HMMT 二月 2022 · 冲刺赛 · 第 19 题

HMMT February 2022 — Guts Round — Problem 19

[11] In right triangle ABC , a point D is on hypotenuse AC such that BD ⊥ AC . Let ω be a circle with center O , passing through C and D and tangent to line AB at a point other than B . Point X is chosen a on BC such that AX ⊥ BO . If AB = 2 and BC = 5, then BX can be expressed as for relatively prime b positive integers a and b . Compute 100 a + b . 20

解析

[11] In right triangle ABC , a point D is on hypotenuse AC such that BD ⊥ AC . Let ω be a circle with center O , passing through C and D and tangent to line AB at a point other than B . Point X a is chosen on BC such that AX ⊥ BO . If AB = 2 and BC = 5, then BX can be expressed as for b relatively prime positive integers a and b . Compute 100 a + b . Proposed by: Akash Das Answer: 8041 2 ′ Solution: Note that since AD · AC = AB , we have the tangency point of ω and AB is B , the reflection of B across A . Let Y be the second intersection of ω and BC . Note that by power of point, 2 ′ 2 2 4 AB we have BY · BC = BB = 4 AB = ⇒ BY = . Note that AX is the radical axis of ω and the BC 2 degenerate circle at B , so we have XB = XY · XC , so 2 2 BX = ( BC − BX )( BY − BX ) = BX − BX ( BC + BY ) + BC · BY. This gives us 2 BC · BY 4 AB · BC 80 BX = = = . 2 2 BC + BY 4 AB + BC 41 20