PUMaC 2020 · 团队赛 · 第 9 题
PUMaC 2020 — Team Round — Problem 9
题目详情
- Consider a regular 2020-gon circumscribed into a circle of radius 1. Given three vertices of this polygon such that they form an isosceles triangle, let X be the expected area of the isosceles 1 triangle they create. X can be written as where m and n are integers. Compute m tan((2 π ) /n ) m + n.
解析
- Consider a regular 2020-gon circumscribed into a circle of radius 1. Given three vertices of this polygon such that they form an isosceles triangle, let X be the expected area of the isosceles 1 triangle they create. X can be written as where m and n are integers. Compute m tan((2 π ) /n ) m + n. Proposed by: Ollie Thakar Answer: 5049 Draw radii from the center of the circumcircle to each vertex of the isosceles triangle. If the 1 central angles thus created are α, α, 2 π − 2 α then the area is simply sin α − sin(2 α ) . This 2 can be seen with law of sines. Let the original side lengths of the triangle be A, B, C, and angles be a, b, c. Then, because the center divides the triangle into three sub-triangles, the 1 1 1 subtriangles have areas RC cos c, RB cos b , and RA cos a, which I found by dividing them 2 2 2 into two congruent right triangles and using base times height. Law of sines tells us, however, that A = 2 R sin a, B = 2 R sin b, C = 2 R sin c. Plugging these relations into our area formula, 1 remembering also that R = 1 and cos a sin a = sin(2 a ) tells us that the total area of the 2 1 triangle is Area = (sin(2 a ) + sin(2 b ) + sin(2 c )) , which is the formula that I used to get 2 1 sin α − sin(2 α ) . 2 For each vertex, the combined area of all of the isosceles triangles whose distinct angle lies at 1 2 π 4 π 2018 π that vertex is simply the sum of sin α − sin(2 α ) where α ∈ { , , ..., } . 2 2020 2020 2020 The sum of sin α for α in the above range is just the height of the regular 2020-gon with 1 side-length 1, which is h = . tan( π/ 2020) 1009 The sum of sin(2 α ) for α in the above range is the imaginary part of the sum 1 + z + ... + z where z is the 1010th root of unity, so is clearly 0. The total number of isosceles triangles is 1009 ∗ 2020 , and the sum of all of their areas, by our 1 1 above logic, is 2020 ∗ , so the expected area of one of the triangles is . 2 tan( π/ 2020) 1009 tan( π/ 2020) 5