HMMT 十一月 2014 · 团队赛 · 第 8 题
HMMT November 2014 — Team Round — Problem 8
题目详情
- [ 3 ] Let H be a regular hexagon with side length one. Peter picks a point P uniformly and at random within H , then draws the largest circle with center P that is contained in H . What is this probability 1 that the radius of this circle is less than ? 2
解析
- [ 3 ] Let H be a regular hexagon with side length one. Peter picks a point P uniformly and at random within H , then draws the largest circle with center P that is contained in H . What is this probability 1 that the radius of this circle is less than ? 2 √ 2 3 − 1 Answer: We first cut the regular hexagon H by segments connecting its center to each 3 vertex into six different equilateral triangles with side lengths 1. Therefore, each point inside H is contained in some equilateral triangle. We first see that for each point inside an equilateral triangle, the radius of the largest circle with center P which is contained in H equals the shortest distance from P to the nearest side of the hexagon, which is also a side of the triangle in which it is contained. √ 3 Consider that the height of each triangle is . Therefore, the region inside the triangle containing all 2 1 points with distance more than to the side of the hexagon is an equilateral triangle with a height of 2 √ 3 − 1 1 . Consequently, the area inside the triangle containing all points with distance less than to the 2 2 ( ) ( ) ( ) √ √ √ √ √ 2 3 3 − 1 3 2 3 − 1 2 3 − 1 √ side of the hexagon has area 1 − = · . This is of the ratio to the 4 4 3 3 3 √ 3 area of the triangle, which is . Since all triangles are identical and the point P is picked uniformly 4 within H , the probability that the radius of the largest circle with center P which is contained in H is √ 1 2 3 − 1 less than is , as desired. 2 3