HMMT 二月 2007 · 冲刺赛 · 第 32 题

HMMT February 2007 — Guts Round — Problem 32

[ 18 ] Triangle ABC has AB = 4 , BC = 6 , and AC = 5 . Let O denote the circumcenter of ABC. The circle Γ is tangent to and surrounds the circumcircles of triangles AOB, BOC, and AOC . Determine the diameter of Γ .

解析

[ 18 ] Triangle ABC has AB = 4 , BC = 6 , and AC = 5 . Let O denote the circumcenter of ABC. The circle Γ is tangent to and surrounds the circumcircles of triangles AOB, BOC, and AOC . Determine the diameter of Γ . √ 256 7 Answer: . Denote by ω, Γ , Γ , and Γ the circumcenters of triangles ABC, BOC, COA, and 1 2 3 17 AOB , respectively. An inversion about ω interchanges Γ and line BC , Γ and line CA , and Γ and 1 2 3 line AB. This inversion also preserves tangency between generalized circles, so the image of Γ is a circle tangent to AB , BC, and CA. It is the incircle of ABC because it is closer to O than these lines and ABC is acute. Now we run a few standard calculations. Where s, r, and R denote the semiperimeter, inradius, and circumradius of ABC , respectively, we have the following: √ √ 15 7 [ ABC ] = s ( s − a )( s − b )( s − c ) = ; 4 √ r = [ ABC ] /s = 7 / 2; abc 8 R = = √ 4[ ABC ] 7 8 2 OI = R ( R − 2 r ) = . 7 Let OI intersect the incircle of ABC at P and Q , with I between P and O. Then OP = r + OI and ′ ′ OQ = r − OI, and P Q is a diameter. Under the inversion, P and Q map to P and Q respectively. ′ ′ Because P, I, O, and Q are collinear in that order, P and Q are diametrically opposed on Γ. It follows that the diameter of Γ is ( ) 2 2 2 R R 1 1 2 rR ′ ′ ′ ′ 2 P Q = OP + OQ = + = R + = . 2 2 OP OQ r + OI r − OI r − OI √ 256 7 We plug in the values found above to arrive at . 17