HMMT 二月 2006 · TEAM1 赛 · 第 8 题
HMMT February 2006 — TEAM1 Round — Problem 8
题目详情
- [15] Given a regular n -gon with sides of length 1, what is the smallest radius r such that there is a non-empty intersection of n circles of radius r centered at the vertices of the n -gon? Give r as a formula in terms of n . Be sure to prove your answer. ◦
解析
- [15] Given a regular n -gon with sides of length 1, what is the smallest radius r such that there is a non-empty intersection of n circles of radius r centered at the vertices of the n -gon? Give r as a formula in terms of n . Be sure to prove your answer. ◦ 1 180 Answer: r = csc 2 n Solution: It is easy to see that, with this r , all the circles pass through the center of the n -gon. The following proves that this r is necessary even if the word “circle” is replaced by the word “disk.” For n even, it is easy to see using symmetry that containing the center point is neces- sary and sufficient. For n odd, there is more work to do. Again, containing the center point is sufficient. To see its necessity, consider three circles: a circle at a vertex A , and the two circles on the segment BC opposite A . Circles B and C intersect in a region R symmetric about the perpendicular bisector of BC , with the closest point of R to A being on this line. Hence, circle A must intersect R at some point on the perpendicular bisector of BC ; and thus we see the entire perpendicular bisector of BC 5 inside of the n -gon is contained in the circles. Now, this bisector contains the center of the n -gon, so some circle must contain the center. But by symmetry, if one circle contains the center, all do. Thus, in any case, it is necessary and sufficient for r to be large enough so the center is contained in a circle. Basic trigonometry gives the answer, which equals the distance between a vertex and the center.