HMMT 二月 2006 · 冲刺赛 · 第 30 题
HMMT February 2006 — Guts Round — Problem 30
题目详情
英文原题
- [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 withdiameter 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 .
IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUNDth
The problems in this batch all depend on each other. If you solve them correctly, you willproduce a triple of mutually consistent answers. There is only one such triple. Your scorewill be determined by how many of your answers match that triple.
解析
英文解析
- 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 circlewith 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 .
Answer:675
XY 44
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 . Wehave h ( AQ ) = DI , and conclude that A, T , and D are collinear. Now since AP is ta diameter of Γ, ∠ P AT is right, implying that DT HP is cyclic. Invoking Power of ta Point twice, we have 225 = AE = AT · AD = AP · AH . Because we are given 210
radius of Γ we can find AP = 10 and AH = = h . If we write a, b, c, s in the usual 45
2 amanner with respect to triangle ABC , we seek BD · DC = ( s − b )( s − c ). But recallthat Heron’s formula gives us
√
s ( s − a )( s − b )( s − c ) = Kr s 2
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 , wea
2 a 2
( )ss h 3 r s 15 67522
a afind = = . Now we compute our answer, = · = .
a 2 r 2 s − a 2 − 1 4 sa