HMMT 二月 2021 · 冲刺赛 · 第 14 题

HMMT February 2021 — Guts Round — Problem 14

[11] In triangle ABC , ∠ A = 2 ∠ C . Suppose that AC = 6, BC = 8, and AB = a − b , where a and b are positive integers. Compute 100 a + b .

解析

[11] In triangle ABC , ∠ A = 2 ∠ C . Suppose that AC = 6, BC = 8, and AB = a − b , where a and b are positive integers. Compute 100 a + b . Proposed by: Freddie Zhao, Milan Haiman Answer: 7303 Solution: Let x = AB , and ∠ C = θ , then ∠ A = 2 θ and ∠ B = 180 − 3 θ . Extend ray BA to D so that AD = AC . We know that ∠ CAD = 180 − 2 θ , and since 4 ADC is isosceles, it follows that ∠ ADC = ∠ ACD = θ , and so ∠ DCB = 2 θ = ∠ BAC , meaning that 4 BAC ∼ 4 BCD . Therefore, we have x + 6 8 2 = = ⇒ x ( x + 6) = 8 8 x √ Since x > 0, we have x = − 3 + 73. So 100 a + b = 7303.