PUMaC 2015 · 几何（A 组） · 第 7 题

PUMaC 2015 — Geometry (Division A) — Problem 7

[ 7 ] Triangle ABC has AB = AC = 20 and BC = 15. Let D be the point in △ ABC such that △ ADB ∼ △ BDC . Let l be a line through A and let BD and CD intersect l at P and Q , respectively. Let the circumcircles of △ BDQ and △ CDP intersect at X . The area of the p locus of X as l varies can be expressed in the form π for positive coprime integers p and q . q What is p + q ?

解析

[ 7 ] Triangle ABC has AB = AC = 20 and BC = 15. Let D be the point in △ ABC such that △ ADB ∼ △ BDC . Let l be a line through A and let BD and CD intersect l at P and Q , respectively. Let the circumcircles of △ BDQ and △ CDP intersect at X . The area of the p locus of X as l varies can be expressed in the form π for positive coprime integers p and q . q What is p + q ? Solution: Let BQ and CP intersect at E , and by Pascal’s converse on the circumcircle of △ BCD , hexagon BBDCCE , and line l , BCDE is concyclic. Let O be the center of the circumcircle of △ BCD , and note that O is also diametrically opposite A in the circumcircle of △ ABC . X is then the Miquel point of cyclic quadrilateral BCDE . By some of Yufei’s lemmas (see his quadrilateral handouts), we know that X lies on l and that OX ⊥ l . Thus, as l varies, abc 20 ⋅ 20 ⋅ 15 80 ” ” X varies along the circumcircle of △ ABC , which has circumradius = = and 15 5 55 4 A 55 4 ⋅ ⋅ 2 2 6400 1280 thus has area = . The desired answer is then 1291 . 55 11 4 Diagram of the case when AP = AQ