HMMT 二月 2009 · 几何 · 第 9 题

HMMT February 2009 — Geometry — Problem 9

[ 7 ] Let ABC be a triangle with AB = 16 and AC = 5 . Suppose the bisectors of angles ∠ ABC and ∠ BCA meet at point P in the triangle’s interior. Given that AP = 4, compute BC .

解析

[ 7 ] Let ABC be a triangle with AB = 16 and AC = 5 . Suppose the bisectors of angles ∠ ABC and ∠ BCA meet at point P in the triangle’s interior. Given that AP = 4, compute BC . Answer: 14 Solution: As the incenter of triangle ABC, point P has many properties. Extend AP past P to its intersection with the circumcircle of triangle ABC, and call this intersection M. Now observe that ∠ P BM = ∠ P BC + ∠ CBM = ∠ P BC + ∠ CAM = β + α = 90 − γ, where α, β, and γ are the half-angles of triangle ABC. Since ∠ BM P = ∠ BM A = ∠ BCA = 2 γ, 3 it follows that BM = M P = CM. Let Q denote the intersection of AM and BC, and observe that 4 AQB ∼ 4 CQM and 4 AQC ∼ 4 BQM ; some easy algebra gives AM/BC = ( AB · AC + BM · CM ) / ( AC · CM + AB · BM ) . 2 Writing ( a, b, c, d, x ) = ( BC, AC, AB, M P, AP ) , this is ( x + d ) /a = ( bc + d ) / (( b + c ) d ). Ptolemy’s 2 2 theorem applied to ABCD gives a ( d + x ) = d ( b + c ) . Multiplying the two gives ( d + x ) = bc + d . We 2 easily solve for d = ( bc − x ) / (2 x ) = 8 and a = d ( b + c ) / ( d + x ) = 14 .