PUMaC 2019 · 个人决赛（B 组） · 第 3 题

PUMaC 2019 — Individual Finals (Division B) — Problem 3

Let M N be a chord of the circle Γ and let S be the midpoint of M N . Let A, B, C, D be points on Γ such that AC and BD intersect at S and A and B are on the same side of M N . Let d , d , d , d be the distances from M N to A , B , C , and D , respectively. Prove that A B C D 1 1 1 1

解析

Let M N be a chord of a circle, and let S be its midpoint. Now let A, B, C, D be points on that circle such that AC and BD both contain S , and A and B are on the same side of M N . Let d , d , d , d be the distances from A, B, C, D respectively to M N . Prove that A B C D 1 1 1 1

= + . d d d d A D B C 1 1 1 1 Solution : It’s natural to convert the expression into − = − . Now we can see that d d d d A C B D 1 1 since B, D can be any chord through S , we need to prove that − = constant , where d d A C that constant only depends on M N and not on the choice of A . Now let’s just compute it. Let O be the center of the circle and let the angle between AC and M N be θ . WLOG let ′ ′ AS ≤ CS . Let the feet of perpendiculars from A and C to M N be A and C . Now from d d 1 1 1 AS 1 AS − CS ′ ′ A C 4 AA S ∼ 4 CC S we have = , so now − = (1 − ) = . Let P be AS CS d d d CS d CS A C A A 1 1 1 SP a point on AC such that AS = CP , now − = . Now we see that ∠ OSC = 90 − θ , d d d CS A C A so SP = 2 OS sin θ . Now from the power of the point S , we have AS · CS = M S · N S , so since 1 1 1 2 OS sin θ 2 OS d = AS sin θ , we have that − = = , which doesn’t depend on A d d AS sin θ CS M S · N S A C A . So now the result follows. Proposed by Igor Medvedev. 1