PUMaC 2012 · 几何（A 组） · 第 5 题

PUMaC 2012 — Geometry (Division A) — Problem 5

[ 5 ] Let 4 ABC be a triangle with ∠ BAC = 45 , ∠ BCA = 30 , and AB = 1. Point D lies on segment AC such that AB = BD . Find the square of the length of the common external tangent to the circumcircles of triangles 4 BDC and 4 ABC .

解析

[ 5 ] Let 4 ABC be a triangle with ∠ BAC = 45 , ∠ BCA = 30 , and AB = 1. Point D lies on segment AC such that AB = BD . Find the square of the length of the common tangent between the circumcircles of triangles 4 BDC and 4 ABC . Solution: We claim that the two circumcircles have same radius. Indeed, the circumcircle of ABC has radius BC/ 2 sin( ∠ BAC ), and the circumcircle of BDC has radius BC/ 2 sin( ∠ BDC ). ◦ ◦ Because AB = BD , we have that m ∠ BDC = 180 − m ∠ BDA = 180 − m ∠ BAC , so ∠ BAC and ∠ BDC have the same sine, from which it follows that the two circumcircles have the same radius. Let O and O be the circumcenters of ABC and BDC . Then, the length 1 2 of the common tangent is the same as the length of O O . The circumradius of ABC is 1 2 ◦ ◦ AB/ (2 sin(30 )) = 1. Note that m ∠ BO C = 2 m ∠ BAC = 90 . If we let M be the intersection 1 √ √ of BC and O O , then O O = 2 O M = 2 O C/ 2 = 2, so the square of the length of the 1 2 1 2 1 1 common tangent is 2 . 2 Problem contributed by Gene Katsevich