HMMT 十一月 2009 · 冲刺赛 · 第 32 题

英文解析

32. [ 20 ] A circle ω of radius 15 intersects a circle ω of radius 13 at points P and Q . Point A is on line P Q
    1 2
    such that P is between A and Q . R and S are the points of tangency from A to ω and ω , respectively,
    1 2
    such that the line AS does not intersect ω and the line AR does not intersect ω . If P Q = 24 and
    1 2
    °
    ∠ RAS has a measure of 90 , compute the length of AR .
    √
    Answer: 14 + 97 Let O be the center of ω and O be the center of ω . Then O O and P Q
    1 1 2 2 1 2
    are perpendicular. Let their point of intersection be X . Using the Pythagorean theorem, the fact that
    P Q = 24, and our knowledge of the radii of the circles, we can compute that O X = 9 and O X = 5,
    1 2
    so O O = 14. Let SO and RO meet at Y . Then SARY is a square, say of side length s . Then
    1 2 1 2
    O Y = s − 15 and O Y = s − 13. So, O O Y is a right triangle with sides 14, s − 15, and s − 13. By
    1 2 1 2
    2 2 2 2
    the Pythagorean theorem, ( s − 13) + ( s − 15) = 14 . We can write this as 2 s − 4 · 14 s + 198 = 0, or
    √
    √ √
    28 ± 388
    s − 28 s + 99 = 0. The quadratic formula then gives s = = 14 ± 97. Since 14 − 97 < 152
    √ √2
    and Y O > 15, we can discard the root of 14 − 97, and the answer is therefore 14 + 97. 1