HMMT 二月 2020 · 几何 · 第 10 题
HMMT February 2020 — Geometry — Problem 10
题目详情
- Let Γ be a circle of radius 1 centered at O . A circle Ω is said to be friendly if there exist distinct circles ω , ω , . . . , ω , such that for all 1 ≤ i ≤ 2020, ω is tangent to Γ, Ω, and ω . (Here, ω = ω .) 1 2 2020 i i +1 2021 1 For each point P in the plane, let f ( P ) denote the sum of the areas of all friendly circles centered at 1 1 P . If A and B are points such that OA = and OB = , determine f ( A ) − f ( B ). 2 3
解析
- Let Γ be a circle of radius 1 centered at O . A circle Ω is said to be friendly if there exist distinct circles ω , ω , . . . , ω , such that for all 1 ≤ i ≤ 2020, ω is tangent to Γ, Ω, and ω . (Here, ω = ω .) 1 2 2020 i i +1 2021 1 For each point P in the plane, let f ( P ) denote the sum of the areas of all friendly circles centered at 1 1 P . If A and B are points such that OA = and OB = , determine f ( A ) − f ( B ). 2 3 Proposed by: Michael Ren 1000 Answer: π 9 Solution: Let P satisfy OP = x . (For now, we focus on f ( P ) and ignore the A and B from the problem statement.) The key idea is that if we invert at some point along OP such that the images ′ of Γ and Ω are concentric, then ω still exist. Suppose that this inversion fixes Γ and takes Ω to Ω of i ′ radius r (and X to X in general). If the inversion is centered at a point Q along ray OP such that √ 2 OQ = d , then the radius of inversion is d − 1. Let the diameter of Ω meet OQ at A and B with A ′ ′ ′ closer to Q than B . Then, ( AB ; P P ) = − 1 inverts to ( A B ; P Q ) = − 1, where P is the point at ∞ ∞ 2 2 ′ ′ ′ r ′ r infinity along line OP , so P is the inverse of Q in Ω . We can compute OP = so P Q = d − d d 2 2 2 d − 1 d − 1 1 − r and P Q = . Thus, we get the equation + x = d , which rearranges to d = x , or 2 2 2 2 r r d − r d − d − d d 2 − 1 2 2 d − x (1 − r ) d − r = 0. Now, we note that the radius of Ω is ( ) ( ) ( ) 2 2 2 2 1 1 d − 1 d − 1 r ( d − 1) r − 1 x AB = − = = r 1 + = r 1 − . 2 2 2 2 2 2 d − r d + r d − r d − r d √ √ 2 2 4 2 2 4 2 2 (1 − r ) ± r − (2 − 4 x ) r +1 1 − r ± r − (2 − 4 x ) r +1 x The quadratic formula gives us that d = , so = − , 2 2 x d 2 r which means that the radius of Ω is √ √ 1 1 2 2 2 r + ± r + − 2 + 4 x 4 2 2 2 r + 1 ± r − (2 − 4 x ) r + 1 r r = . 2 r 2 1 Note that if r gives a valid chain of 2020 circles, so will by homothety/inversion. Thus, we can think r √ 1 1 2 r + ± r + − 1 2 r r 1 of each pair of r, as giving rise to two possible values of the radius of Ω, which are . r 2 This means that the pairs have the same sum of radii as the circles centered at O , and the product √ 2 2 of the radii is 1 − x . (A simpler way to see this is to note that inversion at P with radius 1 − x swaps the two circles.) From this, it follows that the difference between the sum of the areas for each ( ) ϕ (2020) 1 1 5 pair is 2 π − = π . There are = 400 such pairs, which can be explicitly computed as 2 2 2 3 18 2 πk πk 1 − sin 1+sin 1000 2020 2020 , for positive integers k < 1010 relatively prime to 2020. Thus, the answer is π . πk πk 9 1+sin 1 − sin 2020 2020