HMMT 二月 2019 · 几何 · 第 6 题
HMMT February 2019 — Geometry — Problem 6
题目详情
- Six unit disks C , C , C , C , C , C are in the plane such that they don’t intersect each other and C 1 2 3 4 5 6 i is tangent to C for 1 ≤ i ≤ 6 (where C = C ). Let C be the smallest circle that contains all six i +1 7 1 disks. Let r be the smallest possible radius of C , and R the largest possible radius. Find R − r .
解析
- Six unit disks C , C , C , C , C , C are in the plane such that they don’t intersect each other and C 1 2 3 4 5 6 i is tangent to C for 1 ≤ i ≤ 6 (where C = C ). Let C be the smallest circle that contains all six i +1 7 1 disks. Let r be the smallest possible radius of C , and R the largest possible radius. Find R − r . Proposed by: Daniel Liu √ Answer: 3 − 1 The minimal configuration occurs when the six circles are placed with their centers at the vertices of a regular hexagon of side length 2. This gives a radius of 3. The maximal configuration occurs when four of the circles are placed at the vertices of a square of side length 2. Letting these circles be C , C , C , C in order, we place the last two so that C is tangent to 1 3 4 6 2 C and C and C is tangent to C and C . (Imagine pulling apart the last two circles on the plane; 1 3 5 4 6 √ √ this is the configuration you end up with.) The resulting radius is 2 + 3, so the answer is 3 − 1. Now we present the proofs for these configurations being optimal. First, we rephrase the problem: given an equilateral hexagon of side length 2, let r be the minimum radius of a circle completely containing the vertices of the hexagon. Find the difference between the minimum and maximum values in r . (Technically this r is off by one from the actual problem, but since we want R − r in the actual problem, this difference doesn’t matter.) Proof of minimality. We claim the minimal configuration stated above cannot be covered by a cir- cle with radius r < 2. If r < 2 and all six vertices O , O , . . . , O are in the circle, then we have 1 2 6 ◦ that ∠ O OO > 60 since O O is the largest side of the triangle O OO , and similar for other an- 1 2 1 2 1 2 ◦ ◦ gles ∠ O OO , ∠ O OO , . . . , but we cannot have six angles greater than 60 into 360 , contradiction. 2 3 3 4 Therefore r ≥ 2. Proof of maximality. Let ABCDEF be the hexagon, and choose the covering circle to be centered at √ O , the midpoint of AD , and radius 3 + 1. We claim the other vertices are inside this covering circle. First, we will show the claim for B . Let M be the midpoint of AC . Since ABC is isosceles and AM ≥ 1, √ √ CD we must have BM ≤ 4 − 1 = 3. Furthermore, M O is a midline of ACD , so M O = = 1. Thus 2 √ by the triangle inequality, OB ≤ M B + OM = 3 + 1, proving the claim. A similar argument proves the claim for C, E, F . Finally, an analogous argument to above shows if we define P as the midpoint √ √ √ of BE , then AP ≤ 3 + 1 and DP ≤ 3 + 1, so by triangle inequality AD ≤ 2( 3 + 1). Hence √ OA = OD ≤ 3 + 1, proving the claim for A and D . Thus the covering circle contains all six vertices of ABCDEF .