HMMT 二月 2017 · 几何 · 第 3 题

英文解析

3. Let S be a set of 2017 distinct points in the plane. Let R be the radius of the smallest circle containingall points in S on either the interior or boundary. Also, let D be the longest distance between two ofthe points in S . Let a, b are real numbers such that a ≤ ≤ b for all possible sets S , where a is as dlarge as possible and b is as small as possible. Find the pair ( a, b ).R
   Proposed by: Yang Liu
   √
   Answer: ( 3, 2)
   It is easy to verify that the smallest circle enclosing all the points will either have some 2 points in Sas its diameter, or will be the circumcircle of some 3 points in S who form an acute triangle.
   Now, clearly ≤ 2. Indeed consider the two farthest pair of points S , S . Then D = | S S | ≤ 2 R ,D
   1 2 1 2
   as both points S , S are inside a circle of radius R . We can achieve this upper bound by taking S to r
   1 2
   have essentially only 2 points, and the remaining 2015 points in S are at the same place as these 2 points.
   √
   For the other direction, I claim ≥ 3 . Recall that the smallest circle is either the circumcircle of d
   3 points, or has some 2 points as the diameter. In the latter case, say the diameter is S S . Then r
   1 2
   D ≥ | S S | = 2 R , so ≥ 2 in that case. Now say the points S , S , S are the circumcircle. WLOG,D
   1 2 1 2 3
   say that S S is the longest side of the triangle. As remarked above, we can assume this triangle is r
   1 2
   π πacute. Therefore, ≤ ∠ S S S ≤ . By the Law of Sines we have that
   1 3 2
   3 2
   √
   D ≥ | S S | = 2 R sin ∠ S S S ≥ 2 R sin = R 3 .π
   1 2 1 3 2
   This completes the proof. To achieve equality, we can take S to have 3 points in the shape of an 3
   equilateral triangle.