HMMT 二月 2022 · 冲刺赛 · 第 16 题

HMMT February 2022 — Guts Round — Problem 16

[9] Let ABC be an acute triangle with A -excircle Γ. Let the line through A perpendicular to BC intersect BC at D and intersect Γ at E and F . Suppose that AD = DE = EF . If the maximum value of sin B can √ √ a + b be expressed as for positive integers a , b , and c , compute the minimum possible value of a + b + c . c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2022, February 19, 2022 — GUTS ROUND Organization Team Team ID#

解析

[9] Let ABC be an acute triangle with A -excircle Γ. Let the line through A perpendicular to BC intersect BC at D and intersect Γ at E and F . Suppose that AD = DE = EF . If the maximum value √ √ a + b of sin B can be expressed as for positive integers a , b , and c , compute the minimum possible c value of a + b + c . Proposed by: Akash Das Answer: 705 Solution: First note that we can assume AB < AC . Suppose Γ is tangent to BC at T . Let AD = √ 2 2 DE = EF = x . Then, by Power of a Point, we have DT = DE · DF = x · 2 x = 2 x = ⇒ DT = x 2. 2 2 Note that CT = s − b , and since the length of the tangent from A to Γ is s , we have s = AE · AF = 6 x , √ √ √ √ √ so CT = x 6 − b . Since BC = BD + DT + T C , we have BD = BC − x 2 − ( x 6 − b ) = a + b − x ( 2+ 6). √ √ √ Since a + b = 2 s − c = 2 x 6 − c , we have BD = x ( 6 − 2) − c . Now, by Pythagorean Theorem, we have √ √ √ √ √ 2 2 2 2 2 2 2 c = AB = AD + BD = x + [ x ( 6 − 2) − c ] . Simplifying gives x (9 − 4 3) = xc (2 6 − 2 2). This yields √ √ √ √ √ √ x 2 6 − 2 2 6 2 + 10 6 72 + 600 = √ = = . c 33 33 9 − 4 3