HMMT 二月 2018 · 冲刺赛 · 第 26 题

HMMT February 2018 — Guts Round — Problem 26

[ 15 ] Let ABC be a triangle with ∠ A = 18 , ∠ B = 36 . Let M be the midpoint of AB , D a point on ray CM such that AB = AD ; E a point on ray BC such that AB = BE , and F a point on ray AC such that AB = AF . Find ∠ F DE .

解析

[ 15 ] Let ABC be a triangle with ∠ A = 18 , ∠ B = 36 . Let M be the midpoint of AB , D a point on ray CM such that AB = AD ; E a point on ray BC such that AB = BE , and F a point on ray AC such that AB = AF . Find ∠ F DE . Proposed by: Faraz Masroor Answer: 27 Let ∠ ABD = ∠ ADB = x , and ∠ DAB = 180 − 2 x . In triangle ACD , by the law of sines, CD = AD BD · sin 198 − 2 x , and by the law of sines in triangle BCD , CD = · sin x + 36. Combining sin ∠ ACM sin ∠ BCM BD sin 198 − 2 x sin ∠ BCM M B CB sin ∠ BCM the two, we have 2 cos x = = · . But by the ratio lemma, 1 = = , AD sin x +36 sin ∠ ACM M A CA sin ∠ ACM sin ∠ BCM CA sin 36 meaning that = = = 2 cos 18. Plugging this in and simplifying, we have 2 cos x = sin ∠ ACM CB sin 18 sin 198 − 2 x cos 108 − 2 x cos x cos 108 − 2 x ◦ · 2 cos 18 = · 2 cos 18, so that = . We see that x = 36 is a solution sin x +36 cos 54 − x cos 18 cos 54 − x to this equation, and by carefully making rough sketches of both functions, we can convince ourselves that this is the only solution where x is between 0 and 90 degrees. Therefore ∠ ABD = ∠ ADB = 36, ∠ DAB = 108. Simple angle chasing yields ∠ AEB = 72 , ∠ ECA = 54 , ∠ EAC = 54 , ∠ EAB = 72, making D, A, and E collinear, and so ∠ BDE = 36. And because AF = AB = AD , ∠ F DB = 1 / 2 ∠ F AB = 9, so ∠ F DE = 36 − 9 = 27.