HMMT 二月 2021 · 几何 · 第 2 题

HMMT February 2021 — Geometry — Problem 2

Let X be the interior of a triangle with side lengths 3, 4, and 5. For all positive integers n , define X 0 n to be the set of points within 1 unit of some point in X . The area of the region outside X but n − 1 20 inside X can be written as aπ + b , for integers a and b . Compute 100 a + b . 21

解析

Let X be the interior of a triangle with side lengths 3, 4, and 5. For all positive integers n , define X 0 n to be the set of points within 1 unit of some point in X . The area of the region outside X but n − 1 20 inside X can be written as aπ + b , for integers a and b . Compute 100 a + b . 21 Proposed by: Hahn Lheem Answer: 4112 Solution: X is the set of points within n units of some point in X . The diagram above shows X , X , X , and n 0 0 1 2 X . As seen above it can be verified that X is the union of 3 n • X , 0 • three rectangles of height n with the sides of X as bases, and 0 • three sectors of radius n centered at the vertices and joining the rectangles Therefore the total area of X is n 2 [ X ] + n · perimeter( X ) + n π. 0 0 Since X is contained entirely within X , the area within X but not within X is n − 1 n n n − 1 perimeter( X ) + (2 n − 1) π. 0 Since X is a (3 , 4 , 5) triangle, and n = 21, this is 12 + 41 π . 0