HMMT 十一月 2019 · 冲刺赛 · 第 21 题

HMMT November 2019 — Guts Round — Problem 21

[11] A positive integer n is infallible if it is possible to select n vertices of a regular 100-gon so that they form a convex, non-self-intersecting n -gon having all equal angles. Find the sum of all infallible integers n between 3 and 100, inclusive. 1 The A -excircle of triangle ABC is the unique circle lying outside the triangle that is tangent to segment BC and the extensions of sides AB and AC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2019, November 9, 2019 — GUTS ROUND Organization Team Team ID#

解析

[11] A positive integer n is infallible if it is possible to select n vertices of a regular 100-gon so that they form a convex, non-self-intersecting n -gon having all equal angles. Find the sum of all infallible integers n between 3 and 100, inclusive. Proposed by: Benjamin Qi Answer: 262 Suppose A A . . . A is an equiangular n -gon formed from the vertices of a regular 100-gon. Note that 1 2 n the angle ∠ A A A is determined only by the number of vertices of the 100-gon between A and A . 1 2 3 1 3 Thus in order for A A . . . A to be equiangular, we require exactly that A , A , . . . are equally spaced 1 2 n 1 3 and A , A , . . . are equally spaced. If n is odd, then all the vertices must be equally spaced, meaning 2 4 ( ) n n | 100 . If n is even, we only need to be able to make a regular -gon from the vertices of a 100-gon, 2 which we can do if n | 200 . Thus the possible values of n are 4 , 5 , 8 , 10 , 20 , 25 , 40 , 50 , and 100 , for a total of 262 .