A Collection Sangaku Problems

ollection Sangaku Problem. arhall Unger epartment of at ian Language & Literature The Ohio State Univerity Verion of 10 ugut 009 PROL 1: Given two line tangent to circle O at and from a common point, how that the circle pae through the incenter of triangle. 1 P O SOLUTION 1 (U: Since and are tangent, each of the bae angle of the iocele triangle meaure half of O. The um of the angle between the bae and the biector of the bae angle i Q? therefore alo half of O. Hence, wherever Q may be, O the interection of the biector, may be, Q i 180 P O/. Pick any point P on the arc eterior to the triangle; P O/. Since P and Q are upplementary, QP mut be a cyclic quadrilateral. Since,, and P lie on circle O, o mut Q, and it i the incenter of. PROL : What i the relationhip between the radii of three circle of different ize all tangent to the ame line and each eternally tangent to the other two? O 1 O 3 O SOLUTION (&P: The hypotenue of the right triangle i 1 ukagawa & Pedhoe 1989, 1.1.4; lot tablet from Ibaragi, 1896; no olution given. ukagawa & Pedhoe 1989, 1.1.1; well-known; tablet from Gunma, 184. 1

O 1 O 3 O r 1 + r. It hort leg i r 1 r, o the other leg i the quare root of (r 1 + r (r 1 r. I.e., i r 1r (twice the geometric mean of the radii. Likewie, r 1r3 and r r3. dding and dividing through by r 1r r3, we obtain 1 1 1 +. r r r 3 1 PROL 3: Suppoe two circle tangent to a common line at and are tangent to each other eternally; that a third circle i tangent to the line and both of them eternally; and N P Q ( O K that a fourth circle through and i tangent to the third circle internally. Show that the fourth circle and the line define a circular egment uch that, if the radii of the circle

through and are varied, the mall circle tangent to them remain tangent to both the arc and line of the egment. 3 SOLUTION 3 (U after. ogomolny: o the eay cae firt:. Then R R R, 4r R, and R 8r. If R O i the radiu of the fourth circle, we have N O or 16r r(r O r, which lead to 5r R O or R O 5/8. Since thi relationhip depend only on the relative location of,, and O, change in R, R, and r will not affect it. Now conider the hard cae:. Writing m for ½, we have O 5m/4 and O 3m/4 becaue O i a 3:4:5 right triangle. Writing for and y for, we alo have m ( + y/ and ( y/ KO. all the latter n. We now calculate r two way: firt, uing the Pythagorean Theorem in right triangle KOQ, then uing the formulae derived in problem. On the one hand, we have (5m/4 r (3m/4 + r n, which reduce quickly to m 4mr n. oreover, m n y, o 4mr y. On the other hand, we have ² 4R r, y² 4R r, and ( + y² 4R R. ombining thee three equation, we get y r( + y. ut thi i again y 4mr. In other word, for any particular, r y/4m i neceary both for circle Q to be incribed in the circular egment defined by fied,, and O and for all the circle above the line to be tangent a required. PROL 4: Given two unequal circle with concurrent diameter and a hown, tangent from (rep. to O (rep. O 1, O O 1 O 1 O and circle tangent to (rep. and the two tangent from (rep., prove that the radii of thee two circle are equal. 4 SOLUTION 4 (&P: 3 5 ukagawa & Pedhoe 1989, 1.1.; lot tablet from iyagi, n.d.; the hint R 8 i given, but no olution. 4 ukagawa & Pedhoe 1989, 1.3; lot tablet from ichi, 184; olution given. 3

rom ΔT 1 O 1 ~ ΔT O, it follow that r 1 r r O 1 O T 1 O 1 O 1 O T and o r 1 ( + + O r ( r 1. Hence r 1 r 1 + r 1 + r 1 r r 1 r r r 1 r 1 r 1 + r 1 + r 1 r r 1 r r 1 (r 1 + + r r 1 r r1 r r 1. r + + r Thi i algebraically ymmetrical: we would have arrived at the ame right-ide epreion for r if we had tarted at the other end of the figure. Thu r 1 r. 1 P O O 1 O 3 r O(r Q l m PROL 5: O 1, O, and O 3 all have radiu r, center in line m, and form a chain a hown. Line l pae through O and i tangent with O 1 and O 3 on oppoite ide of m. ircle O(r i internally tangent to O 1 and O 3, and i cut by l in P and Q. Prove that PQ r + 3r. 5 5 ukagawa & Pedhoe 1989, 1.3.3; tablet from Ibaragi, 1871; no olution given. 4

SOLUTION 5 (U: The trick i to uperimpoe the implet cae on the general cae. Start with O coincident with O : P Q 6r r + 3r i obviou. Now, for O below O on the perpendicular to m, PQ i longer than P Q by PP QQ. That i, PQ 6r PP QQ. The theorem aert that PP QQ i equal to the change in length of r, which i longer than before by r 3r. ut PQ 6r r 3r if and only if PQ r + 3r. P P Q O 1 O O 3 r O r Q l m O 1 (r 1 PROL 6: Given right triangle and it circumcircle O 1 (r 1, contruct circle O (r tangent eternally to leg a and b and internally to circle O 1. (Such a circle i called mitilinear. Prove that r a + b c. 6 O (r 6 ukagawa & Pedhoe 1989,..7; lot tablet from Hyōgo, n.d.; no olution given. 5

SOLUTION 6 (U: The trick i to add the incircle O 3 (r 3 and firt prove that r r 3, which i preented a a eparate problem. 7 It i eaily done by drawing O T and O T, and noting that O 3 touche all four ide of quare T O T, each ide of which i r 3. Now in any triangle with emiperimeter, the ditance from to the point of tangency of the incircle i c. So in a right triangle uch a, r 3 c (a + b c/, which immediately implie r a + b c. T O 1 O T PROL 7: Right triangle i partitioned into two triangle by the altitude H a hown. Prove that thi altitude i the um of the radii of the three incircle. 8 O (r O 1 (r 1 H SOLUTION 7 (U: ll three triangle are right. We ue the corollary jut tated to calculate r 1 a + b c, r H + H a, and r 3 H + H b. dding thee equation, we get r 1 + r + r 3 H + H + H c H. So r 1 + r + r 3 H. O 3 (r 3 PROL 8: Given two circle of equal radiu incribed a hown below, prove P ( a. 9 7 ukagawa & Pedhoe 1989,.3; tablet from Iwate, 184; no olution given. 8 ukagawa & Pedhoe 1989,.3.; tablet from Iwate, n.d.; no olution given. 6

7 SOLUTION 8 (U: In the figure above, r i the inradiu of Δ, i it emiperimeter, ΔP and ΔP have emiperimeter 1 and, repectively, but the ame inradiu k. Uing for P, oberve that 1 + + (we will ue thi fact immediately and once again later. dding area, r k 1 + k. Hence (r/k. y imilar triangle, c k r b 1. Hence we can calculate two way: c b 1 ( ( pand each equation, and olve their um for : ( ( ( ( ( ( ( ( ( 1 1 1 1 a a a a c b c b + + + + + + + + + 9 ukagawa & Pedhoe 1989,..5; urviving tablet from hiba, 1897; no olution given. P r k k }

ab OROLLRY: If i a right triangle, then k. 10 ab + a + b + c PROO: We have Δ k(b + P + P/ + k(c + P + P/ k(a + b + c + P/, o ab k a + b + c + P. In a right triangle, a r, o P ( a r Δ ab/. That i, 4P ab or P ab. PROL 9: i a quare with ide a and diagonal. The incircle of N and N are congruent. What i their radiu r in term of a? 11 SOLUTION 9 (U: ecaue N i a right triangle, r ( + N N/ (ee problem 6. The congruence of the two incircle implie N (, where i the emiperimeter of (proven in problem 8. We know a, o a + a. N a a a ( + 1 Hence N + a and N a +. Now, ince a ( + 1 a ( 1 N a, a 1 a a a + N. So r + a + + 1. 10 Thi i ukagawa & Pedhoe 1989,..3; lot tablet from iyagi, 1847; equation given, no olution provided. 11 ukagawa & Pedhoe 1989, 3.1.7; urviving tablet from Hyōgo, 1893; the olution given, r 1 (1 + 1a, i clearly an error (indeed, i obviou from the diagram. r < a 8

PROL 10: quare with one diagonal i cut by a line from a third verte to the midpoint of an oppoite ide. circle i incribed in the reulting triangle oppoite the midpoint. What i it radiu? 1 SOLUTION 10 (. ogomolny: Imagine completing the figure a hown below. y congruent triangle, it i eay to ee that the top of the quare biect the ide of the large right triangle. Hence the two line within the quare are median of the large right triangle. The ape of the incribing triangle i it centroid, and divide the two line within the quare in the ratio 1 :. or the ame reaon, if the ide of the quare i a, the altitude of the incribing triangle i a 3 (imagine a line parallel to the top and bottom of the quare running through the ape of the triangle. Now the diagonal of the quare i a and other line in the quare i a 5. The ide of the incribing triangle are 3 of thee length, repectively. ut in any triangle with bae a, altitude thereto h, perimeter p, and inradiu r, Δ pr ha. onequently, a + 3 a (3a + a + + a a r 3+ + 3 a 5 5 a r a 3 5 r a 1 ukagawa & Pedhoe 1989, 3.1.3; urviving tablet from iyagi, 1877; olution given in the form a r 3 + 5 +. 8 9

PROL 11: right triangle ha three circle tangent to it leg and internally tangent to it circumcircle: O 1 i tangent to both leg; O and O 3 are tangent to leg and at their midpoint and N, repectively. Show that r 1 3r r 3. 13 SOLUTION 11 (U: The diameter of O and O 3 are the agittae of chord and : v b r and v a r 3. O 1 N Lemma: In any right triangle, the inradiu r v a v. Proof: b v a R b/ v b R a/ v a c b v b c a 4v a v b ab c(a + b c 4v a v b ab cr v a v b ab/ cr v a v b r cr r( c r. ut r 1 r (problem 6, o r r 1 /4 8r r 3. Thu r 1 3r r 3. PROL 1: In Δ,. If one chooe on and on uch that and the incircle of Δ, Δ, and Δ all have radiu r, then r /4. 14 SOLUTION 1 (U: The trick to olving the problem epeditiouly i to clarify what i given and to prove the convere firt. N Notice that the only connection between and the geometry of Δ i that the incircle of Δ (dahe i tangent to at it midpoint. The two incircle of Δ and Δ cannot be tangent to the hared ide at the ame point, a hown, unle i the point of tangency of the incircle of Δ (dot with ide. 15 Hence. nd the two incircle cannot be congruent unle, in addition, either or, in which cae, Δ Δ and the 13 ukagawa & Pedhoe 1989,.4.6; urviving tablet from Iwate, 1850; no olution given. 14 ukagawa & Rothman 008:194 96, 1 16 15 Given a an eercie Honberger 1995:13, 157. 10

other condition follow. Therefore, given a fied iocele, if the congruence of all three mall incircle implie r /4, a the problem aert, then the convere mut alo be true. Indeed, it i eaier to prove. In the figure above, let a, b, d, and h be the length of,, N, and, repectively. Let r 1 and r be the radii of the incircle of Δ and Δ, repectively. We ue two lemma, which we prove later: 1. a b d. r 1 r ad Given h 4r, quare the equation for the inradiu of a right triangle b h + a/ r r + a/ to get b 4r + ar + a /4, and equate thi with the Pythagorean reult b (a/ + h a /4 + 16r. Thi yield a 6r, which, with h 4r, implie b 5r. Hence, by Lemma 1, d r. Now Δ ha/ 1r r 1, where i the emiperimeter of Δ or 8r. Thu r 1 3r/ 3d. Uing Lemma, we find r a/6 r. rom thi, it i clear that olving the original problem come down to proving, without knowing the value of h, that Δ and Δ are 3:4:5 right triangle. It eay to how in the adjoining figure that ΔTU ~ ΔO U. Thi mean that U i the mean proportional between T and O U T + N. In term of length, we therefore have O 1 S U O T N a/ r + d [r + (a/ r ]/(a/ r (a/ r + d(a/ r r + (a/ r d(a/ r r ad/ dr r r 1 r dr r r 1 d r. (Lemma take the form r 1 r ad ince we are auming r r. Since S r, O 1 S d. Thu O 1 r + d 3d r 1. Hence, by Lemma, rr 1 (d(3d 1d ad, o a 1d 6r. Since a b d (Lemma 1, b 10d 5r. nd becaue i a right triangle, a/ 3r and b 5r imply h 4r. 11

The lemma follow the general cae of any triangle with incircle I. raw a cevian and the incircle O 1 and O of the reulting triangle and. Label the point of tangency of thee incircle a hown. Then H GK N. PROO: y equal tangent, (1, and H I O 1 L O K G N G G L K H K N H N L. Subtracting the equation in econd row from thoe in the firt, we get H GK ( GN (3 H L (4 L K N (5 quating the left and right ide of (4 and (5, + N K + H ( + + N K + ( + H (by 1 + N K + H (by 3 GN + N K + H (GN G + N (K G + H N GK + H (by N GK (6. Linking equation ( and (6, H GK N. H O 1 L K G N O any more inference can be drawn from thi figure, but we need jut two. Lemma 1 follow immediately if and are the midpoint of and, repectively. Then / / H N. In the problem, a eplained earlier, mut be the midpoint of. Requiring that be iocele force to be the midpoint of. Lemma in the general cae i r 1 r L. The dahed line in the figure above how that (r 1 + r + ( L (O 1 O (r 1 r + KN, but KN K + N + L 1

by equal tangent. The ret i jut algebra. The traditional olution prove thi relation for the problem figure, but it i true even for calene Δ. PROL 13: Prove that the um of the radii of the incircle in both triangulation of a (conve cyclic quadrilateral are equal. 16 L I K SOLUTION 13 (U: There are many way to prove thi theorem. I have put together the following equence of reult on the bai of hint from everal different ource. 17 I Lemma 1: The biector from one verte of a triangle, etended, cut the circumcircle at the midpoint of the arc ubtended by the oppoite ide of the triangle, which i the center of the circle defined by the other two vertice and the incenter. Proof: I I + I, that i, half the um of the verte angle at and. I I + I +, the ame um. So I I and ΔI i iocele. y imilar reaoning, o i ΔI. Hence I. oreover, ince the and, which ubtend arc and, repectively, are equal, turn out to be the midpoint of arc. γ If we add another point on the circumcircle a hown, it immediately follow that and I, etended, concur at and that all four line egment, I,, and are equal. omplete the quadrilateral and contruct the eight biector that meet at,, G, and H, the midpoint of arc,,, and, repectively. (The diagonal of the quadrilateral have been omitted. It i eay to prove that H and G are perpendicular: I 16 ukagawa & Pedhoe 1989, 3.5(1; lot tablet from Yamagata, 1800. 17 ot helpful i huja, Uegaki, and atuhita 004. 13

β α G α δ Lemma : If a circle i partitioned into four ector, the line joining the midpoint of the arc are perpendicular. β S H δ Proof: y hypothei, π α + β + γ + δ. dd auiliary line GH. GH ½(α + β. GH ½(γ + δ. So GSH π ½(α + β + γ + δ π π/ π/. γ γ Thi lead to the lat lemma, which i an impreive theorem in it own right: Lemma 3: The incenter of the four triangle formed by the ide of a conve cyclic quadrilateral and it diagonal are the vertice of a rectangle with ide parallel to the line joining the midpoint of the arc ubtended by the ide of the quadrilateral. Proof: In the figure, H and H ubtend equal arc, o H biect. Lemma 1 aure that I L. Thu ΔIL i iocele with bae IL perpendicular to H. pplying the ame reaoning at H, we conclude that K i perpendicular to H, and therefore parallel to L. Likewie, I and LK are parallel and perpendicular to G. Since H and G are themelve perpendicular (Lemma, IKL i a rectangle. β β α I L G α K δ H δ We are now ready to prove the original theorem, which tate: γ γ The um of the radii of the incircle in both triangulation of a (conve cyclic quadrilateral are the ame. PROO: If we draw line through L and parallel to (left and through I and K parallel to (right, the I L perpendicular ditance between each pair of line will be the um of the K radii of the correponding pair of incircle. To prove thee um are equal, it uffice to how that the parallelogram produced by uperimpoing the two et of parallel line i a rhombu, becaue the two altitude of a rhombu are equal. To that end, oberve that G G becaue they ubtend equal arc. G G for the ame reaon. Hence, ΔUG ~ ΔVG with UG VG. That i, and cut G at the ame angle in oppoite direction. 14

α α G Since H and G are perpendicular (Lemma, and likewie cut U N H at W and X at the ame angle in K W X H oppoite direction. Hence all line V L parallel to the diagonal of the quadrilateral cut the ae of γ rectangle KLN (Lemma 3 at the γ ame angle. So the four triangle baed on the ide of the rectangle that, together with it, make up the parallelogram, are all iocele, and we have a rhombu (four ide equal. (nother neceary and ufficient condition for a parallelogram to be a rhombu i that it diagonal be perpendicular: the diagonal of thi rhombu lie on H and G. OROLLRY: The um of the inradii in any of triangulation of a (conve cyclic polygon are all the ame. or eample, here are two of triaangulation of the ame cyclic heagon. There are many other. Yet the um of the radii of the incircle i the ame for all of them. PROO: The previou theorem etablihe thi theorem for cyclic quadrilateral. ume it hold for cyclic n-gon. very cyclic polygon of n + 1 ide can be analyzed a a cyclic n-gon plu a triangle by electing three adjacent vertice of the tarting polygon for the triangle and regarding all the vertice other than the middle one of thee three a a cyclic n-gon. Since the ame triangle i added to every triangulation of the cyclic n-gon, the theorem hold for the larger polygon too. Thi corollary i frequently decribed a a theorem by itelf. 15

PROL 14: In quare, i tangent to emicircle O 1. O i the incircle of. The tangent to O 1 and O meet the ide of the quare in and H and interect in G. O 3 i the incircle of GH. Prove that r /r 3 3/. 18 O O 1 G O 3 H SOLUTION 14 (U: irt, we prove H. tend,, and H and draw the normal KO 1 and LO 1 a hown below. ark equal angle noting where parallel are cut by tranveral, complementary acute angle in known right triangle, vertical angle, and equal angle in imilar triangle. There are two kind of acute angle in each right triangle. oth kind are found at O 1 ; ince they are complementary, KO 1 L mut be a right angle. ll the right triangle containing both kind of acute angle are imilar, and, by the lemma proved preently, have ide in the ratio 3:4:5. Let be the ide of quare and t / be the ide of quare GLO 1 K. Note pair of tangent from the ame point to the ame circle: K, K, and H HL. ecaue of thi lat pair, if we etend LO 1 to meet in, ΔHL ΔHI. or later convenience, ay that a, b, and c are the length of I L, H HL, and HI H, repectively, noting that a:b:c :: 3:4:5. G O O 3 K O 1 t L H I We now prove the key lemma. In the auiliary figure below, we etend KO 1 to meet in N, and add line O 1 and O 1. O 1 and O 1, which form congruent triangle with radii of and equal tangent to circle O 1, biect upplementary angle, o O 1 90 o and KO 1 i the altitude to the hypotenue of right ΔO 1. Hence KO 1 K K, or t K t K. Therefore K t/. Oberve that thi implie ¾, o Δ i a 3:4:5 right triangle. 18 ukagawa & Pedhoe 1989, 3..5, lot tablet of 1838 from Iwate prefecture; no olution given. 16

N Now, returning to the figure above, in Δ, + r ( + ( + K t (by the lemma. Thu r t/. In ΔGH, G + GH H r 3 ( t + (t + HL ( H b. Thu r 3 b. K ut c/b 5/4, o b + c 9b/4. In ΔO 1, t/(b + c 4/3. Thu 3t 9b, or r 3 t/3, r /r 3 3/. O 1 PROL 15: In circumcribed triangle, let and be the midpoint of, repectively, chord and arc. Then v a i the SGITT of the chord a. Prove that the quare of the ditance from a verte of a triangle to it incenter i four time the product of the agittae to the adjacent ide. 19 SOLUTION 15 (&P: Lemma: If r i the inradiu and the emiperimeter of triangle, then ( b( c r + v a r. (Similar tatement hold for the other two ide. Proof: We write a for a, etc. for convenience. Square Heron ormula and divide by : r a b c or a b c r (a + b + c. ivide by r 3 : (a /r(b /r(c /r a /r + b /r + c /r. (Trigonometrically peaking, the product and um of the cotangent of half of each angle in a triangle are equal. biect becaue biector of angle pa through the midpoint of the circumcircle arc they ubtend. (both ubtend arc, o a /r / (½/v a (b + c /v a. Put thi in the equation derived from Heron ormula: ultiply through by v a r : (b + c /v a + b /r + c /r [(b + c /v a ](b /r(c /r. 19 ukagawa & Pedhoe 1989,.; lot tablet of 185 from uahi; olution provided. 17

Now divide by (b + c : r (b + c + v a r(b + c (b + c (b c. r + v a r b c. Likewie, r + v b r a c r + v c r a b. To olve the problem, add and multiply the lat two equation to obtain r r ( r + v ( r + vc + r(v b + v c a (b + c and b a b c a. Uing thee value, the equation r (a + b + c a b c become r r a + + r( v a b + vc r ( r + vb ( r + vc, a which reduce to a + r 4v b v c. ut a + r I. OROLLRY: ince 4v b v c I, 4v a v c I, and 4v a v b I, 4 3 (v a v b v c (I I I, or 8v a v b v c I I I. PROL 16: ind an epreion for the radiu of the mall circle in the figure at the right in term of the ide, inradiu, and/or agitta hown of the partially circumcribed triangle. 0 0 ukagawa &Pedhoe 1989,..8 (1781, n.pl., a hard but important problem. formula for the deired radiu i given, but not proved. 18

SOLUTION: The lat tep are not too hard, but to get to them require proving a difficult lemma and uing ome of it implication. 1 The proof preented here i a retatement of a proof by yetti poted on athlink, 1 anuary 005. Lemma: K Through verte of Δ, draw cevian with on. raw circle 1 tangent to at, at, and the circumcircle of Δ at K. Then the chord pae through the incenter I of Δ. 1. Let and N be the interection of K and K with the circumcircle. Then N becaue 1 i a dilation of with repect to K. We define a the image of under thi dilation. Since i tangent to 1 at, the tangent to through mut be parallel to. ut the tangent parallel to the chord ubtending an arc touche the arc at it midpoint. Thu i the midpoint of, meaning that i the biector of and hence include the incenter I. K N K N. Let be the interection of and, and conider the circle paing through and K. One of them,, contain the chord N. The correponding chord for the circle through mut lie on becaue N. nd jut a i the image of N with repect to the radical ai K, i the image of. Thu the quadrilateral K i cyclic. 1 yme (003. Y. Sawayama, an intructor at the entral ilitary School in Tōkyō publihed the lemma in 1905 coincidental to olving another problem. 19

K N 3. We now apply the iquel Theorem to, electing on, on, and on. The three circle each paing through two of thee point and the verte of not on the line joining them are all concurrent in K. Since lie on the etenion of, it cannot cut (i.e. twice, o the circumcircle of K i tangent to at. 4. ircle 3 centered at and radiu pae through I becaue biect angle. Thi circle i alo orthogonal to the circumcircle of ΔK becaue K, and therefore, by imilar triangle, K. (Indeed, the circumcircle of ΔK i it own invere with repect to 3. Since lie on K, all circle with chord K mut alo be orthogonal to 3. That include the circumcircle of K. Therefore,. ut I, o I. K N Remark: G Q H I P yme (003 ue thi lemma to olve a famou problem of V. Thébault. The Sawayama-Thébault Theorem tate that the center of the two circle P and Q and the incircle I of triangle are collinear. We do not need the theorem itelf (the lat entence of the proof to olve the problem, but we do need the other obervation. To quote yme, ccording to the hypothei, QG, P; o QG P. y [the] Lemma, GH and pa through I. Triangle HG and QGH being iocele, Q i (1 the perpendicular biector of GH, [and] ( the -internal angle biector of triangle HG. utati mutandi, P i (1 the perpendicular biector of, [and] ( the -internal angle biector of triangle. the biector of two adjacent and upplementary angle are perpendicular, we have Q P. Therefore, GH P and Q. oncluion: uing the convere of Pappu theorem applied to the heagon PIGQP, the point P, I and Q are collinear. 0

Unfortunately, the label and H are revered in yme paper, and, ince tranveral cut parallel line in equal correponding angle, the invocation of Pappu i rather gilding the lily. Solution proper: The Sawayama Lemma implie that the incenter of,, and lie on GH and. Hence there are pair of imilar right triangle with inradii for one leg. In the figure, we note five of them and the equation they imply: Q Ib H I I c P v c N (1 P ~ c I c c r c ( c b ( c I c ~ I rr c ( c b (3 c I c ~ b I b r b r c ( c b( b c (4 I ~ b I b r b ( b r( b (5 I ~ N v c ( b (b/r G b c Since i common to both and, we alo know b + c +. Thi allow u to epre b and in (1 and (4 uing only term that occur in the other three equation. Obviouly b c, which i eaily changed to ( b ( c b. rom the figure, ( c. Replacing with b + c, the lat epreion become ( c + ( b c ( c b c ( b c ( c b ( b c. i not o eaily eliminated, but dividing (1 by (, c /r /. Since a jut hown c ( c b ( b c ( c b ( b c, thi mean 1. Uing ( once again, r ( c b ( b c( c b ( c b we have c r. So, by (3, c r rb. Thi rc rc would be good enough if we weren t retricted to term involving only, but the problem require that we eliminate r and r b from thi equation. r[( b ( c b] r( c b We do thi with the modified verion of (4: rb r, or, b b vc ( c b ( c b vc ( c b vcrc by (5, r. Hence c + rc +. b r b b( b c c Thi i equivalent to v ( ( c c c c +, the olution attached to the original bc r angaku problem, becaue, quaring Heron ormula, r c c ( ( ( b. c c c 1

PROL 17: Triangle ha incircle (I, to which (O through and i internally tangent. ircle (P i tangent to and and eternally tangent to (O. ircle (Q i internally tangent to (O and tangent to at it midpoint. If r i the inradiu of, how that r i 4 time the product of the radii of (P and (Q. SOLUTION 17 (U: ircle (Q give the figure a pleaing balance, but all that matter i it diameter d. We mut prove that r dr p, where r p i the radiu of (P. P Our firt tak i to prove that (O i tangent to both (I and (P if and only if one of their intangent i parallel to, which require ome care. Q onider (P with P on anywhere in the egment I. Since the two right triangle formed by I, the two intangent, and radii of (I are congruent, we ee that, G, and I are concurrent in X, X XG, X GX, and G. The equal vertical angle X and GX have meaure G G (an eterior angle of a triangle i the um of the two oppoite interior angle. Hence, onequently if angle G and G, then G and. ut ince < and >, thi equation hold for the figure only if and. Thi prove Lemma: G ( i parallel to if and only if (G i it antiparallel in triangle. P X G Now let be the midpoint of, and conider the coaial ytem Г of circle with center on the perpendicular biector of. The typical circle cut and in two point that, together with and, form a cyclic quadrilateral. If G, then i one of thee becaue, in any cyclic quadrilateral, an eterior angle equal to interior angle at the non-adjacent verte, and we have jut hown that and G. I ukagawa & Pedhoe 1989,.4..

I P X G touche (P and (I in two ditinct point. ecaue all chord cut off by circle in Г are parallel to, the circle (O in Г that pae through or mut alo pa through the other. So by the lemma, there i a circle (O through and tangent to (I internally and to (P eternally, a pecified in the problem, if and only if one of their intangent (viz. G i parallel to. Since G ~, the ret i eay: let H be the point on touched by (I. H a, where i the emiperimeter of. Since (I i the ecircle of G, H i alo the emiperimeter of G. Hence r p /r ( a/. Therefore dr p dr( a/. If dr( a/ r, then d( a r area. Thi i true if and only if d i the radiu of the ecircle touching. Since ecircle (S belong to Σ, point T, where cut, i it contact point on. I i concurrent with the biector of the eterior angle at and in S, and ST jut a L. Therefore, drawing parallel to through N and S, we get rectangle LTUV, LTSW, and SUVW. N d TU LV but we don t yet know the length of US VW. However, T L c; therefore T L. Hence LN NUS: N lie on the diagonal of LTSW, and US VW d. Incidentally, K, L, and N are collinear becaue (I i incribed in the circular egment of (O bounded by arc K and chord. H P K I G T U d N V L S W 3

OROLLRY 3 : In the cae of two triangle, and, if the radii of the two circle tangent to are r 1 and r and the radii of the two mall circle at and are r 1 and r, then r 1 r (r 1 r /. Thi i the reult tated for ukagawa & Pedhoe problem.5.5. or if the diameter perpendicular to meaure d 1 above and d below, d 1 r 1 r 1 and d r r. ultiply thee equation together, noting that d 1 d (/. 3 ukagawa & Pedhoe 1989,.5.5. 4

Reference huja,., W. Uegaki, and K. atuhita. 004. apanee theorem: little known theorem with many proof. Part I and II, iouri ournal of athematical Science 16:7 80, 149 58. yme, ean-loui. 003. Sawayama and Thébault Theorem, orum Geometricorum 3.5 9. ukagawa, Hidetohi, and an Pedhoe. 1989. apanee temple geometry problem. Winnipeg: harle abbage Reearch entre., and Tony Rothman. 008. Sacred mathematic: apanee temple geometry. Princeton: Princeton Univerity Pre. Honberger, Ro. 1995. piode in nineteenth and twentieth-century uclidean geometry. Wahington,..: The athematical ociation of merica. 5