HMMT 二月 2012 · 冲刺赛 · 第 22 题
HMMT February 2012 — Guts Round — Problem 22
题目详情
- [ 13 ] For each positive integer n , there is a circle around the origin with radius n . Rainbow Dash starts off somewhere on the plane, but not on a circle. She takes off in some direction in a straight path. She √ √ √ 5 3 5 moves units before crossing a circle, then 5 units, then units. What distance will she travel 5 5 before she crosses another circle?
解析
- [ 13 ] For each positive integer n , there is a circle around the origin with radius n . Rainbow Dash starts off somewhere on the plane, but not on a circle. She takes off in some direction in a straight path. She √ √ √ 5 3 5 moves units before crossing a circle, then 5 units, then units. What distance will she travel 5 5 before she crosses another circle? √ √ 2 170 − 9 5 Answer: Note that the distance from Rainbow Dash’s starting point to the first place in 5 which she hits a circle is irrelevant, except in checking that this distance is small enough that she does not hit another circle beforehand. It will be clear at the end that our configuration does not allow this (by the Triangle Inequality). Let O be the origin, and let Rainbow Dash’s first three meeting points √ √ 3 5 be A, B, C so that AB = 5 and BC = . 5 Consider the lengths of OA, OB, OC . First, note that if OA = OC = n (i.e. A and C lie on the same circle), then we need OB = n − 1, but since she only crosses the circle containing B once, it follows that the circle passing through B is tangent to AC , which is impossible since AB 6 = AC . If OA = OB = n , Guts note that OC = n + 1. Dropping a perpendicular from O to AB , we see that by the Pythagorean Theorem, 5 121 2 2 n − = ( n + 1) − , 4 20 from which we get that n is not an integer. Similarly, when OB = OC = n , we have OA = n + 1, and n is not an integer. Therefore, either OA = n + 2 , OB = n + 1 , OC = n or OA = n, OB = n + 1 , OC = n + 2. In the first case, by Stewart’s Theorem, √ √ √ √ 24 5 8 5 3 5 2 2 2
- ( n + 1) · = n · 5 + ( n + 2) · . 5 5 5 This gives a negative value of n , so the configuration is impossible. In the final case, we have, again by Stewart’s Theorem, √ √ √ √ 24 5 8 5 3 5 2 2 2
- ( n + 1) · = ( n + 2) · 5 + n · . 5 5 5 Solving gives n = 3, so OA = 3 , OB = 4 , OC = 5. √ 1 2 11 √ √ Next, we compute, by the Law of Cosines, cos ∠ OAB = − , so that sin ∠ OAB = . Let the 3 5 3 5 √ 2 11 √ projection from O to line AC be P ; we get that OP = . Rainbow Dash will next hit the circle of 5 √ √ 2 170 9 5 radius 6 at D . Our answer is now CD = P D − P C = − by the Pythagorean Theorem. 5 5