HMMT 二月 2012 · 冲刺赛 · 第 22 题

英文解析

22. [ 13 ] For each positive integer n , there is a circle around the origin with radius n . Rainbow Dash startsoff 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 inwhich she hits a circle is irrelevant, except in checking that this distance is small enough that she does 5
    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 = .
    Consider the lengths of OA, OB, OC . First, note that if OA = OC = n (i.e. A and C lie on the same 5
    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 ,
    Gutsnote 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, andn 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 firstcase, 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, againby 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