HMMT 十一月 2019 · 冲刺赛 · 第 7 题

HMMT November 2019 — Guts Round — Problem 7

[7] Let S be the set of all nondegenerate triangles formed from the vertices of a regular octagon with side length 1. Find the ratio of the largest area of any triangle in S to the smallest area of any triangle in S .

解析

[7] Let S be the set of all nondegenerate triangles formed from the vertices of a regular octagon with side length 1. Find the ratio of the largest area of any triangle in S to the smallest area of any triangle in S . Proposed by: Carl Schildkraut √ Answer: 3 + 2 2 By a smoothing argument, the largest triangle is that where the sides span 3, 3, and 2 sides of the ◦ ◦ ◦ octagon respectively (i.e. it has angles 45 , 67 . 5 , and 67 . 5 ), and the smallest triangle is that formed by three adjacent vertices of the octagon. Scaling so that the circumradius of the octagon is 1, our answer is √ ◦ ◦ √ sin (90 ) + 2 sin (135 ) 1 + 2 √ = = 3 + 2 2 , ◦ ◦ 2 sin (45 ) − sin (90 ) 2 − 1 where the numerator is derived from splitting the large triangle by the circumradii, and the denominator is derived from adding the areas of the two triangles formed by the circumradii, then subtracting the area not in the small triangle.