HMMT 十一月 2022 · 冲刺赛 · 第 30 题

HMMT November 2022 — Guts Round — Problem 30

[15] Let ABC be a triangle with AB = 8, AC = 12 , and BC = 5. Let M be the second intersection of the internal angle bisector of ∠ BAC with the circumcircle of ABC . Let ω be the circle centered at M tangent to AB and AC . The tangents to ω from B and C , other than AB and AC respectively, intersect at a point D . Compute AD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2022, November 12, 2022 — GUTS ROUND Organization Team Team ID#

解析

[15] Let ABC be a triangle with AB = 8, AC = 12 , and BC = 5. Let M be the second intersection of the internal angle bisector of ∠ BAC with the circumcircle of ABC . Let ω be the circle centered at M tangent to AB and AC . The tangents to ω from B and C , other than AB and AC respectively, intersect at a point D . Compute AD . Proposed by: Eric Shen Answer: 16 Solution: Redefine D as the reflection of A across the perpendicular bisector l of BC . We prove that DB and DC are both tangent to ω , and hence the two definitions of D align. Indeed, this follows by symmetry; we have that ∠ CBM = ∠ CAM = ∠ BAM = ∠ BCM , so BM = CM and so ω is centered on and hence symmetric across l . Hence reflecting BAC across l , we get that DB, DC are also tangent to ω , as desired. 2 2 Hence we have by Ptolemy that 5 AD = 12 − 8 , so thus AD = 16.