PUMaC 2014 · 几何（B 组） · 第 7 题

PUMaC 2014 — Geometry (Division B) — Problem 7

[ 7 ] Consider quadrilateral ABCD . Given that ∠ DAC = 70, ∠ BAC = 40, ∠ BDC = 20, ∠ CBD = 35. Let P be the intersection of AC and BD . Find ∠ BP C .

解析

[ 7 ] Consider quadrilateral ABCD . Given that \ DAC = 70, \ BAC = 40, \ BDC = 20, \ CBD = 35. Let P be the intersection of AC and BD . Find \ BP C . Solution: Take the circumcircle of triangle BCD with center O . Then since CBD and COD both 1 intersect arc CD , we have that m \ CBD = m \ COD ) m \ COD = 70 and similary 2 m \ COB = 40 and we can see that O = A and so A is the center of the circle. Suppose we extend AC to intersect the circle at the opposite side at E , then we have that BC + DE 40 + 180 70 m \ BP C = = = 75 . 2 2