PUMaC 2019 · 几何（A 组） · 第 6 题
PUMaC 2019 — Geometry (Division A) — Problem 6
题目详情
- Let two ants stand on the perimeter of a regular 2019-gon of unit side length. One of them stands on a vertex and the other one is on the midpoint of the opposite side. They start walking along the perimeter at the same speed counterclockwise. The locus of their midpoints traces out a figure P in the plane with N corners. Let the area enclosed by convex hull of P m π sin ( ) A 4038 be , where A and B are coprime positive integers, and m is the smallest possible π B tan ( ) 2019 positive integer such that this formula holds. Find A + B + m + N . Note: The convex hull of a figure P is the convex polygon of smallest area which contains P .
解析
- Let two ants stand on the perimeter of a regular 2019-gon of unit side length. One of them stands on a vertex and the other one is on the midpoint of the opposite side. They start walking along the perimeter at the same speed counterclockwise. The locus of their midpoints traces out a figure P in the plane with N corners. Let the area enclosed by convex hull of P m π sin ( ) A 4038 be , where A and B are coprime positive integers, and m is the smallest possible π B tan ( ) 2019 positive integer such that this formula holds. Find A + B + m + N . Proposed by Jackson Danger Blitz. Answer: 6065 . ( ) ( ) π π 2 Solution: The area of regular a n -gon with circumradius R is nR sin cos . The cir- n n 1 cumradius of the starting n -gon is R = . 0 π 2 sin ( ) n The locus consists of a 2019-star, which has 2019 isosceles triangles affixed to a smaller 2019- gon, for 4038 total corners. The convex hull of this star is a 2019-gon. We can find it’s circumradius by noting that its vertices are precisely the points which are midpoints of a segment connecting a vertex of the 2019-gon and the midpoint of the opposite side. π R 1 − cos 0 ( ( )) 2019 Then a simple calucation gives that R = . Plugging this into the formula we 2 4 π ( ) sin ( ) 2019 4038 π get that the area of the convex hull is . Furthermore, sin is not rational π 4 4038 tan ( ) 2019 when taken to the powers 1 , 2 , 3. Hence the A = 2019 , B = 4 , m = 4 , N = 4038, for a final answer of 6065. 2 Note: We also accepted the interpretation of a corner to not include the self-intersections. This gives us that N = 2019. Following the above logic yields the answer of 4046, which we also accepted.