HMMT 十一月 2011 · 冲刺赛 · 第 33 题

HMMT November 2011 — Guts Round — Problem 33

[ 17 ] Let ABC be a triangle with AB = 5, BC = 8, and CA = 7. Let Γ be a circle internally tangent to the circumcircle of ABC at A which is also tangent to segment BC . Γ intersects AB and AC at points D and E , respectively. Determine the length of segment DE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TH 4 ANNUAL HARVARD-MIT NOVEMBER TOURNAMENT, 12 NOVEMBER 2011 — GUTS ROUND Round 12

解析

[ 17 ] Let ABC be a triangle with AB = 5, BC = 8, and CA = 7. Let Γ be a circle internally tangent to the circumcircle of ABC at A which is also tangent to segment BC . Γ intersects AB and AC at points D and E , respectively. Determine the length of segment DE . 40 Answer: 9 Guts Round A P E D O B Z X M C Y First, note that a homothety h centered at A takes Γ to the circumcircle of ABC , D to B and E to C , since the two circles are tangent. As a result, we have DE ‖ BC . Now, let P be the center of Γ and O be the circumcenter of ABC : by the homothety h , we have DE/BC = AP/AO . − − → Let Γ be tangent to BC at X , and let ray AX meet the circumcircle of ABC at Y . Note that Y is the image of X under h . Furthermore, h takes BC to the tangent line l to the circumcircle of ABC at Y , ̂ and since BC ‖ l , we must have that Y is the midpoint of arc BC . Therefore, AX bisects ∠ BAC . Now, let Z be the foot of the altitude from A to BC , and let M be the midpoint of BC , so that OM ⊥ BC . Note that AP/AO = ZX/ZM . Now, letting BC = a = 8, CA = b = 7, and AB = c = 5, we compute 2 2 2 c + a − b 5 BZ = c cos B = = 2 a 2 by the Law of Cosines, ac 10 BX = = b + c 3 by the Angle Bisector Theorem, and BM = 4 . To finish, ( AP )( BC ) ( ZX )( BC ) (5 / 6)(8) 40 DE = = = = AO ZM (3 / 2) 9 .