HMMT 十一月 2023 · 冲刺赛 · 第 33 题
HMMT November 2023 — Guts Round — Problem 33
题目详情
- [17] Let ω and ω be two non-intersecting circles. Suppose the following three conditions hold: 1 2 • The length of a common internal tangent of ω and ω is equal to 19. 1 2 • The length of a common external tangent of ω and ω is equal to 37. 1 2 • If two points X and Y are selected on ω and ω , respectively, uniformly at random, then the 1 2 2 expected value of XY is 2023. Compute the distance between the centers of ω and ω . 1 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#
解析
- [17] Let ω and ω be two non-intersecting circles. Suppose the following three conditions hold: 1 2 • The length of a common internal tangent of ω and ω is equal to 19. 1 2 • The length of a common external tangent of ω and ω is equal to 37. 1 2 • If two points X and Y are selected on ω and ω , respectively, uniformly at random, then the 1 2 2 expected value of XY is 2023. Compute the distance between the centers of ω and ω . 1 2 Proposed by: Nilay Mishra Answer: 38 2 2 2 2 Solution 1: The key claim is that E [ XY ] = d + r + r . 1 2 To prove this claim, choose an arbitrary point B on ω . Let r , r be the radii of ω , ω respectively, and 2 1 2 1 2 p 2 2 O , O be the centers of ω , ω respectively. Thus, by the law of cosines, O B = d + r − 2 r d cos( θ ), 1 2 1 2 1 2 2 2 2 2 where θ = ∠ O O B . Since the average value of cos( θ ) is 0, the average value of O B is d + r . 1 2 1 2 2 2 2 Now suppose A is an arbitrary point on ω . By the law of cosines, AB = O B + r − 2 r d cos( θ ), 1 1 1 1 2 2 2 where θ = ∠ AO B . Thus, the expected value of AB is the expected value of O B + r which becomes 1 1 1 2 2 2 d + r + r . This proves the key claim. 1 2 2 2 2 Thus, we have d + r + r = 2023. The lengths of the internal and the external tangents give us 1 2 2 2 2 2 d − ( r + r ) = 361, and d − ( r − r ) = 1369. Thus, 1 2 1 2 2 2 2 2 ( d − ( r + r ) ) + ( d − ( r − r ) ) 361 + 1369 1 2 1 2 2 2 2 d − ( r + r ) = = = 865 . 1 2 2 2 2 865+2023 Thus, d = = 1444 = ⇒ d = 38. 2 2 2 2 2 Solution 2: We present another way of showing that E [ XY ] = d + r + r using complex numbers. 1 2 The finish is the same as Solution 1. Let the center of ω and ω be 0 and k , respectively. Select Z and Z uniformly random on unit circle. 1 2 1 2 Then, 2 2 E [ XY ] = E | k + r Z + r Z | 1 1 2 2 = E ( k + r Z + r Z )( k + r Z + r Z ) 1 1 2 2 1 1 2 2 Then, observe that E [ Z ] = E [ Z ] = E [ Z Z ] = E [ Z Z ] = 0 , 1 2 1 2 2 1 so when expanding, six terms vanish, leaving only 2 2 2 2 2 2 E [ XY ] = E [ kk + r Z Z + r Z Z ] = d + r + r 1 1 2 2 1 2 1 2