PUMaC 2011 · 几何（B 组） · 第 4 题

PUMaC 2011 — Geometry (Division B) — Problem 4

[ 4 ] Let ω be a circle of radius 6 with center O . Let AB be a chord of ω having length 5. For 2 2 any real constant c , consider the locus L ( c ) of all points P such that P A − P B = c . Find the largest value of c for which the intersection of L ( c ) and ω consists of just one point.

解析

It is easy to see, by the Pythagorean theorem, that L ( c ) for any c consists of a line perpendicular to AB . Thus, in order for the intersection of L ( c ) and ω to consist of a single point, L ( c ) must be tangent to ω . In this case, define X to be the point on L ( c ) collinear with A, B . If B is between X and A then 2 2 c = ( XA ) − ( XB ) = ( XA − XB )( XA + XB ) = ( AB )(2 r ) = 5 · 2 · 6 = 60 . Note that r above denotes the radius of ω . Otherwise, if A is between X and B then 2 2 ( XA ) − ( XB ) = − ( AB )(2 r ) = − 60 . Thus the possible values of c are ± 60, so our answer is c = 60 .