HMMT 十一月 2024 · 冲刺赛 · 第 19 题

HMMT November 2024 — Guts Round — Problem 19

[11] An equilateral triangle is inscribed in a circle ω . A chord of ω is cut by the perimeter of the triangle into three segments of lengths 55 , 121 , and 55 in that order. Compute the sum of all possible side lengths of the triangle. 4 3 2

解析

[11] An equilateral triangle is inscribed in a circle ω . A chord of ω is cut by the perimeter of the triangle into three segments of lengths 55, 121, and 55 in that order. Compute the sum of all possible side lengths of the triangle. Proposed by: Karthik Venkata Vedula Answer: 410 Solution: A A P 1 Y X Y X P P 1 2 P 2 B C B C Note that the chord splits two of the sides into segments of lengths a, b and c, d , where segments of length a and c is incident to the same vertex of the equilateral triangle. Moreover, a + b = c + d (as the triangle is equilateral) and ab = cd = 55 · 176 by Power of a Point. This means that { a, b } = { c, d } . This means that we have two cases. • Case 1: the chord is parallel to the third side. We must have a = c = 121 and by power of point, b = d = (55 · 176) / 121 = 80, so the side length is 121 + 80 = 201. • Case 2: the chord is not parallel to the third side. In that case, we have that a = d and c = b . Thus, by the Law of Cosines, we have 2 2 2 a + b − ab = 121 . Moreover, ab = 55 · 176 by power of point. Thus, p √ 2 a + b = 121 + 3 · 55 · 176 = 11 121 + 3 · 5 · 16 = 209 , so the side length is 209. This means that the answer is 201 + 209 = 410. (One can check that the two triangles indeed exist, as we can solve for a, b, c, d and see that they are positive real.) 4 3 2