PUMaC 2023 · 团队赛 · 第 13 题

PUMaC 2023 — Team Round — Problem 13

Let △ T BD be a triangle with T B = 6, BD = 8, and DT = 7. Let I be the incenter of △ T BD , and let T I intersect the circumcircle of △ T BD at M ̸ = T . Let lines T B and M D intersect at Y , and let lines T D and M B intersect at X . Let the circumcircles of △ Y BM and △ XDM intersect at Z ̸ = M . If the area of △ Y BZ is x and the area of △ XDZ is y , then the p x ratio can be expressed as , where p and q are relatively prime positive integers. Find p + q . y q 2

解析

Let △ T BD be a triangle with T B = 6, BD = 8, and DT = 7. Let I be the incenter of △ T BD , and let T I intersect the circumcircle of △ T BD at M ̸ = T . Let lines T B and M D intersect at Y , and let lines T D and M B intersect at X . Let the circumcircles of △ Y BM and △ XDM intersect at Z ̸ = M . If the area of △ Y BZ is x and the area of △ XDZ is y , then the p x ratio can be expressed as , where p and q are relatively prime positive integers. Find p + q . y q Proposed by Sunay Joshi Answer: 97 Below, let us relabel points T, D as points A, C , respectively. Let a = BC , b = CA , and c = AB . Since ∠ Y ZB = ∠ Y M B = ∠ XM C = ∠ M ZC and ∠ BY Z = ∠ XM Z = ∠ XCZ , the triangles 2 △ ZY B and △ ZCX are similar. The desired ratio is therefore ( Y B/XC ) . Since triangles △ Y BM and △ Y CA are similar, we have Y B/Y C = M B/AC and Y M/Y A = M B/AC . Using the fact that Y C = Y M + M C and Y A = Y B + AB , we find that Y B = 2 2 M B ( b + c ) M C ( b + c ) . By symmetry, XC = . Since M is the midpoint of arc BC , we have 2 2 2 2 b − M B c − M C 2 2 b − M B 2 M B = M C , and hence the desired ratio reduces to ( ) . 2 2 c − M B a To compute M B , note that M B = . By the Law of Cosines, 2 cos( A/ 2) 2 2 2 a a a bc 2 M B = = = 2 2 2 4 cos ( A/ 2) 2(1 + cos A ) ( b + c ) − a Therefore the desired ratio equals 2 2 2 b ( b + c ) − a b 81 = , 2 2 c ( b + c ) − a c 16 and our answer is 81 + 16 = 97. 6 2