HMMT 十一月 2020 · GEN 赛 · 第 7 题

HMMT November 2020 — GEN Round — Problem 7

In triangle ABC with AB = 8 and AC = 10, the incenter I is reflected across side AB to point X and 2 across side AC to point Y . Given that segment XY bisects AI , compute BC . (The incenter I is the center of the inscribed circle of triangle ABC .)

解析

In triangle ABC with AB = 8 and AC = 10, the incenter I is reflected across side AB to point X and 2 across side AC to point Y . Given that segment XY bisects AI , compute BC . (The incenter I is the center of the inscribed circle of triangle ABC .) Proposed by: Carl Schildkraut Answer: 84 Solution 1: A Q X Y P E F I Let E, F be the tangency points of the incircle to sides AC, AB , respectively. Due to symmetry around line AI , AXIY is a rhombus. Therefore ◦ ◦ ∠ XAI = 2 ∠ EAI = 2(90 − ∠ EIA ) = 180 − 2 ∠ XAI, ◦ which implies that 60 = ∠ XAI = 2 ∠ EAI = ∠ BAC . By the law of cosines, 2 2 2 ◦ BC = 8 + 10 − 2 · 8 · 10 · cos 60 = 84 . Solution 2: Define points as above and additionally let P and Q be the intersections of AI with EF and XY , respectively. Since IX = 2 IE and IY = 2 IF , 4 IEF ∼ 4 IXY with ratio 2, implying that 1 1 IP = IQ = IA . 2 4 IP IP IE 2 ◦ Let θ = ∠ EAI = ∠ IEP . Then = = sin θ , implying that sin θ = 1 / 2 and θ = 30 . From IA IE IA here, proceed as in solution 1.