# Mock Geometry AIME 2011 Problems/Problem 8

## Problem

Two circles have center and radius respectively. The smallest distance between a point on with a point on is . Tangents from to meet at and tangents from to meet at such that are on the same side of line meets at and meets at Q. The length of can be expressed in the form where are relatively prime positive integers. Find

## Solution

First note that the diagram is symmetric about line ; that is, when every point is reflected by , the diagram remains unchanged. This leads to two important observations: , and that . Then the distance from to is half the length of .

Second, note that the smallest distance between two points on two non-intersecting circles lies on the line connecting their centers. This can be demonstrated by letting be on and be on , such that is the smallest distance connecting both circles. Then ; they are both paths from to and the smallest distance between two points is a straight line. This rearranges to with equality iff is on line . Hence coincides with .

Then . By the Pythagorean Theorem on , , and so . Also, from , , and so .

Now let be the foot of the perpendicular from to . Let . Then , where we have used right triangles . Similarly, using right triangles .

These equations rearrange into . We also know . Solving this system of equations for yields . Then . Then and .