PUMaC 2015 · 个人决赛（B 组） · 第 2 题

PUMaC 2015 — Individual Finals (Division B) — Problem 2

On a circle ω , four points A , B , C , D lie in that order. Prove that CD = AC · BC + AD · BD 1 if and only if at least one of C and D is the midpoint of arc AB .

解析

On a circle ω , four points A , B , C , D lie in that order. Prove that CD = AC · BC + AD · BD 1 if and only if at least one of C and D is the midpoint of arc AB . Solution: 1 Without loss of generality, let the diameter of ω be . Notice that the area of quadrilateral 1 2 ABCD can be calculated in two ways: one as the sum of the area of 4 ABC and the area of 1 1 4 ABD , which is AC · BC · sin ∠ ACB + AD · BD · sin ∠ ADB . The other way is to notice 2 2 1 that the area is just AB · CD · sin ∠ AP C where P is the intersection of AB and CD . So we 2 have: 1 1 1 AC · BC · sin ∠ ACB + AD · BD · sin ∠ ADB = AB · CD · sin ∠ AP C 2 2 2 1 1 1 In a circle with diameter , sin ∠ ACB = sin ∠ ADB = AB . Therefore AC · BC + AD · BD = 2 2 2 1 2 1 CD · sin ∠ AP C . So CD = AC · BC + AD · BD iff CD = sin ∠ AP C iff ∠ CAD = ∠ AP C 2 2 or ∠ CBD = ∠ AP C . Since ∠ AP C = ∠ ADC + ∠ BAD , the previous statement is equivalent to AC = BC or AD = BD , as desired. Author: Xiaoyu Xu