HMMT 二月 2006 · 冲刺赛 · 第 30 题
HMMT February 2006 — Guts Round — Problem 30
题目详情
- [10] ABC is an acute triangle with incircle ω . ω is tangent to sides BC , CA , and AB at D , E , and F respectively. P is a point on the altitude from A such that Γ, the circle with diameter AP , is tangent to ω . Γ intersects AC and AB at X and Y respectively. Given XY = 8, AE = 15, and that the radius of Γ is 5, compute BD · DC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND The problems in this batch all depend on each other. If you solve them correctly, you will produce a triple of mutually consistent answers. There is only one such triple. Your score will be determined by how many of your answers match that triple.
解析
- ABC is an acute triangle with incircle ω . ω is tangent to sides BC , CA , and AB at D , E , and F respectively. P is a point on the altitude from A such that Γ, the circle with diameter AP , is tangent to ω . Γ intersects AC and AB at X and Y respectively. Given XY = 8, AE = 15, and that the radius of Γ is 5, compute BD · DC . 675 Answer: 4 XY 4 Solution: By the Law of Sines we have sin ∠ A = = . Let I , T , and Q denote the AP 5 center of ω , the point of tangency between ω and Γ, and the center of Γ respectively. A 1 A Since we are told ABC is acute, we can compute tan ∠ = . Since ∠ EAI = 2 2 2 AE 15 and AE is tangent to ω , we find r = = . Let H be the foot of the altitude 2 2 from A to BC . Define h to be the homothety about T which sends Γ to ω . We T have h ( AQ ) = DI , and conclude that A, T , and D are collinear. Now since AP is T a diameter of Γ, ∠ P AT is right, implying that DT HP is cyclic. Invoking Power of 10 2 a Point twice, we have 225 = AE = AT · AD = AP · AH . Because we are given 45 radius of Γ we can find AP = 10 and AH = = h . If we write a, b, c, s in the usual a 2 manner with respect to triangle ABC , we seek BD · DC = ( s − b )( s − c ). But recall that Heron’s formula gives us √ s ( s − a )( s − b )( s − c ) = K 2 r s where K is the area of triangle ABC . Writing K = rs , we have ( s − b )( s − c ) = . s − a 15 s 1 Knowing r = , we need only compute the ratio . By writing K = ah = rs , we a 2 a 2 s ( ) 2 2 s h 3 r s 15 675 a a find = = . Now we compute our answer, = · = . s a 2 r 2 s − a 2 − 1 4 a