HMMT 十一月 2010 · GEN1 赛 · 第 8 题

HMMT November 2010 — GEN1 Round — Problem 8

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

解析

[ 7 ] Two circles with radius one are drawn in the coordinate plane, one with center (0 , 1) and the other with center (2 , y ), for some real number y between 0 and 1. A third circle is drawn so as to be tangent to both of the other two circles as well as the x axis. What is the smallest possible radius for this third circle? √ Answer: 3 − 2 2 Suppose that the smaller circle has radius r . Call the three circles (in order from left 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)).