HMMT 二月 2006 · 冲刺赛 · 第 25 题

HMMT February 2006 — Guts Round — Problem 25

[9] Points A , C , and B lie on a line in that order such that AC = 4 and BC = 2. Circles ω , 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

Points A , C , and B lie on a line in that order such that AC = 4 and BC = 2. Circles ω , 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 . 2 Answer: 3 Solution: Let the center of ω be O for i = 1 , 2 , 3 and let O denote the center of 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 6 We find r = . Now the key observation is that the circumcircle of triangle CDE is 7 13 20 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