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HMMT 十一月 2010 · GEN1 赛 · 第 8 题

HMMT November 2010 — GEN1 Round — Problem 8

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 7 ] Two circles with radius one are drawn in the coordinate plane, one with center (0 , 1) and the otherwith center (2 , y ), for some real number y between 0 and 1. A third circle is drawn so as to be tangentto both of the other two circles as well as the x axis. What is the smallest possible radius for this thirdcircle?
解析

英文解析

  1. [ 7 ] Two circles with radius one are drawn in the coordinate plane, one with center (0 , 1) and the otherwith center (2 , y ), for some real number y between 0 and 1. A third circle is drawn so as to be tangentto both of the other two circles as well as the x axis. What is the smallest possible radius for this thirdcircle?
    √
    Answer: 3 − 2 2 Suppose that the smaller circle has radius r . Call the three circles (in order fromleft to right) O , O , and O . The distance between the centers of O and O is 1 + r , and the distance
    1 2 3 1 2
    in their y -coordinates is 1 − r . Therefore, by the Pythagorean theorem, the difference in x -coordinates
    General Test
    √
    √ √
    2 2
    is (1 + r ) − (1 − r ) = 2 r , which means that O has a center at (2 r, r ). But O is also tangent
    2 2
    to O , which means that the difference in x -coordinate from the right-most point of O to the center
    3 2
    √
    of O is at most 1. Therefore, the center of O has an x -coordinate of at most 2 r + r + 1, meaning
    3 3
    √
    √ √
    that 2 r + r + 1 ≤ 2. We can use the quadratic formula to see that this implies that r ≤ 2 − 1,
    √
    so r ≤ 3 − 2 2. We can achieve equality by placing the center of O at (2 , r ) (which in this case is
    √3
    (2 , 3 − 2 2)).