HMMT 二月 2021 · 冲刺赛 · 第 5 题

HMMT February 2021 — Guts Round — Problem 5

[9] Let m, n > 2 be integers. One of the angles of a regular n -gon is dissected into m angles of equal size by ( m − 1) rays. If each of these rays intersects the polygon again at one of its vertices, we say n is m -cut. Compute the smallest positive integer n that is both 3-cut and 4-cut.

解析

[9] Let m, n > 2 be integers. One of the angles of a regular n -gon is dissected into m angles of equal size by ( m − 1) rays. If each of these rays intersects the polygon again at one of its vertices, we say n is m -cut. Compute the smallest positive integer n that is both 3-cut and 4-cut. Proposed by: Carl Schildkraut Answer: 14 Solution: For the sake of simplicity, inscribe the regular polygon in a circle. Note that each interior angle of the regular n − gon will subtend n − 2 of the n arcs on the circle. Thus, if we dissect an interior n − 2 angle into m equal angles, then each must be represented by a total of arcs. However, since each m n − 2 of the rays also passes through another vertex of the polygon, that means is an integer and thus m our desired criteria is that m divides n − 2 . That means we want the smallest integer n > 2 such that n − 2 is divisible by 3 and 4 which is just 12 + 2 = 14 .