PUMaC 2023 · 几何（A 组） · 第 7 题
PUMaC 2023 — Geometry (Division A) — Problem 7
题目详情
- Let △ ABC be a triangle with ∠ BAC = 90 , ∠ ABC = 60 , and ∠ BCA = 30 and BC = 4. Let the incircle of △ ABC meet sides BC, CA, AB at points A , B , C , respectively. Let 0 0 0 ω , ω , ω denote the circumcircles of triangles △ B IC , △ C IA , △ A IB , respectively. We A B C 0 0 0 0 0 0 construct triangle T as follows: let A B meet ω for the second time at A ̸ = A , let A C A 0 0 B 1 0 0 0 meet ω for the second time at A ̸ = A , and let T denote the triangle △ A A A . Construct C 2 0 A 0 1 2 √ triangles T , T similarly. If the sum of the areas of triangles T , T , T equals m − n for B C A B C positive integers m, n , find m + n . 7
解析
- Let △ ABC be a triangle with ∠ BAC = 90 , ∠ ABC = 60 , and ∠ BCA = 30 and BC = 4. Let the incircle of △ ABC meet sides BC, CA, AB at points A , B , C , respectively. Let 0 0 0 ω , ω , ω denote the circumcircles of triangles △ B IC , △ C IA , △ A IB , respectively. We A B C 0 0 0 0 0 0 construct triangle T as follows: let A B meet ω for the second time at A ̸ = A , let A C A 0 0 B 1 0 0 0 meet ω for the second time at A ̸ = A , and let T denote the triangle △ A A A . Construct C 2 0 A 0 1 2 √ triangles T , T similarly. If the sum of the areas of triangles T , T , T equals m − n for B C A B C positive integers m, n , find m + n . Proposed by Sunay Joshi Answer: 15 √ We begin by computing the inradius r . The sides of △ ABC are clearly a = 4, b = 2 3, and √ √ c = 2, so that the semiperimeter is s = 3 + 3. The area is 2 3. Since rs = [ ABC ], we have √ 2 3 √ r = . 3+ 3 We now compute the area of T . For convenience we relabel A , B , C by D, E, F . Let B 0 0 0 r r , r , r denote the radii of ω , ω , ω , respectively. Note that by trigonometry, = A B C A B C 2 r A A A sin and EF = 2 cos · r , with similar equalities holding for cyclic permutations of the 2 2 indices. Let us take B = F E ∩ ω WLOG. The key observation is the following spiral 1 C DF DI similarity: △ DF B ∼ △ DII , where I is the center of ω . Therefore = , so that 1 C C C F B II 1 C II C F B = · DF , so that 1 DI r C F B = · 2 r sin B (1) 1 B r 1 B = · 2 cos · r (2) C 2 2 sin 2 Thus EB = F B − EF (3) 1 1 1 B A = · 2 cos · r − 2 cos · r (4) C 2 2 2 sin 2 1 B A C = r · · (cos − 2 cos sin ) (5) C 2 2 2 sin 2 1 A − C = r · · sin | | (6) C 2 sin 2 1 A − C B By symmetry, EB = r · · sin | | . Since ∠ B EB = 90 − , the sine area formula 2 A 1 2 2 2 sin 2 applied to T yields B 1 B 1 A − C 2 2 [ T ] = r · cos · · sin (7) B A C 2 2 2 sin sin 2 2 3 which may be simplified via half-angle and product-to-sum as follows: 1 sin B A − C 2 2 [ T ] = r sin (8) B A B C 4 2 sin sin sin 2 2 2 1 sin B 2 = r (1 − cos( A − C )) (9) A B C 8 sin sin sin 2 2 2 1 1 2 = r (sin B − cos( A − C ) sin B ) (10) A B C 8 sin sin sin 2 2 2 1 1 1 2 = r (sin B − (sin 2 A + sin 2 C )) (11) A B C 8 2 sin sin sin 2 2 2 Summing cyclically, we find 1 1 2 [ T ] + [ T ] + [ T ] = r (sin A + sin B + sin C − sin 2 A − sin 2 B − sin 2 C ) A B C A B C 8 sin sin sin 2 2 2 (12) √ A 1 3 √ Recall A = 90 , B = 60 , C = 30, so that sin A = 1, sin 2 A = 0, sin = , sin B = , 2 2 2 √ √ √ 3 B 1 1 3 C 3 − 1 √ sin 2 B = , sin = , sin C = , sin 2 C = , sin = . Plugging in, the sum becomes 2 2 2 2 2 2 2 2 √ √ √ 1 1 3 1 3 3 2 √ r · (1 + + − 0 − − ) (13) 1 1 3 − 1 8 2 2 2 2 √ √ · · 2 2 2 2 √ √ 2 3 1 3 − 3 2 √ √ = ( ) (14) 2 3 + 3 3 − 1 √ = 2 3 − 3 (15) √ Therefore the sum is 12 − 3 with m = 12 and n = 3, yielding an answer of 12 + 3 = 15. 7