HMMT 二月 2006 · 冲刺赛 · 第 25 题
HMMT February 2006 — Guts Round — Problem 25
题目详情
英文原题
- [9] Points A , C , and B lie on a line in that order such that AC = 4 and BC = 2. Circlesω , ω , and ω have BC, AC , and AB as diameters. Circle Γ is externally tangent to ω and
1 2 3 1
ω at D and E respectively, and is internally tangent to ω . Compute the circumradius of
2 3
triangle CDE .
解析
英文解析
- Points A , C , and B lie on a line in that order such that AC = 4 and BC = 2. Circlesω , ω , and ω have BC, AC , and AB as diameters. Circle Γ is externally tangent
1 2 3
to ω and ω at D and E respectively, and is internally tangent to ω . Compute the
1 2 3
circumradius of triangle CDE .
Answer:2
Solution: Let the center of ω be O for i = 1 , 2 , 3 and let O denote the center of 3
i i
Γ. Then O, D , and O are collinear, as are O, E , and O . Denote by F the point of
1 2
tangency between Γ and ω ; then F, O , and O are collinear. Writing r for the radius
3 3
of Γ we have OO = r + 2 , OO = r + 1 , OO = 3 − r . Now since O O = 1 and
1 2 3 1 3
O O = 2, we apply Stewart’s theorem:
3 2
2 2 2
OO · O O + OO · O O = OO · O O + O O · O O · O O
2 3 1 3 1 2 1 3 3 2 1 2
1 2 3
2 2 2
2( r + 2) + ( r + 1) = 3(3 − r ) + 1 · 2 · 3
We find r = . Now the key observation is that the circumcircle of triangle CDE is 6
13 207
the incircle of triangle OO O . We easily compute the sides of OO O to be , , and
1 2 1 2
7 7
18 27