HMMT 二月 2006 · 冲刺赛 · 第 41 题

HMMT February 2006 — Guts Round — Problem 41

[18] Let Γ denote the circumcircle of triangle ABC . Point D is on AB such that CD bisects ∠ ACB . Points P and Q are on Γ such that P Q passes through D and is perpendicular to CD . Compute P Q , given that BC = 20 , CA = 80 , AB = 65.

解析

Let Γ denote the circumcircle of triangle ABC . Point D is on AB such that CD bisects ∠ ACB . Points P and Q are on Γ such that P Q passes through D and is perpendicular to CD . Compute P Q , given that BC = 20 , CA = 80 , AB = 65. √ Answer: 4 745 Solution: Suppose that P lies between A and B and Q lies between A and C , and let line P Q intersect lines AC and BC at E and F respectively. As usual, we write a, b, c for the lengths of BC, CA, AB . By the angle bisector theorem, AD/DB = AC/CB so ( ) ( ) bc ac ac bc 2 that AD = and BD = . Now by Stewart’s theorem, c · CD + c = a + b a + b a + b a + b 2 2 2 2 ab (( a + b ) − c ) a bc ab c 2

from which CD = . Now observe that triangles CDE and 2 a + b a + b ( a + b ) CA ED F B CDF are congruent, so ED = DF . By Menelaus’ theorem, = 1 so that AE DF BC b ( b − a ) CA AE 2 ab = . Since CF = CE while b > a , it follows that AE = so that EC = . BC F B a + b a + b √ √ 2 2 ab ( c − ( a − b ) ) 2 2 Finally, DE = CE − CD = . Plugging in a = 20 , b = 80 , c = 65, a + b we see that AE = 48 , EC = 32 , DE = 10 as well as AD = 52 , BD = 13. Now let P D = x, QE = y . By power of a point about D and E , we have x ( y + 10) = 676 and y ( x + 10) = 1536. Subtracting one from the other, we see that y = x + 86. Therefore, √ √ 2 x + 96 x − 676 = 0, from which x = − 48 + 2 745. Finally, P Q = x + y + 10 = 4 745.