HMMT 二月 2006 · TEAM1 赛 · 第 8 题

英文解析

8. [15] Given a regular n -gon with sides of length 1, what is the smallest radius r suchthat there is a non-empty intersection of n circles of radius r centered at the verticesof 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 centerof the n -gon. The following proves that this r is necessary even if the word “circle” isreplaced by the word “disk.”
   For n even, it is easy to see using symmetry that containing the center point is necessary and sufficient. For n odd, there is more work to do. Again, containing the centerpoint 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 aregion R symmetric about the perpendicular bisector of BC , with the closest pointof R to A being on this line. Hence, circle A must intersect R at some point on theperpendicular bisector of BC ; and thus we see the entire perpendicular bisector of BCinside of the n -gon is contained in the circles. Now, this bisector contains the center 5
   of the n -gon, so some circle must contain the center. But by symmetry, if one circlecontains the center, all do. Thus, in any case, it is necessary and sufficient for r tobe large enough so the center is contained in a circle. Basic trigonometry gives theanswer, which equals the distance between a vertex and the center.