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
.