Suggestion. — The answer to Ex. 18 is imaginary il k <. 12. 20. Show that the sum of the two values of AF obtained in Ex. 18 is I of AB. 21. Draw a figure to illustrate the two values of ^ F obtained in Ex. 18. Suggestion. — For the first value let F be any arbitrary point on AB so that AF<^AB. The other value of AF may be found from Ex. 20. O A K H F F' G' G Fig. 45. 55. Some of the exercises given in the previous paragraph suggest in- teresting relations in isosceles triangles. 44 PABQUET FLOOR DESIGNS EXERCISES 1. If ABO is any isosceles triangle, and if AF = BG and FX and G Y are parallel to OB and 0/1 , respectively, then A' K is parallel to A B. 2. If ^fi = a and AF=h, find the area of FGYX when ^05 is 2 ah -3 7(2 a right triangle. Ans. 3. If ABO is a right triangle, find points F' and G' so that if F'X' and G'P are drawn parallel to OB and 0^, respectively, the area of F'G'Y'X' shall be equal to that of FGYX. Suggestion. — li AF' = k, the area of F'G'X'Y' 2ak-9k^ Hence, 2aA-3A2 2ak~3k^ 4 4 4. Study the case k^ln. D _G C , and h + k= - a. o Fig. 46a. — Parquet Flooring. 56. In Fig. 46 ABCD is a square with diameters and diagonals drawn as shown. OX = OY = OZ = OW. Lines XY, YZ, etc. intersect the diameters at points M, Q, etc. From points M, Q, etc., lines MK, ML, QR, etc. are drawn parallel to the diagonals. EXERCISES 1. Prove that (1) MK=ML = QR; (2) KE = ME = RF; (3) points K, M, Q, P are collinear; (4) WXYZ is a square. 2. If ^5 = 6 and KE = 1, find the area of the square XYZW, of the trapezoid ABYX, and of the parallelogram AKMX. DESIGNS CONTAINING SQUARES WITHIN SQUARES 45 3. If AB =a a.nd KE= -, find the area of the square XYZW, of n the trapezoid ABYX, and of the parallelogram AKMX. Ans.{l) g(«-2y; (2) J,(«-l); (3) £^("-2). 4. 11 AB = a and iiT^ = ?, find n so that KM = l/X n Ans. 2 +2V2. Suggestion.— li KM = KA, then :^ = J-, from which = i___. ^•6 2 + 2V2' The same result is obtained by solving ^— ^ = ~ (n - 2), for n. n 2n 5. Construct a figure to illustrate the case KM = MX. Suggestion. — Diyide AE in the ratio 1 : V2. See § 54, Ex. 14. 6. li AB = a and KE = -, find n so that iiTlf = ZF. n Ans. 2 + V2. 7. Construct a figure to illustrate the case KM = AT. 8. If the area of ABCD is twice that of A'l'ZITand if AB = 6, find KE. ^ns. .88. Suggestion. ~ Solve the equation [6 - 2(KE)]^ = 18 for KE. 9. If AB = a, find A'E' so that the area of ABCD shall be k times that of XYZW. A ns. "(^^7 ^). 2v'yt 10. If AB = 6, find .fiTi; so that the area of AKMX shall be ^ that of ABCD. Ans. .4 or 2.5 nearly. Suggestion. — Solve for KE in the equation (.3 — KE)KE = ||. 11. Construct a figure to show that both of the results obtained in Ex. 10 have a meaning. 12. If AB = a, find KE so that the area of AKMX shall be - k that of ABCD. Ans. 'il^i^^EEMS. 4/fc 46 PARQUET FLOOR DESIGNS 13. Show that the area of the parallelogram AKMX cannot be more than ^ that of ABCD. Suggestion. — If AKMX were more than ^ ABCD, the result obtained in Ex. 12 would be imaginary. 14. Show that the sum of the two values of KE obtained in Ex. 12 15. Given an isosceles right A AOE with right angle at E. Show how to con- struct KM and K'M' parallel to A 0, and XM and X'M' parallel to AE so that the parallelograms AKMX a,nd AK'M'X' shall be equal in area. Suggestion. — From Ex. 12 and Ex. 14 it follows that KE = AK in Fig. 466. A K' K E Fig, 466. D G C IT v^ t F A E Fig, 47, — Parquet Floorings, 57. In Fig. 47 ABCD is a square with diagonals AC and BD. A£ == BF ■■ CG = DH. £Y and GW are parallel to AC and FZ and HX parallel to BD. EXERCISES 1. Prove that, (1) AEYX is a parallelogram. (2) AEYX^BFZY^CGWZ&DHXW. (3) XYZWiaa. square. 2. If AB = 10 and ^£; = 2, find the areas of EBY, AEYX, and XYZfV. Ans.lQ,8,4:. DESIGNS CONTAINING SQUARES WITHIN SQUARES 47 3. It AB=a and AE = -,find the areas mentioned in Ex. 2 and n the perpendicular distance from Y to AB. Ans. (1) £^n -. ly; (2) £,(n - 1); (3) g; (4) ^^(n - 1). 4. JlAB = a, find ^£ so that the area of XYZ W shaW be — . n Ans. — i:. Vn 5. If 4B = a, find CG? so that Z)G = 2 TFZ. Ans. f. o 12' j4ns. 1.26 or 4.73 nearly. Suggestion. — Solve for n the equation — -; (« — !)= — . Then^S=-. n 7. Construct a figure to illustrate the meaning of the two results obtained in Ex. 6. 8. liAB = a, find ^.B so that the area oi AEYX shall be ^. k Ans. «(^tVF^8¥) 2k 9. Show that the parallelogram AEYX cannot be more than | the square ABCD. 10. Show that the sum of the two values oi AE obtained in Ex. 6 is .45. 11. In connection with Ex. 5 construct the unit upon which the middle figure is based. THIRD POSITION OF THE SECONDARY SQUARE 58. Designs containing secondary squares whose sides are not parallel to either the diagonals or the diameters of the given squares give still another variety. The Arabic unit shown in Fig. 48 gives rise to a great variety of patterns ancient and modern. ^ 1 Day, pp. 45-47. 48 PARQUET FLOOR DESIGNS D G c \ \ >- •^ \ iV ^^ -A H .,-\ ^ \ V ^ \ \ F A E B Fig, 48. — Arabic Design, Fig. 48a. — Parquet Flooring. 59. The design shown in the square ABCD in Fig. 48 may be constructed by either method indicated below. EXERCISES 1. AE = BF = CG = DH and the points are joined as indicated. Prove that (1) AFCH and EBGD ai-e congruent parallelograms; (2) XyZ IF is a square. Suggestion for (2). — Prove that (1) Z 2 + Z 5 = 1 rt. Z ; (2) ZWXY= 1 rt. Z; and (3) XY=XW. 2. As analternative construction let .^1 = Z2 = Z3 = .^4. Prove that XVZW is a square. 3. If ylB = a and^£ = -,showthat ^Z= Jty. 2 Suggestion. — XE is parallel to YB in AABY. 4. If ^^ = a and ^B = -, prove that YF = -YB = -AX =-AF. 2^ 2 2 5 Suggestion. — A ABF is similar to A FB Y. 5. If AB = a and AE = -, showthat AX = ^XY and YF^^AY. 3 2 9 6. If ^B = a and AE , show that AX- ^andi='F = iiZ. 7. If AB= a and AE = -, show that (a) AF = '^V6. 2 (c) AX = XY = BY. (A) AX : 5 .5 DESIGNS CONTAINING SQUARES WITHIN SQUARES 49 8. If ^B = aand^E=-, show that 3' (a) AF = ^ VlO. (J) area of triangle ABF = -. o 6 (c) AX = — VW. (d) EX = ^ = ^. ^ ^ iO ^ ' 3 10 (e) By = j-AF. (/) zr = |^F = ?vio. 9. Ji AB = a and ^E = -, show that (g) Area, of XYZW = ^a^. 9. Ji AB = a and ^ (a) ^F = -Vn2 + 1. n (b) Area of triangle ABF = — . «(«■' +1) Suggestion. — Since AAEX is similar to A ABF, EX : AX : : 1 : n. r /") xf^ = ""(" - ^)'' 10. If ^S = 8, find A. Y and ^F so that the area of A'F^riF shall be32. Ans. h=2(\/6+ V2),c = 2(\/6-V2). Suggestion. — Use the equations (b — c)^ = 32, and i^ + c^ = 61, where h=A Fand c = BY. 11. If .4 B = a, find J[ F and 5 F so that the area oiXYZW shall be a 12. In the right triangle DWC, DC is the hypothenuse and ABCD is the square on the hypothenuse. Draw the figure on paste- board, cut it into pieces as indicated, and show that these can be rearranged so as to form the two squares on the sides DW and CW. Thus illustrate the Pythagorean theorem.^ 1 This proof is of Arabic origin. See Cajori, p. 123. 50 PARQUET FLOOR DESIGNS D K G C Fig. 49. — Tiled Flooring. EXERCISES 60. 1. Draw a diagram showing the construction for the tiled floor design in Fig. 49. Suggestion. — In square HOGD, in Fig. 49, points K,L,M, and iV are the middle points of the sides. The construction in adjacent squares is similar and similarly placed. 2. Prove that the points H, N, and C are oollinear. 3. If, as an alternative construction, points H and C are joined, prove that HC bisects GO. 4. Find the ratio of a side of square XRST to a side of the square VE 5 ■ 5. If four units identical with HOGD are placed as shown in Fig. 49, prove that XWZY is a square. 6. Prove that a side of square XWZY is twice as long as a side of square XRST. 7. If the square XWZY is 4J inches on a side, find the length of one side of square HOGD. Ans. 4| nearly. 8. If a pavement contains 50 square yards, how many square yards are there of tiles like RSTXI Ans. 10. HOGD. Ans. DESIGNS CONTAINING SQUARES WITHIN SQUARES 51 61. The unit on which the de- ■^ sign in Fig. 90 is based is shown in the square ABCD. See Fig. 48. Notice the reversing of the position of the unit in adjacent squares. Fig. 50. — Arabic Lattice. Day, pp. 45 and 46. EXERCISES 1. Prove that ZRSW is a square. 2. n ^5 = a and .4^ = ^, show that XY = — . n n 3. Find the ratio oi XY to ZW under the conditions of Ex. 2. 1 Ans. 2n-l Suggestion. — Solve for n in the equation (2n-l)2 5 Ans. n = J± jV5. 5. Find the value of u in Ex. 2 ii XY^ = —-. Ans. n = — ;- 6. Show that CY, RZ, and AB are concurrent. 7. Find the area of the triangle CYB if AB = a. 62 PARQUET FLOOR DESIGNS Fig. 51. Fig. 5la. — Pompeian Mosaic. Zahn (2), Vol. II, PI. 96 ; Day, pp. 47, 48 62. In Fig. 51 ABCD is a square with its diagonals and diameters. A£=BF = CG = DH. GK and EL are perpendicular to MN and lines from H and F perpendicular to the second diameter. ■ n^ _ 2 » + 2 EXERCISES 1. Prove that EFOII is a square. 2. li AE = ^ AB, prove that the areas of the squares ABCD and EFGH are in the ratio 9 to 5. 3. li AB = a and BE = 2, prove that -^^: 4. If DG = ^ DC, prove that HK = KE. 5. Ji DG =—, prove that HK = ^^^. 6. li AB = a and EB = -.findthe area,oiAEBF. Ans. -3-(n-l). n 2n^ 7. In Fig. 51a, if AB =a and AX = -, find the area of the n Ans. (n — 1). diamond J? W'ZF. Suggestion. — Figure 51a is based on Fig. 51. The construction is shown by the dotted lines. Notice the rpversing of the unit in ad- jacent squares. DESIGNS CONTAINING SQUARES WITBIN SQUARES 53 8. Show that the area of the figure RXSWTZUY is equal to the area of the square ABCD. 9. Construct a design using the square with the inscribed swastika (see square EFGH, Fig. 51) as the unit. Reverse the position of the ■unit in adjacent squares. Remark. — The design suggested in Ex. 9 is from the Alhambra.' n M Q Fig. 52 Arabic Design Unit. Bourgoin (2), Vol. 7, III, PI. 14. Fig. 52a. 63. In Fig. 52 ABCD is a square. AE = BF = CG = DH and AP = BQ = CR = DS. Points E and R, F and S, etc. are joined. EXERCISES 1. Prove that PG and ER are parallel. Suggestion. — ADPG^A EBR and Z 1 = Z 3 = Z .j. 2. Prove that : (a) EXQ is a right angle. Suggestion. — A EXQ is similar to A ERB. (b) WX = XY. Suggestion. — A EXQ s A FYR ^ A P WH and A ERB ^ A HA Q (c) Figures EBFY and AEXH are congruent. \d) XYZW is a square 1 Calvert, (1), p. xxxviii. For the history of the swastika see Thomas Wilson, U. S. National Museum Report, 1894, p. 763. 64 PARQUET FLOOR DESIGNS 3. As an alternative construction, make AE = BF = CG = DH and Al = /.2 = ZZ = Z^. Prove that (1) AP = QB= CR = DS and (2) XYZW is a square. 4. Find Zl so that MN drawn through points X and .Z^ is a diameter of square ABCD. See Fig 52a. Ans. 45°. 5. Show how the design shown in three of the squares in Fig. 52a may be derived from the one in the lower right-hand corner. PART 4. DESIGNS CONTAINING STARS, ROSETTES, AND CROSSES WITHIN SQUARES DESIGN CONTAINING THE RHOMBUS INSCRIBED IN THE SQUARE G Fig. 53a, — Dutch Tile Design, 64. In Fig. 53, ABCD is a square. E and G are the middle points of AB and CD, respectively, and the points are joined as shown. EXERCISES 1. Prove that EFGH is a rhombus. 2. If AB = 6, find the length of EC. 3. Prove that A EBF and BCF are equal in area. Find this area and the area of the rhombus EFGH M AB = Q. DESIGNS CONTAINING STABS, ROSETTES, CBOSSES 55 4. Find the areas mentioned in Ex. 3, if AB = a. Ans. — , — . 8 4 5. Find the perpendicular distance between A G and CE, iiAB = a. , Ans. - Vs. 5 6. Construct the design shown in Fig. 53a. Notice the reversing of the position of the unit in adjacent squares. 7. Prove that in Fig. 53a points /, K, and L, also M, N, and O, are coUinear. 8. In Fig. 53a show that the area of the rhombus FL is f of the area of KLGH. 9. In Fig. 53a prove that ALGH is a right angle. DESIGN CONTAINIKG FOUR-POINTED STARS D N a c Fig. 54a. — Parquet Flooring. 65. In Fig. 54 ABCD is a square, with its diameters and diagonals. Points K, L, M, and N are the middle points of HA, EB, FC, and GD, respectively. KX, LY, MZ, and NW are drawn from the points K, L, M, and N perpendic- ular to the diameters, and the points joined as indicated. EXERCISES 1. Prove that LP OX and KSOW are congruent parallelograms and that A PYO is equal in area to one half the parallelogram LP OX. 2. If AB = a, find the length of LO and KQ, and the area of XLYMZ etc. Ans. — Vo, 4 ?!VI3, ^'. 4 8 56 D PARQUET FLOOR DESIGNS G C /g^ ^w M M X u f s ^1 f Fig, 55a. — Parquet Flooring, Arabic and l\4edjeval, Bourgoin (S), Vol 7, III. PI. 6 ; Wyatt, PI. 12. 66. In Fig. 55 ABCD is a square with diameters and diagonals. EX = FY — GZ = HW, and the points are joined as indicated. EXERCISES 1. Prove that (1) XYZW is a square, (2) ABX and BCY are congruent isosceles triangles, (3) AX is parallel to CZ. 2. If EX = OX, prove that ii AXO and A EX are equal in area. Suggestion. — Triangles having equal bases and equal altitudes are equal in area. 3. If EX = OX, prove that the area of the star AXBY etc. is J that of the square A BCD. , Tj. A T> J T^-i- a 4.V, i OX n — 2 j OX n — 2 4. 11 AB = a a,na EX = -, prove that = and = — - — n^ OE n EX 2 5. If AB = a and EX = -, what is the relation of the area of n A AXO to the area of A AEO, and the area of the star AXBYC etc. to the area of the square ABCD1 Ans. n -2 6. What is the relation of EX to OE if the area of the star is \ of the area of the square ? DESIGNS CONTAINING STAUS, ROSETTES, CB08SES 67 7. What is the relation of EX to OE if the area of the star is - > k of the area of the square ^5C'Z)? Ans. ~ k 8. liAB = a and EX = XO, find the area of the {XAXW and of the square XYZW. Ans. — and — . 32 8 9. \iAB = a and if EX = ^, find lengths of OX, XW, and AK. Ans. (1) "^" - ^) (-2) £^(„ _ 2) ; (3) °^'^<" + ^) . ^'^ 2» ^^2n^ J ^ \ J 4„ 10. Find the areas of t^^AXW, square XYZW, and of the star if Ans. (1) ^,(n^-4); (2) 5^,(«-2)^ (3) °'^" ^) ■ 11. If ^jB = a, what must be the length of EX in order that the area of XYZW shall be jJj of the area of the whole square? 12±2V6 Sugge^ioit. — Solve the equation — ^^ ^^i- = — . 12. Have both results obtained in the answer to Ex. 11 a mean- ing in the figure ? Suggestion. — The values of EX and EZ are obtained. 13. What must be the value of EX if the square XYZW is — of the whole square ? Ans. — i =f^ . 2{k ± \/2k) 14. If AB = a, find EX so that the sum of the areas of the M. AXW, XBY, YCZ, and ZDW shall be ^^ of the area of the whole square. Ans. EX=—- — • 58 PARQUET FLOOR DESIGNS 15. Construct a figure to illustrate the case mentioned in Ex. 14. "v^ 1 /~~ Suggestion. — Since = , £X is the fourth proportional to v5, a EX a, and 1. If a right triangle be constructed with the legs in the ratio 1 : 2, the shorter leg and hypothenuse are in the ratio 1 : V^. 16. Ji AB = a, find EX so that the sum of the areas of the A AXW, XBY, etc., shall be - of the area of the square. k 17. What must be the length of EX if the area of the star is } that of ABCDl Construct an illustrative figure. 18. What must be the length of EX if the area of the star is - of that of ABCDf A ns. °(^~^^ . k 2k 19. If AX, BX, BY, etc. bisect the A OAE, QBE, OEF, etc., prove that OE, AX, and BX are concurrent. Suggestion. — The bisectors of the anglesof a triangle are concurrent. 20. If AX, BX, BY, etc. bisect the A OAE, OBE, etc., prove that OX=OY^OZ =0W. 21. If ^A', BX, BY, etc. bisect A OAE, etc., find the lengths of EX, OX, and .1 A', when AB = a. Ans-Q.) ^(V2-l); (2)5(2-^); (3) |V4-2V2. Suggestion. — The bisector of an angle of a triangle divides the opposite sides into parts proportidnal to the other two sides. 22. Also find the area of the star AXBYC etc., and of A AXB. Ans. (1) a2 (2 - V2) ; (2) ^' ( ^2 - 1). 23. Construct a diagram for a pattern like Fig. 55a in which the octagons formed shall be regular. Suggestion. — Show that in Fig. 55 if AX, BX, BY, etc. bisect the A OAE, OBE, etc., the octagons shown in Fig. 55o will be regulai'. DESIGNS CONTAINING STABS, ROSETTES, CROSSES 59 Fig. 56a. — Parquet Flooring. 67. In Fig. 56 ABCD is a square with diameters and diagonals. OS = OL = OM = ON and the points are joined as shown. EXERCISES 1. Prove that (1) EK = KF = FL, etc. ; (2) Z MHN = Z NEK = /. KFL, etc. ; (3) ONEK and OKFL are congruent. 2. If AB = 12 and OK = 3, find the area of the star EKFLG etc. OB 3. If OK = -^^i compare the areas of A EKO and jBBO, and n prove that the area of the star is - of that of the square A BCD. n 4. li AB= a and OK = ~, find the area of ONEK. Ans. ~. n in ^ „ ., . KN OK 5. Prove that j^ = ^. 6. If j45 = 6 and KP = 1, find the area of the star. Ans. 6 (3 - V2). 7. If ^B = a and KP = — , find the area oiAEFK and of the star. Ans.a) ^; (2) ^ (« - 1). on ^ n 8. Compare A EKF and i^^F and show that the formula for the area of A EFK accords with the theorem : " The areas of triangles having equal bases are to each other as their altitudes." 60 PARQUET FLOOR DESIGN'S 1 9. If the area of the star is - of the whole square, find PK. Ans.PK=OP('-'K k Suggestion. — Use equation a^ ( " ~ ) = — \ 2 n / k- 10. If the sum of the areas of the A EFK, FGL, etc. is of the whole square, find PK. Ans. KP 2 PC k ^ ' k 11. Construct a diagram illustrating the conditions given in Ex. 10 if i = 6 and show that the area of the star is — . 3 68. Figure 57 is based upon a special case of Fig. 56. The unit of which it is composed is shown in square ABCD, where £K, KF, FL, etc. bisect the ^iOEP, OFP, OFQ, etc., re- spectively. Fig. 57. — Arabic Design from Cairo, Bourgoin (2), Vol. 7, III, PI. I and 68. EXERCISES 1. Prove that EK, KF, and OB are concurrent. Suggestion. — The bisectors of the angles of a triangle are con- current. 2. Prove that OK = OL = OM = ON. 3. If AB-a, find the length of OK, the area of EKFLG etc, and of EKFB. Ans. (1) |(^ - 1); (2) f (2 - v^); (3) | V2, DMSIGNS CONTAINING STABS, ROSETTES, CROSSES 61 4. Construct the pattern shown in the figure. 5. Prove that (1) the points KFK' are collinear ; (2) KFK' and V W are parallel ; (3) E WUTF etc. is a regular octagon. 6. Compare the area of EWUTF etc. with the area of EKFLG etc. Ans. Area = (1 + v^) times the area of star. 6y. In Fig. 58 AE, KB, BP, etc. bisect the ^OAE, EBO, OBF, etc., and EM, MF, FR, etc. bisect the ^OEF, OFE, OFG, etc. These bi- sectors are extended until they intersect. EXERCISES 1. Prove that (a) KL ^NP = PQ, etc.; (i) EL = NF= FQ, etc. ; (c) BL = BN = CQ, etc. ; (d) LM = MN = QR, etc. ; (e) BK is the perpendicular bisector of EM; (/) EK = EX ; (^g) DM = EX and EM = XB. 2. Prove that the figures ELKJ and NFQP are congruent ; also figures BNML and CSRQ. Prove that A ELK and EBL are similar. 3. H AB = a, show that (1) EK=-(^y2 - 1)= OM. {Slt?'!»'Vj'5l^-^i' ^S : .. ■'*j"^^*Vv'^^4*T%^'^'-i^'92^'ij**'-jVi- "S^v : ?'it ■■•Si".i.fet:4>*''''»"™"™ *^#-:- ■:*>?$>- ■'$'- 1 $^ E a- if :?''i•^■^ ^^^ ^;p 1 ■■■■ •M^?w-i' St Fig, 66a. — Mosaic Flooring. After the Roman. Barre, Vol, V, 6"" Series, PI. 3. 82. In Fig. 66 ABCD is a square. EFGHKL etc. is a regular octagon in- scribed in the square. The points E and H, F and M, etc., are joined. These lines intersect at P, Q, etc. Lines XY, YZ, etc. are drawn through points P, Q, etc., parallel to the sides of the square. EXERCISES 1. Show how to construct the regular octagon EFGH etc., and give proof. Suggestion. — See § 29. 2. Prove that PQRS is a square. 74 MISCELLANEOUS INDUSTRIES 3. Prove that A' YZW is a, square with vertices on the diagonals of the given square. Suggestion. — Prove that A is the perpendicular bisector of SP ■ and tljat A'.S = XP. Therefore X lies on AO. 4. Prove that .4 EPA', PFBY, BGQYare congruent rhombuses. 5. If AB = a, find the length of AE and EF. Ans. (1)^(2 -V2); (2)a(V2-l). 6. If AB = a, find the length of A' Y and area of WX YZ. Am. (1) a(2 - V2); (2) 0^(6 - 4v^). 7. UAB = a, find the area of A EFP and of the rhombus A EPX. Ans. (1) |'(3-2V2); (2) ^(3^2- 4). 83. In Fig. 67 ABCD is a square with diagonals AC and BD. Fig, 67a. — Mosaic Flooring. EXERCISES 1. Show how Fig. 67 is obtained from Fig. 66. 2. If ^5 = a, find the area of POQZR WX. Ans. -(9-6\/2). 3. li AB = a, find the area of the square BGVF. Ans. ^\3-2v^). 4. Prove that GQOFand RZCK ?ire congruent rhombuses. DESIGNS BASED ON OCTAGONS WITHIN SQUARES 75 Fig. 68. 84. In Fig. 68 ABCD is a square, AR = VB = BU, etc., and the points are joined as indicated. Fig. 68a. — Linoleum Design. EXERCISES 1. Prove that TREN and VUGF are congruent squares. 2. If EFG/^X etc. isaregularootagon, prove tliat(l) OE:EF:FP : : 1 : V2 : 1, and (2) AR: RV : VB : -.l-.^ +yJ^ : 1. 3. Construct Fig. 68 in a given square so that EFGHK etc. is a regular octagon. ' 4. If AB = a, find the lengths of AR, RT, and RV, when EFGH etc. is a regular octagon. Ans. (1) j|(-t-v^); (2) ^(2V5- 1); (3) ^(3+ V2). 5. If AB = a and if EFGH etc. is a regular octagon, find the fol- lowing areas : (a) IS ART; (ft) TREN; (c) RVFE ; (d) EFGHK etc. Ans. (a) ^«i (9 - 4 V2) ; (6) g (9 - iV2) ; (c)g(5V2 + l); (d) ?£(!iV2 + l). Note. — For a different discussion of the designs given in Fig. 68a see § 30. 76 MISCELLANEOUS INDUSTRIES G C -^\Y\ /Z /^-^ D 7\^\ r u E Fig. 72. EXERCISES 90. 1. Construct Fig. 72 from the square ABCD. Suggestion. — • Draw the diameters and diagonals and produce the diameters beyond the sides of the square. Draw BH bisecting Z. ABD and extend to meet OH. Draw CH bisecting Z ACD and extend to meet OH. Continue thus about the square. 80 MISCELLANEOUS INDUSTRIES Fig. 72a. — Cut-glass Design. 2. Prove that BH, FH, and CH are concurrent. 3. Prove that OB = OH, and that AB = GH. 4. Prove that Z GHE is divided into four equal parts by the lines BH, FH, and CH. 5. Prove that lines HB, GD, and AC are concurrent. What other lines are concurrent for the same reason ? Suggestion. — The bisectors of the angles of a triangle are con- current. 6. Prove that BH = EB = GD, etc. 7. Prove that KLMNP etc. is a regular octagon. 8. Prove that (a) KX = KY, (h) DX = DY, (c) DYKX is one quarter of a regular octagon. 9. 'ilAB = a, find the lengths of OL and DK. Ans.(l) ^(2-V^); (2) a(^-l). Suggestion. — Use the theorem : " The bisector of an angle of a triangle divides the opposite side into parts proportional to the other two sides." 10. Prove that Al = HI and that 1-2-3-4-5 etc. is a regular octagon. 11. Prove that (1) DC bisects LE, &nd (2) EL = DY = a(V2 -1). DESIGNS CONTAINING EIGHT-POINTED STABS 81 Fig. 73. — Arabic Lattice Design from Cairo 91. Figure 73 is formed by a repetition of the unit shown in Fig. 72. EXERCISES 1. Prove that (o) EU and BX are parallel; (i) EU, CU, and AB are concurrent; (c) TU = VU = RS = RQ; (rf) PNMLH etc. is a regular octagon. 2. Prove that the area of octagon O is J the area of octagon A and that the side of octagon O is to the side of octagon j4 as 1 . VS. Suggestion. — Consider ratio of OQ to PA. OQ = ^(2-V2), P^ = a(\/2 - 1), when DA = a. PART 2. DESIGNS CONTAINING EIGHT-POINTED STARS 92. Occurrence. — Star polygons in general are dis- cussed in Part 5 of this chapter. The widespread use of the two eight-pointed stars, both in modern industrial design and in historic ornament, entitles them to separate discussion. The use of piaiiy-pointed stars is much more 82 MISCELLANEOUS INDUSTRIES restricted. Besides the illustrations given in this section, the two eight-poiuted stars are very common in medieval mosaics, as in the Duomo de Monreale,i the church of San Lorenzo,^ churches at Ravenna,^ and elsewhere.* THE THREE-EIGHTHS STAR 93. The I star is formed by joining every third vertex of the octagon. When inscribed in a given square, it is related to the octagon obtained by cutting off the corners of that square. (See § 29.) Both the regular and irreg- ular forms are common. Two irregular forms of frequent occurrence are discussed in § 94. A Linoleum Design. A E U F B Fig, 74a. 94. In Fig. 74 ABCD is a square. AE = BF = CH, etc. Points If and G, F and K, etc., are joined; also points E and H, F and M, etc. EXERCISES 1. Prove that (a) MW and ZL are equal and parallel; (J) PL and KS are equal and parallel ; (c) LZ, KZ, and VO are concurrent ; (d) PL, PM, and OD are concurrent; (e) POZL is a parallelogram. 1 Gravina, PI. 85. 2 Wyatt, PI. 10. » Gayet, Vol. Ill, Ravenna, X and XII. * Gruner, PI. 26. DEHIGNS CONTAINING EIGHT-POINTED STAMS 83 2. li AE = , prove that points Y, S, and Z are collinear. Draw o a diagram to illustrate this case. Suggestion. — Points L, Z, aud Y are collinear by construction of the figure. It is necessai-y to prove that points T and S would = - then, = -, and if DC 3 ' OC 3' coincide. To prove this, notice that if VK 1 ,, OS I = - , then — = -. VC 3 OC 3 3. liAE= ^, prove that the points H, S, Z, P, and M, are collinear. Draw a diagram to illustrate this case. Suggestion. — See Fig. 746. 4. Show that Fig. 74a illustrates , ,, AE 1 ,AE-1 both cases = - and = -. AB 4 :ii /\ Uiili /\ i: / ^ ■/A / \ \ V |\ W A ^ \ / 1 .^/X.^ s ~~)>' \ \ / \ A \ X \ / T ^/ '"A \ \ / \ ^x / /^ \^ \ -^^^ Fig. 76a. F 86 MISCELLANEOUS INDUSTRIES Fig. 766. — Mosaic Floor Design. 6. li AB = a and AE = ^ — ~ — ^, find the area of the square AETN, of the rhombus EXZT, and of the entire cross in Fig. 76a. ^«s.(l)g(ll-6V2); (2)g(llV2-12); (3) ||'(1] - 6V2). THE TWO-EIGHTHS STAR 97. The I star is formed by joining every second ver- tex of the octagon and is composed of two superposed squares. When inscribed in a given square, it is related to the octagon discussed in § 88. Figures 77 and 83 show it as it most commonly occurs. Fitted together with cer- tain peculiar crosses, as in Fig. 77a, it forms a very common all-over pattern, which was executed in tiles in Saracenic decoration.! Figure 78 is an Arabic form. The irregu- lar form suggested in § 99, Ex. 4, is from a Roman mo- saic.^ Figure 83 occurs also in both the Arabic and the Roman, and is extremely common in modern ornament. 1 Furnival PI. 81, Cut. 2 Price. DESIGNS CONTAINING EIGHT-POINTED STABS 87 - Parquet Flooring. Fig. 77a. - 98. In Fig. 77 circle O is tangent to the sides of the square ABCD at points E, F, G, and H, and intersects the diagonals at points K, L, M, and N Points E, F, G, and H, also K, L, M, and N are joined. Then the other points are joined as indicated. EXERCISES 1. Prove that (a) EFGH and KLMN are congruent squares; (b) XYZW is a square; (c) DA, KN, and XW are parallel; (rf) IIR = RN = NS; (e) ROhhsBotsANOH; (/) P/JSrO" etc. is a regular octagon. 2. If AB = a, find area of circle that could be inscribed in octagon PRSTU etc. Ans. ^. 3. If AB = a, find the length of HR and RS. Ans. (I) ^(V2-l); (2) |(2- V2). 4. If AB = a, find the area of figure OSNR and of the entire star KPHRN etc. Ans. (1) ^(2 - V2) ; (2) a\^ - V2). 8 5. Compare results in Ex. 4 with those given for Fig. 71a. (See § 88, Ex. 5.) Notice that the area of star KPHRN etc. (Fig. 77) is twice the area of star WKXL Y etc. in Fig. 71. 88 MISCELLANEOUS INDUSTRIES 6. Find area of figure HRNSGD. Ans.^(V2- 1). 7. From results of Exs. 4 and 6 compare the areas of the stars and crosses in Fig. 77a. Ans. area cross _ l area star VS 8. If GH = a, find lengths of HR, RS, and AB. Ans. (1) 3(2-^/2); (2) a(V2-l); (3) aV2- 9. If GH — a, find the area of ORNS and of the entire star KPHRNSG etc. ^„^ ^^^ |^2 - V2) ; (2) 2a\2- V2). 10. In Fig. 77 join P and T, also {7 and F, and extend to meet AD and AB. Extend NK and LK to meet ^D and AB. Prove that (a) ^ii:=i(rP = ^i7'. (i) The figure AH'KE' is i of a star similar to the star KPHRNS etc. 11. If AB - a, find the area of the figure AH'KE'. Ans. ?mO_^l2^. 99. Figure 78 exhibits the star shown in Fig. 77 inscribed in a given square with all of its vertices on the sides of the square. A K -~-E—-'L B Fig. 78.— From Celling of the Infanta's Tower in the Alhambra. Calvert (I), p. 341. EXERCISKS 1. Shovr hovr to construct Fig. 78. Suggestion. — The vertices of the star are determined by using the vertices of the square as centers and the half diagonal as radius. DESIGNS CONTAINING EIGHT-POINTED STARS 89 2. If AB = a, find the length of (1) KM and (2) the radius of the circumscribed circle. Ans. (1) a V2 - Vo ; (O) ?V4-2V2. 3. If AB = a, find the area of the star. Ans. 2 a^ (2 - Vi)". 4. Construct a figure in which points K, L, M, N, etc. shall be the middle points of AE, EB, BF, EC, etc. 5. Show that a circle can be circumscribed about the star con- structed as in Ex. 4, and find its radius, ii AB = a. Ans. -VK D G G S.-' \T ^ V''^/V'' M R (' ^N,/'^N / \ XT / \ '- \ ''"^ / H ■' >(3 >C& >< F P' V / K \ /W L \ J 1 B 1 J Fig. 79. Fig. 79a. — Parquet Flooring. 100. In Fig. 79 ABCD is a square. E, F, G, and H the middle points of the sides. OK = OL = OM = ON. The squares EFGH and ELMN are drawn, and the points are joined as indicated. EXERCISES 1. Prove that HPQR and GSXT are equal squares. 2. Prove that ^A" = PIT. 3. Prove that i^^=^, and f^ = f^. AG AD AG AB 4. If AB = a and ^A'=— , find the area of (1) A PQR, (2) KJ-QKNS etc., (3) square KLMN. ^»s-(l) ^,; (2) -(n-2); (3) "^^{n-iy. in-* n n' 90 MISCELLANEOUS INDUSTRIES 5. If the area of KPQRNS etc. is J that of the square ABCD, find AK. Suggestion. — Use equation — (» — 2) = — - A II' ^O Ans. AA = — — 6. If the area of the cross is - that of the square, find n and AK. Ans. n = -• 7. If OG = ON and the squares EFGH and KLMN are equal, as in Fig. 77, find the area of APQJ? and of the cross KPQRNS etc., if ^^""- ^»s.(l)f(3-2V2); (2)a'(V-2-l). Suggestion. — Fvom § 98, Ex. 3, Hii = ?(V2 -1) and RS = RP = «(2-V2). Fig. 80a. — Steel Ceiling Design. 101. In Fig. 80 ABCD is a square. E, F, G, and H are the middle points of the sides. OK = OL = OM = ON. The squares EFGH and KLMN are drawn, and the points are joined as indicated. Semicircles are constructed on the lines PR, ST, etc. as diameters. 1. K AB = a and AK -■ EXERCISES AO n , find the area of the semicircle PQR. Ans. 2n2 23M8IGN8 CONTAINING EIGHT-POINTED STARS 91 AO ' — ^ 2. If AB = a and AK = , find the area bounded by PK, PQR, RN, NS, SXT, etc. Ans.^J(n-l)^-2 7r-]. AO 3. If AB = a and AK = '^^^, find the area bounded by PQR and n line-segments RN, ND, DA, AK, and KP. in' 4. If AB = a and OG = ON, find the area bounded by PQR and the line-segments RN, NS, etc. Ans. —[4 — ir(6 — 4V2)]. 8 Suggestion. — From § 98, Ex. S, RP = SR = -(2 - V2). 5. If .45 = a and OG = 07V, find the area bounded by AK, KP, ,PQR, RN, ND, AD. Ans. §^ [4 + ir(6 - 4V5)]. D G Fig. 81a. — Parquet Flooring. 102. In Fig. 81 ABCD is a square. E, F, G, and H are the middle points of "the sides. OK = OU = OV = OW. The squares EFGH and KUVW are drawn. KL = UP = UR = VT, etc., and the points are joined as indicated. EXERCISES 1. Prove that (a) KM = NU = US, etc.; (b) EN = FS, etc.; (c) AEMK and EBUN are congruent trapezoids; (d) L, 0, and T are coUinear. 2. If OK = OE, prove that EN = NU. 92 3. If AB=a, AK MISCELLANEOUS INDUSTRIES AO -, and KL = LM, find the areas of (1) KUVW; (2) AEMK; (3) LMENPO. 4. Find these same areas if AK -. AO AB = a, and KL = m ■ LM. Ans. (1) °_;(„_1)- (2) ^,(n-l); (3) ^, ( "'" + "' 7 ■""+ - )■ n^ in' in'\ m + 1 / AO 5. If AB = a and AK= — , what is the length of KL, if 6 LMENPO occupies \ of the square. Ans. KL = f LM. Suggestion. — Use the equation - — ( — I = — and substitute n = 6. 4:n^\ m + 1 / 8 AO 6. If LM = and AK = —— , what part of the square is occupied hy LMENF01 Ans. if. AO 7. If KL = LM and AK = — , what part of the square is occupied n by this figure? Ans. n^-n + 2 DESIGNS COMBINING THE THREE-EIGHTHS AND THE TWO- EIGHTHS STARS Fig. 82a. — Parquet Flooring 103. InFig. 82 ABCDisasquare. E, F, G, and H are the middle points of the sides. OK=OL = OM = ON. The squares EFGH and KLMN are drawn. Points Q and V, P and R, etc., also K and L, V and R, K and N, etc., are joined. DESIGNS CONTAINING MIGBT-POINTEB STABS 93 EXERCISES 1. Prove that (a) HQWP, KSTQ, and XYZW are squares; (A) QTWO is a parallelogram. 2. 'iiAB = a, and C'ii: = -^-^ , find the areas of the squares KSTQ n andHQH^P. in^ 8n2" 3. Find the area of the star HQKSE etc. if AB = a, and 0"K = 0"A Ans. iL^(„2 + i-). 4. Also find the area of the star PWQTSX etc. Ans. -An-1). 5. Also find the area of the trapezoid AESK. Ans. -^^ («2_1). 6. If OiV = OG and 4B = a, find the area of star PWQTSX etc. ^ns. 02(3^2 -4). K B-v / \^ Q -7 ^-7 ■IM /\ N ^ \F JV A E I" B M Fig. 83. r>*<- <% J>*< >'»x ^^ ^st Fig. S3a. — Oilcloth Design. 104. In Fig. 83 ABCD is a square. A£ - BF = BG = CH, etc. Points are joined as indicated. 94 MISCELLANEOUS INDUSTRIES EXERCISES 1. Prove that MFXE, NPQR, and A'i'ZTF are squares. 2. If AB = a and AE = -, find the area of 3 (1) EMFX; (2) XYZW; (3) AEMFBG etc.; (4) EFGHK etc.; (5) AEXFBG etc.; (6) NEMFPG etc. 18' ^»«- (1) T^ ! (2) -s- ; (3) -s- ; (4) -TT ! (5) -^ ! (6) 9 9 9 3 t ^^ r^ fc /^^ ^_^r ^ s ^-y\H ' > t^o© ^ ■(/B ^ 3. Draw a figure to illustrate the case AE = ^ 3 and prove that W. N, and X are collinear. (See § 94, Ex. 2.) 4. If ^B = a and ^£ = ? 4 find the lengtb of EM. Ans ? ^^ * Fig. 836. — Tiled Flooring. Scale J in. = I ft. 5. If ^5 = a and AE =^, find the area of (1) AEMFBG etc.; (2) EFGHK etc.; (3) ^^XF^Gretc.; (4) NEMFPG etc. Ans.(l) ^; (2) If; (3) ^; (4) a=. 6. If AB = a and AE = -, find the above areas. n Ans.(l) ^\„^-2„ + 2); (2) ^> - 2) ; (3) ^(«-l); (4) ^(n-2). 7. If AF=AO = BE = BH, etc., prove that EFGHK etc. is a regular octagon and that the squares WXYZ and NPQR are equal. DESIGNS CONTAINING EIGHT-POINTED STARS 95 8. If AF=AO = BE = BH, etc. prove that AE = EM = MF, etc. 9. ltAB = a and AF = AG, find the area of FBGP. Ans. ^(3-2\/2). 10. If AB = a and AF =A0, find the four areas of Ex. 5, and also that of NEXFPGY &tc. Ans. (1) 2a2(2-V2); (2) 2a2(V2-l); (3) 2a^{^-l); (4) 2a2(V2 - 1); (5) a2(6V2 - 8). 11. If AB = a and jIO = AF, find the length of NS and the area of NSXTPU etc. ^ns. (1) |(3\/2 - 4) ; (2) 0^(20 - 14V2). 12. Verify the areas AEMFB etc., and NSXTPU etc. by applying the theorem : " The areas of two similar figures are to each other as the squares on any two homologous sides." Suggestion. - ^(2-v^)2 4 ^ _ a^(4 - 2V2) — C3V2-4y a^(2{)-li-\/2) 4 13. If AB = a and AF = AO, find the area of ENSX. Ans. ^(5-^/2-7). A X B Fig. 84. — From a Roman Mosaic Pavement. Zahn (2), Vol. II, PI. 79, Vol. Ill, PI. 6. 96 MISCELLANEOUS INDUSTRIES EXERCISES 105. 1. Give the construction for the figure within the square ABCD in Fig. 84. Suggestion. — First inscribe the circle in the square. 2. Prove that GPYZ is a rhombus. 3. If OP = r, find the length of HK. Ans. r (2 - V2). 4. Also find the areas of ZYPG and EFGHK etc. Ans. (1) ^(3v/2 - 4) ; (2) i:i\\^ - 1). PART 3. DESIGNS FORMED FROM ARCS OF CIRCLES 106. Occurrence. — The figures in this section are found in all lines of modern industrial design from all-over lace to iron grills and stamped steel ceilings. In the history of ornament they occur in ancient Assyrian ^ and Egyptian 2, in Chinese' and Japanese, in Roman and medieval, and in modern art. FIGURES FORMED OF SEMICIRCLES AND QUADRANTS A BE F Fig, 85 — Design for Iron Grills and Mosaics, EXERCISES 107. I. Show how Fig. 85 is constructed from a network of squares. 1 Ward (1), Vol. I, p. 126. 2 Ward (2), p. 34 ; Glazier, p. 4. 3 Clifford, pp. 27-35 ; Jones (2), PI. LIX. DESIGNS FORMED FROM ARCS OF CIRCLES 97 2. Prove that the curved figures CEDHC and CHKLC are con- gruent. Suggestion. — Draw the chords CE, DE, etc., and CH, IIK, etc. Prove that (1) the quadrilaterals thus formed are congruent; (2) corresponding arcs as CE and CH, ED and CH are equal; (3) con-e- sponding arcs lie on the same side of corresponding chords. 3. If ^B = 4, find the perimeter and area of the curved figure CEDHC. 4. If j4B = a, find the perimeter and area of the curved figure CEDHC. Ans. (1) 27ra; (2) 2a\ Fig. 86£l. — Roman Mosaic; also Byzantine. See Gayet, Vol. II, PI. 30. 108. Figure 86 is constructed on a network of squares. The semicircles are constructed on the sides of the squares as indicated. EXERCISES 1. Prove that two semicircles are tangent at the intersection of the diagonals of each square. 2. Prove that the curved figures DPKQGLD and HRNSOLH are congruent. 3. If HO = a, find the area of the following curved figures : (a) DPKQGLD; (b) LGMOL; (c) HRNSOPH Ans. (a) — ; (*) y ; (c) ^(8-t). 98 MISCELLANEOUS INDUSTRIES A B Fig. 87. — From a Roman Fig. 88, — Modern Tile. Arabic Design. Prisse d'Avennes, Vol. II, PI. CI. EXERCISES 109. 1. Give the construction for the design unit shown in Fig. 87. 2. If AB = a, find the area and the perimeter of the curved figure. Ans. 8 o^, 4 wa. EXERCISES 110. 1. Give the construction for Fig. 88. 2. If AB = a, find the area and the perimeter of the curved figure. Ans. Tra, — . 2 OVALS FORMED OF mTERSECTIITG QUADSAITTS 111. Occnrrence. — While the illustrations that accom- pany the following figures are all from modern sources, these forms abound in Kon\an, Byzantine, and Gothic work. Some instances are especially noteworthy. In the old Singing School in Worcester Cathedral ^ and in the Bap- tistry at Florence ^ are pavements containing a large num- ber of these designs. In the Basilica Antoniana at Padua is a stone parapet containing numerous different designs based on circles and squares. Figures 89 and 90, as well as trefoils, quadrifoils, six- and eight-pointed rosettes, are 1 Shaw (1), PI. Waring (1), PI. 23 and 25. DM8I6NS FORMED FROM ARCS OF CIRCLES 99 among them.^ These figures are found in churches and cathedrals in Ravenna, Parenzo,^ Venice, and many other places. Figure 90 is found in the details of a window in the Milan Cathedral.^ Figure 92 is a common Gothic design used in tracery windows and inlaid tiles.* See Fig. 217. 112. In Fig. 89 ABCD is a square. The semicircles are constructed within the square on the sides as diameters, as shown in the figure. Fig. 89. EXERCISES 1. Prove that the four leaves formed are congruent. 2. If AB = 12, find the area of each leaf and of the curved figure AHOKB. 3. Find these same areas if AB = a. ^«»-a)f (t-2); (2) |(4-,r). 4. If the area of each leaf is 25, find AB. Am. 13.2+. 5. If the area of each leaf is s, find J. B. Ans.-\i — —. 7r — ^ 6. Show that this figure has four axes of symmetry and also a center of symmetry. 1 Arte Italiana, 1896, PI. 47. 2 Gayet, Vol. Ill ; Ravenna, PI. 10 ; Vol. II, Parenzo, PI. 24. s See § 204, Ex. 4. « Shaw (1), PI. xxx. Hartung, Vol. II, PI. 96. 100 MISCELLANEOUS INDUSTRIES EXERCISES 113. 1. Give the construction for Fig. 90 and prove that the ovals are congruent. •?- 2. If AB = 12, find the area of the four-pointed figure in the center and of one of the ovals. H- 3. If AB = a, find these same areas. 4. If the area of the central figure is s, ns. 2^ /_£_ . c^ ~^~^^ 1*^ /^^^^ E Fig. 90. find AB Ans. '. 5. What per cent of the area of the square is the area of the cen- tral figure? Ans. 21.4% nearly. Fig. 90a. — Iron Grills. EXERCISES 1'14. 1. Draw a network of squares. Draw the diagonals of each square. Prove that these diagonals form two sets of parallel lines that intersect in squares. See Fig. 91. 2. Draw the diameters of the first set of squares. Compare the areas of squares ABCD, ECFD, and AG EH. DESIGNS FORMED FROM ARCS OF CIRCLES 101 F 3. If the circles in column III, Fig. 91, are circumscribed about the small squares, find the area of one of the ovals formed ii AB = a. 4. In column III, prove that the diagonals of the large squares are tangent to tvi^o circles at the vertices of the small squares. 5. Show how the ovals are formed in col- umn I. If AB = a, find the area of each oval. Ans.^(^-2). 6. Compare the area of one of the ovals in column III with the area of one in column I. Explain the relation found. 7. In column I, prove that the diameters of each of the large squares are tangent to two circles at the center of each side of the square. 8. If AB = a, find the area of the crescent XZrOX. Ans. 5-'. 8 9. Draw a diagram showing the construc- tion of Fig. 91a. Suggestion. — See column II, Fig. 91. Fig. 9lfl. — Steel Ceiling Design. 102 MISCELLANEOUS INDUSTBIES Fig. 92a. — Tiled Flooring. Gothic. Shaw, (I), PI. XXX. EXERCISES 115. 1. Construct Fig. 92, given the square ABCD. Suggestion. The arcs are drawn with the vertices A, B, C, and D as centers and intersect in the points X, Y, Z, and W. The inner figure is similarly drawn from points .X, Y, Z, and W. 2. Prove that (a) The points X, Y, Z, and W lie on the diameters of the square ABCD. (i) The lines WB and ZB trisect the angle ABC. (c) The chords A W, WZ, and ZC are equal. (rf) XB Y is an equilateral triangle. (e) XFZ IF is a square. (/) Line .4 C is the perpendicular bisector of WX and Z Y and bisects ZZCF. (gr) Lines BF and BZ trisect Z CBD. (K) The curved triangles BXY axiA. WAX are congruent. 3. In the square XYZW construct arcs similar to those con- structed in the square ABCD. Prove that ZLM is an equilateral triangle and that the points Z, L, B are collinear. 4. Construct in a given circle a figure like the one here shown in a square. DESIGNS FORMED FROM ARCS OF CIRCLES 103 5. If AB = a, find the length of OP, PB, XY, and BQ. ^ns. (1)^(V6-V2); (2)|(3^-V6); (3)^(v^-V2); Suggestion. — Since BXY is an equilateral triangle and XYZW is a square, OB is divided at point P in the ratio —^. va 6. If AB = a, find the area of the following figures : (a) The linear figure XYZW. Ans. a'^(2 - V3) (h) The linear A ZfiF. (c) The linear A £1^^. Ans. — (2V3 -3) 4 (rf) The segment bounded by the arc WZ and the chord WZ. Ans. ^(^-3) a"-h- V3 + 5:> •4 + ! (e) The curved figure JfF^' IT. ^ (/) The curved A BXY. Ans.-(2VS {g) The four-pointed curved figure AXBYCZ etc. Ans. a'^IVS +?^ - 3\ (h) The curved figure AXB. Ans. -{^i - V3 -^) 'y 5^ ^ X / ^ ^ \6 V\ (y / \ \\ / ,\ \ /_, 'XP \\ ' ■-.\ \ / __ — ^ "^-^^ — :::. i-v Fig. 93. Fig. 93u. —Steel Ce.ling Design. 104 MISCELLANEOUS INDUSTRIES EXERCISES 116. 1- Given the square A BCD, construct Fig. 93. 2. Using AT = ^(Vo - ^2) as found in §115, Ex. 5, find the area of the following figures if ^ B = a : (a) The segment bounded by XY and the chord XY. Ans. Y7, (2 IT - 3 V3)(2 - V3). (i) The curved figure XYZW. yl ns. ^ (2 - VS) (3 - 2 tt + 3 V3 ) . o (c) The curved figure BX Y. Ans. - (3 , tV3 - 8). 3. Construct in a given circle a figure like Fig. 93. CUSPED QUADRILATERALS 117. Occurrence. — The cusped quadrilateral shown in Figs. 94 and 95 is the basis of a large number of fan vaulting designs besides those here mentioned. See also Figs. 145 and 146. This type of vaulting is peculiar to the late English or Perpendicular Gothic architecture. The name refers to the manner in which the main ribs ra- diate from the tops of columns forming inverted cones.' 118. In Fig. 94 the arcs HPE, EQF, etc. are drawn with the vertices of the squares as H centers and the half side as radius. The five circles are tangent as shown. K 1 Fletcher, p. 288-290 ; Bond, chapter xxii ; Fergusson, Vol. II, p. 361. DESIGNS FORMES FROM ARCS OF CIRCLES 105 EXERCISES 1. Prove that the sectors HAE, EBF, etc., are congruent. 2. Show how to construct (1) circle X inscribed in the sector HAE, and (2) circle inscribed in the curved figure EFGH. 3. Prove that the diagonals A C and BD pass through the points of tangency P, Q, R, and -^ si/i^— h!^ EXERCISES 133. 1. Show that Fig. 108 is based ^ on Fig. 107. 2. Find the area of HKLMEN etc, if AB = a, (1) if XM = ^, (2) if ZM" = ^,(3) iiXM = — . ^n.9. (1)^(8 + ,r); (2)g(18 + ^); 16 d6 ^ £ B (3)-H!-(2ni' + ,r). i^'S^ '"=• 3. What must be the ratio -— =■ if the semicircle KLM is tangent XE 1 to the lines HA and ^£? ^ns. — r- V2 4. Draw a figure to illustrate this case and find the area of HKLMENPQFR etc. as so formed. V P — Suggention. — In this case XM = ^^ . Substitute V2 for n in the result obtained in Ex. 2, pai't 3. Ans. ~{i + ir)- 8 120 MISCELLANEOUS INDUSTRIES 5. Find the ratio 4^ if 'he area of HKLMENPQFR is | of the "''- ^ XM 1 area of the square. Ans. = — — ■ Suggestion. — Let^M = Solve the equation ° / 2 ~ "7~ for I Then XE n Fig. I09o. EXERCISES 134. 1. Show that Figs. 109 and 109a are based on Fig. 107. 2. Ji AB = a and XL = \ XE, find the area of (1) HAELMN and (2) HNMLEOH. ^ns.(l)^(8+,r); (2)|j(8-,r). 3. If AB = a and XL = J^ Z£, find the above areas. ^n.. (1)^(18 + -); (2) ^(18-.). XE 4. If AB = a and XL = , find the above areas. n ^-- (1) 4::. (2 "^ + -) ; (2) tIt. (^ «^ - -)• 16 n 5. Show that the area of the figure HNMLEPQ, etc. is \ the area of the square ABCD regardless of the length of the radius XL. DJESIGNS FORMED FROM LINE SEGMENTS 121 135. In Fig. 110 the circle is divided into eight equal parts and the points A, C, E, G, and B, D, F, H are Hf joined. Figure KSRQLW etc. is de- rived from BDFH as in Fig. 107. Fig. 110. B - Linoleum Design, EXERCISES 1. Prove that ACEO and BDFH are congruent squares, and that BX = XC = CY = YD, etc. 2. If OH = a, find the area of KSRQLW etc. if (1) PS = ^ PK and (2) PS = PK Ans.O)^a»- .);(2)^^(2»^-.). 3. If HB = a, find this area if (1) PS = ^ KP and (2) PS ■. KP Ans. (1) |(18 - ^); (2) ^^(2 n^ - ,r). 4. Compare the areas of ;& OPL and ONC by using the theorem : " The areas of two similar triangles are to each other as the squares on two corresponding sides." 5. If AC = a, find the area of trapezoid ACLK. Ans. — . Suggestion. — Show that the trapezoid is \ ^ AOC and hence \ of square AC EG. 6. If .4C = a and P5= :^, find the area of AJBXCLQRSK. Ans. -^—(14 «2 - 8 n^\/2 + ir). 16 n^ KP 7. liAO = r and PS = ^^, find the areas of (1) KSRQL WVU n etc. and (2) AJBXCLQRSK. Ans. (1) J1(2»2-^); (2) -!:!-(14„2-8n=V2 + ^). 2n^ on'' 122 MISCELLANEOUS INDUSTRIES DESIGNS RELATED TO THE PRECEDING Fig. Ilia, — Unoteum Design. EXERCISES 136. 1. In Fig. Ill ABCD is a square with diameters EG and HF. Show how to construct the arc MLK so that KE = \ AE and Z KXM is 90°. 2. If AB = a, find the areas of (1) segment MLK, and (2) the figure AKLMBP etc., if Z KXM is 90° and MKE = \ AE. Am. (1) -^(2 + 3^); (2) ^'(10 - ^). 128 3. If AB = a, find the areas as in Ex. 2, except that KE = . n Ans. (1) ^(2 + 3 ,r) ; (2) ^ (2 n^ - 2 - 3 ,r). 4. What must be the ratio — — if circles W and X are tangent to each other and if Z KMX is 90° ? ^ns. AE' Suggestion. — If KT and XU are both perpendicular to AO, AT=TK and Ot/ = UX. If A'Jf = XU, then ^T = TU = UO. Therefore 4i!:=AE. 5. Construct such a figure in which KE shall \)&\AE and Z KXM shaU be 60°. DESIGNS FOBMEJP FROM LINE SEGMENTS 123 H G S T F F E Q Jl V M A w -^K^ L A B Y a D Fig. 112. Fig. Il2a. — Wicker Design. 137. 1. If the distance between the parallels AH, GB, etc. in Fig. 112 is given, show how to construct the figure. 2. If the distance between any two parallels is 4 inches, find the area of the figure KLMNPQRST etc. Ans. 48 sq. in. 3. If the distance between any two parallels is a, find the area of the figure KLMNPQRST etc. Ans. 3 aK Fig. 113. Fig. Ii3a. — Counter Railing Design. 138. In Fig. 113 A ABC and BDC are equilateral. E, F, G, H, etc., are the middle points of the sides AB, BC, AC, BD, etc. The semicircles whose centers are E, F, G, H, etc. are drawn with equal radii. 124 MISCELLANEOUS INDUSTRIES EXERCISES 1. li KE = i AE and AB = 2, find the area of (1) AKLMBN etc. and (2) SPQEDS etc. Arts. (1) (VS+-\ (2) ( V3-^]. 2. If KE = \AE and AB = a, find the same areas. ,1ns. (1) g(6V3 + ^); (2) |J(6Va-^). 3. If KE = and AB = a, find the same areas. n Ans. (1) £i(2n^v^ + 3,r) ; (2) £^,(.2nW3 - 3x). must be the ratio of FN to -FB in order thai with centers F and H be tangent ? Ans. 4. What must be the ratio of FN to FB in order that semicircles FM^l FB 2" 5. If FN = ^ F£, what is the distance between semicircles with centers F and H if AB = a1 Ans. -. 6 6. If JviV = — , find the distance between semicircles with centers n F and H a AB = a. Ans. —(n-2). 2n 7. Construct a figure in which the three semicircles with centers F, H, and T are tangent to each other. If AB = a, find the area of the triangular space inclosed by the three semicircles. Ans. gg (2v3 — tt). 8. If AB = a and the semicircles with centers F and H are tangent, find the area of space inclosed by lines NB and BP and the two arcs PFand NX. Ans. jg (3 VB-t). 9. Upon PR = J BZ) as a chord construct a segment of a circle which shall be tangent to the altitude BT. See § 253 (1). If three such arcs are constructed in the A BCD, will they be tangent to each other? DESIGNS FORMED FROM LINE SEGMENTS 125 D C Fig. Il4a. — Counter Railing Design. 139. In Fig. 114 ABCD is a square with diameters EG and FH and diagonals AC and BD. KE = EM = FN = FQ, etc., and ^EAB = ZEBA = Z FBC = Z FCB = Z GCD, etc. EXERCISES 1. Prove that (1) AE, EB, and OX are concurrent; (2) AAEB ^AFBC, etc. Suggestion for (1), OX is the perpendicular bisector of AB; since AE = EB, E lies in line OX. 2. If Z EAB is 30°, KE is J AE, and AB = a, find the areas of (1) AKLMBA ; (2) AKLMBNPQC, etc. Ans. (1) g^ (27 V3 + 2 ,r) ; (2) g (81 - 27 V3 - 2 ^). 3. Find the areas mentioned in Ex. 2 under the same conditions, AE except that KE = n Ans. (1) -|^^(3n2V3 + 8,r); (2) ^(9„''-3n^VS-8,r). 4. If Zi?-.4B is 30°, construct the figure so that the sectors with centers E and F shall be tangent to each other. Find the radius of each sector in this case. Ans. -z (3v'2 — Vs). Suggestion. — EX = Vs ; therefore OE = ^ (3 - V3) and EF = 2(3V2 - V6). 126 MISCELLANEOUS INJiUSTRIES 5. If AE. EB, BF, etc., bisect A OAB, OB A, OBC, etc., find the length of EX, OE, and EA,iiAB = a. ■Am.(l) |(V2-1); (2) ?(2-V2); (3) ICV^TiV^)- Suggestion. — To get EX, use the theorem : " The bisector of an angle divides the opposite side into parts proportional to the other two sides." 6. If AE, EB, BF, etc., bisect A OAB, DBA, OBC, etc., find the area of (1) AKLMBA and (2) of AKLMBNP etc., ii AB = a, and KE = \AE. ^„^^i) ^(V2-1)(64 + 5W2); (,2 (2) 0^(128 - 64 V2 - lOir + 5 tt V2). 7. If AE, EB, BF, etc. bisect A OAB, OBA, OBC, etc., and AE KE = , find the areas mentioned in Ex. 6. ^«s. (1) -|^ (^/2 - 1)(4 »2 + 5 ,rV2) ; (2) -!^(8n2-4n2V2-107r + 5 7rV2). 4 n^ PART 5. DESIGNS BASED ON REGULAR POLYGONS DESIGNS BASED OR THE REGULAR HEXAGON 140. Occurrence. — All the figures in this group contain the star formed by joining every second vertex of the regular hexagon. It is extremely common in Arabic and medieval^ ornament and in modern designs. Over the bishop's throne in the cathedral at Anagni is a design executed in mosaic and based on this star,^ that is worthy of mention. It seems to be an ancient symbol of Deity,' and is said to be a Jewish talisman. It is in use also in such modern instances as the policeman's star and many common trademarks. 1 Wyatt, PI. IX, Bourgoin (1), Plates. « Arch. Record, XII, p. 217. 3 Waring (2), p. 66. DESIGNS BASED ON REGULAR POLYGONS 127 E D \^ y^~^ V-^ ^^^ A ^*--«.^ ^^' B Fig. 115. Fig. Il5a. — Border, Parquet Flooring. 141. In Fig. 115 the circumference of the circle is divided into 6 equal parts and alternate division points joined. EXERCISES 1. Prove that two equilateral triangles are formed. 2. Prove that points F, 0, Z, and A, O, D are collinear. Suggestion. — Draw radius OH to the center of the arc FA. Prove that OH is the perpendicular bisector of FA. Prove FV = AV. Therefore V lies on OH. 3. Prove that ^AXV, BXY, CYZ, etc. are equilateral and con- gruent. 4. Prove that XYZW etc. is a regular hexagon. 5. Prove that BF is the perpendicular bisector of A 0. 6. If ^0 = ?•, find the length of £F. ^ns. rVJ. 7. If AB = r, find the area of AAVX. Ans. —^ V3. 8. li A0 = r, find the area of (1) the hexagon XYZW etc. ; (2) AAEC; (3) star AXBYC, etc. Ans. (1) |V3; (2) ^VS; (3) rV3. 128 MISCELLANEOUS INDUSTRIES E.-' — £'"'--- i ? \ \ Ft KX, XZ jo Fig. 116. Fig, Il6a. — Tiled Flooring. Scale J in. I ft. Medieval, Wyatt, PI. 3. 142. In Fig. 116 ABCD etc., is a regular hexagon, G, H, K, etc., are the middle points of the sides, and the points are joined as shown in the figure. EXERCISES 1. Prove that GKM and NHL are congruent equilateral triangles with sides parallel to the sides of ABCD etc. 2. Show that star GYHZK etc. is similar to the star shown in Fig. 115. 3. Prove that GBHY and HCKZ are congruent rhombuses. Find the size of the angles in these rhombuses. 4. liAO = r, find the area of (1) rhombus BHYG; (2) A XGY\ (3) hexagon XYZW etc. ; (4) star GYHZK etc. Ans. (1) |V3; (2) j^ V3; (3) ^Z-Vs; (4) ^jf V3. DESIGNS BASED ON REGULAR POLYGONS 129 143. In Fig. 117 ABCDEF is a Tegular hexagon witli each side F\ divided into three equal parts and the points joined as shown. "F ''' ^\ / X*""^- \ Jp V' V/ / 1 \ V--' X-~,.J__l__7„^''Ji Fig. 117. — Register Design. Arabic. Prisse d'Avennc, Vol. 2, PI. 103. EXERCISES 1. Prove that AD, JL, and BC are parallel. Suggestion. — -Extend AB and DC until they intersect. Use the theoi'em : " A line dividing two sides of a triangle proportionally is parallel to the third side." 2. Prove that points P, O, and S are collinear; also points A, X, 0, W, and D\ also points N, P, and G. 3. Prove that BHY.T, CLZG, etc., are congruent rhombuses. 4. Prove that JIP, XPY, GHQ, YQZ, etc., are congruent equi- lateral triangles. 5. Prove that IBGQYP, CKRZQH, XYZWUV, etc., are congru- ent regular hexagons. 6. Prove that star XPYQZR etc., is similar to the star shown in Fig. 115. 7. li AG = r, find the area of (1) each small hexagon; (2) each small equilateral A; (3) the rhombns £7Fi?; (4) the starXPFQZ etc. Ans. (1) r'VS; (2) g ^'3 ; (3) ^^VS; (4) L\/3. 130 MISCELLANEOUS INDUSTRIES rig. 118. - Arabic Design Unit. Bourgoin {2), Vol. 7, II, Pis. 42 and 55. Fig. l|8a. — Linoleum Design. 144. In Fig. 118 the drcumference is divided into 12 equal parts and the points joined as indicated. EXERCISES 1. Prove that (a) ABX, CDY, etc., are congruent equilateral triangles. (b) BCXY, EDYZ, etc., are congruent squares. (c) A'F^PV^f/F is a regular hexagon. 2. Prove that a cii'cle can be passed through the four points ^4, X, Y, and D. 3. Can a circle be passed through A, Y, W, LI Find other sets of four points all lying on the same circle. 4. If OM = r, find (1) the length of AB; (2) the area of A ABX; (.3) the area of hexagon XYZW etc. Ans. (1) rV2 - V3 ; (2) ^(2V3-3); (3) ^(2V3-3). OF and OF: 3rf , 2 VM. Suggestion. — Note that OP ■— j . 5. li ABzza, find (1) length of MO; (2) length of RX; (3) area of the 6-pointed star in the hexagon XYZW etc. Ans. (1) aV2 + V3 ; (2) |V3; (3) a^y/3. DESIGNS BASED ON REGULAR POLYGONS 131 6. Show that Fig. 118a is derived from Fig. 1 18. Draw to scale [J in. = 1 ft.] a diagram for Fig. 118a. Let ^IB represent 2 inches. 7. Show that Fig. 118a can be constructed from three sets of parallel lines. 8. If AB = a, find the distance between the parallels MD and AB in Fig. 118. Ans. ^(Vs + l). STAR POLYGONS 1 145. Construction of Star Polygons. — If a circumference is divided into n equal parts and each division point is joined to the kth one from it, where k is an integer greater than one and less than -, a star polygon is formed. In the discussion which follows these polygons are named from the size of the arcs cut o£E by the chords. Thus, if the circumference is divided into 8 equal parts and every third point joined, the arc cut off is f of the circumference and the star formed is called the | star. If a 4-pointed star is derived from the |- star by omitting part of it, as star AXCZE etc.. Fig. 121, it is called the 4-pointed | star. 146. Occurrence. — The simpler stars are common every- where in modern ornament and are frequently used in advertising. The occurrence of the two 8-pointed stars is discussed elsewhere. The use of many-pointed stars is largely confined to cut glass. Almost every piece contains one or more. The f star is the one used in the United States flag. 147. History. — Star polygons occur in all periods of historic ornament. Their use dates back to the ancient 1 S. Gunther, pp. 1-92. 132 MISCELLANEOUS INDUSTRIES Greeks when the f star was a symbol of recognition among the Pythagoreans.^ Near Winchester, England, are the remains of an old Roman pavement, the design of which includes the ^ star.^ In the church of San Giovanni e Paulo at Rome is a Byzantine mosaic which contains the eight-pointed ^g star.^ Among the mosaics at St. Mark's, Venice, is one that contains the ^ star within the five-pointed -^ star.* On the ceiling of one of the halls in the Alhambra is the | star within the I star.* Stars of various kinds are found in medieval tiled pavements in western Europe ^ and in medieval embroideries.'^ They are extremely common in Saracenic art.* They occur in the ornamentation of primitive peo- ples,' especially in symbols of the sun.'" Fig. 119, 1 Ball, p. 21. 2 Morgan, p. 221. » Wyatt, PI. 5. * Ongania, Folio PI. 30. ' Calvert, (1), p. 124. « Shaw, (1), PI. XXI. ' Shaw, (2). 8 Prisse d'Avennes, Pis. XC, XCI, XCII. 9 Beauchamp, Pis. 1-^. w Waring, (2), pp. 62 and 63. DESIGNS BASED ON REGULAR POLYGONS 133 EXERCISES 148. 1. Construct the five-pointed star shown in Fig. 119. 2. What axes of symmetry has this figure ? 3. Prove that (a) AC= CE = EB, etc. (b) AX =XB = BY, etc. (c) AAUC ^ABVD ^ A CXE, etc. (d) ^BAX = \/.EAB. 4. Prove that points F, 0, C are collinear. Suggestion. — Prove that each point lies on the perpendicular bisec- tor of ^E. 5. Prove that 4 C is divided into mean and extreme ratio at Y and that ^ F is divided into mean and extreme ratio at X. 6. Prove that Z YXB = 2^ YBX. 7. Prove that XYZUV \s a regular pentagon. 10 4"-^, ,'6 Fig. 120. EXERCISES 149. 1. Construct the ^ star and the 5-pointed ^ star. Suggestion. — See Fig. 120. 2. How many 10-pointed stars can be constructed by dividing the circumference .into 10 equal parts? - Every second, third, or fourth point may be joined. 134 MISCELLANEOUS INDUSTRIES 3. Study the composition of each of the 10-pointed stars obtained in Ex. 2. Suggestion. — From Fig. 120 it is seen that the ID-pointed ^ star can be divided into two 5-pointed i% stars. Fig. I2la. — Parquet Flooring EXERCISES 150. 1. Construct the f star in the circle. Suggestion. — See Fig. 121. 2. What axes of symmetry has the figure ? 3. Prove that (a) AN=NB = BP= PC, etc. (6) AB=VA = AX = XC =WB = BY, etc. 4. Prove that (n) ZMAN = Z NBP, etc. (J) /.HMA =ZANB, etc. (c) ZGVA = ZHWB = ZAXC, etc. (d) ZMAN+ZAXC=2 It. A. (e) ZMAN=^ ZHAB. Suggestion. — Turn the figure through an angle of 45.° DESIGNS BASED ON REGULAR POLYGONS 135 5. Find the number of degrees in each angle mentioned in Ex. 4. 6. Prove that points L, 0, and Q, and also H, V, 0, Z, and D, are collinear. 7. Prove that AN = NY, and that GB is perpendicular to AD. 8. Prove that KLMNP etc., and UVWXY etc. are regular octagons. 9. Prove that the chords GD, HC, AF, and BE intersect in a square. 10. Prove that the figures ANWM, BPXN, etc. are congruent. Suggestion. — Turn the figure through an angle of 45°. 11. Prove that the triangles WNX, XPY, YQZ, etc. are congruent. 12. UAO = a, find the lengths of HR and HA. ^___ Ans.HA = a V2 - V2. 13. U AO = a, find the lengths of AM and HC. Arts. (1) °V4-2v'2; (2) a(V2 + l)(V2- V2. 14. li AO = a, find the area of square LQ and of A HMA. Am. (1) n2(2-V2); (2) ^(2- V2). ^■■■p ■■■iD^nr^ ■ ■ z I_ -- ■ - « ■■■-■ ■> Fig, 1216. — Vaulting from Peterboiough Cathedral. 136 MISCELLANEOUS INDUSTBIES 15. Construct a diagram for Fig. 1216. Are all of the lines straight ? 16. Construct the 4-pointed star AXCZE etc. in Fig. 121. 17. If the circumference is divided into 8 equal parts, how many 8-pointed stars are possible? 16 JL- — TV ---15 ■-'% A >^ ,-' / \ // ''V \ \ /\ // > / w 'O' ^. / '' \ ~\ \ Y y b< / \ 3/. / 1 1 \ ^ :/1 1 N^ j-=C ~^^_^ .'K V} -0 iV^v J^^:=^i2 \ V '''.-—-- 7- ^-x / sV / -yii \ . / ^/^ ' \' / '^s. \ / \ /x i' /' \ / \ > ^s \ / i^ 1 •^/ 6v I / \ / \\ >10 \ / \ ' r -» ^-7^- 8 Fig. I2lc. 18. Construct the f star and study its composition. 19. Construct the t star and show that it is composed of two 20. If the circumference is divided into 16 equal parts, how many 16-pointed stars can be constructed? 21. Construct the ^^j and j^j stars. Study the composition of each. DESIGNS BASED ON BEGULAB POLYGONS 137 22. Construct the ^j star. Construct the i-pointed ^j star and the -pointed /j star. 23. Construct the 4-pointed and the 8-pointed ^^ stars. Suggestion. — See Fig. 121c. EXERCISES 151. 1. Construct the regular ^ star. Suggestion. — See Fig. 122. 2. Give the axes of symmetry of this figure. 3. Prove that (a) AE=EB = BF = FC, etc. (A) AQ= QC = BR = ED, etc. (c) AW= WD = BX = XA', etc. 138 MISCELLANEOUS INDUSTBIES t 4. Prove that (a) ZA=ZB = ^C, etc. (J) ZAEB =Z BFC = ZCGD, etc. (c) Z ^ QC = /i^iJi) = Z CSA', etc. (rf) Z ,4 WD = Z £:C^ ' = Z C F£', etc. 5. Find the number of degrees in each angle mentioned in Ex.4. 6. Prove that the points E, V, O, V, and E' are collinear ; also the points B, Q, 0, Q', and D'. 7. Prove that the figures BFQ/i,C(?7iF, etc. are congruent. Prove EQF, FRG, etc. congruent isosceles triangles. 8. Prove that EQVP, FRWQ, etc. are congruent. Prove that PV= FQ=QI'F, etc. 9. Prove that OW = WB = AB, and that AE =^ EB = EW. 10. Prove that (1) EFGHK etc., (2) PQRST etc., (3) UVWXY etc. are regular duodecagons. ' Suggestion. — Turn the figure through an angle of 30°. 11. Prove that PRT etc. is a regular hexagon. How many other regular hexagons can be found in the figure ? 12. Prove that EHE'H' is a square. How many other squares can be found in the figure? 13. Construct the 4-pointed star A WDZC etc. How many such stars are there in the figure ? Compare it with the 4:-pointed | star in Fig. 121. Are the two stars congruent when constructed in equal circles? 14. Show that the 12-pointed star in Fig. 122 may be broken mp DESIGNS BASED ON SEGULAR POLYGONS 139 \\ \ \ \ \ \ \ \ \ \ 3 / 1 5i wS:... f 1 t~ Fig. 123. EXERCISES 152. 1. Construct the f\-star. Suggestion. — See Fig. 123. 2. Construct the 6-pointed star ANCQE etc. How many such stars are in the figure ? 3. Construct the 4-pointed star A UD etc. How many such stars are in the figure ? 4. Construct the 3-pointed star .4Zi?X' etc. How many such stars are in the figure ? 140 MISCELLANEOUS INDUSTRIES 5. If a circumference is divided into 12 equal parts, how many 12-pointed stars are possible ? 6. Explain the composition of the ^ and j^ stars. 7. Construct the /^ star. 8. If a circumference is divided into 24 equal parts, how many 24-pointed stars may be formed? 9. Explain the composition of the j'j, /j, j'j, and ^j stars. 10. Explain the composition of the l^ star. 11. Construct the j\ and JJ stars. By means of each construct a star of 8 points, one of 4 points, one of 12 points, and one of 6 points. 12. In the figure for the | star how many regular octagons occur? 13. In the figure for the -^ star. Fig. 120, how many regular decar gons may be formed by joining corresponding intersections? 14. In the figure for the f^ star, how many regular duodecagons occur? See Fig. 122. 15. In the figure for the -^ star, how many regular duodecagons occur? See Fig. 123. 16. In the figure for the ^ star, how many regular polygons of 16 sides occur? See Fig. 121c. DESIGNS BASED ON STAR-POLYGONS 153. Fig. 124 is the only one of the following designs which is from a modern source. The rest are Mohamme- dan or Arabic. While only extremely simple Moham- medan designs are here given, they are characteristic of that style, which is highly original, strikingly geometrical, and totally unlike anything else in history. The fact that the Koran forbade the use of all living forms in decoration and art, undoubtedly accounts for its peculiar DESIGNS BASED ON REGULAR POLYGONS 141 character. The most complicated interlacing geometrical forms abound in endless variety. It is a style of orna- ment especially adapted to surface decoration. It is executed in wood, stone, plaster, clay, or mosaic, and com- pletely covers pulpits, domes, vaults, and exterior and interior walls. Window lattices are of the same design. This style characterizes Mohammedan art to-day and has strongly influenced Spanish decoration. EXERCISES 154. 1. Construct the design shown in Fig. 124. Suggestion. — ABCD etc. is a regular hexagon with the circle in- scribed. EFGHK etc. is the regular duodecagon. EPGSK etc. is the 6-pointed QR = RS, etc. T^j star. BF = FM, CH = HQ, etc. MN = NP, 142 MISCELLANEOUS INDUSTBIES ^ ' w ^^\ \ ^^0^ 2. Prove that (a) FB=CH= DL, etc. (ft) MF= HQ = LT, etc. (c) PN= SR= VU, etc. (d) EM = iVG= GQ = QK, etc. (e) EN = NG = GR = -RA:, etc. 3. Prove that (a) Figures EPGN, GSKR, etc. are congruent. (6) Figures ENGM, GRKQ, etc. are congruent. (e) Figures £M(?F, GHKQ, etc. are p,g |24«.- Parquet Flooring. congi'uent. (rf) Figures EBFG, GHKC, etc. are congruent. 4. Prove that (a) Z iVCii = Z ieA'[7, etc. (ft) Z MGQ = /. QKT, etc. (c) Z £iVG = Z GRK, etc. 5. If £0 = r, find length of EP. - ^ns. ^ V2. 6. If EO = r, find the area of (1) A EPG, (2) star EPGSK etc. Am. (1) !^'; (2) ^"-(VS- 1). 7. If EO = r, find the area of (1) the hexagon ABCD etc., (2) the figure EPGB. Ans. (1) 2 r^VS; (2) ^(3 + V3). 8. If -BO = r, find (1) the length of FB and (2) the area of LEFB. Ans.O.) r(2V3-3); (2) ^(2V3-3). 9. If EO = r, find the length of (1) PF, (2) PM, (3) il/iV. r 12 Ans. (1) |(3 - V3) ; (2) ^(15 - 7V3) ; (3) ^-^(15 - 7V3). 10. If EO = r, find the area of figure ENGM. ^ns. g(15-7V3). DESIGNS BASED ON BEGULAB POLYGONS 143 Fig, 125. — Arabic All-over. Bourgoln (1), PI. I, EXERCISES 155. 1. Construct the pattern shown in Fig. 125. Suggestion. — Construct an all-over pattern of regular hexagons. In each hexagon construct the | star shown in Fig. 116. 2. Prove that the points A, M, and iVare collinear. Prove that RS = XY. 3. Show that Fig. 125 can be constructed from three sets of par- allel lines. 4. If RS = a, And the perpendicular distance between the paral- lels and the angle of intersection between the sets of parallels. ,4 ns. "y/3: 60°. 5. If RS = a, find the area of the hexagon ABCDEF. Ans.^V%. 8 6. If RS = a, find the area of the star CLKH, etc. ^ns.ir'VB. 7. Show that Fig. 125 may be constructed from an all-over of equilateral triangles each of whose sides is equal to ZN. 144 MISCELLANEOUS INDUSTRIES / \ / \ Y Fig. 126. —Arabic All-over. Bourgoin (I), PI. 12. EXERCISES 156. 1. Construct the pattern shown in Fig. 126. Suggestion. — Construct an all-over pattern of regular hexagons. In each hexagon inscribe a circle. In each circle construct the 6-pointed ^ star (see § 152, Ex. 2), so that its points shall be the midpoints of the sides of the hexagon. Why is this possible ? 2. Prove that points H, C, a.nd K are coUinear. 3. Prove that lines AB and HK are parallel. 4. Prove that figure CDEFGH is equilateral and that it has two sets of equal angles of three each. Find the size of each. 5. Prove that all figures of the type CDEFGH are congruent. 6. Prove that if GE is drawn A GEO is equilateral. 7. If the radius of one of the circles is r, find the length of GH. Ans. -y/2. 2 8. If the radius of one circle is /•, find the area of (1) the star ; (2) the figure CDEFGH. Am. (1) ?|^'(V3-1); (2) J(V3 + 3). DESIGNS BASED ON REGULAR POLYGONS 145 9. If a side ZF of one of the hexagons is a, find the length of GH. Ans. Ve. 4 10. If XY=a, find the area of (1) the star; (2) the figure CDEFGH. Ans. (1) ~{y/Z-\); (2) i^(-v/3 + 3). 8 16 E Fig. 127. —Arabic All-over. Bourgoin (I), PI. 42. EXERCISES 157. 1. Construct the pattern shown in Fig. 127. Suggestion. — Construct an all-over pattern of squares. In each square inscribe a circle. In each circle construct the f star, so that every second vertex will coincide with the middle points of the sides of the square. Construct the small star O by extending lines in stars I, II, III, and IV until they intersect. 2. Prove that points A, B, and S, also C, D, and E, are collinear. 3. Prove that EF, CH, and ZO are concurrent. 4. Prove that (1) EG = GH = HK, etc. (2) Z GHK = Z KLM, = etc. (3) Star FGHKL etc., is the regular 4-pointed f star. 146 MISCELLANEOUS INDUSTRIES 5. Prove that (1) LR = RQ = QP = PN; (2) ML = MN. G. Find the number of degrees In each angle of figure LMNPQR. 7. Prove that the figures LMNPQR and BIHK etc. are con- gruent. G EXERCISES 158. 1. Show how to construct Fig. 128. Suggestion, — Construct the 6-pointed /j star AKCME etc. in a given circle. Construct the star KYMZP etc. by joining the proper points. DESIGNS BASED ON BEGULAB POLYGONS 147 2. Prove that (a) AB = AK = KC = CM, etc. (c) Z AKC = Z CME, etc. (J) Z KCM = Z il/£:P, etc. 3. Find the number of degrees in each angle mentioned in Ex. 2. 4. Prove that the points B, K, O, R, and B' are collinear, also the points A, H, 0, Q, and G. 5. Prove that HKLMN etc. is a regular duodecagon. 6. Prove that KCMY and MEPZ are congruent squares. 7. Prove that the star whose points are KLMN etc. is a y\ star. 8. Prove that the points 0, X, A, and 0, Y, C, etc., are collinear. 9. Prove that the circle through points A, K, and C is equal to the giveh circle. 10. If OA = r, find the length of (1) AB; (2) OX; (3) KB. Ans. (1) rV2 - V3; (2) r(2 - V3) ; (3) r-(2 - V3). 11. If 04 =r, find the area of (1) the square CMYK; (2) the rhombus ABCK; (3) the star AKCME etc.; (4) the star XKYMZ «'^<=- 4ns. (1) r2(2 - V3); (2) |?(2 - V3); (3) 3r\V3~ 1); (4)3r2(3>/3-5). Fig, i28a. — Arabic All-over. Bourgoin, (I) P). 13. 12. Construct the pattern shown in Fig. 128a. Suggestion. — Construct an all-over pattern of regular hexagons. About each circumscribe a circle. 148 MISCELLANEOUS INDUSTRIES Fig. 129. —Common Unit in Arabic Design. Jones (3), Vol. II, PI. XLV. EXERCISES 159. 1. Show how to construct Fig. 129. Suggestion. — Construct the | star in a given circle. Let K be any arbitrary point on the radius OA, where A is one vertex of the | star. Draw the circle whose center is O and whose radius is OK. In this circle construct the f star with its vertices on the radii OA, OE, OB, etc. Extend the lines as indicated. 2. Prove that (a) XK = KY = ZM = MW, etc.; (6) HX = YE= AZ = WB, etc. ; (c) YR = RZ = WS, etc. 3. Prove that points R, L, 0, L', R' are collinear. 4. Find the number of degrees in each angle of figure RZMLKY. 5. Prove that figures I, II, and III are congruent. DESIGNS BASED ON REGULAR POLYGONS 149 6. Prove that YY', RR', and ZZ' are parallel. 7. Point K may be determined by bisecting Z ORH. Let the bisector intersect OA at K. Prove that in this case YK = YR. 8. Is it possible to make YR = KL ? Suggestion. — Is it possible to make Z KLR = Z LRK1 9. Construct figures similar to Fig. 129, by dividing the circum- ference into sixteen equal parts, joining every second point, and con- structing any sixteen-pointed star desired in the inner circle. 10. Construct a figure like Fig. 129 but of 12 parts. Suggestion.. — See Fig. 129a. Fig. IS9a.j— A«bkC Dscign Unit. P:i3Fe d'Avemes, PI. 50.. CHAPTER IV GOTHIC TRACERY: FORMS IN CIRCLES 160. Origin and Development.^ — The origin of Gothic tracery is evident from a study of English buildings. During the latter part of the twelfth and during the thirteenth century, lancet windows which formerly were used singly were often grouped under a common drip stone. When two lights were so grouped, the stone be- tween them and the arch above was sometimes pierced by a round window. This was the beginning of plate tracery. True bar tracery came later. In it the open- ings are separated by extensions of the mullions with no solid masses of stone between the tracery bars. Tracery is usually associated with Gothic windows, but during the middle and later Gothic, the interior and ex- terior of buildings are,. prof u,se],y^ decorated with it. It is cut in stone .jiad.par.^ e,d. in. wjxid,, It adorns altars, screens, seats, pulpits, and all the interior furnishings of churches and .cathedrals.,. The .design?, at first very simple, became, iat-^v -m.ost i:ompl;cated and elaborate. 1 Bond, Chap. X'XxlV E"ergus8on, 'Vol; H-J p. 163 ; Sturgis (1), p. 268 ; Sharpe (1). 150 FORMS CONTAINING CIRCLES IN TRIANGLES 151 PART 1. FORMS CONTAINING CIRCLES INSCRIBED IN TRIANGLES 161. With one exception, all of the illustrations used in this section are from decorated rafters. History and Occurrence. — Timber roofs have been in use whenever and wherever lumber was abundant enough to be used for building. In the early Christian and in the Romanesque buildings the use of the circular arch left rectangular spaces between the arch and the rafters or ceilings above, which often received ornament more or less elaborate. In the Middle Ages highly ornamented open-timber roofs abounded in En gland. ^ The hammer-beam roof of the fifteenth century, especially, afforded opportunity for the use of tracer3% (See Fig. 132a.) The modern methods of artificial heating in winter render it impossible to use the main structural features of the roof for ornamental purpose as was done in the four- teenth and fifteenth centuries. The ornamented rafters seen in modern buildings, therefore, are not the rafters that support the roof. 162. In Fig. 130 AFEC is a rectangle cir- cumscribed about the semicircle AFB. A " O F Fig. 130. — Decorated Rafter. Church of San Minrato, Florence. Fletcher, PI. 93. Fletcher, p. 290; Brandon (2), Pis. 18, 19, 20. 152 GOTHIC TRACERT: FORMS IN CIRCLES EXERCISES 1. Find the relation between AC and AF. 2. If is the center of the semicircle, prove that Z A OC is one half a right angle. 3. Construct the square with CD as diagonal. 4. li AO = r, find the area of the square CD. Ans.'^(Z -2V2). 5. If ^C = a, find the area of the figure bounded by the arc ADB and the tangents A C and BC. Ans. - (4 - ir). E 163. In Fig. 131 ACB is a right angle with AC = CB. Circle ADB is tangent to AC and CB at A and B j/ respectively. Circle O' is tangent to AC, CB, and to the arc ADB. Fig. 131. — Decorated Rafter or Tracery Win- dow Design, EXERCISES 1. Show how to find the center of the circle ADB. Suggestion. — Erect perpendiculars to .4C and CB at A and B, respectively. 2. Prove that A OBC is a square. 3. Is the problem, " To construct a circle tangent to the sides of an angle at given points on these sides," generally possible? Why is it possible in this case ? 4. Show how to find the center of circle 0'. FORMS CONTAINING CIRCLES IN TRIANGLES 153 5. Show that the problem given in Ex. 4 is a special case of the problem, " To construct a circle tangent to two given intersecting lines and to a given circle." Why is this special case capable of a simple solution ? CO' V^ 6. Show that the point 0' divides CD so that = ^ O'D 1 7. If CA = a, find the length of CD, CO', and O'D. Am. (1) a ( V2 - 1) ; (2) a (3 V2 - 4) ; (3) a (3 - 2\%. 8. li AC = a, find the area of the small circle. Ans. -ra^ (17 -12V2). 9. li AC = a, find the area of the figure bounded by EHF and the line-segments CE and CF. Ans. -j (17 - 12V2)(4 - w). 10. If ^ C = a, find the area of the figure bounded by E WD, AD, and line-segment AE. Ans. ^ (12v'2 - 16 - IBir + 9 7rV2). Fig. 132a. — Hammet-beam Rafter. Bran- don (2), Pis. 18 and 19. EXERCISES 164. 1. In Fig. 132 Ji CA = CB and A ^ CB is a right triangle with right angle at C, construct the circles 0, 0' and 0". 2. If AB = a, find the length of OD. Am. | ( V2 - 1). 3. If AB = a, find the area of circle 0. Ans. -j- (3 - 2^/2). 4. If AB = a, find the area bounded by EGF and the tangents CE and CF. Ans. jg (4 - tt) (3 - 2 V2). 154 GOTHIC TRACERY: FORMS IN CIRCLES 5. If AB = a, find the area of the quadrilateral AEOD. Ans. -(Vl-l). 6. If AB = a, find the area bounded by EHD and the tangents EAsiuAAD. ^ns. —(8 V2 -8-97r + 6irV2). 7. If CA = a, find the area of circle 0. Ans. ^(3_2V2). 8. If AC = a, find the area of (1) the quadrilateral AEOD; (2) the figure bounded by EGF and the tangents CE and CF; (3) the figure bounded by EHD and the tangents EA and AD. Ans. (1) ?i (V2 - 1) ; (2) '^ (4 - ,r)(3 - 2Vl) ; (3) j^(8V2-8-97r + 67rV^). 9. Construct circles 1, 2, and 3 in any right triangle as shown in Fig. 132a. EXERCISES 165. 1. In Fig. 133 circles E and F are inscribed in the isosceles right triangles ADC and ABC. Show that arcs HG and MN can be drawn tangent to the circles as indicated. 2. Prove that EHC, FGC, etc., are straight lines. ROUNDED TREFOILS 155 3. If AB = a, find (1) the area of circle E ; (2) the length of AE ; (3) the length of AM. Am. (1) ^(3-2V2); (2) aV2-V2; (3) aV2- V2-?(2- V2). 4. Make a drawing for a tiled floor composed of tiles like that shown in Fig. 133. The tiles are to be so placed that the corner quadrants of four tiles meet to form a circle. The tiles are six inches on a side. Make the drawing on a scale 4 inches to 1 foot. PART 2. ROUNDED TREFOILS 166. History and Occurrence. — The names trefoil and quadrifoil are applied to various three-leaved and four- leaved forms, respectively. Those considered in the two following sections are the rounded forms and are con- structed from equal circles which are tangent or which intersect. They abound in Byzantine and Gothic archi- tecture, especially in the cathedrals of the thirteenth and fourteenth centuries^ where they form small windows in clerestories and gables, and occur as details in the tracery of large windows, in painted glass, in the wood carvings of interior furnishings, in sculptured stone, and in orna- mental iron. They are supposed to symbolize the Trinity and the four Gospels, respectively. ^ Trefoils have been used since the Middle Ages in nearly every variety of ornament. Examples may be found in almost every town. Illustrations of their occurrence in tracery windows may be found in Figs. 208a, 246a, and 256a. Figure 137a shows two different kinds of trefoils. Many more or less complicated designs have been based on the simpler ones here given. ^ 1 A. Pugln, Illustrations ; A. and A. W. Pugin, Illustrations. 2 Womum, p. 106. 8 Billings (3) ; Day, p. 36 ; Brandon (2), PI. 40. 156 GOTHIC TRACERY: FORMS IN CIRCLES THE TREFOIL FORMED OF TANGENT CIRCLES E C_ 167. Figure 134 shows a trefoil foiuned of the three circles X, Y, and Z tangent to each other at the points T, S, and R. It is inscribed in the circle as shown. EXERCISES 1. Show how to construct the figure. Solution. — Circumscribe an equilateral triangle about the circle. Connect each vertex with the center. Inscribe a circle in each of the triangles FOG, GOE, and EFO. 2. Prove that the small circles are tangent to the large circle and to each other. 3. Prove that the curved figures A TB and BSC are congruent. 4. 'iiOA= r, find EF. 5. If XF=6, find OA. Suggestion.-^ — Find XS, OX, then OA. 6. If J:r=2a, findO^. 7. If OA = r, find XT. Am. 1 rVS. Ans. (2V3 + 3). ^ns. |(2\/3 + 3). Ans. ?-(2\/3 - 3). 8. If Xr = 6, find the area of (1) the curved figure RST ; (2) the trefoil TBSCR etc. ; (3) the curved figure CSB. Ans. (1) 1.45+ ; (2) 86.27+; (3) 14.99+. 9. li XY ='2a, find the areas mentioned in Ex. 8. Ans. (1) t(2VZ-ir); (2) ^(5,7+2 V3); (3) ?i(8TV3-6V3-7r). 2 2 18 ROUNDED TREFOILS 157 10. If OA = r, find the areas mentioned in Ex. 8. Ans. (1) r%2lV3 + 6 ttVS - ^ - 36) ; (2) ^ (35 IT - 20 ttVS + 14 V3 - 24) ; (3) - (60 IT V3 - 42 V3 - 103 TT + 72). 168. Figure 135 is based on Fig. 134. Figure 13Sa shows the working drawingfoT i certain parts of Fig. 1356 and is based on Fig. 135. Fig. 1356. —Scottish Rights Church, St. Louis. EXERCISES 1. How many degrees in the arc TRA of Fig. 135 ? Ans. 210°. 2. Prove that the curved figures AEBTR and BFCST are con- gruent. 158 GOTHIC TRACMBT: FORMS IN CIRCLES 3. liAO = r, find the area of the figure AEBTR. Ans. — (23 TT - 12 tt Va - 42 V3 + 72). 4. Show how to construct Fig. 135a if the radius of the outer circle is given. Suggestion. — Points X, Y, and Z are the centers for the small circles. They are determined by the A HKL. 5. If KM= 2 a in Fig. 135a, find the lengths of (1) HK, (2) XR, (3) OA, and (4) AM. Ans. (1) aV3; (2) |VB; (3) |(2 + Vj) ; (-1) ^(2 - V3). EXERCISES 169. 1. Inscribe a trefoil in the equilateral triangle ABC. Suggestion. — See Fig. 136. Draw the medians AD, BF, and CE. Inscribe circles in the figures OEBD, ODCF, etc. 2. \iAB = a, find the length of GF and OT. ^ns.(l)|(V3-l); (2)^(-3_V3). VC 1 Suggestion. — Since BY =2 YG, we have — — = — -■ OB V3 SOUNDED TREFOILS 159 3. If AB = a, find the following areas : (a) The curved figure RST. Ans. ^(2 - V3)(2V3 - tt). (J) The trefoil TGHSK etc. ^ns. :^ (2 V3 + 5 w) (2 - >/3) . (c) The figure bounded by GH and the line-segments GB and 5//^. Ans. ^(2-V^)(3V^-7r). (rf) The figure bounded by NT and Tff and the line-segment NG. ^„s. g(2-V3)(4-,r). ROUNDED TREFOILS FORMED OF INTERSECTING CIRCLES C Fig. 137a. — From St. Mary's Church, Speenhamland, Newbury, England. 170- In Fig. 137 EFG is an equilateral triangle inscribed in a circle and X, Y, and Z, are the centers of the semicircles constructed as shown. EX£RCISES 1. If OA = a, find the lengths of OZ, ZH, and GH. Ans. GH = ~{^-V). 2. Prove that the curved figures GHC and KFC are congruent. 3. Find the ratios (1) ||; (2) ||; (3) ||. Ans.{\)^; (2) V3; (.3) i- 160 GOTHIC TRACEBT: FORMS IN CIRCLES 4. H AO = a, find the area of the figure bounded by the arcs GC and CH and the line-segment GH. Am. j^i5n-6Vl). 5. Ji.AO = a, find the area of the figure GHCKFLB, etc. Ans. 5^(2-n/3+t). o Fig. 138a. — From Monk's Church, St. Louis. 171. In Fig. 138 the radii OA, OE, OB, etc., divide the circle into six equal parts, and X, S, Y, T, etc., are the middle points of these radii. EXERCISES 1. Prove that XSYT etc. is a regular hexagon. 2. If X, Y, and N are the centers of RAS, SBT, and TCR, re- spectively, prove that these arcs meet at the points R, S, and T. 3. Prove that the curved figures ASB and BTC are congruent. 4. If 40 = a, find the area of the trefoil RASBT etc. Ans. ^(3\/3 + 47r). 8 5. Find the area of the curved figure AEBS. 6. What fraction of the circle is within the trefoil? BOUNDED qUADBIFOILS 161 Fig. 139. Fig. 139a. — From South Park Con- gregational Church, Chicago. 172. In Fig. 139 the circumference of the circle is divided into three equal parts by the points A, B, and C. the trefoil ATBSC etc. is composed of semicircles tangent to the circle at points A, B, and C. EXERCISES 1. Show how to construct the trefoil. Suggestion. — Construct A OCR and OCS each 45°. 2. Prove that these semicircles intersect on the radii which bisect AB, BC, and CA. 3. Prove that the curved figures A TB and BSC are congruent. vlns. ^(3-V3). 4. liAO=r, find the length of AX. XA Vs' 5. If ^0 = r, find the area of (a) the trefoil ATBSC etc. and (ft) the curved figure A TBA. ^'"'(2-^/3)(3,r-|-2^/3); (6) ^(18- 14,7-12 V3 + 9irVj). Suggestion. — Show that - Ans. (a) • PART 3. ROUNDED QUADRIFOILS 173. Occurrence. — While the quadrifoil is essentially a Gothic form, it is extensively used in modern industrial ornament. This may be due to the very general use of 162 GOTHIC TBACERY: FORMS IN CIRCLES the square as the basis for modern design. With the exception of the eight and sixteen foiled figures none of the others are well adapted to such designs. In Gothic bnilding^s they are extremely abundant. The following illustrate their use in tracery windows: Fig- ures 161a, 193a, 205a, 207a, 216a, 229a, 230a, 237a, 262a. When used in stone cutting and wood carving, the two forms are frequently combined in one figure, as in Figs. 149 and 151. Some special instances of the use of the quadrifoil in medieval tiled pavements are noticeable. In the pavement in Chertsey Abbey, England, is one based on Fig. 140. The complete design is a square covering four tiles, each 7| inches on a side. The circle containing the quadrifoil is surrounded by a handsome scroll border, and all spaces are fully decorated with a fine all-over pattern.^ In Gloucester Cathedral and in Great Malvern, Worcestershire, England, are tiled pavements containing designs based on Fig. 142. The first dates back to 1455.^ There are a multitude of designs more or less closely related to quadrifoils which are common not only in medieval and modern work, but also in the Chinese and Japanese work and in the decorations of primitive people.* Among them is the Chinese Monad used as a charm by the Chinese and as a trademark by the Northern Pacific Railroad (§ 186, Ex. 5, and Remark). (See also Figs. 142 and 152.) The quadrifoil is also found in American Indian decoration.* 1 Shaw (1), PI. XX. 2 Shaw (1), Pis. XXXVIII, XXXIX, XLI. s " Wonderland," 1901. Northern Pacific R. R. Diefenbach, Series B, Pis. VII, XXIII, XXX. * Beauohamp, PI. 3. BOUNDED QUADBIF0IL8 163 THE ROUNDED QUADRIFOIL FORMED OF TANGENT CIRCLES D^ __S_ ^ EXERCISES 174. 1. Construct a quadrifoil of tangent circles in a square, so that each circle shall be tangent to one side of the square at its middle point and to two of the other circles. Suggestion. — See Fig. 140. Draw the diagonals of the square. Inscribe a circle in each triangle thus formed. 2. Prove that the centers of the circles lie on the diameters of the square and that the lines joining their centers form a square. 3. Prove that the curved figures EQF and FRG and also BEQF and CFRG are congruent. 4. Inscribe a quadrifoil of tangent circles in a given circle. Suggestion. — Circumscribe a square about the circle and inscribe the quadrifoil in the square as in Ez. 1. 5. If AB = a, find the length of EY. Ans. - ( V2 - 1). 6. If AB = a, find the following areas : (1) the quadrifoil EQFR etc. ; (2) the small curved figure PQRS; (3) the cusped figure HPE ; (4) the cusped figure HAEP. Ans. (1) ^ (3 - 2 v^) (3 ,r + 4) ; (2) ^ (4 - ,r) (3 - 2 V2) ; (3) ^(3tv'2+4 V2-4^-6); (4) ^ (6ir V2 + 8 V^-97r-8). 8 16 164 GOTHIC TRACEST: FORMS IN CIRCLES 7. If the area of the quadrifoil EQFRS etc. is 144 square inches, find AB. 8. If the area of the quadrifoil is s, find AB. Ans. 2(1 + ^2) sj- '37r + 4 9. If the area of the cusped figure HPE is s, find AB. 10. If a circle is inscribed in the square X YZ W, find the area of the oval PQ and of the cusped figure bounded by PEQ and PQ,. Let AB = a. Ans. (1) |(3-2V2) (,r - 2) ; (2) | (3 - 2 V2) (,r + 2). 11. If XY=b, find the following areas: (a) the quadrifoil PEQFR etc.; (V) the oval PQ; (c) the cusped figure bounded by PEQ and PQ. Ans. (a) i' (3 ,r + 4) ; (ft) | (^ - 2) ; (c) |(,r + 2). a: £ i Fig. 141. Fig. I4la. — Inlaid Tile Design. Scale J in. <= I ft. EXERCISES 175. 1. Construct the quadrifoil of tangent circles inscribed in a square so that each circle is tangent to two sides of the square and to two other circles. Suggestion. — See Fig. 141. 2. Prove that the lines joining the centers of the circles form a square. ROUNDED QUADBIFOILS 165 3. If ^S = a, find the following areas : (a) Of the quadrifoU isTQiAfiJ etc. , a" ,o .^ Ans. — (3 IT + 4). (6) Of the small curved figure PQRS. . a^ ,. , 16 ^ -" (c) Of the figure bounded by the arcs KQ and QL, and the line- :ment KL. 4. If the area of the quadrifoil is s, find AB. Ans. 4 -yl 4 + 37r Fig. 142. — From Westminster Abbey Cloisters. Parker (!), p. 177. EXERCISES 176. 1. Show how to construct Fig. 142. 2. How many degrees are there in the arc APQI Ans. 225°. 3. Prove that the figures APQB, BQRC, etc., are congruent. 4. If ^ = r, find the area of the figure APQB. Ans. ^y(27r-6 + 4V2-TV2). 5. If j4 = 2, find the area of the curved figure PQRS. 6. If the area of the figure PQRS is .s, find A 0. 166 GOTHIC TBACERT: FOBMS IN CIRCLES 177. 1. Show how to construct Fig. 143. Suggestion. — The construction is effected by means of two quadri- foils of tangent circles ; w, x, and y are centers for three of the circles. 2. Prove that xA = OK = OL = OM = ON, etc. Fig. 143, — Rose Window, Strassengal, Germany. Fig. i43a. — Stamped Steel Ceiling Design. 3. Prove that line OE bisects /.GO A. Suggestion. — Draw O// the bisector of /.GO A. Prove that «)0 = Ox; OH is the perpendicular bisector of xw ; Ew = Ex. Therefore, E lies on OH. 4. If OA = a, find the area of (1) the cusped figure EAHG etc. and (2) the quadrifoil KEAFM etc. 5. Show that K, L, M, N, etc. are the vertices of a regular octagon. 6. If OA = a, find the area of the octagon KLMN etc. Suggestion. — OK in Fig. 143 is J ZF in Fig. 144. See also § 88. 7. Draw a diagram for the design shown in Fig. 143a, assuming that all arcs are arcs of circles. BOUNDED QUADRIFOILS 167 (From Diefenbach, Series B. PI. XI.) Fig. I44a. 178. In Fig. 144 the circles X, Y, Z, and W are tangent, as in § 174, Ex. 4. The lines HK, KF, FG, and GH are perpendicular to OA, OB, OC, and OD, respectively, at the points X, Y, Z, and W, respectively. The small circles are tangent as indicated. EXERCISES 1. Prove that HKFG is a square and that vertex K is on the radius through the point of contact of the circles X and Y. 2. Prove that (1) OP = KM = KL ; (2) OP = KN. Suggestion for (2). — XY= 0K = PB. 3. Prove that circle K is tangent to the circles X and Y and to the large circle O. 4. If OB = a, find the length of OY, YP, and OP. Ans. (1) a(2 - V2); (2) a(V2 - 1); (3) a(.3 -2V2.) 5. If 0P= b, find OB and PY. Ans. (1) 6(3 + 2v'2) ; (2) A(V2 + 1). 6. If P F = c, find OB and OP. Ans. (1) c(V2 + 1) ; (2) c(V2 - 1). 7. What per cent of the area of the large circle is contained within the five small circles 0, F, G, H, and Kt Ans. About 15 %. 168 GOTHIC TRACERY: FORMS IN CIRCLES 8. What per cent of the area of the large circle is contained within the four circles X, Y, Z, and W ? Ans. 68.62 %. 9. Give the construction for Fig. 144a. 179. In Fig. 145 ABCD is a rectangle with AD = i AB. £ and F are the middle points of the sides AB and CD, re- spectively. Semicircles are constructed with E and F as centers and quadrants with A, B, C, and D as centers, each with radius = i A£. A E Fig. 145. — Fan Vaulting from Gloucester Cathedral, Fletcher, PI. 112. B England. EXERCISES 1. Lines are drawn tangent to the arcs at the points where they cut the diagonals of the squares AEFD and EBCF. Prove that these tangents form the squares WXYZ and KLMN. 3. Show how to construct the entire figure. 3. li AD — a, find the radius of one of the small circles. Ans. ^(3-2V2"). 4. li AD = a, find the following areas : (a) The large cusped figure bounded by four quadrants. Am. j^i-T). (i) The quadrifoil formed by the four small circles. Ans. ^(17 - ISV^XStt + 4). (c) The small cusped figure included by the four small circles. Ans. j(4-7r)(17-12V2). (d) The four-sided curved figure bounded by two quadrants and two small circles. Ans. j(9 7rV2 + 12 ^2 - 13 ir - 16). ROUNDED QUADBIF0IL3 169 180. In Fig. 146 ABCD is a square, EF, FG, GH, and HE are drawn with A, B, C, and D as centers and the half side AE as radius. The four circles are tangent, as indicated. D G ^^"-\k/ H -^ >^, find the radius of circle O. Ans. -. 6. If DH = J CD, find the radius of circle 0. .^«s-f|(8+ V3). Suggestion. — Let h ='— y/3 in the formula found in Ex. 4. 7. Make a drawing for the window shown in Fig. 200a. 244. In Fig. 201 the line AB is given, and the arcs CAE and CBE are drawn with points B and A as centers and the arcs ADB and AHB with points E and C as centers, respectively. Fig. 201. — From First Presbyterian Church, Fort Wayne, Indiana, 224 GOTHIC TRACERY: POINTED FORMS EXERCISES 1. Construct circle C tangent to arcs ADB and AHB. 2. Construct circle tangent to arcs AC, BC, and ADB. 3. What loci must be used to find the centers of circles 3 and 4? Suggestion. — The locus of the centers of circles tangent to any two given circles. This problem does not come within the province of elementary geometry. Fig. 202a. —St Andrews M. E. Church, New York City. 245. In Fig. 202 D is the middle point of AB. ACB is a semicircle. Circle O is tangent to the line AB and to the semicircle ABC. Circles O' and O" are tangent, as indicated. EXERCISES 1. Show how to construct circle 0. 2. If circle 0' is tangent to circle at .X and to the semicircle A CB at N, prove that O, X, 0' and also D, 0', N are coUinear. 3. li AB = s, find the radius of circle O' tangent to circle O, to the semicircle ACB, and to line AB. Ans. Suggestion. — Drop perpendiculars from 0' to AB and CD. In AO'DS,T)S^ ^TJO'^ -TFS'^. U{:^0'GO,W(? = '0a^ -'0(?. Since DS = O'G, DO'^ - O'S^ = 00'^ - If radius of circle 0' hs r, 0' D -■ s 2~ 00' =-+ r. 4 OG=--i i The above equation becomes [5 — '') ""'"''= (j + '') ~(i~''') 1 Hanstein, PI. 19. FOBMS CONTAINING TANGENT CIRCLES 225 4. Construct circle 0'. 5. What fraction of the area of the semicircle A CB is occupied by the three included circles ? Ans. |. Remark. — Figure 202a shows a design related to Fig. 202. The construction of the two small circles on either side of the oval does not come within the province of elementary geometry. Fig. 203. EXERCISES 246. 1. Problem. — To construct a circle tangent to a given line at a given point and to a given circle. Suggestion. — See Fig. 203. Let be the given circle, AB the given line, and P the given point. From and P draw lines perpendicular to AB. Let the perpendicular through cut the circle in points C and D. Draw CP cutting the circle at E. Draw OE meeting the perpendicular through P at 0'. Prove A COE and PEO' similar. Hence, O'E = O'P, since EO = OC. Show that a second circle is possible. Give construction and proof for circle 0". 226 GOTHIC TRACERY: POINTED FORMS 2. In the preceding construction show the application of the theorem, " If two circles have a common point in the line of centers, the circles are tangent." 3. What locus is used in this construction, and where? 4. Perform this construction in case point P lies within circle 0. 5. For another solution for this problem, see § 119, Ex. 7. CIRCLES INSCRIBED IK EQUILATERAL CURVED TRIANGLES • 247. Occurrence. — The equilateral circular triangle occurs throughout Gothic architecture. It was used for small windows at an early date.^ Fine examples are found in Westminster Abbey, where the inscribed circle contains an eight-foil, in Hereford Cathedral, where it contains a six-foil, and in Lichfield Cathedral (Fig. 208). That in Westminster Abbey dates from 1250. When used as details of tracery, circular triangles are an indication of later geometrical work. Its modern use is indicated in Figs. 188a, 189a, 204a, 205a, 206a, 258a. Fig. 204a. — From Second Uni- ver&alist Church, Boston. 1 Parker (2), pp. 139-140 ; Rickman, pp. 121, 190-192 ; Sharpe (1), p. 71. FOBMS CONTAINING TANGENT CIRCLES 227 EXERCISES 248. 1. If ABC is an equilateral curved triangle with A, B, and C the centers for the arcs, show how to inscribe the circle in the tri- angle, as shown in Fig. 204. Suggestion. — Let DC, EB, and^li^be the medians of the linear triangle ABC. Extend them until they meet the arcs at G, H, and K, respectively. Let them intersect at 0. is the center for the re- quired circle. Prove OG — OH = OK, and that the circle is tangent to the arcs AB, BC, and A,C. 2. If CG extended is considered the locus of a certain class of circles tangent to each of the two given equal circles of which AC and CB are arcs, prove by intersecting loci that point is the center of a circle tangent to the three arcs AB, AC, and CB. 3. If AB = s, iind the length of GO and the area of the circle. An,. (1) ^(3- V3); (2) 1^(2 - VS). 4. If GO = a, find AB. Ans. -(3 + V3). 5. Prove that the curved triangles AHMG and BGNK are con- gruent. 6. liAB = s, find the areas of the curved triangles ABC and A GH. 18' Ans. (1) fy(fl- - \/3) ; (2) f; (4 7rV3 - 5 tt - 3V3). Fig. 205a From Boyton Church, England. 228 GOTHIC TBACERY: POINTED F0BM8 EXERCISES 249. 1. If the circumference of a circle is divided into six equal parts at points A, B, M, N, F, and E, show that it is possible to con- struct the equilateral curved triangles OAB, OMN, and OEF, with centers at these points as shown in Fig. 205. 2. If circle Q is constructed tangent to arcs OM, OB, and BM, prove that its center must lie on the radius perpendicular to AN. r 3. Find the length of QT if OT = r. 4. Give the complete construction of Fig. 205. Ans. Fig. 206. Fig. 206a. EXERCISES 250. 1. Give complete construction for Fig. 206. 2. Show that Fig. 206 can be inscribed within an equilateral curved triangle ABC, if A, B, and C are the centers for the arcs BC, A C, and AB, respectively. Suggestion. — In Fig. 206o ABC is an equi- lateral triangle with the medians AE, BF, and CD. The angles COF, FOA, A OD, etc., are bisected by the lines OM, ON, OG, etc. Prove that OM = ON = OG. Use superposi- tion, folding along any median. Then prove that linear A GKM and HLN are equilateral and congruent. Fig. 2066. — From St Bride's R. C. Church, Chicago. FORMS CONTAINING TANGENT CIRCLES 229 3. Construct the curved A GKM and HLN and prove that a circle can be inscribed in the curved hexagon XYZ etc. Suggestion. — Prove that OX = OY = OZ, etc., and that the circle and the arcs NH, GK, HL, etc., are tangent. / / / / yfK E'\ Fig. 207. 251. In Fig. 207 ABC is an equilateral triangle. L, M, and IT are arbitrary points on the medians A£, BF and CD, respectively, so chosen that AL = BM — CN. L, M, and N are the centers and LC is the radius for the circles that intersect at A', B', and C. EXERCISES 1. Prove that these circles also intersect at points A, B, and C, and that the points A', B', and C are on the medians extended. Suggestion. — In order to prove that circles with centers at Z and M intersect at point C prove that LC = MC. 230 GOTHIC TRACERT: POINTED FOEMS 2. Prove that a circle can be inscribed in the curved triangle ABC. 3. Show that Ex. 2 may be proved by intersecting loci. 4. Braw a diagram for one of the windows shown in Fig;. 207a. F.g. 207(1. — St. Anne R C. Church, Chicago. 5. Draw a diagram like Fig. 207 with points L, M, and N on the lines EA, FB, and DC ex- tended. 6. Prove Exs. 1 and 2 for the figure called for in Ex. 5. 7. If A LMX is given equilateral and circles are constructed with L, M, and N as centers and with any equal radii, intersecting in points A, B, and C, and A', B', and C", prove that (a) A circle can be inscribed in the curved AABC. (6) A circle can be circumscribed about the three circles. (e) The points E', A, 0, A' are collinear. (d) AL = BM= CN. (e) Linear ^5C is equilateral. Fig. 208a. — From Lichfield Cathedral. 252. In Fig. 208 ABC is an equilateral curved triangle with A, B, and C as centers for the sides. AF, BE, and CD are the medians of the linear A ABC, extended to the arcs as indicated. The circles O, O', and O" are tan- gent to each other and to the sides of the triangle. FORMS CONTAINING TANGENT CIRCLES 231 EXERCISES 1. If circle is tangent to lines EX and XD, prove that its center must lie on line AX. If circle is tangent to arcs AC and AB, prove that its center must lie on AX. 2. Construct circle tangent to lines EX and DX and to the arcs AC a.ndBC. 3. Make O'B = OA and construct a circle with 0' as center and a radius equal to that of circle O. Prove that this circle is tangent to lines XD and XP and to arcs BA and BC. 4. It AB = s, find the length of the radius of circle 0'. Ans. About .2is. Suggestion. — In A A KG', AK^ = AO'^ - K0'\ If OK' = x, XK = 5 Vij, and ^ K = ^ v^ + * Vs. Hence get the equation {s-xy-x'^ = \{s + xy. 5. What is the per cent of error if O'K is called '-? EXERCISES 253. 1. Problem. — To construct a circle tangent to the sides of an angle and to a given circle. Suggestion. — • The solution may be accomplished by the three fol- lowing steps. 232 GOTHIC TRACERY: POINTED FORMS (1) Construct a circle passing through two given points and tan- gent to a given line. See Fig. 209a. Solution. — Let ABhe the line and C and D the given points. Let CD intersect AB at P. Find the mean proportional between PC and PD. Lay off PQ and PR equal to the mean proportional. Erect perpendiculars to ^fi at Q and R. Draw the perpendicular bisector of DC, meeting the perpendiculars at and O'. Prove that O and 0' are the required circles. Note that the problem is impossible if the two points are on oppo- site sides of the given line. Study the case when one of the points is on the given line. What locus is used in this construction? Fig. 2096. (2) Construct a circle tangent to the sides of an angle and passing through a given point. See Fig. 2096. Solution. — Let EBA be the given angle and C the given point. Draw O'B the bisector of the angle. Draw CD so that O'B is the per- pendicular bisector of CD. Construct a circle passing through points C and D and tangent to AB. Prove that and 0' are the required circles. Note that the problem is impossible if the given point is outside of the given angle. When is only one circle possible ? What locus is used in this construction ? FORMS CONTAINING TANGENT CIRCLES 233 (3) The main problem. — To construct a circle tangent to the sides of angle EBA and tangent to circle 0. Solution. — Drav! A' B' and E'C parallel to the sides of ^ ABE, and at a distance from them equal to the radius of circle O. Con- struct circle 0' tangent to A'B' and E'C, and passing through point 0. The circle with ff as center, and O'Q as radius, is the required circle. 2. Study the following special cases : (a) When the given circle cuts both sides of angle EBA. Suggestion. — The parallels A'B' and E'C may be inside the given angle. In this case four circles are possible. (A) When it cuts one side of the angle and is tangent to the other. (c) When it cuts one side of the angle only. {d) When it is wholly within the angle. (e) When it is wholly without the angle. (/) When it is tangent to one side only and within the given angle. (g) Whsn it is tangent to one side only and without the given angle. (A) When it is tangent to both sides of the given angle. 234 GOTHIC TRACERY: POINTED FORMS TWO-LIGHT WINDOWS Fpg. 210a. — First Presbyterian Church, Chicago. Fig. 210. 254. In Fig. 210 the arch ACB is equilateral, and the circle O is constructed on AB as diameter. The equilateral arches GEH and HFE are drawn with the half span AO as radius and with HE and HF tangent to circle O. EXERCISES 1. Show how to find points G, H, and K, the centers for the arcs EG, EH, FH, and FK. Suggestion. — This involves the problem, " To construct a circle of given radius, tangent to a given circle and to a given line." See § 265. Notice thart the arcs are tangent to the lines A G, CH, and BK, as indi- cated. 2. How many circles can be constructed that will fulfill the condi- tions of the figure ? Is there any difficulty in determining which circle is intended ? Suggestion. — See § 265. 3. Prove that the circle O and the arc EH are tangent at the apex of the arch GEH. Suggestion. — Draw GO and KG. Prove that the line GO passes through the point of contact and that A GOK is equilateral. Since ZEGH = 60°, line EG falls on OG. FORMS CONTAINING TANGENT CIRCLES 235 4. If ^fi = 8, find the area of the curved figures: (a) AHEF; (b) AGEA; (c)_AACBNA. Ann. (o) 8(2 V3 - tt) ; (c) KSr-eVS). 6. li AB = i, find the area of the figures mentioned in Ex. 4. Ans. (a) i'(2v^-^); (A) ii(9V3-4^); (c)^{5^-QV3). 48 6. Ji AB = s, find the length of A G. C 24' ^ns. - V3. Fig. 21 Fig. 2lla. — From Notre Dame, Pans. 255. Figure 211 shows a design like that of Fig. 810, with AQ and PB for radii of AC and CB, respectively, each | of AB, and the radii for EG, EH, Sf, and VS., each | of AO. EXERCISES 1. Show how to construct the figure. Suggestion. — Let Ox' and Ow' be | of OA and OB, respectively. Draw x'x and w'w perpendicular to AB. The centers for arcs EH and ii"^ must lie on these lines. Why? Determine points x and w the centers for the arcs EH and FH. Draw line wx and complete the construction. 2. Is A Oxw equilateral ? 3. If AB = s, find the length of A G. Ans. - V6. 236 GOTHIC TRACERY: POINTED FORMS C Fig. 212. — Northfleet Church, Kent, England. Brandon, (I), p. 41. Fig. 213, — From Chapter House, York Cathedral. RIcknnan, p. 182. 256 In Fig. 21 2 the equilateral curved triangle ABC is constructed with A, B, and C as centers. The circle O is inscribed in this triangle. Arches GEH and HFK are equilateral, erected on the half span GH, so that the arcs HE and HF are tangent to circle O and to line CD extended. EXERCISES 1. Construct the figure, making AB three inches. Show all work in detail. Suggestion. — See § 248 and § 265. 2. Is the point of contact the summit of the arch GEH1 Suggestion. — Draw GO and OK. Is A GOK equilateral? 3. If AB = a, find the radius of circle and the length of BK. Ans. (1) I (3 - V3) ; (2) | V2I - 9 V3 - ^ V3. FORMS CONTAINING TANGENT CIRCLES 237 257. In Fig. 213 the circle O is constructed as in Fig. 212. All arches are equilateral except MDN. Arcs EM and FN are tangent to circle O and con- structed as in Fig. 210. Arcs MD and DN are constructed to pass through point D and be tangent to arcs EM and FN at points M and N respectively. EXERCISE Construct the figure, making AB three inches. Show all ■work in detail. 258. In Fig. 214 arch ACB is equi- lateral, with A and B as centers. Circle O is inscribed in the curved triangle ABC. Arches GEH and HFE are equilateral and tangent to circle O, as in Fig. 210. Fig. 214. EXERCISES 1. Construct a figure like Fig. 214, in which AB is five inches. Show all work in detail. 2. If AB = s, find the radius of circle and the length of BK. Ans.{l) ^; (2) |(V33-3). Suggestion. — For, the radius of circle O see § 241, Ex. 3. 238 GOTHIC TRACEBT: POINTED FORMS C 259. The design in Fig. 215 is like that in Fig. 214 with the radii of the arcs AC and CB each | of the span AB, and the radii -^ of the arcs EG, EH, HF, and FK each | of AD. ^ Fig. 215. - From Lady Chapel, St. Patrick's Cathedral, New York City. Am. Arch, and BIdg. News, March 24, 1906. EXERCISES 1. Consti-uct a figure like Fig. 215 with AB = 4. Show all work in detail. 2. If AB = s, find the length of RK. Ans. .2935 .?. Suggestion. — For the radius of circle O, see § 242, Ex. 4. Fig. 216. Fig. 216a,— From Cologne Cathedral, FORMS CONTAINING TANGENT CIECtES 239 260. In Fig. 216 the arch AXB and the curved square WXYZ are con- structed as shown in § 115. The arches GEH and HFE are equilateral. The arcs EH and HF are tangent to the arcs WEZ and ZFY, respectively. EXERCISES 1. Show how points G and K are determined, and prove GK parallel to AB. 2. Make a drawing for Fig. 216, showing all work in detail. Let AB\)& five inches. ^ns. i(V5-2). 3. If AB = s, find the length of BK. 4. Prove that X, Y, Z, and W are the vertices of a square. 5. Prove that Y is the midpoint of arc XB. 6. li AB =: a, find the area of the square whose vertices are X, Y, Z, and W. Suggestion. See § 115. 261. In Fig. 21 T the circle is in- scribedinthe curved square constructed as in Fig. 216. In the equilateral arches GEE and HFK the arcs EE and FH are tangent to circle O. 240 GOTHIC TBACERY: POINTED FOBMS EXERCISES 1. Construct a figure similar to Fig. 217, so that AB shall be four inches. Show all work in detail. 2. Prove that it is possible to inscribe circle in the circular square xyzw. 3. If AB = s, find the radius of circle O and the length of BK. Ans.(l) •^(2-V2); (2) i(VlO - 6V2 - 1). 262. InFig. 218 ACBisanegui.. lateral arch. Point O is an arbitrary point on the altitude CD. The equi- lateral arches GEH and HFK are tangent to circle O. EXERCISES 1. Construct circle tangent to the arcs A C and CB. Suggestion. — Draw BO and extend to the arc A C. Use the theorem, " If two circles have a common point on the line of centers, the circles are tangent." 2. Construct the equilateral arches GEHsmd HFKon the half span AD and DB, so that the arcs EH and HF will be tangent to circle 0. FORMS CONTAINING TANGENT CIRCLES 241 3. If ^5 = 12 and Z)0 = 5, find the length of DM and DH. Ans. (1) 4.189 = (12 -Vei) ; (2) 3.2. 4. li AB = s and DO = 6, find the length of DM and i/0. Ans. (1) s-l Vs2 + 4 62 ; (2) ^V9 s^ + 4 6^ - 6 sVs^ + 4 J^. 5. If DiZ^ = a and ^S = s, find the length of DO. 263. Figure 219 shows a de- sign like Fig. 81 8. AQ = PB={AB. P and Q are the centers for the arcs CB and AC, respectively. O is an arbitrary point on the altitude CD. The arches GEH and HFK are tangent to circle O and drawn with radius j of AD. Fig. 219. — From Reims Cathedral, France, Sturgis (I), p. 267. EXERCISES 1. Construct circle tangent to the arcs CB and CA. 2. Construct arches GEH and HFK on the half span with a radius f of AD, so that the arcs EH and HF are tangent to circle and to the line CD extended. Suggestion. — Find the points x, z, y, and w, the centers for the arcs EH, EG, FK, and FH, respectively. 3. If ^B = 8 and DO = IJ, find OM and DH. _ Ans. (1) 33 ; (2) J (5V33 - 7). 242 GOTHIC TRACEBT: POINTED FORMS 4. If AB = s,DO = b, and AQ = iAB, find CM and OH. Ans. (1) gj-^S^'+lSft.'. (2) jV^T+li^^^^lsTi^^^l^+TlT^ 5. If ^£ = s, ^ Q = fcs, and Z)0 = 6, find DM and OH. G 264. In Fig. 220 ABC is an equilateral curved triangle with A, B, and C as centers. The trefoil is inscribed in this triangle as in Fig. 208. Points E and F are obtained by dropping perpendiculars from the middle points of AD and DB. Arches GEH and HFK are equilateral arches on the half span GH = AD, passing through the points £ and F. ^K Fig, 220. — From a German Church, Hartung, Vol. 2, PI. 97. EXERCISE Show how to locate points G, H, and K. Suggestion. — See § 266, Ex. 1. Fig. S2I. FORMS CONTAINING TANGENT CIRCLES 243 EXERCISES 265. Pkoblem. — Given the circle with radius r and the line AB. To construct a circle of given radius r' tangent to circle and to AB. Suggestion. — Solve by intersecting loci. Use the locus of centers of circles of given radius and tangent to a given line, and the locus of centers of circles of given radius and tangent to a given circle. How many circles are possible ? 1. What must be the distance d of line AB ivova center in order that circle X'Y'Z'W may (a) be tangent to line CD'i Ans. d = 2 r' + r. Suggestion for (a). — The radius of circle X'Y'Z'W'= r' + r, and the distance from to CD is d — r'. In order that X'YiZ'W may be tangent to CD, r' -\- r must equal d — r' . Therefore, solve the equation r' + r = d — r' for d. (A) cut the line CD'i Ans. 2 r' -k- r > d. (c) be tangent to line EFt Ans. d = r. AB \i tangent to the given circle, (rf) cut line EF? Ans.d<,r. j4fi cuts the given circle. 2. If the radius of the required circle is less than the radius of the given circle, what must be the distance of line AB from the center 0, in order that circle XYZW ratiy (a) be tangent to line CDI Ans. d = r. Line AB is tangent to the given circle. Suggestion. — As in Ex. 1 the radius of r. Ans. (a) d = 2r' -r; (b) d<2r' - r; (c) d = -r; that is, AB is tangent to circle O but on the opposite side of the center from line EF ; (d) d<, — r\ that is, AB is outside the circle. 244 GOTHIC TRACERT: POINTED FORMS 4. If 2 r* < r, show that there will be No solution to the given problem, if r/ > 2 ?^ + r. One solution, if d = 2 r' + r. Two solutions, if2 r' + r>rf>r. Four solutions, if d = r. Six solutions, ifr>d>r — 2r'; Seven solutions, if rf = r — 2 r'. Eight solutions, ii d r' > r, show that there will be No solutions to the given problem, ii d>2r' + r. One solution, if d = 2 r' + r. Two solutions, if 2 r' + r > rf > 2 r' — r. Three solutions, if rf = 2 r' — r. Four solutions, if d < 2 r' — r. 6. Discuss the cases (1) 2 r' = r, (2) r' < r < 2 r', (3) r' -. (4) r' > 2 r, (5) r = r'. ■ 2r THREE-LIGHT WINDOWS 266. In Fig. 222 the equilateral arch ACB and the curved triangles are con- structed as in § 226. The arches MPH and KQN are equilateral, constructed with \ AB as radius, so that their apexes will lie on the arcs APD and D QB, respectively. The arcs DH and DK are so constructed that their centers lie on the line MN and that they will pass through the points H and D, and K and D, respectively. Fig. 222. FORMS CONTAINING TANGENT CIRCLES 245 EXERCISES 1. Show that the arc PM may be constructed by means of either of the following problems : (1) To construct a circle passing through a given point, of given radius, and tangent to a given line. (2) To construct a circle passing through a given point, of given radius, and with its center on a given line. 2. What loci would be used in each of the above constructions? 3. Show that the arc DH may be constructed by means of either of the following problems : (1) To construct a circle passing through a given point and tan- gent to a given line j^t a given point. (2) To construct a circle passing through two given points with its center on a given line. 4. What loci would be used in each of the above constructions ? 267. In Fig. 283 ACB is an equi- lateral arch constructed from A and B as centers. AG = GH = HB. AGE, HBF, and EEC are equilateral curved triangles constructed from their ver- tices as centers. The arch RDS is also equilateral. Fig. 223. — From St. Nazaire Cathedral, Carcas- sonne, France, Sturgis (I), p. 278, 246 GOTHIC TRACERY: POINTED FORMS EXERCISES 1. Show how to construct the three curved triangles A GE, HBF, and EFC from the linear triangle ABC. 2. If B is the center of the arc EDF, show how to locate points R and S so that DR = RS = DS. Construct the arcs DR and DS. 3. Construct a complete design like Fig. 223. Let AB be four inches. Suggestion. — The trefoils in 1 and 2 are described from the middle points of the straight-line triangles as centers ; those in 3 and 4 from the middle points of the sides of the circular triangles as centers. See §§ 232 and 233. 4. If AB = s, find the area of the curved figures : (o) the A CEF; (b) the A CFB ; (c) the A KFB ; (d) the trefoil HMBKF etc. ^„,. (a)2A\^_V3); (b) ^(2, W^(2' a/3). 3V3); (c)^(2,r-3V3); Fig. 224. Fig. 224a. — From Exeter Cathedral. EXERCISES 268. 1. Construct a design like that given in Fig. 223 with the span of arches 1 and 2 twice the span of arch 5. FORMS CONTAINING TANGENT CIRCLES 247 2. Ji AB = s, find the area of the curved figures : (a) A CEF ; (4) A CFB; (c) window head AEG; (d) window head liDS. Am. (a) 1^' (,r - V3) ; (6) |;^(2 tt - 3 V3) ; (c) f! (4 tt - 3 V3) ; W3^(4^-3V3-). 25 75' 269. In Fig. 225 ABC is an equilat- eral curved triangle with A, B, and C as centers. Circles O, O', and O" are con- j^ structed as in Fig. 208. Fig. 225. EXERCISES 1. If the equilateral arch GEH drawn with \ AB as radius is so constructed that arc EH is tangent to circle 0, show how point G is determined. Suggestion. — See § 265. 2. Determine points G and L so that arcs EH and FK, drawn with a radius equal to ^ AB and with G and L as centers, will be tangent to circles O and 0', respectively. Prove GL parallel to AB. 3. Let AN = NM = MB. Draw NH and MK perpendicular to AB from points iV and Af, respectively, meeting GL at H and K. Complete the construction of arches GEH and KFL. 4. If arc HD is tangent to line NH at point H and to circle 0, how is the center for arc HD to be determined V 248 GOTHIC TRACERY: POINTED FORMS 5. Construct arcs DK and DH by the same method and prove that they will have the same radii. Fig. 226.— From Milan Catliedral. 270. In Fig. 226 the center of the large central circle is probably arbitrary. The two small side arches are constructed with one third of the main span as radius and with one side of each tangent to the circle. The small central arch is equilateral and made to touch the circle. EXERCISES 1. What circle constructions are used for the small arches ? 2. Make a diagram for a window like Fig. 226 with the span 4 inches and the radius of the sides of the main arch { of the span. PART 3. FORMS CONTAINING TANGENT CIRCLES: ALGE- BRAIC ANALYSIS 271. Most of the diagrams given in this Part are treated as types. By means of the formulae given it is easy to obtain the data for any special cases under these types. The cut that accompanies the diagram is usually one of these possible special cases. Windows like those given TANGENT CIRCLES: ALGEBRAIC ANALYSIS 249 and many others abound. The relation between the span and the radius of the arcs forming the main arch may be obtained roughly from even small photographs. If tlie length of the span and the altitude are known for any particular case,i the radius may be found as in § 275, Exs. 20 and 21. It is therefore possible to study with more or less accuracy any particular window desired. ONE-LIGHT WINDOWS 272. Figures 228, 236, and 247 furnish illustrations of the following type of one-light window. The formula for their solution is given in Ex. 7 below. M E N ') ^ 1 ^l \ff^o^ i 1 v^ 1 l\ 1 / \ . 1 / \ / L '^"'^Nv \ / /-^^^i^J f- -f---^/ „__„P?__J^ D ' 1 \ 1 ;_ 1 j A H B Fig. 227. — From Norwegian Legation, Wasilington, D. C. Arch. Annual, 1907. In Fig. 227 CED is an equilateral arch constructed with C and D as centers. The rectangle CDNM is constructed with DN parallel to the altitude EK. ABDC is a rectangle with DB = \ AB. AFB is a semicircle on AB as diameter. Circles O' and O" are tangent as shown. EXERCISES 1. Find the ratio of DN to MN. 2. li AB = s, find the radius of circle 0'. Ans. Ans. V3. 2 lis 40 ' ^ E. Sharpe (2), gives the dimensions for a large number of windows from English churches and cathedrals. 250 GOTHIC TRACERY: POINTED FORMS Suggestion. — Vse ADaK. 0'D = s-r, KD = ^-, aK=^- + r. Solve for r in the equation (s — J")^ — j=l7 + ''l • 3. If DB = hAB = hs, find the radius of circle 0'. 2 3-2A Suggestion. — In A DO'K, O'K = (f, - ''•'•" + '")• 4. Find the radius of circle 0' if DB is zero. Ans. -. Suggestion. — Let h be zero in the formula found in Ex. 3. See Fig. 228. 5. If AMNB is a square, (a) Find the ratio of ^ Ans " ~ "^ AB ' ' •> ' Suggestion. — JtNB = s and ND=iV3,DB = '^ (2 - v^). (6) Find the radius of circle O'. Ans. - (7-aVTi) or about — . 8^ ^ 40 6. Construct circle 0". Suggestion. — See § 2.o3 and § 332. 7. If the radius of CE and DE is Jcs and DB = hs, find the radius of circle 0' ■when AB = s. , s /l + 2h^-2h~ 2k\ Ans. - [ — ■ . 2\ 2h-2k- 1 / 8. Show how the formula obtained in Ex. 3 may be derived from that obtained in Ex. 7. Suggestion. — Let k = 1. Why? 9. Show how the result in Ex. 2 may be derived from that in Ex. 7. 10. Draw a figure to illustrate the case mentioned in Ex. 5. TANGENT CIBCLES : ALGEBUAIG ANALYSIS 251 Fig. 228a. — From St, Pol de Leon, Franpe. 273- In Fig. 228 the diameter of the semicircle AFB is the span of the equilateral arch ACB. CD is the perpendicular bisector of AB. The circle O is tangent to the semicircle and to the arcs as indicated. EXERCISES 1. Prove that the points B, O, and Z are coUinear. 2. If AB = s, find the radius of circle 0. Ans. -. Suggestion. — A ODB gives the equation (s — r)^ = ( - ] + ( ^ + '' ) where r is the radius desired. 3. Construct circle 0. 4. What per cent of the area above the line AB is occupied by the circle and semicircle ? Ans. About 78 %. 5. Construct a figure like Fig. 228 in which the radius oi AC and CB is I of the span AB. Show that the radius of circle O in ,, . „ . AB this figure is ^ 8 6. Find the radius of circle 0, if the radius oi AC and CB is ks. . s 2i - 1 Ans. - ■ —- -. 2 2k + \ 7. Show that the formulae obtained in Exs. 2 and 6 may be derived from that obtained in § 272, Ex. 7. 1 Hanstein, PI. 19. 252 GOTHIC TBACERT: POINTED FORMS TYPES OF TWO-LIGHT WINDOWS 274. Type I. —Figures 229-234, 237, 238, 256, and 258 furnish illustrations of this type. The formulae for their solution are given in Exs. 9, 28, and 25 in § 275. The form of this type shown in Figs. 229 and 230 is the com- mon design of early geometrical windows. Its use is universal. Figure 193a shows an illustration from St. Louis. By repetition of this design, windows of four and eight lights are formed as in Fig. 231. The ratio of the radius to the span varies greatly. It is said to be ^ in the choir at Lincoln Cathedral, ^ in Salisbury Cathe- dral, and ^ in the nave at York Cathedral.^ Q B Fig. 229a. — From the Chapter House, Salisbury Cathedral. ^^275. In Fig. 229 AQ = BP = | AB. P and Q are the centers for BC and AC, respectively. The arches AED and DBF are constructed on the halves of the span AB, with radius f AD and points S', R', R, and S as centers. Circle O is tangent to the arcs as shown. 1 Enc. Brit., Arches and Masonry in article on Building. TANGENT CIRCLES: ALQEBBAIC ANALYSIS 253 EXERCISES 1. Prove that (1) CD extended is the common chord of the two intersecting circles of which AC and BC are arcs ; (2) CD is perpen- dicular to ^£; (3) the linear A ^ CB is isosceles. 2. Prove that points z, 0, Q, and also 0, x, S are collinear, when 2 and X are the points of tangency of circle and CA and FD. 3. If circle is tangent to the two equal intersecting circles A C and CB, or to the two equal tangent circles ED and FD, prove that its center lies on the line CD. 4. Find the radius of circle O, it AB = s. Ans. — 30 Suggestion. — Solve for r in the equation &-')'-H%*'y-{ur 5. Find the center of circle O. Suggestion. — Use Q as center and {AQ~ r) as radius, and strike an arc cutting line CD at 0. 6. If AB= s, what is the length of the radius used in Ex. 5 ? Ans. 5i. 5 7. If S is used as a center what radius must be used to obtain Voir^tO? Ans. SD + r, 01-^. 20 8. IiAQ=^AB and DS = | DB, find the radius of circle 0, if 36' ^B=s. ^„,.7. 9. If ^iJ = s and AQ = ks, find the radius of circle O. . s 4/t-l Ans. — • 12 k Suggestion. — Let r be the radius of the circle. Then OQ — ks — QD=ks--, OS = — + r, DS = -■ Get the equation '(--)'-{'-?r=(i-)'-(f)' 254 GOTHIC TRACERY: POINTED FORMS 10. liAQ = i AB = f s, find the radius of circle O by substituting 4 for k in the formula obtained in Ex. 9. Ans. 7_s 36' 11. If AQ = iAB, find the radius of circle Oby substituting J for k in the formula obtained in Ex. 9. Draw illustrative figure. See Fig 234. > s ° Ans. — 6 12. If AQ= AB, find the radius of circle by substituting 1 for k in the formula obtained in Ex. 9. Draw illustrative figure. See Fig. 230. Ans. 13. li AQ = ^ AB and AB = s, find the radius of circle by sub- stituting in the formula obtained in Ex. 9. Ans. 30 14. If .1 Q = l-s, and Ox = r, find value of ^B in terms of k and r. 12 rk Ans. 4k- 1 AS'PK D Q B Fig. 2296. 15. li AQ = %AB and OX = 4, find the lengths of AB, CD, EK, and DO. For EK see Fig. 2296. Ans. (1) 17^; (2) ^v^; (3) V^-v^; (4) yVIi. 16. If ^Q = is, S'D = k. AD, and 4B = s, find the value of CD and E2i:. See Fig. 2296. Ans. (1) - V4yi:-1 ; (2) - V4/fc- 1. 17. Prove that S'E and PC are parallel. Suggestion. — From the computations in Ex. 16, A S'EK and PCD have their sides proportional and are therefore similar. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 255 18. Prove that M^AED and ACB are similar. Suggestion. — Since S'E and CP are parallel, the isosceles A S'ED, and PCB may be proved similar and Z CBP may be proved equal to Z EDS'. For another discussion of this problem see § 328. 19. Prove that the points A, E, and C are collinear, and that E is the center of A C. 20. Find the value of 4 Q in terms of CD and AB. Ans. AQ = ■ i^AB Suggestion. — Draw SQ. Prove A j4C-D and ^2? Q similar. Hence AQ^AE AC ad' 21. If CD is 8 feet and AB 9 feet 10 inches, find the relation between AB and AQ. Ans. A Q is about j?^ of AB. 22. Construct a figure in which AQ = BP = ^ AB, and the arches AED and DFB are equilateral. Show that the radius of circle O in this figure is — . ® 32 23. li AQ = k-AB and DS = h ■ DB, find the radius of circle 0. . s ik-l Ans. - ■ 4 2k+ h 24. Show that the formula obtained in Ex. 23 becomes that ob- tained in Ex. 9 ii h = k. Remark.— Th& center of circle in Figs. 230, 232, and 234 may be found by the formula obtained in Ex. 23, as will be shown later. 25. Show that if the arch A CB is stilted by an amount equal to n, and ii AQ= k ■ AB and DS' = h ■ DB, the radius of circle may be found fi'om the equation, in which AB = s, ^(^ks-.y-[ks-|y=^j{hi+.y-{h|y-n. Suggestion. — For an illustration of stilted arches see Fig. 233 in which AH represents the n of Ex. 25. Is there any limit to the value of n in a figure of the general type suggested in Ex. 25 ? 256 GOTHIC TRACERY: POINTED FORMS 276. Figure 230 represents a special case under the pre- ceding, and may be solved by formulae obtained in § 275, Exs. 9 or 23. It is of common occurrence, and is more readily solved by methods suggested below. Fig. 230a. — From the Union Park Churcli, Chicago. In Fig. 230 D is the middle point of AB, A is the center for the arcs CB and ED, B for the arcs AC and DF, and D for the arcs A£ and BF. Circle O is tangent to the arcs as shown. EXERCISES 1. Prove that ED and DF, FB and CB, AE and AC are tangent in pairs, and that CD extended is the common chord of the two inter- secting circles of which A C and CB are arcs. 2. Prove that points A, E, C, also B, F, C are collinear. 3. If circle O is tangent to arcs A C and CB or to ED and FD, prove that its center must lie on the line CD. Suggestion. — Use the theorem, "The perpendicular bisector of the line of centers of two equal circles is a part of the locus of the centers of circles tangent to the two given circles." What two cases of this theorem occur in this figure ? 4. If circle is tangent to the two concentric arcs ED and CB, prove that its center must lie on a circle drawn with 4 as a center and .4 Q as a radius, if Q is the middle point of the line DB. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 257 5. If circle is tangent to the two concentric arcs FD and A C, where must its center lie ? 6. If circle is tangent to the arcs CB, AC, ED, and DF at the points Y, Z, W, and X, respectively, prove that the points Z, O, X, B, and also Y, 0, W, A are collinear. Suggestion. — Use the theorem, " If two circles are tangent, the line of centers passes through the point of tangenoy." 7. Construct circle 0. Suggestion. — The construction may be performed by any two of three loci. It is necessary to make OX = OW = OY = OZ, and to prove the circle tangent to the arcs A C, BC, ED, and FD. The sim- plest proof is obtained by constructing the arcs drawn with A and B as centers and with A Q and BP as radii, respectively. 8. Show that the radius of circle O may be obtained from the formula in § 275, Ex. 9. s 4 ifc — 1 Suggestion. — Let i = 1 in the formula — • . 9. Show that the radius of circle may be obtained from the formula in § 275, Ex. 23. s 4 it — 1 Suggestion. — Let i = A = 1, in the formula - • r-r • 10. If AB = 12, find the length of DO and the area of circle O. 11. If AB = s, find the length of DO and the area of circle 0. Ans. (1) |V5; (2) ^. 12. If the radius of circle is 6 ft., find (1) the span AB; (2) the altitude DC; (3) the altitude of the arch AED. Ans. (1) 24 ft.; (2) 12\/3; (3) 6V3. 13. Find the lengths of the segments mentioned in Ex. 12 if the radius of circle O is r. Ans. (1) 4 r; (2) 2 rVS ; (3) rVE. 14. If the altitude of the arch ^ EC is 4 ft., find (1) the span AB ; (2) the altitude CD ; (3) the radius of circle 0. 4ns. -VV3; (2)8; (3) |V3. 258 GOTHIC TRACERY: POINTED FORMS 15. If the altitude of the arch AED is r, find the lengths of the 3 segments mentioned in Ex. 14. .4ns. (1) -»V3 ; (2) 2r; (3) - VS, 16. TiAB= 12, find the areas of the following figures : (a) the curved A A CB. Ans. 12(4 ir - 3 V3). Suggestion. — The sector CBA (B as center of circle) is J of the area of the circle or 24 ir. The area of the equilateral AACB = d6 Va. The area of the segment bounded by the arc AC and the chord ^ C =24 5r-36 V3. The area of the curved A .4 CB is the area of the sector CAB (A as center) and the segment AC or 48 7r — 36\/3 = 12(4 7r-3V3). (6) The curved triangle AED. Ans. 3(4 t - 3 V3). 17. Show that the area of the curved AACB is equal to the area of a segment whose arc is 120°. 18. li AB = s, find the areas mentioned in Ex. 16. Ans. i^(4,r-3V3); (2) |!(4,r-3V3). 12 48' Fig. 231. Fig. 231a. — From Lincoln Cathedral, England. EXERCISES 277. 1. Construct Fig. 231. Suggestion. — This figure involves Fig. 230. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 259 2. If AB= s, find (1) the radii of circles O, 0', and 0", and (2) the altitudes of arches ASF, A ED, and ACB. Ans. (1) -,-,-; (2) ?- Va, - V3, - VS. ^ '^ 4 8 16^ '^ 8 4 2 3. If the radius of circle O is 5 ft., find (1) the radii of circles 0' and 0", and (2) the altitudes of arches ASF, A ED, and ACB. Ans. (1) 2i ft., li ft; (2) 2iv'3, 5V3, 10\/3. 4. If the altitude of arch A CB is 12 ft., find (1) the span AB and (2) the radii of circles 0, 0', and O". Am. (1) 8V3; (2) 2V3, Vs, ^V3. 278. Subtype. — Figures 232-234 show a subtype which may be explained by reference to Fig. 229, (see § 275, Ex. 23 and 25), or may give rise to the formulse obtained below. Fig. 232«. — From Amiens Cathedral. In Fig. 233 AQ = BP = | AB. P and Q are the centers for the arcs CB and CA, respectively. DA = DB. Semicircles are constructed on AD and DB. EXERCISES 1. Find the radius of circle tangent to the arcs AC, CB, AwD, 7 s and DxB, ii AB = s. Ans. —. 2. Construct circle 0. 3. li AQ = kAB = ks, find the radius of circle 0. , S 4:k-l Ans. - • 2 4 i; + 1 260 GOTHIC TRACERY: POINTED FORMS 4. Show that the radius of circle O may be derived from the formulae obtained in § 275, Ex. 23. 5. IiAQ = AB = s, find the radius of circle O. Ans. — - . Suggestion. — Substitute i = 1 in the result obtained in Ex. 3. 6. Draw a figure to illustrate the case k = 1. Suggestion. — See Fig. 282 h. Fig. 2326. 7. In the figure obtained in Ex. 6, what per cent of the total area above the line AB is occupied by the included circle and semicircles? Ans. 78% about. 8. In Fig. 232 if ^ Q = J ^B = J «, find the radius of circle and draw an illustrative figure. Ans. Suggestion. — See Fig. 234. 9. If ^ Q = ks, what value of k will make the radius of circle O equal the radius of one of the semicircles ? 6 Suggestion. — Solve the equation s(ik-l) _s 8k+ 2 4 10. Verify the result obtained in Ex. 9, by a figure. 11. Construct a figure in which' the arch A CB is stilted by J the span and AQ = ^ AB. Find the radius of circle in this figure. 12. Show that if the arch ACB is stilted by an amount equal to n, and ii AQ = ks, the radius of circle O may be obtained from equation, TANGENT CIRCLES: ALGEBRAIC ANALYSIS 261 Fig. 233. Congregation, Chicago, III. 279. In Fig. 233 HK is the diameter of the semicircle HCK. AB is par- allel to HK. HA and KB are straight lines perpendicular to HK and equal to i HK. The semicircles AED and DFB are constructed on i AB as diameter. EXERCISES 1. Find the radius of circle O tangent to the three semicircles HCK, AED, and DFB, if AB = s. Ans. — . 32 Suggestion. — Let the radius of the circle O be r. A ODS gives the equation (| + .^ = (|)% (^ _ .y. 2. Construct circle 0. 3. What per cent of the figure above the line AB is occupied by the included circle and semicircle ? Ans. about 69 %. 4. What loci would be needed to find the centers of the two small circles shown in the photograph ? Suggestion. — The locus needed is the locus of centers of circles tangent to each of two tangent circles. Its construction is beyond the province of elementary geometry. 5. If HA = ks, find the radius of circle 0. i S (1 + 2 i!)2 Ans. - • ' „ ^ , / 2 3 + 4i 262 GOTHIC TRACERY: POINTED FORMS Suggestion. — A CDS gives the equation {ic--)-T=(i-r-(i)' 6. Find the radius of circle O if HA is zero. ' Ans. ' Suggestion. — Substitute it = in the result obtained in Ex. 5. 7. Draw a figure to illustrate Ex. 6. See Fig. 234. 8. If HA = ks, what value of k will make the radius of circle equal to the radius of the semicircle AED'i Suggestion. Ans. k ■■ - Solve the equation, - • i — ^-^ — '- = '- • ^ 2 3 + 4/fc 4 V3 - 1 280. In Fig. 234 AD=DBandthe three semicircles ACB, AED, and DFB are drawn on AB, AD, and DB, respec- tively, as diameters. The circles O, O', and O" are tangent as indicated. Fig. 234. EXERCISES 1. Find the radius of circle tangent to each of the three semi- circles, if ^.B = s. Ans.-' 6 Suggestion. — Let r be the radius of circle O. A ODS gives the equation, (I -r)' +(!)'= (I + r)l 2. Show how to construct circle as indicated. 3. Erect ffR perpendicular to AD a,i, K the middle point of AD. Find the radius of circle O' tangent to semicircle AED at E and to semicircle ACB, it AB = s. Ans. — ■ 12 Suggestion. — Let the radius of circle C be r'. A CRD givef the equation(| + r')%(|)^=(|-.')^ TANGENT CIRCLES: ALGEBRAIC ANALYSIS 263 4. Prove that this circle will be tangent to circle 0. Suggestion. — O'R is parallel to OD. O'M = i^ = OD. Hence o ao is parallel and equal to RD. But r + r'= — + 1 = - = RD. ^ 12 6 4 Therefore the line of centers of circles O and O' equals the sum of the radii, and the circles are tangent. 5. Show how to construct circle 0'.^ Suggestion. — Since 00' is parallel to RD, perpendiculars erected at R to line A D and at to line CD intersect at 0'. 6. What fraction of the semicircle ACB is occupied by the in- cluded circles and semicircles ? Ans. |. 281. Type II. —Figures 196, 235-241, 245, 250-252, and 257 furnish illustrations of this type, the formula for which is obtained from Ex. 3 below. D NQ Fig. 235. Fig. 235a. — From Union Parl< Church, Chicago. In Fig. 235 AQ = PB = ^ AB. P and Q are centers for the arcs CB and AC, respectively. Arcs ED and FD are drawn with R and S as centers and AQ as radius. 1 Hanstein, PI. 19. 264 GOTHIC TBACEBT: POINTED FOBMS EXERCISES 1. Find the radius of circle inscribed in the curved quadrilateral CEDF, iiAB=s. Ans. ^■ 64 Suggestion. — Let r be the radius of circle 0. Solve for r in the —•(¥-')'-&T-(^'-)'-(¥)" 2. TiAB = s, find the lengths of OD and FN. Ans. (1) 320 V385; (2) g^VlS. 3. Jl AQ = kAB = ks, find the radius of circle O. . s /ik-l\ Ans. — I |. 16\ k J Suggestion. — A ODQ and ODS give the equation (ks - r)2 - f its - 1 y = (ks + ry - kh^ 5 s 4. If AQ= ^ AB ='—, find the radius of circle by substi- tuting fc = I in the result obtained in Ex. 8. Ans. —. 40 Draw a figure to illustrate this case. Let AB = 5 in. 6. If ^Q = AB = s, that is, if ^ = 1, find the radius of circle 0, and draw an illustrative figure. Ans. —■ 16 6. If AQ = ^ AB =-, find the radius of circle O, and draw an illustrative figure. (See Fig. 240.) Ans. -• 8 7. Give construction for circle 0". Suggestion. — See § 242, Ex. 1. 8. nAQ = ^AB = is, find the radius of circle 0". Ans. — . 128 Suggestion. — Join 0" with the center for the arc EA . Use A 0" QM. 9. If ^ Q = kAB = ks, find the radius of circle 0"- Ans.±.^Ji^. 32 k TANGENT CIRCLES: ALGEBRAIC ANALYSIS 265 10. Construct the entire figure for the case fc = 1. Suggestion. — By substituting in the necessary formulae, the radii of circles and 0" are — and — , respectively. 16 32 ^ ^ 11. a AQ = kAB, find the ratio between the radii of circles 8fc-2 and 0" Ans. • \k-l 12. Show that if AQ = kAB, it is impossible to find a value for k that will make circle equal to circle 0". P A B Q EXERCISES 282. 1. Give the construction for Fig. 236. The radius for the arcs CA and CB is J AB. Suggestion. — For the construction of circles 0' and O", see § 273, Ex. 6. 2. Find the radius of circles and 0', if the radius for the arcs ^Cand CB is kAB = ks. Ans. (2) J ik-1 4 4kk + l 3. Show that the result obtained in Ex. 2 may be derived from that in § 273, Ex. 6. 4. Construct the entire figure if the arch ACB is equilateral. Let AB = 5 in. 266 GOTHIC TRACERY- POINTED FORMS Fig. 237a.— From Cath- olic Church, Evans- vitle, Indiana. 283. In Fig. 237 AQ = BP = | AB. P and Q are the centers for the arcs CB and AC, respectively. Arcs ED and DF are drawn with AQ as radius and R and S as centers. Arches AHM, MKD, etc. are equilateral. AM = MD = DN = NB. EXERCISES 1. Find the radius of circle 0" tangent to the arcs EA, ED, HM, 17 s a,udKM,MAB = s. 2. li AQ = kAB = ks, find the radius of circle 0". Ans. 104 s 8 jfc - 1. S ik + 1 3. Show that the formula obtained in Ex. 2 may be derived from that in § 275, Ex. 23. 4. Construct the entire figure ii AQ = AB= 5 in. Suggestion. — For the radius of circle see § 281, Ex. 3. 5. li AQ = kAB, for what value of k will the radius of circle equal the radius of circle O'? Ans. k = \. Suqqeslion. — Solve the equation = . ■^^ ^ IQk 8(4^ + 1) 6. Make a drawing to illustrate the case k = \. 7. Show how to construct the quadrifoils shown in the circles in Fig. 237 a. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 267 C /\( '^ 1/ 284. In Fig. 238, arch ACB is equi- lateral. Arcs ED and FD are drawn with AB as radius, and centers on AB extended. Arches AGK, EHD, etc., are equilateral. Fig. 238. — From Choir, Tintern Abbey, Eng- land. Bond, p. 475. EXERCISES 1. If AB = s, find the radius of circle and circle 0'. Ans. (1) ^; (2) ^. ^ ^ 16 ^ ^ 40 2. Construct the entire figure. 3. Show that the radius of circle 0' may be derived from the for- mula obtained in § 275, Ex. 23. D S Fig. 239. M 268 GOTHIC TRACERY: POINTED FOBMS 285. IiiFig.239AQ=BP=JAB. P and Q are the centers for the arcs BC and AC, respectively. AR=RD= DS=SB. Arcs GH, ED, KS, etc., are drawn with AQ as radius, and points on line PM as centers. M is the cen- ter for the arc FD. Fig. 2390. — From Wells Cathedral Chapter House. EXERCISES 1. Find the radius of circle inscribed in the quadrUateral CEDF. Ans. Suggestion. — See § 281, Ex. 3. 2. Ji AQ = ^ AB = ^s, find the radius of circle O'. Ans. 3. H AQ = ks, find the radius of circle 0' where AB = s. , s Sk- Ans. — ; • 5 80 ' 64 k 4. If 4 Q = I s, find the radius of circle O' from the formula ob- tained in § 281, Ex. 3. Suggestion.— From § 281, Ex. 3, r = — • iA^ where k is the ^^ 16 k relation between DM and DB. Hence k = -, DB = '-. Make the nec- 2 2 essary substitutions. 5. Show that the formula obtained in Ex. 3 may be derived from that in § 281, Ex. 3. Suggestion. — Sizice ~ =^^^ change k ix> 2 k. Since DB 1 s = - AB, change « to - in formula obtained in § 281, Ex. 3. 6. li AQ = iAB = ^s, find the radius of circle and of circle O' and draw an illustrative figure. Let AB = 8 in. TANGENT CIRCLES : ALGEBRAIC ANALYSIS 269 7. li AQ= lAB = is, find the radius of circle and of circle 0' 5s 48' 8. It AQ = AB = s, find the radius of circle and of circle 0' and draw an illustrative figure. AB = Q in. Ans. (1) (2) and draw an illustrative figure. Ans. (1) — ; (2) — . ^ ^ 16' ^ -^ 64 9. If ^Q = IcAB = ks, for what value of k will circles O and 0' have the same radius ? Suggestion. Ans. ^ = f . Solve the equation 'I^A^zll = s(ik-l) _ ^ 64 /fc 16 i 10. Draw a figure to illus- trate the case i = |. (See Fig. 239 6.) 11. If ^Q = ».4Band the radius of circle 0' is 18, find AB, and the altitudes CD and FS. Ans. (1) 172f; (2) ; (3) ?|^V5. 12. liAQ = -^AB and the radius of circle O is 2, find span AB and the altitudes Ci3 and J^S. Ans. (1) llf; (2) 4{v'2i; (3) fjVsi. 286. In Fig. 240 AB is the diameter of the semicircle ACB. A and B are the centers and i AB the radius for the arcs DE and DF. Circle O is tangent to the arcs as shown. A D B Fig. 240.— From Great Waltham Church, Essex, Eng. Brandon (I), Sect. II, Woodwork, PI. 2. EXERCISES 1. Prove that the points C, 0, and D are coUinear. 2. Prove that the arcs ED and FD are tangent at point D. 3. If circle is tangent to the arc DF at point x, prove that the points O, X, and B are collinear. 4. If is the center of the circle, prove that OD is perpendicular io AB. 270 GOTHIC TRACERY: POINTED FORMS 5. If AB = s, find the radius of circle O. Ans. -. Suggestion. — Let the radius of circle O be r. A ODB gives the equation, (|y- + (I -,y=(|+r)'- 6. Show how to construct the entire figure. 7. Show that Fig. 240 is a special case of Fig. 235. 8. If AB = s, find the area of the curved figures : (a) triangle A ED; (b) triangle EFD. Ans. (a) g(4x-3V3); (6) g(3V3-,r). nr Fig. 241a. — Ospedale Maggiore, Milan. 287. In Fig. 241 a series of semicircles is inscribed between the two parallel lines AB and CD as shown, and circle O is tangent to arcs CE and EQ. EXERCISES 1. Prove that the center of circle O must lie on a perpendicular to AB at E. Suggestion. — Use the theorem : "The common tangent to two equal tangent circles is a part of the locus of centers of circles tangent to the two given circles." 2. If circle O and arc EQ are tangent at Q, prove that the points H, Qa,nd O are coUinear. 3. Find the radius of circle tangent to the arcs CE and EQ and to the line CD, \i EP = s. Ans. -■ 4 Suggestion. — In A EHO, OH^ = OE^ + EH'^. Let r be the radius of circle 0, and s = EH = EP. Since OH = s -^ r and OE = s — r, the equation becomes (s + ry = (s — r)^ + s^. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 271 4. Show how to construct the entire figure. 5. If EP = s, find the area of the curved figure of which PCE is half. Ans. I (4 - ^). 6. Can the answer to Ex. 3 be obtained from that of § 286, Ex. 5 ? TYPES OF THREE-LIGHT WINDOWS Fig. 242a. — From Lincoln Cathedral, England. 288. Type I. — in Fig. 242 AQ = I AB. AQ is the radius for the arcs CB and CA. Arches AER and SFB are equilateral and AR = RS = SB. EXERCISES 1. Find the radius of circle 0' tangent to the arcs CA and ER and 5s the line CD. A ns. 24 Suggestion. — Dvaw O'Q and O'A. In ^AO'Q, 0'A^-AM^ = O'Q^- MQ'or MD = r? 2. Construct circles 0" and 0'. Prove that tliey are tangent to each other. 272 GOTHIC TBACEBT: POINTED FORMS 3. M AB = s oaA. AQ = —, find the length of O'M. Am. :j^V30. 4. Construct arcs GR and GS tangent to circles O' and O'', respec- tively, and to lines Rx and Sy at iJ and S respectively. Suggestion. — This involves the problem, " To construct a circle tan- gent to a given line at a given point and to a given circle." See § 246. Point N is the center for the arc RG. 5. Find the radius of circle tangent to arcs CA and CB and to circles O" and 0', TiAB = s. Ans. .208 s. Suggestion. — OD = KD + KG, KG =-y|/5i+ :c]^- (^']\ DG = ^OCt-DCt- Hence get the equation, 6. What is the per cent of error if the radius of circle is said to be equal to the radius of circle O" 1 7. H AQ= kAB = ks, find the radius of circle O'. Ans. i . 9illi. 6 6/fc+l Suggestion. — Get the equation, 8. If ^ Q = kAB, for what value of k will the radius of circle 0' be -g- 7 Ans.k = l Suggestion. — Solve the equation - . = £.. 6 6 i + 1 6 9. liAQ = kAB = ks, find the radius of circle O. Suggestion. — This involves the equation, where p is the radius of circle 0' or - • . 6 6^ + 1 TANGENT CIRCLES: ALQEBBAIC ANALYSIS 273 10. If AQ = kAB-- ks and the radius of the arcs AE, ER, SF, and FB is — , find the radius of circle 0'. o Am. ^ (9 - k). Suggestion. — Get the equation 11. If AQ = kAB = ks, and the radius of the arcs AE, ER, SF, and FB is — , find the radius of circle 0. 289. Type II. Figures 243-247 and 260 furnish illus- trations of this type, the formula for which is obtained in Ex. 4 below. c In Fig. 243 AQ = BP = f AB, P and Q are the centers for the arcs CB and AC, respectiTely. AR = RS = SB. RN = BN'=|RB. N and S' are centers for the arcs RF and BF. Arch AES is similarly drawn on AS. EXERCISES 1. Find the radius of circle O tangent to the arcs CA, CB, ES, and RF, if AB = s. Ans. ^■ 150 Suggestion. — ii OQD and OND give the equation, {¥-')'-(l)"=(¥-)'-(!i)" 2. If AQ = kAB = Ics and RN = SM = f fc, find the radius of circle O. Ans. —[ — ; — )• 30V k I Suggestion. — Solve for r in the equation, 274 GOTHIC TRAGERT: POINTED FORMS 3. If it = 1, find the radius of circle O by substituting in the result obtained in Ex. 2. Am. -■ Note. — This is Fig. 244. 4. If ^Q =ifcs and RN = h • — find the radius of circle O. ^ . s (iik-2h-2\ Ans. - •[ )■ 2 V 9;i: + 6A / 5. Show how the formula in Ex. 2 may be obtained from that in Ex.4. 290. Figure 244 is a special case under Fig. 243. It is readily solved by the special method indicated below, or by the formula in § 289, Ex. 2. In Fig. 244 the arch ACB is equilateral. AR = RS = SB. Arches AES and RFB are equilateral on AS and RE as spans. Fig. 244. — From Church in Hert- fordshire, England. Brandon (I), p. 28. EXERCISES 1. If D is the center of AB, prove that the points C, K, and D are colliiiear. Suggestion. — Use the theorem, " If two equal circles intersect, the common chord is the perpendicular bisector of the line of centers.'' 2. If circle O is tangent to the arcs A C and BC or to the arcs ES and RF, prove that its center must lie on line CD. 3. If circle O is tangent to the concentric arcs A C and FR, where must its center lie ? 4. If circle O is tangent to the concentric arcs BC and ES, where must its center lie ? TANGENT CIRCLES: ALGEBRAIC ANALYSIS 275 AB 5. Show that the radius of circle is -— , and construct circle O. 6 Suggestion. — It is necessary to prove that OX = OY = OZ = OW, and that circle is tangent to the four arcs A C, BC, ES, and FR. 6. Show that the radius of circle may be obtained from the formula found in § 289, Ex. 2. 7. Ji AB = s, find the length of the line OD, the altitude of the arch AES, and the length of KB. Ans. (1) ?1; (2) |V3i (3) ^V7. 291. Figure 245 shows a six-light window which is a combination of the three-light type shown in Fig. 243, and the two-light type shown in Fig. 235. Fig. 245a. — From the West Front, Lichfield Cathedral, England. In Fig. 245 AQ is the radius of AC and BC. AQ = | AB. AR = RS = SB. PR is the radius of EA, ES, FR, and FB. PR = | RB. All of the smaller arcs are drawn with PR as radius. EXERCISES 1. Find the radius of circle 0'. 34 2. Show that the answer to Ex. 1 may be derived from the for- mula in § 289, Ex. 4. 276 GOTHIC TRACERY: POINTED FORMS 5 s 3. Find the radius of circle 0"- Ans. — 4. Show that the answer to Ex. 3 may be derived from the formula in § 281, Ex. 3. 5. Find the radius of circle 0'". Ans. —■ 14 6. Show that the answer to Ex. 5 may be derived from the for- mula in § 281, Ex. 3. 292. Figure 246 shows a sub-type which may be ex- plained by reference to V'lg. 243 (see § 289, Ex. 4), or may give rise to the formuUe obtained below, c Fig. 246a. — From Masonic Hall, 91st St., Chicago, III. In Fig. 246 AQ = BP == | AB. P and Q are the centers for the arcs CB and AC, respectively. AR = RS = SB. Semicircles AWS and RXB are drawn on AS and RB as diameters. EXERCISES 1. Find the radius of circle O tangent to the arcs A C, CB, WS, and RX, if AB = ». Ans. -. 2. Show how to construct the entu-e figure. 3. If -4 Q = kAB = ks, find the radius of circle 0. . s Sk -1 Ans. - ■ . 2 Sk+l 4. Show that the formula obtained in Ex. 3 may be derived from that in § 289, Ex. 4. 5. If ^Q = AB, find the radius of circle O. Ans. -. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 277 Suggestion. — Substitute fc = 1 in the result obtained in Ex. 3. 6. Draw a figure to illustrate Ex. 5. (See Fig. 246 a.) 7. In the figure obtained in Ex. 6, what per cent of the entire area is not occupied by the in- cluded circle and semicircle ? Ans. about 22%. 8. Ill Fig. 2i6, a AQ = kAB = ks, what value of k will make the radius of circle equal the radius of the semicircle RXB V Draw an illustrative figure. Suggestion Fig. 2461). Ans. k : To obtain the value of k, solve the equation, s 3k -1 ^s 3" 2 3yi; + l 9. What value of k will make the diameter of circle equal to the line iJS? Ans. k = i. 10. If AQ = i AB, find the radius of circle and draw an illustrar tive figure. ^^_ ± 10 11. Find the radius of circle ii AQ = kAB and AS = mAB, where ^B = s. Ans. ^ ( „ ~ "' )• 2 \2k + ml 12. Draw a figure to illustrate the case k = \,m= \, and find the radius of circle for this case. Ans. ■ 14 13. Show that the formula obtained in Ex. 3 may be derived from that in Ex. 11. 14. Can the formula obtained in Ex. 11 be derived from that in §289, Ex. 4? 15. Show how to construct the trefoil shown in the circle in Fig. 246a. 278 GOTHIC TRACERY: POINTED FORMS 293. Figure 247 shows a window which is a combination of that suggested in § 292, Ex. 12, with the one-light type shown in Fig. 228 and with the two-light type shown in Fig. 235. D E A F G Fig. 247. Fig. 247a. — From Or Cjan Michela, Florencoi Italy, In Fig. 247 CNB is a semicircle on CB as diameter. CD — DA = AG = GB. Semicircles are constructed on the line segments CG, DB, CD, DG, and GB as diameters. The circles are tangent to the arcs as shown. Arcs DM and L6 are drawn with DF as radius and centers on the line KH. EXERCISES I 1. Find the radii of circles and O'". Show that they may be derived from the formula obtained in § 273, Ex. 6. Arts. (1) — ; (2) —. ^ -' 20^ ^ 16 2. Find the radius of circle 0". Show that it may be derived from the formula in § 292, Ex. 11. An. -. 3. Find the radius of circle 0' . Show that it may be derived irom the formula in § 281, Ex. 3. Ans. 12 4. Construct the entire figure, using AB = 7 in. 5. What per cent of the total figure is occupied by the six circles ? 6. Construct a figure similar to Fig. 247, using instead of the semicircles on CB, CG, and DB pointed arches with the radius for the arcs | of the span. TANGENT CIRCLES: ALGEBMAIC ANALYSIS 279 294. Type III. Figures 248-252 and 257-261 furnish illustrations of this type. The formula to which they may be referred is obtained in Ex. 3 below. -M L__P__i_-_L._.I..j^...l._..__.::i-.#_ A R D S B Fig. 248. — From Merton College Chapel, Oxford, Eng. A. and A. W. Pugin, Vol. I, PI. 5. In Fig. 248 AQ = BP = | AB. P and Q are the centers for the arcs CB and AC, respectively. AR=SS=SB. Arcs ER and FS are drawn with M and N as centers and AQ as radius. EXERCISES 1. Find the radius of circle O tangent to the arcs AC, CB, ER, and F5, if 4B = s. Ans.^. 15 2. Construct the entire figure, using AB = 5 in. 3. If AQ,= kAB = ks and AR = mAB — ms, find the radius of . 1 ^ A s (1 k — m) (1 — m) circle O. Ans. - ■ ^ ^ ' . 4 k Suggestion. — The equation is (ifcs _ ry -1^2 yfc - 1)2 = (ks -\- r)" - ~(2 k + 1 - 2 m). 4. Let ifc = 1 and m = \. Find the radius of the circle by substitut- ing in the formula obtained in Ex. 3. Draw an illustrative figure. A 5s Ans. — . 18 5. If ik = 1 and m = J, find the radius of the circle by substituting in the formula obtained in Ex. 3. Draw an illustrative figure. Compare with Fig. 235. Compare the result obtained with that in § 281, Ex. 5. 280 GOTHIC TEACEBY: POINTED FORMS C A 1 \ / / /\ / ■ 1 /o B , V G K 1 1 Fig. 249. — From Hyde Park Presbyterian Church, Chicago, III. 295. Figure 249 is identical with Fig. 248 with the addition of the arch GHK. The arcs GH and HK are constructed with AQ as radius, tangent to lines RG and SE, respectively, and passing through point H. EXERCISES 1. Show how to construct the arcs GH and HK. Suggestion. — This involves the construction of a circle by intersecting loci. What loci are required? 2. Show that in general two circles can be constructed to fit the given conditions. (See Fig-. 249a.) 3. Are the arches GHK and SFB congruent if GK is made to coincide with SB ? 4. Construct a figure similar to Fig. 249 with the arch ACB equilateral. ^ 5. Construct a figure similar to Fig. 249, with AQ= ^ AB. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 281 296. Figure 250 is a special case under the types shown in Figs. 248 and 235. c X MAPRDSQBN Y Fig. 250.— From Durham Cathedral. E. S. Prior, p. i95. In Fig. 250 P and Q are the centers for the arcs CB and CA, respec- tively. AQ = BP = I AB. AR = RS = SB. The arcs KR, GS, HR, and FS are drawn with radius AQ and centers on line AB. EXERCISES 1. Find radius of circle inscribed in quadrilateral CGKH, if AB = s. Ans.—. 10 2. Verify the result obtained in Ex. 1 by substituting k = |, m = I in the result shown in § 294, Ex. 3. 3. Show how to construct circle 0. 4. If ^ Q = BP — kAB = ks, find the radius of circle 0. Ans. J- . M^l. 18 k 5. Verify the result shown in Ex. 4 by substituting m = | in the result obtained in § 294, Ex. 3. 6. IiAQ = BP = iAB = ^s, find the radius of circle 0" Ans. — 15 7. Construct the entire figure when AQ = ^AB. Let AB = 5 in. 8. If ^ Q = kAB = ks, find the radius of circle 0". i s Qk-\ Ans. — • 36 k 282 GOTHIC TRACERY: POINTED FORMS 9. Show that the formula found in Ex. 8 can be obtained from that in § 281, Ex. 3. Suggestion. — From § 281, Ex. 3, r = ii^ ■ ^^'~^ , where k' = — . Since -1^ = 5 ^, k'=-lc, AS=- s. Make the necessary substi- AS 2 AB' 2 ' 3 ■' tutions. 10. Show that the formula found in Ex. 8 can be obtained from that in § 294, Ex. 3. Suggestion. — From § 294, Ex. 3,r = ^ (2 k' - m'){l - m') ^^^^^ 4 k' AT} Y7? m' and k' are the ratios -— and '——, respectively. First substitute A!^ AS m' = I, and reduce to ^ • i^lnJi 16 k' 11. If AQ = AB, find the radius of circle and of circle 0' by substituting k = lia the results obtained in Exs. 4 and 8. Ans. (1) - ; (2) — . ^ ^ 9 ' *■ '^ 36 12. Construct the entire figure when k = 1. 13. What value of k will make the radius of circle O one half the radius of circle 0' ? Ans. i = i. Su^^es^ion. — Solve the equation -i • f^AziI] ^2s/ 3fc-i y A B Fig. 251. — From R. C. Church, Oyster Bay, N. Y. See Arch. Reir., Vol. V, p. 76. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 283 EXERCISES 297. 1. If the arch ACB is equilateral, construct the diagram shown in Fig. 251. Suggestion. — To find the center of circle 0', use the problem, "To construct a circle tangent to a given circle and to a given line at a given point." See § 246. 2. Show that the radius of circle may be derived from the formula in § 281, Ex. 3. 298. Figure 252 shows a four-light window which is a combination of the types discussed in §§ 281 and 294. In Fig. 252 the arch ACB is equilat- eral. AR = RD = DS = SB. RG, eB, ESi etc., are drawn with radius AB and centers on AB extended. EXERCISES 1. Find the radii of circles 0" and 0"' if AB = s, and show that these may be obtained from the formula found in § 294, Ex. 3. Ans. (1) ^; (2) — . ^ '^ 32 ^ '^ 64 2. Find the radius of circle 0' and show that it may be obtained from the formula found in § 281, Ex. 3. 3. If AB = s, find the altitude of the arches RLB, DFB, and SHB. 4. What per cent of the entire figure is occupied by the six circles ? 5. Construct the entire figure, using ^B = 8 in. 284 GOTHIC TBACEBT: POINTED FORMS TYPES OF FIVE-LIGHT WINDOWS 299. Type I. Figures 253-256 furnish illustrations of this type, the formula for which is obtained in Ex. 5 below. For other illustrations see Fig. 137a and the note under Ex. 4, § 300. In Fig. 253 AQ = BP = fAB. P and Q are the centers for the arcs CB and AC, respectively. AM = NB = |AB. SN = RB=6BN. R and S are the centers for the arcs FB and FN, respectively. Arch AEM is similarly drawn. A P M D N Q B Fig. 253, — From Exeter Cathedral Aisle and Choir. Sharpe (2). EXERCISES 1. Find the radius of circle O tangent to the arcs AC, CB, EM, and FN. if AB = s. Ans. 2. If AQ= kAB and SN = kNB, find the radius of circle O, if 3s Qk-2 AB = s and NB = 1$. Ans. 70 3. Prove A ACB and AEM similar and the points A, E, and C collinear. Suggestion. — See § 275, Exs. 16-19. 4. Ji AQ= AB, find the radius of circle 0. Ans. Ss 10 ■ 2s 5. If AQ = kAB = ks and SN = hNB = A • — , find the radius of circle 0. 10 V 5 i + 2 A / TANGENT CIRCLES: ALGEBRAIC ANALYSIS 285 300. InFig.254AQ = BP = f AB. P and Q are the centers lot the arcs CB and AC, respectively. Arches A£M and NFB are equilateral. AM = NB = |AB. M D N Q Fig. 254. EXERCISES 1. Find the radius of circle tangent to the arcs A C, CB, EM, and FN, if AB = s. 2. liAQ = kAB: Ans. 61s 220' and BN = | AB, find the radius of circle 0. Ans. _s_ 25^-4 10 ' ok+2' 3. Show that the results obtained in Ex. 1 and 2 may be derived from the formula in § 299, Ex. 5. 4. liAQ^iAB and BN = i AB, find the radius of circle O. Note. — The First Congregational Church, Portland, Oregon, con- tains a window of the proportions given in Ex. 4. The muUions in the arches AEM and NFB are of the type shown in Fig. 235. 5. Make a diagram for the window in the church referred to in Ex. 4, note. 6. Show how to construct the small arch erected on the span MN, if the arcs that form it are tangent to the arcs EM and FN at M and N respectively, and if its apex just touches the circle 0. 7. What locus is needed iu the construction called for in Ex. 6 ? 8. Find the radius of circle if A; = 1 and make an illustrative drawing. . 286 GOTHIC THACEST: POINTED FORMS C Fig. 255a. — From Union Congre- gational Church, Boston, Mass. 301. In Fig. 255 D is the middle point of AB. AH = BK^§ofAB. A is the center for the arcs EH and CB, B for the arcs EF and CA, H and K for the arcs £A and FB, respectively. EXERCISES 1. Prove that arcs CA and AE are tangent; also FB and CB. 2. If circle is tangent to the arcs CA and CB or to arcs EH and FK, prove that its center must lie on the line CD. 3. If circle O is tangent to the concentric arcs EH and CB, where must its center lie V 4. If circle O is tangent to the concentric arcs CA and FK, where must its center lie ? 5. Show how to construct circle O. Suggestion. — It is necessary to prove that Ox = Oy = Oz = Ow, and that circle O is tangent to the four arcs AC, CB, EH, and KF. 6. Show that the radius of circle O may be derived from the formula obtained in § 299, Ex. 5. 7. Prove that the points j4 , to, 0, and y, also z, O, z, and £ are coUinear. 8. Construct the arc GK, passing through the points G and K with its center on line AB. 9. Show that the construction required for Ex. 8 is equivalent to constructing a circle passing through a given point and tangent to a given line at a given point. TANGENT CIBCLE8 : ALGEBRAIC ANALYSIS 287 10. Show how to construct the complete figure. 11. If AB = 15, find (1) the radius of circle 0; (2) the lengths of OD and GD; (3) the area of curved triangle A EH. 12. li AB = s, find the lengths and areas mentioned in Ex. 11. ^ns. (1) J-'; (2) -V(Jand^(2VU-3); (3) i^(4x-8V3). 13. If R is the center for the arc GK, show that RD = 2KD Tile ^K Suggestion. — A RSK and GDK are similar, therefore = — — or RK = \^„ ■ S is the mid-point of line GK. RD = RK- DK. DK 14. If ^ B = s, find DR. ^ns. 1(8-3 V6). 15. If AH= KB = kAB = Tcs, find (1) the radius of circle 0; (2) the length of OD, DG, and DR ; (3) the area of curved tri- angle A EH. Ans. (1) ,^ (1 - k) ; (2) i VWT2l, ^ (i- - 1 + VF+Tfc), Fig. i:56a. — Lincoln Calhedrai. R D s Fig. 256. 302. InFig. 256 AQ = BP = f AB. P and Q are the centers for the arcs CB and AB, respectively. AB is divided into five equal parts. Arcs EA, ER, FS, and FB are drawn from S, P, Q, and R as centers and radius | AB. 288 GOTHIC TBACEBT: POINTED FOBMS EXERCISES 1. Find radius of circle tangent to arcs A C, FS, CB, and ER. Ans. 10 Suggestion. — Notice that the arcs A C and FS, also CB and ER are concentric. 2. Show that the result found in Ex. 1 may be derived from the formula in § 299, Ex. 5. 3. Find the radius of circle 0' tangent to the arcs jFjS and FB, and to the arcs HT and TK, if the arches SHT and l^RB are equilateral. Ans. — 8 4. Show how the answer to Ex. 3 may be derived from the formula obtained in § 275, Ex. 23. 5. If the arches BKT, THS, etc., in Fig. 256 had been semi- circles, what would have been the radius of circle 0', if AB = si A S Ans. _■ 7 6. Derive the answer to Ex. 5 from the results obtained in § 278, Ex.3. Suooesfton. — From §278, Ex.3, r=i2£ • — -; in Fig. 256 SB „ „ 2 4i + l 2 s 3 SB = — ; k = —- — Make the necessary substitutions. 7. Draw a diagram to illustrate the case suggested in Ex. 5. 303. Type II. Figures 257 and 258 are illustrations of Type II. They may be referred to Type III of three- light windows (§ 294) as indicated in Ex. 2 below. A M P D Q N B Fig. 257. — From Chester Cathedral Presbytery. Bond, p. 524. TANGENT CIBCLES : ALGEBRAIC ANALYSIS 289 In Fig. 257 AQ = BP = I AB. P and Q are the centers for the arcs CB and AC, respectively. AB is divided into 5 equal parts. All arcs shown except zP and zQ are drawn with AQ as radius and centers on Une AB or AB eztended. EXERCISES 1. Find radius of circles and 0' placed as shown in the figure, itAB = s. Ans.(l) |; (2) A. 2. Show that the radius of circle can be obtained from the formula in § 294, Ex. 3, and the radius of circle 0' from the formula in § 281, Ex. 3. 3. Construct the entire figure. Suggestion. — Arcs xP and xQ are constructed to pass through points X and P, and x and Q, respectively, and to have centers on the line AB. 4. If AB = s, find the length of OD and Dx. Ans.(l) ^V13; (2) ^(^15-2). 5. li AB = s, find the distance RD, when -B is the center for the a,rcxQ. Ans. j- {9 ~2Vl5). Suggestion. — As in § 301, Ex. 13, RD = ^^ ~ ^^ • 6. If ^ Q = kAB = ks, find the radius of circle 0'. . s 10 yfc - 1 Ans. • 100 k 7. If AQ, = AB = s, find the radii of circles and 0', and con- 6 s 9 s struct the entire figure. Ans. {X) — ; (2) — — ■ Suggestion. — For the radius of circle 0, the formula obtained in § 294, Ex. 3, may be used. 8. Ji AQ = kAB = ks, find the radii of circle 0. . 3 s /5k~l\ A ns. — • I 50 \ k / 9. Construct a figure to illustrate the case k = I. 290 GOTEIC TRACERY: POINTED FORMS C P D Q Fig. 258, Fig. 258a. — From Medinah Temple, Chicago. 304. Figure 258 is like Fig. 257 except that arches AGM, MHP, QKN, NLB are equilateral. EXERCISES 1. Find the radius of circle 0' ii AQ = BP = ^AB = ^s. Ans. -. • 8 2. Construct the entire figure. Suggestion. — The radius of circle can be derived from § 294, Ex.3. 3. H AQ = l-AB = ks, find the radius of circle 0'. Ans. s 10 fc - 1 5 10 k + 2 4. Show that the formula obtained in Ex. 3 may be derived from that in § 275, Ex. 23. 5. If .4 Q = I AB = is, find the radius of circle and 0', and draw an illustrative figure. Ans. (1)—; (2) — . 6. If the arches A GM, MHP, etc., are semicircles, and ii AQ = AB 7 ■ ^AB, what is the radius of circle 0'? Ans 7. Draw a figure to illustrate Ex. 6, using .4B = 7 in. 8. If the arches A GM, MHP, etc., are semicircles, and if ^ Q = o lOifc — 1 kAB = ks, find the radius of the circle 0'. Ans. '- ■ 5 lOk + 1 9. Show that the results obtained in Exs. 6 and 8 may be found from § 278, Ex. 3. TANGENT CIRCLES: ALGEBRAIC ANALYSIS 291 SEVEN-LIGHT WINDOWS C Fig. 259. 305. In Fig. 259 AQ = BP = f AB. P and Q are the centers for the arcs CB and AC. AB is divided into ^ equal parts. The arcs shown passing through the various points of division have radii equal to AQ and centers on AB or AB extended. • EXERCISES 1. Find the radius of circle tangent to the arcs A C, CB, ER, and FS, ii AB = s. Ans. — . 14 Fig. 259a. — From Merton College Chapel, Oxford. 292 GOTHIC TRACEBT: POINTED FORMS 2. Construct the entire figure. 3. Verify the result derived in Ex. 1 by substituting ^ = ?, m = ?, in the result obtained in § 294, Ex. 3. 4. If AB = s, find the length of XD and the altitude of the arch A GP. Ans. (1) ^ (2 v/n - 3) ; (2) ^ V23. 5. 'liAQ = kAB = ks, find the radius of circle O. Ans. s lik-3 49 k 6. Verify the above result by substituting m = ^ in the formula obtained in § 294, Ex. 3. C A- — \0' -V-r'~7T— B 306. In Fig. 260 AQ = BP = ^ AB. P and Q are the centers for arcs CB and CA respectively. AB is divided into 7 equal parts. Arcs ER and FS are drawn with radius AQ and center on line AB extended. Arches NLB, SKQ, etc., are equilateral on the lines NB, SQ, etc. respectively. Fig. 260. — From Fxeter Cathedral, Lady Chapel. See Bond, p. 473. EXERCISES 1. Find the radius of circle 0' tangent to arcs FS, FB, KQ, and K A ._j ^ 1 1 ._L_ I A 1 p R S"" n'" Q~~ LN, if AB = s. Ans. 8 2. Construct the entire figure, using AB = 7 in. Suggestion. — The radius of circle can be found from formula given in § 294, Ex. 3. Arcs XS and XR are constructed to pass through the points A' and R, and S and R, respectively, with centers on line AB. 3. If ^Q=fc^.B = i«, find the radius of circle 0'. Ans. — • ^• 14 7 k + 2 4. Show that the answer to Ex. 3 may be derived from the formula obtained in § 289, Ex. 4. VENETIAN TRACERY 293 Fig. 261. 307. In Fig. 261 the arch ACB is equilateral. AB is divided into 7 equal parts, and the arcs drawn as shown. EXERCISES 1. Find the radius of circle and of circle 0', if AB = s. Ans. 15 s 6 s 98 49 2. Construct the entire figure. 3. Show that the radius of circle O may be found from the formula given in § 294, Ex. 3. 4. Show that the radius of circle 0' may be found from the formula given in § 294, Ex. 3. 5. Show that the radius of circle 0' may be found from the formula given in § 281, Ex. 3. 6. If the radius for the arcs CA and CB is kAB, find the radii of circles O and 0'. PART 4. VENETIAN TRACERY 308. Venetian tracery is peculiar in that it is intended for indefinite horizontal extension, and not designed to fit a definite closed space. ^ For a modern example of the same see the Montauk Club, Brooklyn, N. Y.^ ^ Brochure series, January, 1895, p. 7. 2 Arch. Record, Vol. II, p. 136. 294 GOTHIC TRACERY: POINTED FORMS .-^.- +0- — \N n-\ ^ M /\ r\ G F K Fig. 262. 309. In Fig. 262 circles L, M, TH, etc., are tangent to the line AB, to each other, and to the semicircles drawn with C, G, F, K, etc., as centers. EXERCISES 1. Prove that AX = XC and that AL = LX, if AB and CD are parallel and E, H, etc., are midpoints of the arcs CEF, GHK, etc. 2. Given the parallel lines AB and CD. Construct the figure. 3. If ^ C = 6, find the area of the following curved figures : (a)AGYF. Am. 6.52+. (b) Triangle YFHZ. An.t. 3.08+. (c) Quadrilateral EOHY. Arts. 1.35+. 4. If AC = h, find the areas mentioned in Ex. 3. Ans. (a) g(4 7r-3%/3); (J) |!(3V3_^); (c) — (36 — 7 T — 6v3). Fig. 262o. — From Franchetti Palace, 9o Venice, Italy, VENETIAN TRACERY 295 •^'.'i, I \ Fig. 263. 310. In Fig. 363 AB is parallel to CD, the equal circles M and N are tan- gent to each other at H and to the line AB, and the semicircle with Q as its center is tangent to circle N at V and to circle M at U. EXERCISES PH 1. If line Q,H iS extended to meet AB at P, find the ratio — —. Suggestion. — Prove A iViVf Q equilateral with the points of con- tact H, U, and V on the sides of the triangle, and HN = HM = HP. Therefore ■—— = . HQ VB 2. Given the two parallels AB and CD. Construct the series of circles and semicircles as indicated. Suggestion. — Divide PQ at H in the ratio of the altitude of an equilateral triangle to one half its side. Through H draw EF parallel to AB. Construct the circles and semicircles as indicated. Prove the semicircles tangent to the circles. 3. If PQ = 6, find the area of the curved figure RVHUS._ Ans. 18(2V3 - 3). 4. If PQ = a, find the same area. Ans. - (2v3 - 3). 5. If PQ = 6, find the length of the curve RVH. 6. If PQ = a, find the length of the same curve. 7. If PQ = 6, find the area of the cusped triangle VUH. 8. If PQ = a, find the same area. 296 GOTHIC TBACEBY: POINTED F0B3IS A P S_ B \H K 7A\ -pi N\- 7 ^" r X V i ] / \ / \ G Q R D Fig. 264. miLp Fig. 264a. — From the Doge's Palace, Venice, 311. In Fig. 264 lines AB and CD are parallel and the circles E, H, and E are tangent to each other and to the equilateral arches, CLQ, QMR, etc. 1. Prove that (1) the radius of circle H is equal to \ CQ ; (2) PQ is perpendicular to AB and CD; (3) Q, H, and P are coUinear. 2. Find the ratio PH HQ PH = HL. HC = S HL. CQ = 2 HL. Hence Suggestion. PH ^ 1 HQ, V5' 3. Given the parallel lines AB and CD. Construct the series of tangent circles and equilateral arches as indicated. CHAPTER VI TRUSSES AND ARCHES PART 1. RAFTERED ROOFS AND TRUSSES 312. Definition. — Raftered roofs are classified accord- ing to construction into single-framed or untrussed roofs, and double-framed or trussed roofs. In double-framed roofs the ordinary rafters are carried by sets of supporting timbers or by steel frames called trusses. These are placed eight or nine feet apart, and carry horizontal bars called purlins which support the common rafters. 313. Development of the Truss. — The isosceles triangle is the simplest form of truss and contains all elements really essential. In spanning large spaces the tie beam {AB, Fig. 265) may be so long and heavy as to sag of its own weight. The king-post ( Ci), Fig. 265) is then added as support to the tie beam. Rafters over 12 feet long are also liable to sag. Hence the addition of struts or braces {^DE and BF, Fig. 265). As the space to be spanned becomes still larger, trusses become more and more com- plicated. Figure 266 shows the queen posts EF and HK and 4 braces. Figure 268 shows the princess posts, KL, PQ, etc., additional braces, and an attic space between the queen posts ECr and FH. For a complete study of the problem of the truss, graphic statics is necessary. All 297 298 TRUSSES AND ARCHES parts must be in equilibrium under the weights to be supported. The forms shown, with the exception of Figs. 270 and 271, may be made wholly or partly of wood. Those shown in Figs. 270 and 271 are steel types. The earliest steel types were like the wooden ones ; later, more economical designs were introduced, many of which were originally suggested by the stress and strain diagrams. 314. In Fig. 265 ABC is an isosceles triangle. CD is the median to the base AB. DE and DF are drawn from D to the middle points of the lines AC and BC, respectively. ^ A D B Fig. 265. — The King-rod Truss.' EXERCISES 1. Prove that (1) CD is perpendicular to AB, (2) DE is ^ AC, and (3) A ADE is ^ the A ABC. 2. If CD = iAB and AE is 12 ft, find the length of AB.. Ans. 42.93+. 3. If the pitch of the roof is 7 in. to the foot and AE is 12 ft., find AB. 315. In Fig. 266 ABC is an isosceles triangle. CD is drawn from C to middle point of AB. AG = GE = EC and CH = HL = LB. DF = ^AD = DK. The points are joined as indicated. D K Fig. 266, — Queen-ioa Truss, EXERCISES 1. Prove that (a) ,EF is perpendicular to ,45 ; (b) EF=HK; (c) GF = GA; (rf) AFDE has same area as AFEG; (e) the area of AFDE is I the area of A DCE. 1 The names of trusses foUov? Kidder (2). RAFTERED ROOFS AND TRUSSES 299 Suggestion for (e). — Consider EF and DC as the bases; then the triangles have equal altitudes and are to each other as their bases. Prove that EF = i DC. 2. If GE = 12 and CD = iAB, find (1) AB, (2) CD, (3) EF, (4) ED. Ans. (1) 59.90; (2) 19.96; (3) 13.31; (4) 16.64. Suggestion ioi (4). — By use of radicals FD = ^VlS and EF = tlVia. Squaring and adding ED^ = &ff2. 316. In Fig. 867 ABC is an isosceles triangle. AG = GF = FC and CL = LK = KB. FD = DL. FE and LH are perpen- dicular to AB, and the points are joined as ^ ^ x P" B indicated. Fig. 267. — Combination King-rod and Queen-rod Truss. EXERCISES 1. Prove that (1) CD is perpendicular to FL and, if extended, to AB. (2) CD if extended would bisect AB. (3) GE = AG. (4) FL is parallel to AB and equal to J /IS. (5) FE = LH. (6) The areas of A AEG and GEF are equal, and equal to J the area of EHLF. 2. Find the positions of points F and L so that EHLF is a square. 3. What must be the relation between CX and ^B if EHLF is a square and CL is J C5 ? Ans. CX = ^ AB. Suggestion. — If Ci = J CB, CD is i CX. Therefore CD = DL, and hence CX = XB. 4. What must be the relation between CX and AB if EHLF is a square and CL = J CB ? 300 TRUSSES AND ABCHES 5. If LH=l FL and CL = \ AB, find the relation between CX and AB. Ans. CX = \ AB. 6. If CX = \ AB, find the relation between FL and LH, so that CL = \ CB. Ans. FL = '2 LH. 7. Given the A ABC; find points F and L so that FL = ZLH. 317. In Fig. 268 ABC is an isosceles triangle. AK = KE = EP = PC and CR = RF = FM = MB. ED = DF. KL.EG.FH, MN are perpendicular to AB. PQ and RS are perpendicular to EF. A L a JC H N B Fig. 268. — Combination Kmg-rod and Queen-rod Truss. EXERCISES 1. Prove that (1) CD is perpendicular to EF. (2) GK and DP are equal and equal to ^ AC. (3) EF is parallel to .4B and equal to i AB. (4) KG and DP are parallel. (5) The area of A EEC is equal to J that of EGHF. (6) The area oi A ALK is ^ the area of A ABC. 2. If CX = XB, prove that HE = ED. 3. If CF = EB and EE. CX_ XB — , find the relation between EH and 4. If CF = EB and EH = ^ EF, find the ratio — . Ans. }. £ P / X \ 4 i: /> £ Fig. 269, —Mansard Roof Truss. BAFTEBED B00F8 AND TBUSSES 301 EXERCISES 318. 1. In Fig. 269, if EF is parallel to AB and CE = CF, prove that (a) ZA = ZB. (6) The line bisecting Z C is the perpendicular bisector oi AB and EF. (c) The lines BE and AF are equal. 2. The outline for the truss may be determined as follows : Draw semicircle ABC and divide it into five equal parts at points F, H, G, and E. Let C be the center of arc GH. Join points as indicated.^ In this case prove that (a) EF is parallel to AB. (b) CK is perpendicular to EF and bisects Z C,ii K is center of EF. (c) CK extended bisects AB sA, right angles. 3. If the outline of the truss is constructed as in Ex. 2, find the number of degrees mZEAB,ZA EC, Z CEK, Z ECF. Ans. (1) 72°; (2) 135°; (3) 27°; (4) 126°. 4. A second method for constructing the outline of the truss for any given height and span is as follows : At the ends of the span construct A A and B each 60° (60° to 70° in practice). Draw the line EF parallel to AB at the desired height above AB. Construct A CEF and CFE at points E and F on line EF each 30°.^ Construct the figure according to the above method to scale, ^ inch to foot. AB = Z0 ft., EX = 10 ft. 5. Can the figure constructed as in Ex. 4 be inscribed in a semi- circle ? 6. If AB = 30, EX = 10, ZA = 60°, and Z CEK = 30°, find the length of AE, EF, CK, and EC. Am. (1) ^^3 = 11.54; (2) 18.45; (3) 5.32; (4) 10.65. 7. If EM = MC = 8, find the length of AB, ii ZA= 60°, Z CEK = 30°, and EX = 10. Ans. 39.26. 1 Cassell's, p. 134. 302 TBUSSMS AND ABCSM8 D H -Fan Truss, Steel. EXERCISES 319. 1. In Fig. 270 ABC is an isosceles triangle. Show how to construct the figure so that AF=FC = CH = HB and AE = EF = FG= GC. — = — and .4 B = 40, find the length of CF AD 12 ' ^ 2. If the ratio and EF. Suggestion for TF. — Show that ^Z)=20 and CD Solve for X in the equation x^ - (20 - x)2 = (y)2. Ans. (1) 11.73 ; (2) 6.3. Let^F=; equation, x^ - (V - a:)" = (W)^- 3. Prove that ^ ABC, AFC, and AFE are similar and verify the results obtained in Ex. 2 by means of the similar A. 320. In Fig. 271 ABC is an isosceles triangle. CH andCQ are so constructed that AH =HC = CQ = QB. F and K are the middle points of AH and HC, respectively, andAE = EG=GL = LC. A F H D Q B B Fig. 271. — Compound Fink Truss, Steel. EXERCISES 1. Prove that (1) FE, HG, and KL are perpendicular to ^ C. (2) GiC is parallel to ylD. (3) FHKG is a rhombus. 2. If P and iV are the middle points of CB and CQ, respectively, prove that the points G, K, N, and P are collinear. CT) 1 3- If Jo ^ i ^""^ ^^ "^ ^' ^^^ ^^^ ^^^S^D. of CH, HG, and EF. Ans. {I) 15f; (2) i^VS; (3) f|V5. 303 A B. Fig. 272. — Scissors Truss EXERCISES 321. 1. In Pig. 272, if CA = CB and CF = CE, prove that AE = FB and that DA = DB. 2. Join C and D. Prove that CD bisects Z C and is the perpen- dicular bisector of AB. 3. If CA = CB, SinAAE and BF are perpendicular to CB and CA, respectively, prove that AE = BF and that AD - DB. 4. If Z C = 60°, if AE and BF are perpendicular to CB and ^ C, respectively, and if CA = 16, find AE, AD, and CD. Ans. (1) 8V3; (2) i/V3; (3) yVs. PART 2. ARCHES 322. History. — The arch first appeared as an archi- tectural feature in the valleys of the Tigris and Euphrates rivers. Here the alluvial soil, which contained no stone and bore no trees, forced the use of brick for building material and made the use of stone and wooden lintels im- possible. The discovery of the arch naturally followed. Semicircular and pointed forms are said to date as far back as the eighth or ninth century b.C.^ No form of the arch was in general use in Europe until Roman times. In Greece stone available for building material was abun- dant, and consequently Greek architecture is based on the lintel. The general introduction of the arch, the vault, and the 1 Fergusson, Vol. I, p. 215 ; Fletcher, p. 43. 304 TRUSSES AND ARCHES dome as architectural features is due to the Romans. These forms had two immense advantages over the lintel : they were adapted to all kinds of building material ; they did away at once with the crowd of pillars that filled rooms and halls when short stone lintels were used. Because of these advantages carried throughout the Ro- man dominions, these new architectural features were soon established as essential to all building. The outline of all Roman forms was semicircular. Later other forms of the arch came into use as necessity demanded. At first these were formed of the arcs of cir- cles as shown in the following sections. In the best modern construction, elliptical, hyperbolic, parabolic, and cycloidal arches have replaced the earlier and cruder forms. As the drawings necessary to determine the shapes of the stones when the arches are true mathematical curves are much more difficult than is the case when they are com- binations of arcs of circles, the latter are still used in all but the very best work. POINTED OR GOTHIC ARCHES 323. History. — The general introduction of the pointed arch was due to the medieval architects of western Europe. While this arch may possibly have been borrowed from the Saracens, who up to this time had used it for orna- mental purposes, yet the problems in vaulting encountered by the Gothic builders made its independent discovery and use almost certain, i In the earliest buildings in Eng- land in which it is found it is used for structural purposes only. Among these buildings is Gloucester Cathedral.^ 1 Fletcher, pp. 272 and 283 ; Fergusson, Vol. II, p. 45. 2 Bond, p. 266. ABCHE8 305 324. Gables and canopies over doors and windows abound in Gothic buildings. Pierced and filled with tracery they were quite characteristic of fourteenth-century work. Later they became very elaborate.^ N M B I C D O E F B Fig. 273. — Common in Pullman Cars. 325. In Fig. 273 AB is the diameter of the semicircle AHB. AB is divided into six equal parts. NM = MA = AC = BR, etc. The arcs DP, LO, KE, etc. are drawn with M, A, C, etc. as centers and ^ AB as radius. EXERCISES 1. Prove that AQC, CXD, DYO, etc. are congruent isosceles triangles. 2. Prove that the curved triangles AQC, CXD, DYO, etc. are congruent. 3. Prove that the linear AAPD, CIO, DJE, etc. are congruent and isosceles ; also that the curved A APD, CIO, DJE are congruent. 4. Prove by superposition that the curved quadrilaterals PQCX, IXDY, etc. are congruent. 5. If this design is to be executed in leaded glass, how many differ- ent shaped patterns must be cut? Prove. G E G Fig. 273a. — From Canterbury Cathedral. ^ Sturgis (1) pp. 273, 330. 306 TRUSSES AND ARCHES 6. Show that the equilateral Gothic arch may be derived from a series of intersecting semicircles. (See Fig. 273a.) Suggestion. — Frore ^ABC, CDE, EFG ■■■ etc. equilateral. Fig. 273(>. 7. Show that designs like those in Figs. 273 and 2736 may be considered as based on a series of intersecting equilateral arches, or on a series of intersecting semicircles. 326. In Fig. 274 AB and CD are of given lengths. CD is the perpendicular Ht bisector of AB. q Fi E EXERCISES 1. If AC and CB are so constructed that their centers are on line AB and that they pass through the points A and C, and B and C, respectively, prove that they have equal radii. 2. If AC and BC are constructed as in Ex. 1 and if ^ is the center for BC, prove that A ABC is equilateral. ARCHES 307 3. Construct the drawing for eight stones on the side ^ C so that the extrados EF, FG, GH, etc. shall be equal. Suggestion. — EP and AC must be concentric and the sutures FP, GG', etc., must be on radii. Therefore join P and B, and divide the angle EBP into eight equal parts. 4. How far are problems like Ex. 3 possible by elementary geometry ? Ans. When the number of stones required is 2" Remark. — The true Gothic arch has no keystor^ . Fig, 275. 327. In Figs. 275 and 275a, CD is the perpendicular bisector of AB. P and Q are the centers of the arcs CB and AC, respectively. EXERCISES 1. If ^Q < AB, prove that AC < AB. li AQ> AB, prove that AC> AB. Suggestion. — Let E be the center of line AC. A AEQ, and CAD A O A F are similar. Therefore ^^^ = '^^^ ■ AC AD From this it follows that i.^ = A? or AQ-AB = ACK Hence AC AB the relations given above. 2. Draw a figure for a pointed arch in which the altitude is | the span. Draw a iigure in which the altitude is less than J the span. What must be the relation between the altitude and span in a true Gothic or pointed arch ? 308 TRUSSES AND ARCHES 3. JiAB = s and CD = h, find the length of A Q. Ans. — ■ -• 4« Suggestion. — Use the similar k^AEQ and ACD. 4. If ^B = 8 and CD = 5, find the length of A Q. Ans. 5\. 5. If ^Q = 6|, and CD = 6, find the length of AB. Ans. 9 or 16. 25 s'^ + 144 Suggestion. — Solve the equation — = — — for s. 4 4s • 6. Construct a figure to scale to illustrate each answer obtained in Ex. 5. 7. 11 AB = s snA AQ = r, find CD. 8. ltAQ = r and CD = h, find AB. Ans. ^ v'4 rs — s^. Ans. 2r± 2Vr^-h^ B Fig. 276. 328. In Fig. 276 P and Q are respectively centers for the arcs BC and GF, in the arches ACB and EGF. EXERCISES 1. If :5^ = 3E prove that A ABC and EFG are similar. AB EF ^ Suggestion. — Prove A ABC and PBC similar. Hence AB ■ PB = BCK Let 4^ = m. Therefore BP = mAB and mAB^ = BC\ AB Similarly mEF = GF . Hence -—— = -— —. Therefore, since the EF FG triangles are isosceles, they are similar. 2. Draw a figure in which PB is greater than AB and prove Ex. 1 for this figure. ARCHES 309 GABLES AND CANOPIES 329. In Fig. 277 and 277a, ACB is a Gothic arch formed of the two equal intersecting circles whose /'" centers are H and K. DC /' is the common chord. / MPL is a gable over the .' arch with its sides PL and ! PM drawn from a point in \ DC extended and tangent \ to the sides of the arch at ^"'-^ L and M, respectively. ^^~ -- Fig. 277. EXERCISES 1. Show how to construct the gable over a given Gothic arch, if point P is given. 2. Prove that PL = PM. 3. Prove that /. OPL = Z. OPMyvheu (a) AB, the span of the arch, is less than HK, the line of centers, (6) AB > HK, (c) AB = HK. Suggestion for (a). —See Fig. 277. Prove AHPM = APLK and AHNP = A KNP. Construct a figure to illustrate case (J). P 310 TRUSSES AND ARCHES 4. Show how to construct the gable over a given arch, if the length of one side of the gable is given. Suggestion. — With the radius of the circle and the given length as legs construct a right triangle. The hypothenuse will be the length of KP. Locate point P. 5. Show how to construct the gable over a given arch, if its angle LPJM is given. Suggestion. — At any point in the line PN construct an angle equal to \ the given angle. Then draw a tangent to the circle which shall be parallel to one side of this angle. Is this problem always possible? Is there any ambiguity in the construction ? 6. Construct a figure in which PH and PK fall upon PM and PL, given the radius of the circles that form the arch and the span of the arch. Suggestion. — See Fig. 277a. Is the problem always possible ? 330. In Fig. 278 ACB is an equilateral Gothic arch. MPL is a gable above the arch. Circle O is tangent to PL and to PM and passes through point C. A O B Fig. 278.— Rouen Cathedral. Porter, Vol. II, p. 266. EXERCISES 1. Show iow to construct circle X. 2. Construct a figure similar to Fig. 278, making the arch equi- lateral. Let AB = i: in., CF = 3 in. 3. Construct a figure similar to Fig. 278, making AB = 5 in., CO = 4^ iu., and PL = 5 in. ARCHES 311 Fig. 279a. — From Church of Our Lady of Lourdes, New York Cify. 331.— In Fig. 279 ACB is an equilateral arch with the gable MPL. CFE is an equilateral circular triangle struck from the centers C, F, and E. EXERCISE Show how to construct the entire figure. THE POINTED ARCH IN THE SQUARE FRAME B C_ 332. In Fig. 280 AC and CB are tan- gent to AE and FB, respectively. C is the center of line £F. Circles O and O' are tangent to the sides of the rectangle and to the arcs as indicated. EXERCISES 1. Show how to find P and Q, the centers for the arcs BC and A C, respectively. Suggestion. — See § 266, Ex. 3. J I '"? D- Q i 3 Fig. 280. — Gothic Doorway, Parlter (2), p. 179. 312 TRUSSES AND ARCHES 2. Prove that ^ Q = PS. 3. Construct circles and O'. Suggestion. — This involves the problem, "To construct a circle tangent to the sides of an angle and to a given circle." As the arc A C -would cut the line EC if extended, three circles are possible in this particular application of the problem, and considerable care must be exercised in making that construction which gives the circle desired. See § 253. 4. ItAB = 8 and PB = 6, find the length of CD. Suggestion. — Draw the lines CB, CA, and CP. Use the similar ^ACB&nA CPB. OGEE ARCHES 333. Occurrence. — The ogee arch is much used on Turkish buildings, and is common as an ornamental fea- ture in the late Gothic. It is very generally used in canopies and gables ^ in the masses of ornament that cover the inside and the outside of the later Gothic cathedrals. Half of it is sometimes used for veranda rafters, and for roofs of bay and oriel windows. THREE-CENTERED OGEE ARCHES F K O 334. In Fig. 281 the semicircle AGHB is constructed on AB as di- ameter, and CD is perpendicular to AS at its middle point. E ARCHES 313 EXERCISES 1. Construct arcs CH and CG tangent to the line CD at point C, and to the semicircle. Suggestion. — Make KC = AD. Draw NE, the perpendicular bi- sector of KD, meeting CK extended at E. E is center for the arc CH. What general problem in construction of circles is involved here? 2. If the arcs CH, drawn with £^ as a center, and HB drawn with Z) as a center, are tangent at H, prove that the points D, H, and E are col linear. 3. Prove that C, H, and B are collinear. Suggestion. — Join C and H and B and C, and prove that each is parallel to DK. 4. JiAB = 8 and CD is 8, find the length of CE. Ans. 6. Suggestion. — KD = 4V5. Compare the sides of the similar trian- gles KCD and KNE. Find KE and hence CE. 5. UAB = s and CD = h, find the length of CE. 4 A2 _ s2 6. For what value of CD will CE = DB1 Ans. CD = is abVs o Suggestion. — Let AB = s, then CE = -■ Solve the equation for A, » 4A^-.^ 2 is 7. Construct a figure to illustrate the case given in Ex. 6. Suggestion. — Construct A CB an equilateral triangle. 8. If CE = DB, prove that CH = HB. Suggestion. — See Ex. 3. Compare the similar A CEH and DHB. 9. If CE = DB and AB = s, find the area of the curved figui'e A GCHB. Ans. - v/3. 4 314 TRUSSES AND ARCHES 10. If the span AB is given and if CD is of indefinite length, construct Cfl so that it shall be tangent to CD and to the semicircle . A GHB and have a given radius. Suggestion. — Point E may be found by the intersection of two loci, namely: (1) the locus of centers of circles of given radius and tangent to a given circle ; (2) the locus of centers of circles of given radius and tangent to a given line. 11. liAB = s and CE = r, find the length of CD. Suggestion. — Solve for h the equation, r = Ans. Jv's(4r + s). 4h^-s^ is FOUR-CENTERED OGEE ARCHES R O D Fig. 282. ,335. In Fig. 282 AB and CD are of given lengths. A is the center for the arc HB, B is the center for the arc AG. CD is the perpendicular bisector of the span AB. EXERCISES 1. Construct the arcs CH and CG tangent to the arcs HB and A G, respectively, and to the line CD at C. 2. Draw lines AE and CB. Prove that they intersect at the point of tangency H. 3. liAB = s and CD = h, find the length of CE. Ans. 4A2_3«a ARCHES 315 Suggestion. — Find the value of KA and compare the sides of the similar ^ KRA and KEN. 4. For what value of C£»will CE = ABt Ans. CD = "^-^ . Suggestion. — Let CE = AB = s. Solve for h in the equation 4s 5. If CE = AB, prove that H is the mid-point of the line CB. 6. What must be the length of CE ii CH = IHB'> Ans. CE = \ AB. Suggestion. — Compare the sides of the similar A CEH and HAB. 7. Find the value of CD ii CE = i AB. Ans. CD = i^^. 8. What must be the length of CD if CE ^^ABl Ans.^V¥9. 6 9. If the span AB is given and if CD is of indefinite length, construct CH so that it shall be tangent to the line CD and to HB and have a given radius. Suggestion. — Find point E by intersecting loci as in § 334, Ex. 10. 10. By the construction suggested in Ex. 9, draw a figure in which CE = AB. 336. In Fig. 283 AB and CD are of given lengths. P and Q are fixed points on the line AB such that AQ = PB. P and Q are the centers for the arcs HB and GA, respectively. CD is the perpendicular bisector of AB. 316 . TRUSSES AND ABCHES EXERCISES 1. Construct the arcs CH and GC tangent to the line CD at C, and to the arcs HB and GA. 2. Draw lines PE and CB. Prove that they intersect at the point of tangency H. 3. If AB = s,PB=: ks, and CD = h, find the length of CE. Arts. ■ ^^ . is 337. In Fig. 284 AB and CD are given lengths. H and G are fixed points on the sides of A ABC. CD is a per- pendicular bisector of AB. BS is per- pendicular to AB at B. Fig. 284. EXERCISES 1. Construct the arcs CH and HB so that they shall be tangent to CD at C and to BS at B, respectively, and so that they shall pass through a given point H in the line CB. 2. If E and P are the centers for the arcs CH and HB, respec- tively, prove that the points E, H, and P are collinear and that the arcs CH and HB are tangent at H. . Suggestion. — To prove that the points E, H, and P are collinear, prove that Z CHE = Z BHP. 3. If CG= CH, construct the entire four-centered arch AGCHB, given the span AB, the altitude CD, and the points of inflection G and H on the sides of AABC such that CH = CG.^ 1 Hanstein, PI. 20, gives a construction which is a special case of the above. ASCHES 317 4. If P and Q are centers for the arcs ffB and GA, respectively, and E and F for CH and C<5, respectively, prove that (1) EC = CF. (2) PD= DQ. (3) £P, 7<'Q, and CD are concurrent. BASKET-HANDLE ARCHES 338. History. — Basket-handle arches, which were espe- cially common in France,^ seem to have arisen in late Gothic times, as arches lower than semicircles came to be used. They had one distinct advantage over arches formed of an arc of a single circle ; in these latter forms, weight upon the top of the arch results in considerable outward pressure at the point where the arch springs from the supporting pillars, while in basket-handle arches a large component of the pressure at this point is downward. 339. Of the constructions given in this section, § 341 and § 342 contain those that are most generally applicable, as in them the center for one of the circles is arbitrary when the altitude and span are given. It may be readily seen by experiment that the relative position of the centers determines the slope of the arch for any given span and rise. Mi' 340. In Fig. 285 PQ, the line of centers of the two equal non-intersect- ing circles P and Q, is extended to meet the circles at A and B, respec- tively. CD is the perpendicular bisector of AB. 1 Bond, p. 267. 318 TRUSSES AND ABCHES EXERCISES 1. Prove that a circle which is tangent to circle P and has its cen- ter on line CD is tangent also to circle Q. 2. If A'F is perpendicular to CD at C, construct a circle tangent to XY at C and to circle P. Show that two circles are possible. 3. Construct a figure for each of the following cases and note the relative position of circle to circles P and Q. (n) CD>AD, (6) CD = AD, (c) CD = AP, {d) CD < AP, (e) AP-- Suqqestion. — In the formula AP = : , if 8 r/i — 4 A^ — s^ is negative, a figure is obtained as suggested in Ex. 8. 10. Show that in the figure OA must be less than CO or no arch of the type here discussed is obtained. 11. If DO = r-h and AD = ^, find AO and show that Ex. 9 states algebraically the condition that is stated geometrically in Ex. 10. Suggestion. — AO'^ = ^ + (r - hy. From Ex. 10 AO^ AB. 3. Construct Fig. 288 when CD > \AB. 4. Study the exercises in §§ 343 and 344, for a figure in which AP = AD; for a figure in which AP '^AD. 5. Ji AP is equal to or greater than AD, can a basket-handle arch be constructed by methods suggested in §§ 341, 342? TUDOR ARCHES 345. History. — In origin and advantage the Tudor arch is similar to the basket-handle arch.^ While the latter is more common in France, the use of the Tudor arch was largely confined to England. The Tudor arch is seen on many university and college buildings in this country. 1 Bond, p. 266. ABOHES 323 346. The Tudor arch may be constructed if the length of the span and the altitude, the positions of the centers of the two end circles, and the direction of the tangent at the apex are given. ^ C F o" X Fig. 290. -l' In Fig. 290 CD and AB are of fixed length. CD is the perpendicular bisector of AB. P and Q are fixed points so that AP = BQ. C£ is a straight line mail- ing any arbitrary acute angle with CD. EXERCISES 1. Construct an arc tangent to CE at C and to the circle P. Show that two such circles are possible. 2. Let CE intersect BA or BA extended at F. Construct a figure to illustrate each of the following special cases : (a) CD > AP and point A between F and P. (b) CD > AP and point F between A and P. (c) CD > AP and point F between P and D. (d) CD= AP. (e) CD < AP. I Enc. Brit. . Building. 324 TRUSSES AND ARCHES 3. What are the conditions for forming a Tudor arch as shown in Fig. 290? 4. Construct the Tudor aveh as shown in Fig. 290, making the half CBB symmetrical with CDA. Lines AB, CD, and CE and point P are given in position. 5. Prove that lines PR, QR', and CD are concurrent. 6. Prove that line O'O" is parallel to line AB and ia bisected at right angles by CD extended. 7. Prove that lines PO', QO", and CX are concurrent ; also lines SO', S'O", and CX. 8. Construct the figure so that CG and CH are straight lines. Let P and Q be fixed points. 9. In Fig. 290 let lines AB, CD, and CE be given in position. Construct a perpendicular to AB at A, intersecting CE at E. If point P is arbitrary, for what positions of point P is the arch possible? 10. For what position of point P will CE be tangent to circle P, if lines CE, CD, and AB are fixed in position ? 347. The following construction for Tudor arches has been given. It is of limited application as the height of the arch is fixed by the construction. c In Fig. 291 AB is of given length. CD, of indefinite length, is the perpendicu- lar bisector of AB. AP = PD = DQ = QB. ABO'O" is a square erected on AB. Lines O'P and 0"Q are drawn and extended. Points P, O', O", and Q are the centers for AG, GC, CH, and HB, respectively. Fig. 291. ARCHES 325 EXERCISES 1. Show how to construct the complete figure. 2. Prove that A G and CG are tangent ; also BH and HC. 3. Prove that CG and CH and the line CD are concurrent. 4. If AB = s, find the length of GO' and CD. Ans.a) ^■, (2) s(V2-l). 5. Construct a figure similar to Fig. 291, dividing AB into three equal parts. 6. Find the length of GO' and CD for the figure constructed in Ex.5. 348. The following construction from Becker, page 80, like the preceding construction, is of limited application. c In Fig. 216 AB is of given length. CD, of indefinite length, is the perpendicular bisector ofAB. AP=PD = DQ = OB. The rectangle PO"0'Q is so constructed that PO" is f of PQ. P, O', 0"Q are the centers for PG, GC, CH, and HB, respectively. EXERCISES 1. Show how to construct the complete figure. 2. Prove that A G and GC are tangent ; also HB and HC. 3. Prove that GC and CH and the line CD are concurrent. 4. If AB = s, find the length of GO' and CD. Ans.(l) ^(l + VlS); (2) ^( VlS + 2Vl3 - 3). 5. Construct a figure similar to Fig. 292, dividing AB into five equal parts, and making PO"=i PQ and PQ = ^ AB. 326 TRUSSES AND ARCHES 349. Becker, page 80, gives also the following special construction for a Tudor Arch. '^ \ \ (\i/'' / / / \ \ \ t ',/ / / H /i\ > / ---'■Ji--'— J— 5ik-- 0"'x 0' Fig. 293. In Fig. 293 AB is of given length. AP = PD = DQ = QB. AO"0'B is a semicircle constructed on AB as diameter. P and Q are the centers and PQ the radius for PO' and QO"- With P as center and AP as radius draw arc AG; with O' as center and O'G as radius draw arc CG. Similarly Q and O" are centers for BH and HC. EXERCISES 1. Prove that PO' = O'B = AO" = QO". 2. Prove that A G and GC are tangent ; also BH and CH. 3. Draw CDX, the perpendicular bisector of AB. Prove that O'O" is parallel to AB and equal to J AB, and that CX is the perpendicular bisector of O'O". 4. Prove that the arcs GC and CH and the line CD are concurrent. 5. If ^5 = s, find the altitude of A ^ 0" Q. Ans. - vTS. 8 Suggestion. — Draw 0"D which equals - • Draw altitude from 0" meeting AB at K. Use A K0"£>. ^ ALPHABETICAL BIBLIOGRAPHY In this book the following works have been referred to under the names of the authors. Different books by the same author are dis- tinguished by the number preceding the title. American Architect and Building News. Boston. Architectural Annual. Philadelphia. Architectural Record. New York. — Cosmati Mosaic. Vol. XII, p. 203. Architectural Review. Boston. Arte Italiana Decorativa e Industriale. Milan. Ornamenti Geometrici. 1896, p. 62. II Mosaioo nei Pavimenti Romani. G. Patroni. 1903, pp. 29 and 41. Audsley, W. J. and G. A. — Handbook of Christian Symbolism. London, 1865. Ball, W. W. R. — A Short Account of the History of Mathematics. London, 1901. Barre, Louis. — Herculanum et Pompei. Paris, 1863-70. Vol. V contains the mosaics. Baudot, A. de. — Les Cathedrales de France. Paris. See illustra- tions. Beauchamp, W. M. — Metallic Ornaments of New York Indians. Bulletin of New York State Museum, 1903. Bulletin 73. Archaeology, 8. Becker, H. — Geometrisches Zeichnen. Leipzig, 1907. Billings, R. W. — (1) Illustrations of Carlisle Cathedral. London, 1839-1840. (2) Illustrations of Durham Cathedral. London, 1843. (3) Power of Form as Applied to Geometric Tracery. London, 1851. See illustrations. 327 328 ALPHABETICAL BIBLIOGRAPHY Bond, Francis. — Gothic Architecture in England. London, 1906. See text and illustrations. Borrman, Richard. — Die Keramik in der Baukunst. In Hand- buch der Architektur. Erster Theil, Vol. 4. Bourgoin, Jules. — (1) Les Elements de 1' Art Arabe. Paris, 1879. See illustrations. (2) Precis de I'Art Arabe. Paris, 1892. See illustrations. (3) Les Arts Arabes. Paris, 1873. Brandon, J. R. and J. A. — (1) An Analysis of Gothic Architecture. London, 1858. Treats of English architecture. See text and illustrations. (2) Open Timber Roofs of the Middle Ages. London, 1849. See illustrations. Brochure Series of Architectural Illustrations. The Alhambra, Details of Ornament. August, 1898, p. 129. Cosmati Mosaics. Dec, 1902, p. 267. Itahan Renaissance Ceilings. June, 1896, p. 83 ; March, 1898, p. 39. Specimens of Gothic Wood-carving. July, 1900, p. 115. Illus- trations, pp. 109-115. Bum, R. S. — Ornamental Drawing and Architectural Design as Applied to Industrial and Decorative Design. London, 1894. Cajori, F. — A History of Elementary Mathematics. London and New York, 1907. Calvert, A. F. — (1) The Alhambra. London and Liverpool, 1904. (2) Moorish Remains in Spain. London and New York, 1906. Clifford, C. R. — The Decorative Periods. New York, 1906. Coleman, Caryl. — Cosmati Mosaic. Architectural Record, Vol. XII, p. 203. Contains pictures of medieval mosaics in place. Corroyer, E. — Gothic Architecture. New York, 1893. Cummings, C. A. — A History of Architecture in Italy. Boston, 1901. Daniels, F. H. — The Teaching of Ornament. New York, 1900. Contains chapters on the history of ornament. Day, Lewis F. — Pattern Design. London and New York, 1903. Contains analysis of some geometric patterns. ALPHABETICAL BIBLIOGRAPHY 329 Dehli, Arne and Chamberlin, G. H. — •Norman Moniiments of Palermo and Environs. Boston, 1892. Contains measured drawings and mosaics. Dictionary of Architecture. Architectural Publication Society. London, 1892. Dief enbach, L. — Geometrische Ornamentik. BerUn. A collec- tion of geometric designs. Elis, Carl. — Handbuch der Mosaik-und Glasmalerei. Leipzig, 1891 . Contains list of references. Encyclopaedia Britannica. See articles on Building, Mosaics, and Mural Decorations. Essenwein, August von. — Die Romanisohe und die Gothische Baukunst. In Handbuch der Architektur. Zweiter Theil, Vol. 4, parts 1 and 2. Fairbairns, A. — Cathedrals of England and Wales. London, 1905. Fergusson, James. — A History of Architecture in all Countries. Vol. II : Ancient and Medieval. London, 1893. Fletcher, B., and Fletcher, B. F. — A History of Architecture on the Comparative Method. London, 1905. See text and illustrations. Ferrer, Robert. — Geschichte der Eiiropaischen Fliesen-Keramik vom Mittelalter bis zum Yahre 1900. Strassburg, 1901. See illustrations. Fowler. — Colored Engravings of Roman Mosaic Pavements. Winterton, 1769-1829. Franz, Julius. — Die Baukunst des Islam. In Handbuch der Architektur. Zweiter TheU, Vol. 3, part 2. Furnival, W. J. — Leadless Decorative TUes, Faience, and Mosaic. Stone, Staffordshire, 1904. Contains chapters on the history of tiles and mosaics. Gayet, Al. — L'Art Byzantin d'apres les Monuments de I'ltalie, de I'lstrie, et de la Dalmatie. Paris. Contains illustrations of mosaics. Gerspach, E. — La Mosaique. Paris, 1881. Glazier, R. — Manual of Historic Ornament. Part I : History of Ornament. Part II : The Applied Arts. London, 1906. Contains list of references. 330 ALPHABETICAL BIBLIOGRAPHY Gonse, Louis. — L'Art Gothique. Paris, 1890. Gravina, Domenico Benedette. — II Duomo de Monreale. Palermo, 1859. Contains illustrations of mosaics. Gruner, W. H. L. — Specimens of Ornamental Art Selected from the Best Models of the Classic Epochs. London, 1850. Giinther, S. — Vermischte Untersuchungen. Leipzig, 1876. Pages 1-92 contain a detailed history of star-polygons and star- polysedra. Gusman, Pierre. — La ViUa Imperiale de Tibur. Paris, 1904. See p. 219, chapter on mosaics. Hamlin, A. D. F. — A Text-book of the History of Architecture. London and New York, 1900. Handbuch der Architektur. Stuttgart, 1885. Previous to 1897 published in Darmstadt. Ausbildung der Fussboden-, Wand-, und Deckenflachen. Dritter TheU, Vol. 3, part 3. Die Baukunst in Islam. Zweiter Theil, Vol. 3, part 2. Die Keramik in der Baukunst. Erster TheU, Vol. 4. Die Romanisehe und die Gothische Baukunst. Zweiter Theil, Vol. 4. Hanstein, H. — • Constructive Drawing. Chicago, 1904. Hartel, A. — Architectural Details and Ornaments of Church Build- ing in the Styles of the Middle Ages. New York. Contains illustrations of GermajU churches. Hartung, H. — Motive der Mittelalterlichen Baukunst in Deutsch- land. Berlin, 1896-1902. See illustrations. Hasak, M. — Die Romanisehe und die Gothische Baukunst. Einzel- heiten des Kirchenbaues. In Handbuch der Architektur. Zweiter Theil, Vol. 4, part 4. See p. 137, windows; p. 237, floors. Hasluck, P. W. — CasseU's Carpentry and Joinery. London and New York, 1908. Heideloffs, Carl. — • Ornamentik des Mittelalters. Niirnberg, 1838- 1852. See illustrations ; mostly from German buildings. International Library of Technology. Volume on Historic Omar- ment and AppUed Design. Scranton. ALPHABETICAL BIBLIOGRAPHY 331 Jacobsthal, J. E. — SM-Italiemsche Pliesen-Ornamente. BerKn, 1886. See illustrations. Jeypore Portfolio of Architectural Details. 1890. See illustra- tions. Jones, Owen. — (1) Designs for Mosaic and Tessellated Pave- ments. London, 1842. (2) Grammar of Ornament. London, 1868. See illustrations. (3) Plans, Elevations, and Sections of tie Alhambra. London, 1842-1845. Kent, W. — - Mechanical Engineer's Pocket-Book. New York. Kidder, F. E. — (1) Architect's and Builder's Pocket-Book. New York. (2) Building Construction and Superintendence. New York, 1906 — . Koch, Hugo. — AusbUdung der Fussboden-, Wand-, und Dec- kenflachen. In Handbuch der Arohitektur. Dritter Theil, Vol. 3, part 3. See p. 40 for Roman mosaics. LefSvre, Leon. — Architectural Pottery. London, 1900. Leighton, J. — Suggestions in Design. London and Glasgow, 1881. See text and illustrations. Loth, A. — Les CathSdrales de France. Paris, 1900. See illus- trations. Liibke, W. — Ecclesiastical Art in Germany during the Middle Ages. Edinburgh, 1877. Meyer, F. S. — Handbook of Ornament. New York, 1894. Moore, C. H. — Development and Character of Gothic Architec- ture. New York, 1906. Morgan, Thos. — Romano-British Mosaic Pavements. London, 1886. Morris, I. H., and Husband, J. — Practical Plane and Solid Geome- try. London and New York, 1903. Ongania, Ferd. — La Basilica di San Marco in Venezia. Venice, 1881-1895. See illustrations. Otzen, J. — Ausgefiihrte Bauten. Berlin, 1894. See illustrations. Parker, J. H. — (1) ABC of Gothic Architecture. London, 1903. (2) An Introduction to the Study of Gothic Architecture. Lon- don, 1902. 332 ALPBABETICAL BIBLIOGRAPHY Patroni, G. — II Mosaico nei Pavimenti Romani. Arte Italiana, 1903, pp. 29 and 41. Poole, Stanley Lane. — The Art of the Saracens in Egypt. London, 1886. See illustrations. Porter, A. K. — Medieval Architecture. New York, 1909. Price, J. E. — A Description of the Roman Tessellated Pavement found in Bucklersbury. 1870. Prior, E. S. — A History of Gothic Art in England. London, 1900. Prisse d'Avennes. — L'Art Arabe d'aprfes les Monuments du Kaire, depuis le Vile si&cle jusqu'a la fin du XVIIle. Paris, 1877. See illustrations. Pugin, A. — - Specimens of Gothic Architecture ; Selected from Various Ancient Edifices in England. London, 1823. See illustrations. Pugin, A. and A. W. — Examples of Gothic Architecture ; Selected from Various Ancient Edifices in England. London, 1850. See illustrations. Racinet, A. C. A. — L'Ornement Polychrome. Paris, 1873. See illustrations. Redtenbacker, R. ^ Leitfaden zum Studiiim der Mittelalterlichen Baukunst. Leipzig, 1881. Rickman, Thos. — An Attempt to Discriminate the Styles of Architecture in England. Oxford, 1817. 7th edition, 1881. Romussi, Carlo. — II Duomo di Milano. Milan, 1902. See illustrations. Rossi, Giovanni Battista de. — Musaici Cristiani e Saggi dei Pavementi deUa Chiese di Roma, Anteriori al Secolo XV. Rome, 1872-1896. See illustrations. Ruskin, J. — The Stones of Venice. Sarre, Friedrich. — Denkmaler Persischer Baukunst. Berlin, 1901 . See illustrations. Schaef er, C. — Die Mustergiltigen Kirchenbauten des Mittelalters in Deutschland. Berhn, 1892-1901. See illustrations. Sharpe, E. — (1) A Treatise on the Rise and J*rogress of Decorated Window Tracery in England. London, 1849. ALPHABETICAL BIBLIOGEAPHT 333 (2) Decorated Windows, A Series of Illustrations. London, 1849. Shaw, Henry. — (1) Specimens of Tile Pavements. London, 1858. See illustrations. (2) Decorative Arts of the Middle Ages. London, 1851. See illustrations. Simpson, F. M. — A History of Architectural Development. London and New York, 1905. Smith, D. E. — The Teaching of Geometry. Boston, 1911. Speltz, Alexander. — Styles of Ornament shown in Designs. Berlin, Paris, and New York, 1906. See illustrations. Statz, v., und Ungewitter, G. — The Gothic Model Book, The Archi- tecture of the Middle Ages. London, 1862. See Ulustra^ tions. Strack, Heinrich. — Ziegelbauwerke des Mittelalters und der Renaissance in Italien naoh Original-Aufnahmen herausgege- ben. Berlin, 1889. See illustrations. Sturgis, Russel. — (1) European Architecture. London and New York, 1896. (2) A History of Architecture. Vol. I : Antiquity. Vol. II : Romanesque and Oriental. New York, 1906. Terzi, Andrea. — La Cappella di S. Pietro nella Reccia di Palermo. Palermo, 1889. See illustrations of mosaics. Trautwine, J. C. — Civil Engineer's Pocket-Book. New York, 1909. Ward, James. — (1) Historic Ornament. New York, 1897. (2) Principles of Ornament. London, 1890. Waring, J. B. — ■ (1) The Arts Connected with Architecture. Illus- trated by Examples in Central Italy from the 13th to the 15th Century. Examples of Stained Glass, Fresco Ornament, Mar- ble and Enamel Inlay, and Wood Inlay. London, 1858. (2) Ceramic Art in Remote Ages, with Essays on the Symbols of the Circle, the Cross and Circle, the Circle and Ray Orna- ment, the Fylfot, and the Serpent. London, 1874. See illustrations. Watt, J. C. — Examples of Greek and Pompeian Decorative Work. London, 1897. See mosaic pavement, PI. 40. 334 ALPHABETICAL BIBLIOGRAPHY White, Gleeson. — Praotioal Designing. London, 1894. Contains chapters on designing for tiled floors, printed fabrics, and linoleums. Wilmowsky, G. — (1) Die Romische Villa zu Nennig. Bonn, 1865. (2) Die Romischen Moselvillen zwischen Trier und Nennig. Trier, 1870. Wilson, E. — Cathedrals of France. New York, 1900. See illustrations. Wilson, Thos. — The Swastika, the Earliest KInown Symbol and its Migrations. Report of U. 8. National Museum. 1894, p. 763. Contains bibliography. Womum, R. N. — Analysis of Ornament. London, 1896. Wyatt, Matthew Digby. — Specimens of Geometric Mosaics of the Middle Ages. London, 1848. See illustrations. Young, J. W. A. — The Teaching of Mathematics. London and New York, 1907. Zahn, Wilhelm. — Ornamente aller Klassischen Kunst-Bpochen nach den Originalen in ihern Eigenthiimlichen Farben. Ber- lin, 1870. (2) Die Schonsten Ornamente und Merkwiirdigsten Gemalde, aus Pompeji, Herkulanum, und Stabiae. Berlin, 1829. CLASSIFIED BIBLIOGRAPHY References on special subjects may be found in the preceding Alphabetical Bibliography under the names of the authors as indicated below. Architecture General : Dictionary of Architecture ; Fergusson ; Fletcher ; Hamlin ; Simpson ; Sturgis. Gothic : Brochure Series 1900 ; Corroyer ; Essenwein ; Gonse ; Hasak ; Moore ; Parker ; Porter ; Redtenbacker ; Statz ; Chapters in works on the history of ornament and archi- tecture. English: Billings (1) and (2); Bond; Brandon; Fairbairns; Parker ; Prior ; Pugin, A. ; Pugin, A. and A. W. ; Riokman ; Sharpe. French : Baudot ; Loth ; Wilson, E. German: Hartel; Hartung; Heideloffs; Liibke; Otzen; Schaefer; Statz. Italian: Arte Italiana; Brochure Series 1896, 1898, 1902; Cum- mings ; Gravina ; Gusman ; Ongania ; Romussi ; Ruskin ; Strack ; Terzi. United States : Architectural Magazines. Saracenic : Calvert ; Franz ; Jones (3) ; Sarre. India : Jeypore PortfoUo. Decoration General and Historical (mostly illustrated) : Chfiord ; Daniels ; Glazier; International Library of Technology; Jones (2); Meyer ; Racinet ; Speltz ; Ward (1) ; Wornum (no illustra- tions) ; Zahn. Geometrical: Arte Italiana 1896; Billings (3); Burn (Sees. I and II); Diefenbaeh; Day (chap. II-VI). 335 336 CLASSIFIED BIBLIOGRAPHY Symbolism : Audsley ; Waring (2) ; Wilson, Thos. ; Wornum (pp. 10&-107). Theory and Teaching : Burn ; Daniels ; Day ; International Library of Technology ; Leighton ; Ward (2) ; White. Classic : Gruner. See Mosaics and Tiles. Byzantine: Dehli; Gayet; Gravina; Ongania; Terzi; Waring (1). Gothic : Heideloffs ; Shaw ; chapters in works on history of archi- tecture and ornament. Saracenic : Brochure Series 1898 ; Bourgoin ; Calvert ; Jones (3) ; Poole ; Prisse d'Avennes ; chapters in works on history of architecture and ornament. See Saracenic architecture. Primitive : Beauchamp ; Waring (2) ; Wilson, Thos. Mathematics Drawing : Becker ; BUlings (3) ; Diefenbaeh ; Hanstein ; Morris. History: Ball; Cajori. Theory: Giinther. Teaching : Smith ; Young. Mosaics and Tiles General : Borrman ; Dictionary of Architeetiu:e (Pavements) ; EHs ; Encyclopaedia Britannica (Mosaics) ; Forrer ; Furnival • Gerspach; Glazier; Jones (1); Koch; LefSvre; White (one chap.). Byzantine : Brochure Series 1902 ; Coleman ; Dehli ; Gayet ; Gravina ; Jacobsthal ; Ongania ; Rossi ; Terzi ; Wyatt ; Zahn (1). Roman: Barr6; Fowler; Gruner, PI. 27; Gusman; Morgan; Patroni; Price; Watt; WUmowsky; Zahn (2). Gothic : Shaw (1). Technical Carpentry and Building : Hasluok ; Kidder : Encyclopaedia Britan- nica (Building). Engineering : Kent ; Trautwine. COMMERCIAL CATALOGUES The following firms very kindly furnished information or supplied catalogues that have been freely used in preparation of notes and figures. American Encaustic Tiling Co. (TUes and Mosaics), Zanesville, O. Architectural Decorating Co. (Stucco and Plaster), Chicago, 111. Darby, Edw., & Sons Co., Pennsylvania Wire Co., Philadelphia, Penn. Dunfee, J., & Co. (Parquetry), Chicago, lU. EUer Manufacturing Co. (Stamped Steel Ceilings), Canton, 0. Flanagan and Biedenweg Co. (Ornamental Glass), Chicago, lU. Hawes and Dodd (Tiles), Chicago, 111. Illinois Mosaic TUe Co., Chicago, lU. Interior Hardwood Co. (Parquetry), Indianapolis, Ind. Marshall Field & Co. (General Merchandise), Chicago, lU. Maw and Co. (Tiles and Mosaics), England. McCully & Miles Co. (Ornamental Glass), Chicago, lU. Moore, E. B., & Co. (Parquetry), Chicago, 111. Mosaic Tile Co., Zanesville, O. Mott, J. L. Iron Works, New York City. New York Metal Ceiling Co., New York City. Northern Pacific R. R., Minneapolis, Minn. Northwestern Parquet Floor Co., Minneapolis, Minn. Rice, J. H. Co. (Ornamental Glass), Chicago, lU. Roberts, E. L., and Co. (Ornamental Glass), Chicago, lU. Smith, P. P., Wire and Iron Works, Chicago, 111. United States Encaustic Tile Works, Indianapolis, Ind. Wild, Joseph, & Co. (Linoleum and Oilcloths), Chicago, 111. Winslow Bros. Co. (Ornamental Iron), Chicago, lU. Wood Mosaic Flooring & Lumber Co. (Parquetry), Rochester, N.Y. 337 INDEX TO PROBLEMS AND THEOREMS Numbers refer to paragraphs and exercises ; thus, 94 (10) means § 94, Ex. 10. Algebraic analysis. (See Constructions.) Altitudes of a triangle: 235 (2). Angles, equality of, proved by means of theorems concerning: I. Congruent triangles: 67 (1); 98 (1); 318 (2); 321 (2); 329 (3). II. Parallel lines and transversals: 90 (4). III. Equilateral triangles : 115(2 6). IV. Measurement of arcs: 88 (9); 115 (2 j) ; 148 (3, 6); 150 (4) ; 151 (4) ; 156 (4) ; 158 (2). V. The angles of similar triangles : 63 (2) ; 341 (2). VI. Symmetry: 150(4); 151(4); 154(4). VII. Miscellaneous relations : 177(3). Angles made by Parallels and transversals, used in connection with theorems concerning : I. Congruent triangles : 63 (1) and similar problems : 10 (1) ; 29 (10) ; 66 (1) 68 (5) ; 74 (1). II. Isosceles triangles : 54(1,3); 98(1); 129(1). Angles, measures of, computed by means of : I. Divisions of a right angle : Figs. 57 and 68. II. Theorems concerning the sum of the angles of a triangle : 41 (3) ; angles in Figs. 57 and 58; 162 (2). III. Angles of an equilateral triangle or regular hexagon : 15(1); 39 (1) ; 115 (angles in triangles in Fig. 92) ; 142 (3). IV. Theorems concerning measurements of arcs : 88 (3) ; 115 (2g); 150 (5); angles in Figs. 119-123; 151 (5); 156 (4) ; 157 (6) ; 158 (3) ; 159 (4) ; 318 (3). Arcs, equality of, proved by. means of theorems concerning: I. Equal chords: 203 (3); 216 (5); 217 (9); 218 (3); 219 (5) ; 220 (3). 338 INDEX TO PROBLEMS AND THEOREMS 339 II. Measurement of angles: 216 (4). III. Symmetry : 203 (3) ; 208 (4) ; 216 (5) ; 217 (9) ; 218 (3) ; 219 (5); 220 (3). (See Curved figures, congruence of.) Arcs, measures of, computed by means of : I. Divisions of a right angle : 176 (2) ; 186 (8). II. Angles of an equilateral triangle : 127 (1) ; 168 (1) ; 260 (5) ; Ares in Fig. 92. III. Simultaneous equations : 198(3). Areas, comparison of. (See Equivalent areas and Ratios of areas.) Areas, computations of. (See Circles, Curved figures. Equilateral triangles, Hexagons, Isosceles triangles. Octagons, Ovals, Paral- lelograms, Polygons, Quadrilaterals, Ratios of areas. Rectangles, Rhombuses, Right triangles. Sectors, Segments, Semicircles, Square roots. Squares, Star-polygons, Trapezoids, Triangles.) Areas, constructions for figures with given areas, requiring : I. Algebraic analysis : 54 (9, 12, 14, 17, 21) ; 56 (11) ; 57 (7) ; 66 (15) ; 67 (11) ; 71 (17, 20) ; 74 (9, 10) ; 94 (11). II. Theorems concerning areas : 66(17); 73(3). Bisectors of angles, used for : I. Computations of length of lines : 11(8); 66(21); 68(3) 69 (3) ; 88 (11) ; 90 (9) ; 139 (5). II. Concurrent lines: 66 (19); 68 (1); 69(6, 7); 90 (5) 91 (d: III. Construction of equal lines : 43 (1, 3) ; 45 (2) ; 46 (2) 159 (7). IV. Equahty of lines: 43 (3, 7) ; 44 (2). V. Construction of circles. (See Inscribed circles I, II, III, and Circles, construction of, V.) Chords, equality of: 115 (2) ; 122 (1) ; 123 (1). Circles, areas of : I. Radius evident : 14 (5) ; 36 (2, 3) ; 118 (Circle O, Pig. 94) ; 132 (1-3); 192 (4). II. Radius to be computed : 98 (2) ; 101 (1) ; 118 (5, 7) ; 163 (8) ; 164 (3) ; 165 (3) ; 178 (7, 8) ; 203 (5, 7) ; 204 (3, 5) ; 245 (5) ; 279 (3) ; 280 (6) ; 293 (5). 340 INDEX TO PROBLEMS AND THEOREMS Circles, construction of: I. Passing through three given points: 198 (5) ; 216 (1). II. Passing through two given points and a. With center on a given line : 266 (3) with applications : 301 (8) ; 303 (3) ; 306 (2) ; 326 (1). 6. Tangent to a given line: 253 (1) with applications: 138 (9). c. Of given radius: 18 (2). III. Passing through a given point, of given radius and a. Tangent to a given line: 266 (1) with applications: 264; 267 (2); 268 (1) ; 270; 295 (1). 6. With center on a given line : 266 (1). IV. Tangent to two parallel lines : 36 ; 37 ; 38. V. Tangent to two given intersecting lines and a. Passing through a given point : 253 (2). 6. Passing through a given point in the bisector of the angle: 121 (1). c. Passing through a given point on one of the lines : 163 (1, 3); 201 (1); 202 (1); 203 (1, 7); 204 (1) ; 205 (1). d. Tangent to a given circle : 253 with applications : 1.63 (5) ; 180 (2) ; 198 (2) ; 252 (2) ; 272 (6) ; 332 (3). VI. Tangent to a given line at a given point and a. Of given radius : 18 (1). 6. Passing through a second given point : 266 (3) with applications: 301 (9); 332 (1); 337 (1). c. Tangent to a given circle: 119 (7) and 246 with appli- cations : 119 (3) ; 230 (3) ; 237 ; 238 (1) ; 241 (2) 242 (1) ; 244 (2) ; 269 (4) ; 288 (4) ; 297 (1) ; 334 (1) 335 (1) ; 336 (1) ; 340 (2, 3) ; 341 (1) ; 342 (1-3) 346 (1, 2). VII. Tangent to a given line, of a given radius and tangent to a given circle: 265 with applications: 254 (1, 2); 255 (1); 256 (1); 257; 258 (1); 259 (1); 260 (1) ; 261 (1) ; 262 (1) ; 263 (2) ; 269 (1) ; 334 (10) ; 335 (9). VIII. Tangent to two equal intersecting circles, and (special eases only), a. Tangent to two concentric circles : 276 (7) ; 290 (5) ; 301 (5) ; 302 (1). INDEX TO PROBLEMS AND THEOUEMS 341 6. Tangent to a second pair of equal intersecting circles : 290 (5). c. Tangent to two equal tangent circles : 275 (3) ; 276 (7) ; 302 (3). d. Tangent to two equal noninterseeting circles : 301 (2) ; 302. IX. Miscellaneous : 136 (1, 5) ; 144 (2, 3) ; 207 (1) ; 234 (1). Circle, divisions of: 186 (4-6); 190 (note); 148-152; 191-202; 200(6) ; 201 (3) ; 202 (5, 6). (See Regular polygons.) Circumscribed circles : 99 (5) ; 192 (4) ; 251 (7) ; 318 (5). Collinear points. Proofs depending upon theorems concerning: I. The perpendicular bisector of a line : 51 (1) (see sugges- tion) ; 70 (1); 82 (3); 115 (2); 141 (2); 143 (2); 148 (4) ; 150 (6) ; 151 (6) ; 158 (4, 8) ; 159 (3) ; 184 (2) ; 205 (2) ; 241 (1) ; 311 (1). II. Tangent circles : 165 (2) ; 241 (1) ; 245 (2) ; 273 (1) ; 275 (2) ; 276 (6) ; 286 (1, 3) ; 287 (2) ; 301 (7) ; 334 (2). III. Intersecting circles: 290 (1) (see suggestion); 208 (3) (see suggestion) with applications : 183 (4) ; 185 (3) ; 217 (1-3) ; 218 (1) ; 219 (2) ; 233 (1) ; 234 (3) ; 235 (2); 251 (7). IV. Miscellaneous relations : 38 (2) ; 41 (8) ; 45 (4) ; 46 (3) ; 49 (2) ; 56 (1) ; 60 (2) ; 64 (7) ; 68 (5) ; 88 (12) ; 94 (2, 3) (see suggestion) and 104 (3); 102 (1); 115 (3); 129 (1); 131 (3) (see suggestion); 155 (2); 156 '(2) ; 157 (2) ; 185 (1, 5) ; 216 (6) ; 217 (8) ; 275 (19) ; 276 (2) ; 320 (2) ; 334 (3) (see suggestion) and 335 (2) ; 336 (2) ; 337 (2). Complementary angles: 341 (3). Concentric circles, construction of circles tangent to : 276 (7) ; 301 (5) ; 302 (1). Concurrent lines, proved by means of theorems concerning : I. The perpendicular bisector of a line : 82 (3) ; 94 (1) ; 139 (1) ; 157 (3) ; 208 (1) ; 337 (4) ; 346 (5, 7). II. The bisector of an angle: 44 (2) ; 94 (1) ; 131 (5). III. The bisectors of the angles of a triangle : 66 (19) ; 68 (1) ; 69 (6, 7) ; 90 (5) ; 91 (1). IV. The altitudes of a triangle : 235 (2). 342 INDEX TO PROBLEMS AND THEOREMS V. Diameters and diagonals of squares : 49 (1). VI. MisceUaneous relations : 61 (6); 90 (2); 211 (1); 221 (2). Congruent figures. (See Curved figures. Equilateral triangles. Isosceles triangles, Parallelograms, Polygons, Quadrilaterals, Rhombuses, Right triangles. Squares, Star-polygons, Trapezoids, Triangles.) Construction of figures having given data. (See Circles, Circum- scribed circles. Equilateral triangles. Escribed circles, Hexagons, Inscribed circles. Intersecting circles. Isosceles right triangles. Lines, Octagons, Points, Polygons regular. Polygons similar. Proportionals, Right triangles. Segments, Semicircles, Squares, Star-polygons, Tangent circles, Tangent lines.) Construction of given figures, requiring: I. Straight Une constructions only : 45 (3) ; 46 (3, 4) ; 62 (9) ; 64 (6) ; 68 (4) ; 86 (1) ; 90 (1) ; 99 (4) ; 105 (1). II. Circle constructions : 109(1); 110(1); 113(1); 114(9); 115 (1, 4) ; 116 (1) ; 168 (4) ; 178 (9) ; 182 (1) ; 186 (7) ; 188 (4) ; 203 (7) ; 204 (4) ; 220 (1) ; 226 (4) ; 228 (6) ; 229 (1) ; 233 (3) ; 238 (1) ; 330 (2, 3). III. Regular polygon constructions : 122(1); 123(1); 154(1); 155 (1) ; 156 (1) ; 157 (1) ; 201 (3) ; 202 (5, 6) ; 209 (1) ; 216 (7) ; 217 (2, 3, 10) ; 218 (5) ; 219 (7). (See Drawings to scale.) Constructions requiring algebraic analysis: I. Circles: 245 (3, 4); 273 (5); 275 (22); 278 (6); 279 (7) ; 281 (4-6, 10) ; 282 (4) ; 283 (4^6) ; 285 (6-8, 10) ; 292 (6, 10, 12) ; 293 (4, 6) ; 294 (2, 4, 5) ; 295 (4, 5) ; 296 (7, 12) ; 297 (1) ; 298 (5) ; 300 (5, 8) ; 302 (7) ; 303 (3, 7, 9) ; 304 (5, 7) ; 306 (2) ; 334 (7) ; 341 (8) ; 342 (7). (See Formulce, algebraic.) II. Figures with given area: 54 (9, 12, 14, 17, 21); 56 (11); 57 (7) ; 66 (15) ; 67 (11) ; 71 (17, 20) ; 74 (9, 10) ; 94 (11). III. Lines: 56 (5, 7). Curved figures, areas of, composed of : I. Equilateral triangles and a. Sectors : 167 (8) ; 254 (4 a, 4 6, 5 a, 5 6). Harder prob- INDEX TO PROBLEMS AND THEOREMS 343 lems : 138 (7, 8) ; 167 (10) ; 168 (3) ; 169 (3) ; 191 (4) ; 203 (5, 6). 6. Segments : 37 (4) (see suggestion) ; 37 (3-6) ; 122 (3) 123 (3) ; 124 (2, 3) ; 126 (2-7) ; 127 (3-7), ; 211 (3) 212 (2, 4); 225 (5-7) (see suggestion); 226 (3) 228 (4, 5, 7); 229 (4, 12); 232 (1, 2); 248 (6) 267 (4); 268 (2); 276 (16, 18) (see suggestion) 286 (8) ; 301 (11, 12, 15) ; 309 (3, 4) ; 310 (3, 4) 334 (9). Harder problems: 213 (4, 5); 214 (3) 227 (2). c. Semicircles : 138 (1-3) ; 170 (4, 5) ; 172 (5) ; 200 (4, 5) ; 254 (4 c, 5 c). li. Hexagons and sectors : 171 (4^6) ; 192 (2, 3) ; 194 (3). III. Squares and semicircles or quadrants :' a. Radius evident : 36 (2) ; 107 (3, 4) ; 108 (3) ; 109 110; 112 (2, 3) ; 113 (2, 3, 5) ; 114 (3, 8) ; 118 (8) 120 (2); 121 (7); 132 (1-3); 133 (2); 134 (2^) 137 (2, 3); 175 (3); 181 (6); 182 (3); 186 (3) 187 (2) ; 204 (3) ; 218 (6) ; 287 (5). 6. Radius not evident: 18 (4); 52 (2-4); 101 (2-5) 118 (8 c, 10) ; 121 (4, 5) ; 133 (4) ; 135 (2, 3, 6, 7) 163 (9, 10) ; 164 (4) ; 174 (6, 10, 11) ; 176 (4, 5) 177 (4) ; 179 (4) ; 183 (6, 7) ; 184 (8) ; 185 (7, 8) 188 (3, 6). IV. Miscellaneous figures : 15 (4, 5) ; 16 (3) ; 115 (6) ; 116 (2) 136 (2, 3) ; 139 (2, 3, 6, 7) ; 164 (6, 8) ; 193 (4) ; 196 (2) 201 (2) ; 202 (4). Curved figures, congruence of: 16 (1) ; 107 (2) ; 108 (2) ; 112 (1) 113 (1) ; 115 (2) ; 121 (3) ; 122 (2) ; 123 (2) ; 124 (1) ; 126 (1) 127 (2) ; 167 (3) ; 168 (2) ; 171 (3) ; 172 (3) ; 174 (3) ; 176 (3) 181 (4) ; 182 (2) ; 186 (2) ; 198 (4) ; 202 (2) ; 212 (3) ; 221 (5) 233 (2) ; 248 (5) ; 325 (2-4). Curved figures, perimeters of: 109; 110; 124 (3); 126 (4, 5) 127 (9, 10). Decagons, regular. I. Determination of: 152 (13). II. Length of side: 209 (2). Discussions of special cases : 54 (8, 9, 12) ; 55 (4) ; 71 (13) ; 217 (10); 243 (4, 5); 253 (2); 265; 276; 286 (7); 290; 296; 301. (See Formulm, substitution in.) 344 INDEX TO PROBLEMS AND THEOREMS Drawing instruments, use of: 22 (8). Drawings to scale : 22 (6, 7) ; 23 (4) ; 24 (4, 6) ; 25 (4) ; 26 (3, 4) ; 27 (1, 3) ; 28 (4, 5) ; 29 (6) ; 30 (2) ; 318 (4) ; 327 (6). Duodecagons : I. Determination of : 151 (10) ; 152 (14, 15) ; 158 (5). II. Length of side : 115(5); 144(4); 158(10). Equations, indeterminate: 55 (3); 74 (8-11); 75 (6). Equations, linear: I. Containing binomial and trinomial squares: 119 (6); 121 (2 a) ; 195 ; 230 (5) ; 234 (2) ; 238 (4) ; 241 (3) ; 242 (3, 4) ; 243 (4) ; 245 (3) ; 249 (3) ; 258 (2) ; 259 (2) ; 272 (2, 3, 7) ; 273 (2, 6) ; 275 (4, 8, 9, 22, 23) ; 278 (1, 3) 279 (1, 5) ; 280 (1, 3) ; 281 (1, 3, 8, 9) ; 282 (2) ; 283 (1 2) ; 284 (1) ; 285 (2, 3) ; 286 (5) ; 287 (3) ; 288 (1, 7, 10) 289 (1, 2, 4) ; 291 (1, 3, 5) ; 292 (1, 3, 11) ; 293 (1, 2, 3) 294 (1, 3) ; 296, (1, 4, 8) ; 298 (1) ; 299 (1, 2, 5) 300 (1, 2) ; 302 (1, 3, 5) ; 303 (1, 6, 7, 8) ; 304 (1, 3, 6, 8) ; 305 (1, 5) ; 306 (1, 3) ; 307 (1) ; 319 (2) ; 320 (3) II. Miscellaneous : 54 (9, 10) ; 56 (4, 6) ; 67 (9, 10) ; 71 (5, 18) 74 (7) ; 75 (5) ; 100 (5, 6) ; 102 (5) ; 242 (6, 7) ; 278 (9) ; 283 (5) ; 285 (9) ; 288 (8) ; 292 (8) ; 296 (13) ; 341 (9). Equations, quadratic : I. Complete: 51 (9-11); 54 (11, 13, 16, 18); 56 (8-10, 12); 57 (6, 8) ; 61 (4, 5) ; 66 (11, 13) ; 71 (9, 12, 14, 16, 19) ; 94 (10) ; 279 (8) ; 327 (5) ; 341 (7) ; 342 (6). Contain- ing binomial squares : 180 (6) ; 252 (4) ; 262 (3, 4) ; 263 (3, 4). II. Incomplete : 29 (4, 5) ; 37 (7, 8) ; 51 (7, 8) ; 66 (14, 16) ; 112 (4, 5); 113 (4); 118 (6); 121 (6); 132 (4, 5); 133 (5); 174 (7-9); 175 (4); 176 (6); 334(6, 11); 335 (4, 7, 8). Equations, radical: 121 (2); 239 (2); 275 (25); 278 (11, 12); 288 (5, 9, 11). Equations, simultaneous: 59 (10, 11); 198 (3). INDEX TO PROBLEMS AND THEOREMS 345 Equilateral ' triangles : I. Angles of, problems involving : 15 (1) ; 39 (1) ; 115 (2) ; 127 (1) ; 142 (3) ; 168 (1). II. Area of: 14 (7); 15 (2); 22 (4, 5). Harder problems: 115 (6) ; 141 (7, 8) ; 142 (4) ; 143 (7) ; 144 (4) ; 229 (10). (See Curved figures, areas of.) III. Congruence of: 141 (3); 142 (1); 143 (4); 144 (1); 211 (1) ; 213 (1) ; 225 (2). IV. Construction of: 221 (1); 331. V. Determination of : 14 (4) ; 24 (3) ; 37 (2) ; 115 (2, 3) ; 141 (1, 3) ; 142 (1) ; 143 (4) ; 144 (1) ; 156 (6) ; 211 (1) ; 213 (1) ; 225 (2) ; 236 (1) ; 251 (7) ; 325 (6) ; 326 (2). Equivalent areas : 6 (1) ; 7 (2) ; 24 (3) ; 48 (5) ; 65 (1) ; 66 (2, 3) ; 131 (4) ; 314 (1) ; 315 (1) ; 316 (1) ; 317 (1). Ratios of areas III.) Escribed circles : 199. Extreme and mean ratio : 148 (5). Formulae : I. Solution of : 41 (13) ; 144 (5) ; 178 (5, 6) ; 184 (7) ; 248 (4) ; 327 (8). Involving incomplete quadratics : 29 (4, 5) ; 37 (7, 8) ; 112 (4, 5) ; 113 (4) ; 118 (6) ; 121 (6) ; 174 (7-9) ; 175 (4) ; 176 (6) ; 327 (7). II. Substitution in : 54 (8) ; 71 (6) ; 74 (6) ; 81 (3) ; 94 (7, 8) 102 (7) ; 243 (6) ; 272 (4, 5, 8, 9) ; 273 (7) ; 275 (10-13 21, 24) ; 276 (8, 9) ; 278 (4, 5, 8) ; 279 (6) ; 281 (4-6, 10) 282 (3) ; 283 (3) ; 284 (3) ; 285 (1, 4-8) ; 289 (3, 5) 290 (6) ; 291 (2, 4, 6) ; 292 (4, 5, 10, 12-14) ; 293 (1-3) 294 (4, 5) ; 296 (2, 5, 6, 9-11) ; 297 (2) ; 298 (1, 2) ; 299 (4) ; 300 (3, 4, 8) ; 301 (6) ; 302 (2, 4, 6) ; 303 (2, 7) ; 304 (2, 4, 5, 9) ; 305 (3, 6) ; 306 (2, 4) ; 327 (4) ; 341 (6) ; 342 (5, 8). Formulae, algebraic, for solution of tangent circle questions in church window problems. I. One-light windows : a. Type I: 1. General formula : 242 (4). 2. Special case : 281 (8, 9). 346 INDEX TO PROBLEMS AND THEOREMS b. Type II : 1. General formula : 272 (7). 2. Special cases : 273 (7) ; 282 (3) ; 293 (1). II. Two-light windows : o. Type I : 1. General formulae : 275 (23, 25) ; 279 (5). 2. Special cases: Under 275 (23): 275 (9); 276; 278 (4) ; 280 (1) ; 283 (3) ; 284 (3) ; 302 (4) ; 304 (4). Under 279 (5) : 280. 6. Type II : 1. General formula : 281 (3). 2. Special cases : 282 ; 283 (4) ; 284 ; 285 (5) ; 286 ; 291 (4, 6) ; 293 (3) ; 296 (9) ; 297 (2) ; 298 (2) ; 303 (2) ; 307 (5). III. Three-light windows : o. Type I : General formula : 288 (7, 9). 6. Type II : 1. General formulae : 289 (4) ; 292 (11). 2. Special oases : Under 289 (4) : 290 (6) ; 291 (2) ; 292 (4) ; 306 (4). Under 292 (11) : 293 (2). c. Type III : 1. General formula : 294 (3). 2. Special cases: 296 (2, 5, 10); 298 (1); 303 (2); 304 (2) ; 305 (3, 6) ; 306 (2) ; 307 (3, 4). IV. Five-light windows : a. Type I : 1. General formula: 299 (5). 2. Special cases : 300 (3) ; 301 (6) ; 302 (2). 6. Type II : 303 ; 304 (special cases of 294). (See Windows pointed, in Index to Notes and Illustrations.) Fourth proportional : 66 (15). Fractions, algebraic: I. Combined with surds : (See Square roots III.) II. Without surds : 51 (6(3)) ; 56 (3) ; 57 (3) ; 62 (3, 6, 7) ; 66 (4) ; 71 (4) ; 74 (3) ; 75 (3) ; 96 (2) ; 112 (3) ; 113 (3) ; 342 (4). Geometric interpretation of : I. Negative results : 341 (8) ; 342 (8-11). II. Solutions of equations : INDEX TO PROBLEMS AND THEOREMS 347 a. Linear : 242 (8) ; 342 (9-12). 6. Quadratics: 54 (17, 19-21); 56 (11, 13-15); 57 (7, 9, 10) ; 66 (12) ; 71 (10) ; 327 (6) ; 341 (8) ; 342 (7). Hexagons, irregular : I. Areas of : 10 (2) ; 27 (2, 3) ; 51 (5). II. Congruence of : 10 (4) ; 75 (1) ; 156 (5) ; 157 (7) ; 159 (5). III. Similarity of: 10 (5, 6). Hexagons, regular : I. Areas of : 14 (7) ; 24 (2) ; 39 (7) ; 40 (4, 5) ; 141 (8) ; 142 (4) ; 143 (7) ; 144 (4) ; 154 (7) ; 155 (5). (See Curved figures, areas of.) II. Congruence of : 143 (5). III. Construction of: 39 (3). IV. Determination of: 14 (2); 40 (2); 141 (4); 143 (5); 144 (1); 151 (11); 171 (1); 191 (5). V. Length of side: 39 (4). Indeterminate equations. (See Equations.) Inequalities : 327 (1). , Inscribed circles : I. In triangles: 164 (1); 165 (1); 236 (3). Applications: 118(2); 163(4); 164(1,9); 167(1); 174 e< se?. II. In squares : 14 ; 88 ; 98 ; 105. III. In quadrilaterals: 169 (1); 193 (1); 209 (1). IV. In ovals: 203 (7); 204 (4); 244 (1). V. Incurved triangles: 230 (1); 248 (1); 249 (4); 256 (1); 257. VI. In Gothic arches : 241 (2) ; 242 (1, 2) ; 243 (1) ; 258 (1) ; 259 (1). VII. In curved polygons : 118(2); 191(2); 196(4); 197(3); 203 (3); 204 (2); 210 (1); 214 (2); 218 (4); 250 (3); 261 (2). Intersecting circles, circles tangent to : I. Circles equal : 230 (4) ; 248 (2) ; 251 (3) ; 252 (1) ; 262 (1) ; 263 (1) ; 275 (3) ; 276 (3, 7) ; 290 (2, 5) ; 301 (2, 5) ; 302 (1, 3). II. Circles not equal : 244 (3). 348 INDEX TO PROBLEMS AND THEOREMS Intersecting loci, used in construction of circles : 248 (2) ; 251 (3) ; 265 ; 266 ; 276 (7) ; 290 (5) ; 295 (1) ; 300 (6) ; 334 (10) ; 335 (9). (See Loci.) Isosceles right triangles: I. Areas of: a. Not requiring binomial surds : 5 (2) ; 54 (5, 6) ; 57 (2, 3) ; 94 (6-8) ; 100 (4) ; 131 (6, 7) ; 154 (6). 6. Requiring binomial surds : 29 (3) ; 43 (11) ; 45 (5, 7) ; 82 (7) ; 84 (5) ; 100 (7) ; 150 (14). II. Congruence of : 150 (11). III. Construction of : 5 (1). Isosceles triangles : I. Angles in, problems involving : 41 (3) ; 162 (2). II. Areas of : 5 (2) ; 54 (5, 6) ; 57 (2, 3) ; 64 (3, 4) ; 67 (7) ; 71 (4) ; 94 (6-8) ; 100 (4) ; 131 (6, 7) ; 154 (6). Harder problems: 29 (3); 43 (11); 45 (5, 7); 66 (8, 10, 22); 71 (7, 8, 11) ; 82 (7) ; 84 (5) ; 100 (7) ; 150 (14). III. Congruence of: 66(1); 71 (1); 75(1); 139 (1); 151 (7); 325 (1, 3). IV. Detemination of: 66 (1); 71 (1); 151 (7); 275 (1); 325 (1, 3). (See Lines, equality of, II.) Linear. (See Equations, linear.) Lines, construction of : 43 (1) ; 45 (2) ; 46 (2) ; 56 (5, 7) ; 316 (7) ; 319 (1). Lines, equality of, proved by means of theorems concerning : I. Congruent triangles combined with theorems involving a. Straight Unes : 54 (1) ; 56 (1) ; 60 (3) ; 66 (20) ; 67 (1) 68 (2) ; 69 (1) ; 90 (6) ; 94 (1) ; 235 (1) ; 236 (1) 317 (1, 2); 321 (1, 3). Harder problems : 44 (2) 46 (5) ; 63 (2, 3) ; 70 (2) ; 81 (4) ; 90 (8) ; 94 (5 h) 315 (1). 6. Circles: 88(3); 98 (1); 135 (1); 148(3); 150(3, 7); 151 (3) ; 154 (2) ; 156 (4) ; 158 (2) ; 318 (1). Harder problems: 38 (5); 217 (4, 5). II. Isosceles triangles : 43 (1-3) ; 45 (2) ; 46 (2) ; 54 (2) ; 90 (3) ; 98 (1) ; 135 (1) ; 148 (3 6) ; 150 (3) ; 151 (3) ; 158 (2). III. Bisectors of angles : 43 (7). INDEX TO PR0SLEM8 AND THEOREMS 349 IV. ParaUelograms : 6 (1) ; 8 (4); 9 (1) ; 35 (4); 49 (3); 54 (4) ; 57 (5) ; 85 (1 c, le) ; 94 (1 6) ; 178 (2) ; 316 (1(5)). V. Equal chords : 115 (2 c). VI. Segments made by parallels : 59 (3) ; 62 (4). VII. Proportionallines : 329 (2) ; 334 (8). VIII. Symmetry: 150 (3); 151 (3); 154 (2); 203 (3); 208 (4, 5) ; 216 (5) ; 217 (6) ; 218 (2) ; 219 (4) ; 220 (2). IX. Miscellaneous: 85 (1 d) ; 86 (3); 90 (10, 11); 91 (1) 98 (10) ; 102 (1, 2) ; 104 (8) ; 151 (8, 9) ; 154 (2) 155 (2) ; 157 (4, 5) ; 159 (2, 7, 8) ; 177 (2) ; 208 (1) 211 (1 6) ; 221 (3) ; 251 (7) ; 332 (2) ; 337 (4) ; 343 ; 344 (1) ; 349 (1). (See Equilateral triangles, determination of; Isosceles tri- angles, determination of; Chords.) Lines, lengths of, computed by means of : I. Pythagorean theorem : a. Relations simple : 1. Involving square root of 2 : 3 (2, 3) ; 4 (2, 5) 5 (2, 5) ; 6 (3) ; 7 (3) ; 8 (1-3) ; 10 (2) ; 17 (2) 18 (5) ; 26 (1, 2) ; 29 (3) ; 34 (7) ; 41 (4, 5, 9) 49 (6, 7) ; 50 (3) ; 61 (5) ; 74 (3) ; 104 (4) 118 (4); 121 (4, 5, 7, 8); 154 (5); 156 (7) 183 (5). Harder problems : 51 (6); 66 (9). 2. Involving square root of 3 : 15 (2) ; 23 (3) ; 39 (4) 141 (6); 144 (8); 154 (8); 155 (4); 200 (5) 202 (3) ; 235 (3) ; 243 (3) ; 254 (6) ; 277 (2) 318 (6, 7) ; 321 (4). 3. Involving other square roots : 64 (2, 5) ; 65 (2) ; 99 (5) ; 156 (9) ; 255 (3) ; 298 (3) ; 314 (2, 3) ; 315 (2); 347 (4, 6); 348 (4). 6. Relations more elaborate, including facts concerning: 1.' Squares: 150 (12, 13). 2. Equilateral triangles : 167 (4-7) ; 170 (1) ; 194 (2) ; 203 (4) ; 213 (3) ; 229 (5) ; 248 (3). 3. Octagons: 88 (4); 95 (1); 99 (2). 4. Circles inscribed in curved figures : 275 (15, 16) ; 276 (10-15) ; 277 (3, 4) ; 281 (2) ; 285 (11, 12) ; 288 (3) ; 290 (7) ; 301 (11, 12) ; 303 (4) ; 305 (4). 360 INDEX TO PROBLEMS AND THEOREMS II. Equal ratios made by : o. Similar triangles : 59 (4-9) ; 69 (3, 5) ; 70 (6) ; 275 (20, 21) ; 301 (13, 14) ; 303 (5) ; 315 (2) ; 319 (3) ; 334 (4, 5, 8) ; 335 (3, 6) ; 341 (4, 5) ; 342 (4). 6. ParaUels : 59 (3-9). c. The bisector of an angle : 11 (8) ; 66 (21) ; 68 (3) ; 69 (3) ; 88 (11) ; 90 (9) ; 139 (5). III. Division of a given line into parts proportional to lines in : a. The isosceles right triangle : 43 (8-10) ; 96 (5) ; 118 (4) ; 163 (7) ; 164 (2) ; 174 (5) ; 178 (4) ; 183 (5) ; 184 (6); 185 (6). 5. The equilateral triangles : 172 (4) ; 193 (3) ; 200 (4). c. The regular octagon: 29 (13); 30 (5); 82 (5, 6); 84 (4) ; 85 (2) ; 86 (4) ; 98 (3, 8) ; 104 (11) ; 105 (3). IV. Miscellaneous theorems : a. Easy : 25 (3) ; 34 (5, 6, 8) ; 39 (5) ; 40 (6) ; 138 (5, 6). 6. Harder: 11 (7); 30 (4); 115 (5); 119 (4); 139 (4) 144 (4, 5) ; 154 (9) ; 158 (10) ; 165 (3) ; 168 (5) 169 (2); 179 (3); 188 (2, 5); 209 (2); 210 (2) 230 (2) ; 256 (3) ; 258 (2) ; 259 (2) ; 260 (3) ; 261 (3); 262 (3, 4); 263 (3, 4, 5). Loci of centers of circles (see Inlersecting loci) : I. Passing through two given points : 18 (2) ; 138 (9) ; 266 (3(2)) with appUoations : 301 (8) ; 303 (3) ; 306 (2) ; 326 (1). II. Of given radius passing through a given point : 266 (1(1)) with apphcations: 264; 267 (2) ; 268 (1) ; 270; 295 (1). III. Of given radius tangent to a given line : 265 with appli- cations: 254 (1, 2); 255 (1); 256 (1); 257; 258 (1) ; 259 (1) ; 260 (1) ; 261 (1) ; 262 (1) ; 263 (2) ; 269 (1) ; 334 (10); 335 (9); 266 (1(1)) with applications. (See II above.) IV. Of given radius tangent to a given circle : 265 with appli- cations. (See III above.) V. Tangent to two intersecting lines: 121 (1); 163 (1, 3) 201 (1) ; 202 (1) ; 203 (1, 7) ; 204 (1) ; 205 (1) ; 252 (1) 253 with applications : 163 (5) ; 180 (2) ; 198 (2) 252 (2); 272 (6); 332 (3). VI. Tangent to a given hne at a given point : 18 (1) ; 163 (1, 3) INDEX TO PROBLEMS AND THEOREMS 351 201 (1) ; 202 (1) ; 203 (1, 7) ; 205 (1) ; 266 (3(1)) ; 119 (7) and 246 with applications : 119 (3) ; 230 (3) ; 237 ; 238 (1) ; 241 (2) ; 242 (1) ; 244 (2) ; 269 (4) ; 288 (4) ; 297 (1) ; 334 (1) ; 335 (1) ; 336 (1) ; 340 (2, 3) ; 341 (1) ; 342 (1-3) ; 346 (1, 2). VII. Tangent to two concentric circles : 276 (4) ; 290 (3, 4) ; 301 (3, 4) ; 302 (1). VIII. Tangent to two given equal circles: 230 (4); 248 (2); 249 (2) ; 251 (3) ; 252 (1) ; 262 (1) ; 263 (1) ; 275 (3) ; 276 (3) ; 287 (1) ; 290 (2) ; 301 (2). IX. Tangent to any two given circles : 244 (3) ; 279 (4). Loci of points : I. Equally distant from two given points, used for : a. Construction of circles. (See Circumscribed circles, and Circles, construction of, II.) b. Determination of coUinear points: 51 (1) (see sug- gestion) ; 70 (1) ; 82 (3) ; 115 (2) ; 141 (2) ; 143 (2) ; 148 (4); 150 (6); 151 (6); 158 (4, 8); 159 (3); 184 (2) ; 205 (2) ; 241 (1) ; 311 (1). c. Determination of concurrent lines : 82 (3); 94 (1) ; 139 (1) ; 157 (3) ; 208 (1) ; 337 (4) ; 346 (5, 7). II. Equally distant from the sides of an angle, used for : o. Construction of circles. (See Inscribed circles I, II, III, and Circles, construction of V.) b. Determination of equal lines: 43 (7). Mean proportional : 253 (1). Measurement of arcs and angles: 88 (3, 9); 115 (2); 148(3,6); 150 (4, 5) ; 151 (4, 5) ; 156 (4) ; 158 (2, 3) ; 159 (4) ; 216 (4) ; 318 (3) ; Figs. 119-123. Medians: theorems concerning the intersections of the medians of a triangle used in : I. Conputations of areas of curved figures : 203 (5 b). II. Computations of lengths : 167 (4-7) ; 168 (5) ; 169 (2) ; 191 (3) ; 194 (2) ; 203 (4) ; 248 (3). III. Construction of circles : 169 (1). Octagons, irregular: I. Areas of: 11(3); 12(1); 27(2); 28(5). II. Similar: 11 (4). 352 INDEX TO PROBLEMS AND THEOREMS Octagons, regular: ' I. Areas of: 29 (3); 41 (6); 84 (5); 88 (6); 104 (10) 177 (6). II. Construction of: 29 (1); 41 (1,9); 66 (23); 88 (1,10) 90 (7); 150 (8). III. Determination of: 29 (8); 30 (3); 66 (23); 68 (5) 69(6, 7) ; 70 (7) ; 71 (15) ; 73 (9) ; 88 (1, 10) ; 90 (7, 8, 10) ; 91 (1) ; 94 (5) ; 98 (1) ; 104 (7) ; 177 (5) ; 197 (2) IV. Length of side : 41 (11, 12) ; 66 (21) ; 88 (11). Ovals, areas of: 18 (4); 37 (3-4); 112 (2, 3); 113 (2, 3); 114 (3, 6, 6) ; 122 (3) ; 123 (3) ; 174 (10) ; 201 (2) ; 202 (4) ; 203 (5); 204 (3); 211 (3); 213 (4). Parallel lines, proved by means of theorems concerning : I. Alternate interior or corresponding angles with theorems concerning : o. Congruent triangles: 10 (1); 29 (10); 63 (1) (see suggestion) ; 66 (1) ; 68 (5) ; 74 (1). 6. Isosceles triangles: 54 (1, 3); 98 (1); 129 (1). II. Parallelograms : 11 (6) ; 35 (2) ; 41 (2). Harder problems : 39 (2); 229 (3). III. Chords and tangents : 14(3); 208(1); 318(2). IV. Proportional segments: 51 (1) ; 54 (1, 3) ; 81 (1) ; 98 (1) ; 129 (1) ; 131 (2) ; 142 (1) ; 143 (1) (see suggestion) ; 316 (1(4)); 317 (1); 320 (1). V. MisceUaneous relations : 44 (1); 55 (1); 85 (1); 86(3); 88 (2) ; 94 (1) ; 96 (1) ; 156 (3) ; 159 (6) ; 269 (2) ; 346 (6) ; 349 (3). Parallelograms. (See Rectangles, Rhombuses, and Squares.) I. Areas of : 56 (2, 3) ; 57 (2, 3) ; 73 (4, 10). II. Congruence of: 69 (1); 65 (1); 73 (1). III. Determination of : 57 (1); 59 (1); 94 (1); 103 (1). Pentagons, irregular: I. Areas of: 8 (1-3); 9 (4); 13 (1). II. Congruence of: 8 (5); 74 (1). III. Similarity of : 8 (6, 7) ; 9 (2) ; 13 (2, 3, 4). Pentagons, regular: I. Determination of: 148 (7). II. Length of side : 209 (2). INDEX TO PROBLEMS AND THEOREMS 353 Perimeters of curved figures: 109; 110; 124 (3); 126 (4, 5); 127 (9, 10). Perpendicular bisectors : I. Determination of: 11 (2); 69 (1); 115 (2); 141 (5); 150 (7) ; 208 (1 d) ; 236 (1) ; 318 (1) ; 321 (2) ; 346 (6) ; 349 (3). II. Used for : a. Collinear points: 51 (1) (see suggestion); 70 (1) 82 (3) ; 115 (2) ; 141 (2) ; 143 (2) ; 148 (4) ; 150 (6) 151 (6); 158 (4, 8); 159 (3); 184 (2); 205 (2) 241 (1) ; 311 (1). 6. Concurrent lines : 82 (3); 94 (1); 139 (1); 157 (3); 208 (1) ; 337 (4) ; 346 (5, 7). c. Other purposes : 177 (3). Perpendicular lines, proved by means of theorems concerning : I. Congruent triangles : 11 (2) ; 54 (3) ; 64 (9) ; 69 (1) ; 275 (1) ; 286 (4) ; 314 (1) ; 316 (1) ; 317 (1) ; 318 (1, 2) ; 320 (1). II. Perpendiculars and parallels : 35 (3) ; 38 (3) ; 236 (1) ; 315 (1). III. Perpendicular bisectors : 115 (2/) ; 141 (5) ; 150 (7) ; 208 (1 d) ; 321 (2). IV. Similar triangles : 63 (2). V. Miscellaneous relations : 18 (3) ; 63 (4) ; 346 (6) ; 349 (3). Points, determination of: 216 (1). (See Circles, construction of, for determination of centers of circles.) Polygons, irregular, areas of, figures derived from : I. Squares and triangles: 8 (1-3); 9 (4); 10 (2); 11 (3); 12 (1) ; 13 (1) ; 27 (2, 3) ; 28 (5) ; 45 (6) ; 51 (5, 6) ; 65 (2) ; 67 (2, 6, 7) ; 72 (3) ; 73 (6) ; 74 (2, 4, 5) ; 75 (2, 3) ; 81 (2, 3) ; 94 (6-8) ; 104 (2, 5, 6) ; 130 (2-4) ; 131 (6, 7). Harder problems : 45 (8) ; 95 (2) ; 100 (4) ; 102 (3, 4, 6) ; 103 (3, 4). II. Regular hexagons : 141 (8) ; 142 (4) ; 143 (7) ; 144 (5) ; 154 (6, 7, 10) ; 155 (6) ; 156 (8, 10) ; 158 (11). III. Regular octagons : 31 (3) ; 68 (3, 6) ; 73 (10) ; 81 (5) ; 83 (2) ; 84 (5) ; 85 (3) ; 88 (5, 6) ; 94 (13) ; 98 (4, 9, 11) ; 99 (3) ; 100 (7) ; 103 (6) ; 104 (10, 11) ; 105 (4). 354 INDEX TO PROBLEMS AND THEOREMS Polygons, irregular, congruence of : I. Four-sides: 13 (5); 63 (2); 67 (1); 69 (2); 70 (4) ; 72 (1) ; 73 (1) ; 74 (1) ; 150 (10) ; 161 (7, 8) ; 154 (3). 11. Many-sides: 8 (5); 10 (4); 74 (1); 75 (1); 156 (5); 157 (7) ; 159 (5). Polygons, regular: I. Areas of: o. Hexagons: 14 (7); 24 (2); 40 (4, 5); 141 (8); 142 (4) ; 143 (7) ; 144 (4) ; 154 (7) ; 155 (5). 6. Octagons : 29 (3) ; 41 (6) ; 84 (5) ; 88 (6) ; 104 (10) ; 177 (6). II. Congruence of: Hexagons : 143 (5). III. Construction of: a. Hexagons : 39 (3). b. Octagons: 29 (1); 41 (1, 9); 66 (23); 88 (1, 10); 90 (7) ; 150 (8). IV. Determination of : 14 (2) ; 29 (8) ; 30 (3) ; 40 (2) ; 66 (23) 68 (5) ; 69 (6, 7) ; 70 (7) ; 71 (15) ; 73 (9) ; 88 (1, 10) 90 (7, 8, 10) ; 91 (1) ; 94 (5) ; 98 (1) ; 104 (7) ; 141 (4) 143 (5) ; 144 (1) ; 148 (7) ; 151 (10, 11) ; 152 (13, 14, 15) ; 158 (5) ; 171 (1) ; 177 (5) ; 191 (5) ; 197 (2). V. Length of side : a. Decagons: 209 (2). 6. Duodecagons: 115 (5); 144 (4); 158 (10). c. Hexagons: 39 (4). d. Octagons: 41 (11, 12); 66 (21); 88 (11). e. Pentagons: 209 (2). Polygons, .similar: I. Areas of: 8 (6); 10 (5); 11 (4). II. Construction of : 8 (7); 10 (6); 13 (2-4); 14 (11, 12). III. Determination of : 8 (6) ; 9 (2) ; 10 (5) ; 98 (10) ; 142 (2) ; 143 (6). Proportional lines, obtained by : I. Bisectors of angles: 11 (8); 66 (21); 68 (3); 69 (3); 88 (11) ; 90 (9) ; 139 (5). II. Parallels, used for : o. CoUinear points : 94 (2, 3). INDEX TO PROBLEMS AND THEOREMS 356 6. Lengths of lines : 59 (3-9). c. Determination of ratios : 51 (2, 4) ; 59 (3-9) ; 61 (2, 3) ; 62 (4, 5). III. Ratios derived from given data, used to form parallel Knes : 51 (1); 54 (1 3); 81 (1); 98 (1); 129 (1); 131 (2); 142 (1) ; 143 (1) (see suggestion) ; 316 (1) ; 317 (1) ; 320 (1). IV. Similar triangles, used to find lengths of lines : 59 (4-9) ; 69 (3,5) ; 70 (6) ; 275 (20,21) ; 301 (13-15) ; 303 (5) ; 315 (2) ; 319 (3) ; 334 (4, 5, 8) ; 335 (3,6) ; 341 (4, 5) ; 342 (4). Proportionals, construction of: I. Fourth proportional : 66 (15). II. Mean proportional : 253 (1). III. Proportional segments : 54 (14) ; 56 (5, 7) ; 71 (17, 20) ; 84 (3) ; 96 (4) ; 272 (10) ; 310 (2) ; 311 (3). Pythagorean theorem: I. Proof for: 59 (12). II. Used for: a. Lengths of lines. (See Lines, problems: 258 (2); 259 (2); 263 (3, 4). 6. Formation of equations. (See and Equations quadratic, I, squares.) Quadratics. (See Equations, quadratic.) Quadrilaterals : I. Areas of : 74 (2) ; 88 (5) ; 98 (4, 9). (See Parallelograms, Rectangles, Rhombuses, Squares, Trapezoids.) II. Congruence of: 13 (5); 63 (2); 67 (1); 69 (2); 70 (4); 72 (1) ; 73 (1) ; 74 (1) ; 150 (10) ; 151 (7, 8) ; 154 (3). Radical equations. (See Equations.) Radicals. (See Square roots.) Ratios of areas, involving : I. Arithmetical computations, figures derived from : o. Parallelograms and triangles : 44 (4) ; 45 (8) ; 102 (6). lengths of.) Harder 260 (3) ; 262 (3, 4) ; Equations linear, I, containing binomial 356 INDEX TO PROBLEMS AND TBEOREMS b. Hexagons: 39 (7). c. Octagons: 41 (16) ; 68 (6) ; 95 (2) ; 98 (5, 7). d. Parts of circles: 36 (2, 3); 113 (5); 118 (7, 10); 126 (6, 7) ; 171 (6) ; 178 (7, 8) ; 181 (7) ; 183 (7) ; 185 (8) ; 213 (5) ; 245 (5) ; 273 (4) ; 279 (3) ; 280 (6) ; 292 (7) ; 293 (5) ; 298 (4). II. Algebraic computations : 54 (19) ; 55 (3, 4) ; 56 (14, 15) ; 57 (9) ; 59 (7 d) ; 59 (8 g) ; 59 (9 /) ; 70 (3) ; 102 (7). III. Theorems concerning areas of triangles and parallelograms : 6 (1) ; 7 (2) ; 24(3) ; 48 (5) ; 65 (1) ; 66 (2, 3,5) ; 67 (3,8) ; 71 (2) ; 72 (2) ; 73 (2) ; 75 (4) ; 86 (3) ; 131 (4) ; 314 (1) ; 315 (1); 316 (1); 317 (1). IV. Theorems concerning similar figures : 8 (6) ; 9 (3) ; 10 (5) ; 11 (4) ; 13 (2) ; 91 (2) ; 104 (12) ; 114 (2, 6) ; 135 (4, 5) ; 229 (11). V. MisceUaneous methods : 9 (5) ; 62 (8) ; 64 (8) ; 130 (note) ; 134 (5). Ratios of lines, obtained from : I. Algebraic equations: 54 (10, 11, 13); 74 (11); 75 (6). II. Computations of length : 29 (2) ; 66 (4) ; 119 (5) ; 243 (2). III. Known ratios : a. The side to diagonal of a square : 5 (3, 4) ; 9 (3) ; 11 (1) ; 12 (2) ; 25 (2) ; 2g (1) ; 34 (4) ; 43 (5, 6) ; 86 (3) ; 96 (3) ; 118 (9) ; 133 (3) ; 163 (6). 5. Lines in an equilateral triangle : 39 (6) ; 170 (3) ; 193 (2) ; 242 (5) ; 272 (1) ; 310 (1). c. Lines in a regular octagon: 29 (2); 30 (1); 31 (2); 41 (9, 10) ; 84 (2) ; 85 (1) ; 91 (2). IV. Theorems concerning : a. Similar figures : 59 (4-9) ; 67 (5) ; 100 (3) ; 316 (1, 3-6) 317 (3, 4) ; 335 (5, 6). 6. ParaUel lines : 61 (2, 4) ; 59 (3) ; 61 (2, 3) ; 62 (4, 5) 316 (3-6) ; 317 (3, 4). c. The* bisector of an angle: 11 (8); 66 (21); 68 (3) 69 (3) ; 88 (11) ; 90 (9) ; 139 (5). V. MisceUaneous: 136 (4); 138 (4); 272 (5); 311(2). Rectangles, areas of : 6; 7; 51(6). {See Squares.) Regular polygons. (See Polygons, regular.) INDEX TO PB0BLEM8 AND THEOREMS 367 Rhombuses : I. Angles of: 142 (3). II. Areas of : 11 (6) ; 40 (4, 5) ; 62 (7) ; 64 (3, 4, 8) ; 73 (4, 5) ; 82 (7) ; 86 (5) ; 96 (2, 6) ; 105 (4) ; 142 (4) ; 143 (7) ; 158 (11). III. Congruence of: 29 (9); 73 (1); 82 (4); 83(4); 86 (3); 88 (1) ; 96 (1) ; 142 (3) ; 143 (3). IV. Determination of : 29 (9) ; 40 (3) ; 64 (1) ; 73 (1) ; 82 (4) ; 83 (4); 86 (3); 88 (1); 96 (1); 105 (2); 142 (3); 143 (3) ; 320 (1). V. Diagonals of : 11 (7). Right angle, triseotion of : 115 (2 6). Right triangles: I. Areas of: 5 (2); 54 (5, 6); 57 (2, 3); 61 (7); 62 (6); 94 (6-8) ; 100 (4) ; 131 (6, 7) ; 154 (6). Harder prob- lems: 29 (3); 43 (11); 45 (5, 7); 82 (7); 84 (5); 100 (7) ; 150 (14). II. Congruence of : 70 (4) ; 150 (11). III. Construction of: 5 (1). IV. Proportional lines in : 69 (3, 5) ; 70 (6). Rotation about a point, used for : I. Equal angles : 150 (4) ; 151 (4) ; 154 (4). II. Equal arcs : 203 (3) ; 208 (4) ; 216 (5) ; 217 (9) ; 218 (3) ; 219 (5) ; 220 (3). III. Equal lines :■ 150 (3) ; 151 (3) ; 154 (2) ; 203 (3) ; 208 (4, 5) ; 216 (5) ; 217 (6) ; 218 (2) ; 219 (4) ; 220 (2). IV. Congruent figures : 150 (10, 11); 151 (7, 8). V. Regular polygons : 150(8); 151(10,11). Sectors : I. Areas of : 15 (3). {See Curved figures, areas of .) II. Congruence of : 118(1). Segments : I. Areas of: 17 (1); 115 (6); 116 (2); 124 (2); 136 (2). (See Curved figures, areas of.) II. Construction of segment on a given line to contain a given angle : 136 (5). 358 INDEX TO PROBLEMS AND THEOREMS Semicircles : I. Areas of: 101 (1). (See Curved figures, areas of.) II. Construction of: 52 (1); 133 (4); 138 (7); 184 (5); 200 (1). Similar polygons. (See Polygons, similar.) Similar triangles. (See Triangles, similar.) Simultaneous equations. (See Equations.) Square roots: (See Lines, lengths of; Squares, areas of; Trapezoids; Triangles.) I. Numerical and monomial: a. Areas : 3-10 ; 14 (7, 10) ; 15 (2) ; 22 (4, 5) ; 24 (2) ; 27 (2) ; 40 (4) ; 48 (4) ; 74 (2) ; 94 (7, 8) ; 104 (2, 5) ; 129 (2, 3) ; 130 (2, 3) ; 141 (7, 8) ; 142 (4). 6. Lengths: 3-10; 23 (3); 26 (2); 34 (7); 39 (4); 59 (7, 8); 64 (5); 70 (6); 104 (4); 314 (2, 3); 315 (2). c. Ratios: 11 (1); 12 (2); 25 (2); 28 (1); 34 (4); 39 (6) ; 133 (3) ; 170 (3). II. Numerical and binomial : a. Areas : 29 (3) ; 31 (3) ; 41 (6, 7) ; 43 (8-11) ; 45 (5-7) ; 46 (7) ; 66 (22) ; 67 (6) ; 68 (3) ; 82 (6, 7) ; 84 (5) ; 104 (10) ; 129 (5) ; 164 (3) ; 177 (6). 6. Lengths : 11 (7) ; 29 (13) ; 30 (4, 5) ; 66 (21) ; 68 (3) ; 69 (3) ; 82 (5, 6) ; 88 (11) ; 90 (9) ; 115 (5) ; 144 (4) ; 158 (10) ; 163 (7) ; 167 (7) ; 209 (2) ; 248 (3). c. Ratios: 29 (2); 30 (1). III. Combined with literal or incommensurable expressions : a. Areas : 51 (6) ; 52 (3, 4) ; 54 (6) ; 66 (10) ; 67 (7) 71 (11) ; 74 (4, 5) ; 81 (2) ; 94 (6) ; 104 (6) ; 122 (3) 124 (2, 3) ; 126 (2, 3) ; 127 (3-6) ; 129 (4) ; 130 (4) 132 (1-3); 133 (2); 134 (2-4); 171 (4); 226 (3) 6. Lengths : 51 (6) ; 52 (2) ; 59 (9) ; 66 (9). Squares : I. Areas of : a. Without radicals : 49 (9) ; 50 (2, 4) ; 54 (5, 6) ; 56 (2, 3) ; 57 (2, 3) ; 94 (6, 7, 8). b. Involving monomial radicals : 3 ; 4; 27 (2) ; 48 (3, 4) ; 49 (10); 104 (2); 129 (2-4). Harder problems: INDEX TO PROBLEMS AND THEOREMS 359 69 (7-9) ; 66 (10) ; 94 (9) ; 100 (4) ; 102 (3, 4) ; 103 (2). c. Involving binomial radicals : 31 (3) ; 44 (3, 4) ; 46 (7) 82 (6) ; 83 (3) ; 84 (5) ; 86 (5) ; 96 (6) ; 104 (9) 119 (2); 120 (3); 129 (5); 150 (14); 158 (11) 162 (4) ; 260 (6). II. Congruence of : 6 (1) ; 35 (6) ; 70 (3) ; 84 (1) ; 94 (5) ; 96 (1) ; 98 (1) ; 100 (1) ; 104 (7) ; 135 (1) ; 144 (1) ; 158 (6). III. Construction of: 34 (1, 3); 35 (1); 50 (1); 162 (3); 316 (2-4). ly. Determination of : 29 (7) ; 30 (3) ; 44 (2) ; 46 (6) ; 48 (1, 2) ; 49 (1, 3, 5) ; 51 (1) ; 54 (1, 3) ; 56 (1) ; 57 (1) 59 (1, 2) ; 61 (1) ; 62 (1) ; 63 (2) ; 66 (1) ; 82 (2, 3) 86 (3) ; 103 (1) ; 104 (1) ; 114 (1) ; 115 (2) ; 119 (1) 120 (3) ; 129 (1) ; 131 (5) ; 150 (9) ; 151 (12) ; 163 (2) 174 (2) ; 175 (2) ; 178 (1) ; 179 (1) ; 181 (2) ; 184 (4) 185 (1) ; 260 (4). Star-polygons : I. General : 145. II. Three-pointed : Construction of : 152 (4). III. Pour pointed : a. Areas : 1. Irregular forms : 65 (2) ; 66 (3, 5, 10) ; 67 (7). 2. Regular forms : 66(22); 68(3). b. Congruence of : 151(13). c. Construction of : 1. Irregular forms : 65 ; 66 ; 67. 2. Regular forms : 68 ; 150 (16, 22, 23) ; 151 (13) ; 152 (3, 11). IV. Five-pointed : Construction of : 148(1); 149(1). V. Six-pointed : a. Areas : 24 (2) ; 141 (8) ; 142 (4) ; 143 (7) ; 144 (5) ; 154 (6) ; 155 (6) ; 156 (8) ; 158 (11). b. Construction of : 141 ; 151 (14) ; 152 (2; 11). c. Determination of : 142 (2) ; 143 (6). VI. Eight-pointed : u,. Areas of : 360 INDEX TO PROBLEMS AND THEOREMS 1. Irregular forms : 94 (6-8, 13) ; 103 (3, 4). 2. Regular forms : 98 (4, 9) ; 104 (10). 6. Construction of : 1. Irregular forms : 93 ; 94 (2, 3) ; 97. 2. Regular forms : 93; 97; 150(1,18); 152(11). VII. Ten-pointed : Construction of : 149(1). VIII. Twelve-pointed : a. Composition of : 151 (14) ; 152 (6). 6. Construction of : 151 (1); 152(1,5). IX. Sixteen-pointed : a. Composition of : 150(19,21). 6. Construction of : 150 (19, 20). X. Twenty-four-pointed : a. Composition of : 152 (9, 10). h. Construction of : 152 (7, 8). Sum of the angles of a triangle, used for : I. Proofs of equality of angles : 38 (3) ; 63 (2) ; 341 (2). II. Computations for measures of angles : 41 (3) ; 162 (2) ; angles in Figs. 57 and 68. Superposition, proofs by: 250 (1); 325 (4). (See Curved figures, congruence of.) Supplementary angles: 150 (4). Surds. (See Square roots.) Symmetry : 112 (6) ; 148 (2) ; 150 (2, 4, 10) ; 151 (2, 10) ; 208 (4) ; 217 (6) ; 219 (3-5) ; 229 (9) ; 232 (3, 4). (See Rotation.) Tangent circles : I. Construction of : a. Circles equal : 1. Two tangent circles : 36; 37; 38; 139(4). 2. Three tangent circles : 138(7); 165(1); 167-169. 3. Four tangent circles: 174-180; 183; 185; 188. 4. Many tangent circles : 191-199. 6. Circles unequal. (See Inscribed circles, IV-VII, and Chap. V, parts 2 and S.) II. Determination of : 108 (1); 118 (3); 136 (4); 138 (4) • 225 (4) ; 236 (2) ; 280 (4) ; 286 (2) ; 301 (1) ; 337 (2) '; 340 (1) ; 343 ; 344 (1) ; 347 (2) ; 348 (2) ; 349 (2). INDEX TO PROBLEMS AND THEOREMS 361 III. Theorems concerning, used for: a. Collinear points : 276 (6) (see suggestion) ; 165 (2) ; 241 (1) ; 245 (2) ; 273 (1) ; 275 (2) ; 286 (1, 3) ; 287 (2) ; 301 (7) ; 334 (2). 6. Construction of circles : 276 (3) ; (see suggestion) ; 230 (4) ; 248 (2) ; 249 (2) ; 251 (3) ; 252 (1) ; 275 (3) ; 287 (1) ; 290 (2) ; 301 (2) ; 262 (1) and 263 (1) ; 244 (3) and 279 (4). (See Loci of Centers of Circles, VII-IX ; Circles, construction of, VIII ; Intersecting circles, I.) Tangent lines : I. Construction of: 329 (4, 5) ; 330 (2, 3) ; 346 (8). II. Determination of: 229 (3); 235 (4). III. Equality of : 329 (2). Trapezoids : I. Areas of : a. Requiring monomial surds only: 11 (5); 14 (10); 40 (4, 5) ; 48 (4) ; 49 (8) ; 54 (5) ; 56 (2) ; 102 (3) ; 129 (2, 3) ; 131 (6). Literal problems : 54 (6-8) ; 55 (2) ; 56 (3) ; 102 (4) ; 103 (5) ; 129 (4) ; 131 (7) ; 135 (5). 6. Requiring binomial surds : 41 (6, 7) ; 43 (8-10, 12) ; 84 (5); 85 (3); 96 (2); 129 (5); 229 (6). II. Congruence of : 43 (4); 102 (1). III. Determination of : 11 (5) ; 14 (4) ; 49 (4) ; 85 (1) ; 96 (1) ; 229 (6). Triangles, areas of : I. Requiring monomial surds only : 5 (2) ; 14 (7) ; 15 (2) ; 22 (4, 5) ; 54 (5, 6) ; 57 (2, 3) ; 61 (7) ; 62 (6) ; 64 (3, 4) ; 67 (7) ; 71 (4) ; 94 (6-8) ; 100 (4) ; 131 (6, 7) ; 154 (6). Harder problems : 66 (8, 10) ; 71 (7, 8, 11) ; 141 (7, 8) ; 142 (4) ; 143 (7). II. Requiring binomial surds: 29 (3); 43 (11); 45 (5, 7); 66 (22) ; 82 (7) ; 84 (5) ; 100 (7) ; 115 (6) ; 144 (4) ; 150 (14); 229 (10). Triangles, congruent: I. Determination of : 66 (1) ; 70 (4) ; 71 (1) ; 75 (1) ; 139 362 INDEX TO PROBLEMS AND THEOREMS (1); 141 (3); 142 (1); 143 (4); 144 (1); 150 (11); 151 (7) ; 213 (1) ; 325 (1, 3). II. Used for: a. Equality of angles : 67 (1) ; 98 (1) ; 318 (2) ; 321 (2) ; 329 (3). 6. Equality of lines: 54 (1); 56 (1); 60 (3); 66 (20); 67 (1) ; 68 (2) ; 69 (1) ; 88 (3) ; 90 (6) ; 94 (1) ; 98 (1) ; 135 (1) ; 148 (3) ; 150 (3, 7) ; 151 (3) ; 154 (2) ; 156 (4) ; 158 (2) ; 235 (1) ; 236 (1) ; 317 (1, 2) ; 318 (1); 321 (1,3). Harder problems : 38(5); 44 (2); 46 (5) ; 63 (2, 3) ; 70 (2) ; 81 (4) ; 90 (8) ; 94 (5 6) ; 217 (4, 5) I 315 (1). c. MisceUaneous purposes: 11 (2); 54(3); 69 (1) ; 229 (3); 275 (1); 286 (4); 314 (1); 316 (1); 317 (1); 318 (1, 2); 320 (1) ; 63 (1) and similar problems: 10 (1) ; 29 (10) ; 66 (1) ; 68 (5) ; 74 (1). Triangles, similar: I. Determination of : 69 (2) ; 70 (5) ; 275 (18) ; 328 (1) ; 341 (2). II. Used for: a. Equality of angles : 63 (2) ; 341 (2). b. Lengths of lines : 59 (4r-9) ; 69 (3, 5) ; 70 (6) ; 275 (20, 21); 301 (13,14); 303 (5); 315 (2); 319 (3); 334 (4, 5, 8) ; 335 (3, 6) ; 341 (4, 5) ; 342 (4). c. Determination of ratios : 59 (4r-9) ; 67 (5) ; 100 (3) ; 316 (1, 3-6) ; 317 (3, 4) ; 335 (5, 6). d. MisceUaneous purposes : 327 (1). Verification by paper cutting and folding : 59 (12) ; 127 (8). INDEX TO NOTES AND ILLUSTRATIONS Numbers in Koman type refer to paragraphs and exercises, as in the previous index. Numbers in italics refer to illustrations. Alhambra, 62 (9), 147, 215; 6Za, 78, 106a. Amiens Cathedral, France, 23S a. Anagni, Italy, 140. Arabic art (Mohammedan, Moslem, Saracenic) : characteristics of, 2, 19, 140, 147, 153. designs from, 97, 215; 4S, 60, 62, 66a, 67, 71a, 73, 88, 102, 103, 117, 118, 125, 126, 127, 128a, 129, 129a. use of pointed arch, 323. (See AlhaTnbra, Cairo, Moorish, Spanish art.) Arabic proofs, 59 (12). Arches : Basket-handle : construction of, 340-344; 286- 289. history of, 338, 339. Gothic : gables over, 324, 329-331; 277-279a. (See Canopy.) geometry of, 325-328. history of, 323. use in window heads, 184-281a. History of, 322. Modern use of, 322, Ogee : [284. construction of, 334-337; 281- occurrence, 333. Semicircular : designs, 169, 202, 233, 234, HO, 241, 27 Sa. occurrence, 161, 223. Trefoiled : construction of, 233 (3), 234 (8) ; 190, 190b, I94. occurrence, 231. Tudor : construction of, 346-349; 290- 293. history of, 345. Assyrian, 106. Berlin, Germany, 189. Boston, Eng., 197a. Boston, Mass., 189; 166a, 190b, 197a, 204a, 266a. Boyton, Eng., 206a. Brooldyn, N.Y., 308. Byzantine architecture, trefoils and quadrifoils in, 166. Byzantine mosaics. (See Mosaics.) Cairo, Egypt, 33; 73. Canopy, 324 ; ie4b, 277-279a. Canterbury Cathedral, Eng., 101, Carcassonne, France, 363 364 INDEX TO NOTES AND ILLUSTRATIONS Carlisle Cathedral, Eng., 223 ; 185b. Ceiling patterns, 19; S4a, 80a, 91a, 9Sa, 97a, 14Sa, IBOa, 178a. Chartres Cathedral, France, 100, 162. Chertsey Abbey, Eng., 173. Chester Cathedral, Eng., 166a, 267. Chicago, m., lS9a, 151, 171a, 188a, 189a, 200a, 206b, 207a, 210a, 230a, 2SSa, 236a, £46a, 249, 258a. Chinese art, 33, 106, 173. Circles, designs based on circles or parts of circles: Arabic, 88, 102, 103. ceilings, 91a, 9Sa, 97a. embroidery, 99a. glass, 29, SO, 100. Gothic, 92a, 94, (p. 105, remark), 96, 130-264a. iron work, 31, 90a, 100, 108b, 112a, 113a, 114a. linoleum, 110, Ilia. plaster, 96. Roman mosaics, 4^, 86a, 87. tiles, 13, 14, 14a, 14b, 16, 16, 92,92a. Clay products, 76. (See also Mosaics, Tiles, and Tiled floors.) Cleveland, Ohio, 169. Cologne Cathedral, Germany, 216a. Curved triangles, 184, 186, 185a, 185b, 186, 187, 187a, 188, 188a, 189, 189a, 190a, 191, 192, 193, 204-208a, 212, 213, 220, 222, 226. Cusps, 231, 232, 237; 190-196. Decagons, designs based on: five-pointed rosettes, 171— 173a. ten-pointed stars, 120. Doorways, lS6b, lS8a, 164b, 193a, 200a, 202a, 230a, 2S5a, 280. Duodecagons, designs based on: Arabic, 118, 128, 128a, 129a. foiled figures, 200 (6), 202 (5); 18Sa, 189a. star-polygons, 122, 123. Durham Cathedral, Eng., 250. Dutch design, 53a. Early Christian architecture, 161. Egyptian art (see Cairo), 106. Eight-pointed stars : Construction of, 93, 97, 150 (1, 17- 19, 22, 23). Designs based on: Arabic and Moorish, 78, 129. ceilings, 80a. linoleum, 74a, 110, Ilia. mosaic floors, 76b. oilcloth, 8Sa. parquetry, 77a, 79a, 81a, 82a, 121a. Roman mosaics, 84- tUed floors, 7Jfi, 76a, 83b. vaulting, 121b. Occurrence, 92, 93, 97, 146-147. Ely Cathedral, Eng., 94 (p. 105, remark). Embroidery, 147; 99a. Equilateral trianglss: Designs based on: Arabic, 102. curved triangles, 184-194, 204- 208a, 212, 213, 220, 222-225. embroidery, 99a, foiled figures, 167, 167a, 183. iron work, 100, 113a. pointed arches, 184, 187, 198, ZOO, 201, 210a, 212-214, 216- INDEX TO NOTES AND ILLUSTRATIONS 365 Z18, 22Z-Z^B, M7, 228, 230, 231, 23Sb, 238, 244, U8b, 251, 252, 261, 2e2-264h, 273a, 274, 278, 279. pointed trefoils, 190-193. (See Trefoils.) rounded trefoils, 134-137. star-polygons, 115, 116, 122. tiled floors, 17, 19. Occurrence in ornament: Arabic art, 19, 140. Gothic art, 224. Evansville, Ind., 148, 237a. Exeter Cathedral, Eng., 160a, 187a, Fabrics, 19. Flag, U. S., 146. Florence, Italy, 111, 215; ISO, 247a. Fort Wayne, Ind., 201. Gables. (See Canopy.) Glass: cut, 146; 72a.. leaded, 215, 216 (8); 29, SO, 162a, 165, 166, 166a, 179a. painted, or stained, 76; 100, I48. Gloucester Cathedral, Eng., 173, 323; 14s. Great Malvern, Worcestershire, Eng., 173. Great Waltham, Essex, Eng., 240. Greek architecture, 322. Greek designs, 33. Gothic architecture and ornament: EngUsh Gothic, 117, 160, 161, 189, 222, 223, 323, 345. French Gothic, 189, 223, 338, 345. General : characteristics, 166, 173, 189, 222, 223, 231, 323. nature of designs from, 111, 166, 173, 189, 231, 240, 247, 323, 324, 333. origin and development of trac- ery, 160. German Gothic, 223; 220. Gothic designs, 92a, 94 (p. 105, remark), 95, 100, 130- 264a, 27S-29S. (See Arches, Curved triangles. Cusps, Doorways, Equilateral triangles, MuUifoils, Pavements, Quadrifoils, Trefoils, Vaulting, Windows.) Grills. (See Iron.) Henry VII Chapel, 95. Hereford Cathedral, Eng., 247. Hertfordshire, Eng., 196, 244. Hexagons : Designs containing: Arabic designs, 117, 125-127. foiled figures, 138, 156-158, 160a, 164, 16B, 166, 166b, 174-177, 180, 204a, 206b, 210a, 231a, 239a, 255a. glass, 179a. , parquetry, 32a, SSa, 115a, 124a. tiled floors, 19, 116a. tiles : hexagons irregular, 8b, 20, 22, 25a, 26; hexagons regular, 12. Occurrence in ornament, 19, 140. Indian, American, 173. Iron, ornamental, 76; 61, 85, 366 INDEX TO NOTES AND ILLUSTRATIONS 90a, 100, 108b, 112a, llSa, 114a, 117, 181a. Isosceles triangles, designs based on : tiles, 3. parquetry, SS, 38. rafters, 132. trusses, 265-B7S. Japanese art, 33, 106, 173. Jervaulz Abbey, Eng., 190. Jewelry, 215. Jewish talisman, 140. Kent, Eng., 172. Lace pattern, 18a. Lichfield Cathedral, Eng., 247; Lincoln Cathedral, Eng., 274; 164b, 231a, 24£a, 266a. Linoleum: designs for, 26a, 26, 44a, 68a, 69a, 74a, 110, Ilia, 118a. manufacture of, 19, 78. Mansard roofs, 318(2, 4); 269. Milan, Italy: Cathedral, 111, 204 (4); 226. . Ospedale Maggiore, 241a. Mohammedan art. (See Arabw.) Monreale, Sicily, 92. Moorish art (see Alhambra), S$, 69. Mosaics: Byzantine (medieval) : designs from, 40, 66a, 86, 86a, 116a. general character of, 21, 79. history and occurrence, 21. special cases mentioned, 111, 140, 147. Modem, 66a, 67a, 76b, 86, 16Sa. Pompeian, 202 (6, remark); 61a, 71a, 84, 94 (see p. 105, re- mark), 174, 176. Roman: designs from, 4S, 61a, 70, 71a, 84, 86a, 87, 94 (see p. 105, remark), 163, 174, 176. history and occurrence, 79. Multifoils : Designs : pointed forms, 171-183. rounded forms, 139a, 166-166b, 189a, 193a, 200a, 210a, 2S9a, History and occurrence, 189, 223. Special cases mentioned. 111, 190. Needlework, 215. Newbury, Eng., lS7a. New York City, 149, 160, 167a, 202a, 216, S79a. Northfleet, Kent, Eng., 212. Notre Dame Cathedral, Paris, 221 (5, remark) ; 18Sa, 211a. Oakland, Cal., 215. Octagons : Designs based on: Arabic, 73. ceilings, 34a. foiled figures, 200(6), 202(5, 6); 139a, 143a, 144a, 161a, 168, 169, 178, 179, 181, 182, 189a, 200a. INDEX TO NOTES AND ILLUSTRATIONS 367 linoleum, S5a, Z6, 68a, 69a. mosaics, 66a, 67a,. parquetry, 71a. tiles: octagons, irregular, 9, 10, 22; octagons regular, f^a. Uses of (see Eight-pointed stars), 19, 76. OUcloth, 78; 8Sa, U7a. Opus ,Alexandrinum, 21. Oxford, Eng., 215; 148, 243, 259a. Oyster Bay, N.Y., 251. Padua, Italy, 111. Parallel lines: Designing on, 19. Designs based on: one system, 112a. two systems, 85, 86, 103. three systems, 17-20, 100, 101, 102, 118a, 125. four systems, 21, 22, 24a, 25, 26. Parallelograms, designs in, 27- S4, 145. Parenzo, Austria, 111. Paris, France, 221 (5, remark) ; 18Sa, 211a. Parquet floors: designs, 19; 18, 27-64a, 6Sa, 71a, 77a, 79a, 81a, 82a, llBa, 121a, 124a. history and manufacture of, 32, 33. Pavements : medieval tiled, 2, 147, 173. mosaic, 21, 79. parquetry, 32. special cases mentioned, 111, 147, 173. tiled, 20. (See Mosaics, Parquet floors. Tiled floors.) Pentagons, design based on : foiled figures, 160, 164b, 193a, pentagram star, 146, 147; 119. tiles, pentagons irregiilar, 6, 6a, 7, 7a. (See Decagon.) Pentagram star, 146, 147 ; 119. Peterborough Cathedral, Eng., 121b. Plaster design, 96. Pompeian. (See Mosaics.) Portland, Oregon, 300 (4). Primitive art, 147, 173. Pullman cars, 273. Pythagoreans, 147. Quadrifoils : designs, 140-154, 161a, 193a, 237a, 262a. history and occurrence, 166, 173, 223. special cases mentioned. 111, 173. Rafters, 161, 312, 313; 130- RaUings, 31, 61, 108b, 113a, 114a. Ravenna, 92, 111; 75. Register design, 117. Regular polygons, used for: many-pointed rosettes, 178- 183. rounded multifoils, 155-166b. star-polygons, 145; 119-129a. Reims Cathedral, France, 219. Renaissance designs, 223, 368 INDEX TO NOTES AND ILLUSTRATIONS Roman : arch, 322. mosaics, 79. Romanesque, 161, 223. Rome, Italy, 92, 147. Roofs : hanmier-beam, 161; ISSa. open timber, 161. (See Trusses.) Rosettes : circular forms. 111; 171-177, 178-183. straight line forms, 60, 61, 6S. (See Star-polygons.) Rouen Cathedral, France, 168, 195a, 278. St. Alban's Abbey Church, Eng., 166h. St. Louis, Missouri, ^356, lS8a, 161a, 193a. St. Mark's, Venice, 75 (9), 147, 215. St. Pol de L€on, France, Z28a. Salisbury Cathedral, Eng., 274; Salt Lake City, Utah, 185. San Lorenzo, Rome, 92. San Miniato, Florence, Italy, 215; 130. Saracenic art. (See Arabic.) Semicircular designs. (See Arches and Windows.) Sicily, 21. Six-pointed stars : Designs containing : Arabic, 1Z6, 126, 128, 128a. medieval, 116a. parquetry, llSa, 124a. tiled floors, 19, 116a. Occurrence, 140. Spanish art, 153. (See Alhambra and Moorish art.) Squares : Designs in: modern industrial designs, 24, 35a-84, 85-97, 103-109, 111, 133. Gothic forms, I40, I4I, I46- 147a, 149, 150, 151, 153, 154. Occurrence, 19. Star-polygons : Construction, 145. Definition, 145. History, 147 : Occurrence, 92, 140, 146, 153. Special stars, 147: four-pointed, 150 (16, 22, 23), 151 (13), 152 (3, 11); 54, 65, 56, 67. five-pointed, 119, 120, 171, 172. six-pointed, 151 (14), 152 (2, 11); 115-117, 124-126, 128, 128a. eight-pointed, 150 (17, 18, 22, 23), 152 (11); 74-75a, 77, 78, 81, 83-83b, 121-121C, 127, 129. ten-pointed, 120. twelve-pointed, 151 (1), 152 (1, 11); 122,123, 128, 129a. (See Eight-pointed, Six-pointed, and Pentagram stars.) Stonecutting, 173; 143. Strassburg Cathedral, Germany, 169. Swastika, 62 (9 and footnote). Symbols, 140, 147, 166. Talisman, Jewish, 140. Tiled floors: designing for, 20. INDEX TO NOTES AND ILLUSTRATIONS 369 designs, 7a, 8b, 14b, 17a- S4a, 27a, 39a, 4I, 43, 53a, 74b, 75a, 9Za, 116a, 133, 152a. history of designs for, 21. medieval, 2, 173. Tiles : designs, 190; 141a, ie4a. history, 2, 76. occurrence, 1. shapes and sizes, 1, 2; 1- 16, 88, 106. Tintern Abbey, 238, 261a. Toledo, Spain, 59. Tracery : designs (see Windows and Gothic) . origin and development, 160. Venetian, 308; 262-26 4a. Trademarks, 140, 173, 186 (5 and remark) . Trefoils : Pointed : designs, 232 (5), 234(6); 190a- 193a, 223. history and occurrence, 223, 231. Rounded : designs, 134-139, 208a, 220, history and occurrence, 166, 223. Trusses : combination king-rod and queen- rod, 267, 268. compound Fink, 271. definition, 312. development of, 313. fan, 270. king-rod, 285. mansard roof, 269. queen-rod, 266. scissors, 272. Turkish architecture, 333. Vaulting, 117; 94 {p. 105, remark), 95, 121b, 14s, 146, 176. Venetian tracery: characteristics, 308. designs, 2e2-264a. Venice, Italy, 75 (9), 111, 147, 190, 215; 262-264a. Washington, D.C., 227. Wells Cathedral, Eng., 184, 2S9a. Westminster Abbey, 247; 95, 14£. Winchester, Eng., 147. Windows pointed: Classification and characteristics, 222, 223. Designs for: one-li;ht, 198-201, 227, 228. two-lights, 196, 197, 210-220, 229, 230, 232-236, 240. three-lights, 222-226, 242-244, 246-251, 279. four-lights, leOa, 197a, 237, five-lights, lS7a, 161a, 187a, 213, 224a, 253-258a. six-lights, 21 ea, 24Ba. seven-lights, 259-261 a. eight-lights, 156a, 231a, 238. nine-lights, ITSa. Method of study, 271. Windows, semicircular: designs for, 159, 165-ieea, 202a, 2S3a, 234, HO, 24I. occurrence, 223. 370 INDEX TO NOTES AND ILLUSTRATIONS Windows, triangular curved: designs for, 171a, 185, 185b, 188a, 189a, 20ia~208a. occurrence, 247. Windows, wheel : designs for, 143, 155-162, 164, 166b-173a, 183, 183a. occurrence, 189. Woodcarving, 173; 151, 154, 175, 177. Worcester Cathedral, Eng., 111. York Cathedral, Eng., 274; 213. INDEX TO RBPERBNCBS Numbers refer to pages. Am. Arch, and Building News, 238. Arch. Annual, 249. Arch. Record, 126, 293. Arch. Review, 282. Arte Italiana, 99. Ball, 132. Barre, 73. Beauchamp, 132, 162. Becker, 325, 326. Billings, 155, 206. Bond, 104, 150, 177, 206, 207, 267, 288, 292, 304, 312, 317, 322. Bourgoin, 53, 56, 60, 112, 126, 130, 143, 144, 145, 147. Brandon, 151, 153, 155, 192, 206, 218, 236, 269, 274. Brickbuilder, 181. Brochure Series, lOf, 176, 178, 194, 196, 293. Burn, 202. Cajori, 49. Calvert, 18, 53, 78, 88, 132. Cassell, 301. Clifford, 96. Day, 47, 51, 52, 155. Dictionary of Architecture, 70. Diefenbach, 157, 162, 167, 187, 189, 194, 197, 198, 201. Enc. Brit., 72, 252, 323. Fergusson, 104, 150, 303, 304. Fletcher, 104, 151, 168, 177, 204, 206, 303, 304. Furnival, 86. Gayet, 36, 82, 84, 97, 99. Glazier, 96. Gravina, 14, 82. Gruner, 82, 176. Giinther, 131. Handbuch der Architektur, 195. Hanstein, 217, 220, 224, 251, 263, 316, 821, 322. Hartung, 99, 188, 242. Jones, 66, 96, 117, 148. Kidder, 298. Leighton, 115, 209. Morgan, 72, 77, 132. Northern Pacific Railroad, 162, 175 (remark) . Ongania, 14, 132, 197. Otzen, 178. Parker, 112, 165, 213, 226, 311. Porter, 188, 310. Price, 86, Prior, 281. 371 372 INDEX TO REFERENCES Prisse d'Avennes, 98, 129, 132, 149. Pugin, A., 106, 155, 279. Pugin, A. and A. W., 155, 169, 197. Rickman, 226, 236. Komussi, 189. Sharpe, 150, 205, 213, 226, 284. Shaw, 98, 99, 102, 111, 132, 162, 178. Sturgis, 150, 241, 245, 305. Ward, 14, 96. Waring, 98, 126, 132, 197. Wilson, Thos., 53. Wornum, 155. Wyatt, 56, 82, 126, 128, 132. Zahn, 52, 72, 78, 95, 105, 186, 193, 195. MATHEMATICS Plane Geometry with Problems and Applications By H. E. SlaUGHT, Associate Professor of Mathematics in the Uni- versity of Chicago, and N. J. Lennes, Instructor in Mathematics in Columbia University, New York City. i2mo, cloth, 288 pages. Price, $1.00. THE two main objects that the authors have had in view in preparing this book are to develop in the pupil, gradually and imperceptibly, the power and the habit of deductive reasoning, and to teach the pupils to recognize the essential facts of elemen- tary geometry as properties of the space in which they live, not merely as statements in a book. These two objects the book seeks to accomplish in the follow- ing ways: — 1 . The simplification of the first five chapters by the exclusion of some theorems found in current books. 2. By introducing many applications of special interest to pupils, and by including only such concrete problems as fairly come within the knowledge of the average pupil. Such problems may be found in tile patterns, parquet floors, linoleums, wall papers, steel ceilings, grill work, ornamental windows, etc., and they furnish a large variety of simple exercises both for geometric construction and proofs and for algebraic computation. 3. The pupil approaches the formal logic of geometry by natural and gradual processes. At the outset the treatment is informal ; the more formal development that follows guides the student by questions, outlines, and other devices, into an attitude of mental independence and an appreciation of clear reasoning. The arrangement of the text is adapted to three grades of courses : — (a) A minimum course, providing as much material as is found in the briefest books now in use. (b) A medium course, fully covering the entrance require- ments of any college or technical school. (c) An extended course, furnishing ample work for those schools where the students are more mature, or where more time can be given to the subject. 72 MATHEMATICS Solid Geometry with Problems and Applications By H. E. Slaught, Associate Professor of Mathematics of University of Chicago, and N. J. Lennes, Instructor in Mathematics, Columbia University, New York City. i2mo, clolh, 196 pages. Price, 73 cents. IN the Solid Geometry the authors carry forward the two main purposes of their Plane Geometry : the development of the reasoning powers of the pupil, and the presentation of the essen- tial facts of elementary geometry as properties of the space in which we live. The important features by which the Solid Geometry seeks to accomplish these two objects are : 1. The concepts of three-dimensional space are made clear by many simple illustrations and exercises, and the pupil is led gradually to a comprehension of space forms and relations. 2. A large number of interesting concrete applications are interspersed throughout the text, emphasizing the practical sig- nificance of certain fundamental theorems. 3. The logical structure is made more complete and more prominent than in the Plane Geometry, thus following up the knowledge and appreciation of deductive reasoning which the pupil has already gained. 4. Throughout both parts of this Geometry a consistent scheme is followed in the presentation of incommensurables and the theory of limits. A complete and scientific treatment leads up to the final chapter, in which the theory of limits is presented in such a way as to leave nothing to be unlearned or compromised in later mathematical work. Plane and Solid Geometry By H. E. Slaught, Associate Professor of Mathematics, University of Chicago, and N. J. Lennes, Instructor in Mathematics, Columbia University, New York City, izmo, cloth, 483 pages. Price, $i.z$. THIS is the Plane Geometry and the Solid Geometry bound in one volume. It will be found convenient for pupils who buy their own books and keep them from one class to the next. 73 MATHEMATICS Plane and Spherical Trigonometry By President ELMER A. LYMAN, Michigan State Normal College, and Professor Edwin C. Goddard, University of Michigan. Cloth, 149 pages. Price, go cents. MANY American text-books on Trigonometry treat the solu- tion of triangles fully ; English text-books elaborate ana- lytical Trigonometry. No book seems to have met both needs adequately. It is the aim of Lyman and Goddard to find the golden mean between the preponderance of practice in the one and that of theory in the other. The book is a direct outgrowth of the classroom experience of the authors, and it has no great resemblance to the traditional text- book in Trigonometry. Two subjects that have called for origi- nal treatment or emphasis are the proof of the formulae for the functions of a ± ;8 and inverse functions. By dint of much prac- tice extended over as long a time as possible it is hoped to give the pupil a command of logarithms that will stay. Computation Tables By Lyman and Goddard. Price, 50 cents. THESE are five-place logarithmic tables which have been prepared to accompany Lyman and Goddard's Trigonome- try. They have been rendered exceedingly convenient by the use of distinctive type, which makes it easy to find the numbers sought. The natural and logarithmic series are given side by side in the tables. The column of difference is in unusually direct proximity to the logarithms themselves. Plane and Spherical Trigonometry, with Compu- tation Tables By Lyman and Goddard. Complete Edition. Cloth, 214 pages, Price, ^1.20. 75 MATHEMATICS Introduction to Geometry By William SCHOCH, of the Crane Manual Training High School Chicago. i2mo, cloth, 142 pages. Price, 60 cents. THIS book is intended for pupils in the upper grades of tlie grammar schools. It is a book of exercises and problems by means of which the pupil will convince himself of a number of fundamental geometric truths. Practical problems are given and the pupil's knowledge is constantly tested by his power to apply it. Principles of Plane Geometry By J. W. MacDonald, Agent of the Massachusetts Board of Educa- tion. i6mo, paper, 70 pages. Price, 30 cents. Logarithmic and Other Mathematical Tables By William J. Hussey, Professor of Astronomy in the Leiand Stan- ford Junior University, California. 8vo, cloth, 148 pages. Price, Ji.oo. VARIOUS mechanical devices make this work specially easy to consult; and the large, clear, open page enables one readily to find the numbers sought. It commends itself at once to the eye, as a piece of careful and successful book-making. Plane Trigonometry By Professor R. D. BoHANNAN, of the State University, Columbus, Ohio. Cloth, 379 pages. Price, ^2.50. Calculus with Applications By Ellen Hayes, Professor of Mathematics at Wellesley College. i2mo, cloth, 170 pages. Price, jSi.20. THIS book is a reading lesson in applied mathematics, in- tended for persons who wish, without taking long couises in mathematics, to know what the calculus is and how to use it, either as applied to other sciences, or for purposes of general culture. 76