HMMT 二月 2013 · 团队赛 · 第 8 题

HMMT February 2013 — Team Round — Problem 8

[ 35 ] Let points A and B be on circle ! centered at O . Suppose that ! and ! are circles not A B containing O which are internally tangent to ! at A and B , respectively. Let ! and ! intersect at A B C and D such that D is inside triangle ABC . Suppose that line BC meets ! again at E and let line EA intersect ! at F . If F C ? CD , prove that O , C , and D are collinear. A

解析

[ 35 ] Let points A and B be on circle ω centered at O . Suppose that ω and ω are circles not A B containing O which are internally tangent to ω at A and B , respectively. Let ω and ω intersect at A B C and D such that D is inside triangle ABC . Suppose that line BC meets ω again at E and let line EA intersect ω at F . If F C ⊥ CD , prove that O , C , and D are collinear. A Answer: N/A Let H = CA ∩ ω , and G = BH ∩ ω . There are homotheties centered at A and B B taking ω → ω and ω → ω that take A : F 7 → E , A : C 7 → H , B : C 7 → E and B : G 7 → H . In A B particular CF || EH || CG , so C, F, G are collinear, lying on a line perpendicular to DC . Because of the right angles at C , DF and DG are diameters of ω , ω , respectively. Also, we have that A B the ratio of the sizes of ω and ω , under the two homotheties above, is CF/EH · EH/CG = CF/CG . A B Therefore, DF/DG = CF/CG , but then △ DCF and △ DCG are both right triangles which share one side and have hypotenuse and other side in proportion; it is obvious now that the two circles ω and A ω are congruent. B Therefore, O has the same distance to A and B , and so the same distances to the centers of the two circles as well. As a result, O lies on the radical axis CD as desired.