PUMaC 2017 · 几何（A 组） · 第 7 题

PUMaC 2017 — Geometry (Division A) — Problem 7

Let ACDB be a cyclic quadrilateral with circumcircle ω . Let AC = 5 , CD = 6 , and DB = 7. Suppose that there is a unique point P on ω such that P C intersects AB at a point P and 1 P D intersects AB at a point P , such that AP = 3 and P B = 4. Let Q be the unique point 2 1 2 on ω such that QC intersects AB at a point Q , QD intersects AB at a point Q , Q is closer 1 2 1 p to B than P is to B , and P Q = 2. The length of P Q can be written as where p and q 1 2 2 1 1 q are relatively prime positive integers. Find p + q .

解析

We will make use of projective geometry. Q P ( A, P ; P , B ) = ( A, D ; C, B ) = ( A, Q ; Q , B ) 2 1 2 1 1 AP P B AC DB AQ Q B 1 2 1 2 · = · = · P P AB DC AB Q Q AB 2 1 2 1 3 5 AQ 1 · 4 = · 7 = · Q B 2 P P 6 Q Q 1 2 2 1 72 implying P P = . Note Q B = P B − P Q = 4 − 2 = 2. Thus, 1 2 2 2 2 2 35 AQ 35 1 = Q Q 12 1 2 Furthermore, AQ + Q Q + Q B = AP + P P + P B 1 1 2 2 1 1 2 2 12 72 AQ + ( AQ ) + 2 = 3 + + 4 1 1 35 35 72 5 + 106 35 P Q = AQ − AP = − 3 = 1 1 1 1 12 47 1 + 35 so the answer is 106 + 47 = 153 . Problem written by Jackson Blitz