HMMT 十一月 2009 · 冲刺赛 · 第 32 题

HMMT November 2009 — Guts Round — Problem 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 .

解析

[ 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 2 s − 28 s + 99 = 0. The quadratic formula then gives s = = 14 ± 97. Since 14 − 97 < 15 2 √ √ and Y O > 15, we can discard the root of 14 − 97, and the answer is therefore 14 + 97. 1