HMMT 十一月 2010 · 团队赛 · 第 7 题
HMMT November 2010 — Team Round — Problem 7
题目详情
英文原题
- [ 7 ] ABC is a right triangle with ∠ A = 30 and circumcircle O . Circles ω , ω , and ω lie outside
1 2 3
ABC and are tangent to O at T , T , and T respectively and to AB , BC , and CA at S , S , and S ,
1 2 3 1 2 3
′ ′ ′
respectively. Lines T S , T S , and T S intersect O again at A , B , and C , respectively. What is
1 1 2 2 3 3
′ ′ ′
the ratio of the area of A B C to the area of ABC ?
Linear? What’s The Problem?
A function f ( x , x , . . . , x ) is said to be linear in each of its variables if it is a polynomial such that
1 2 nno variable appears with power higher than one in any term. For example, 1 + x + xy is linear in x
2 2
and y , but 1 + x is not. Similarly, 2 x + 3 yz is linear in x , y , and z , but xyz is not. 1
解析
英文解析
- [ 7 ] ABC is a right triangle with ∠ A = 30 and circumcircle O . Circles ω , ω , and ω lie outside
1 2 3
ABC and are tangent to O at T , T , and T respectively and to AB , BC , and CA at S , S , and S ,
1 2 3 1 2 3
′ ′ ′
respectively. Lines T S , T S , and T S intersect O again at A , B , and C , respectively. What is
1 1 2 2 3 3
′ ′ ′
the ratio of the area of A B C to the area of ABC ?
√
3+1
Answer: Let [ P QR ] denote the area of 4 P QR . The key to this problem is following fact:
[ P QR ] = P Q · P R sin ∠ QP R .12
√2
√
°3
Assume that the radius of O is 1. Since ∠ A = 30 , we have BC = 1 and AB = 3. So [ ABC ] = .
′ ′ ° ′ ° ′2
Let K denote the center of O . Notice that ∠ B KA = 90 , ∠ AKC = 90 , and ∠ B KA = ∠ KAB =
° ′ ′ ′ ′ ° ′ ′ °