2005 AIME II Problems/Problem 8
Problem
Circles and
are externally tangent, and they are both internally tangent to circle
The radii of
and
are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of
is also a common external tangent of
and
Given that the length of the chord is
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution
Let be the centers and
the radii of the circles
,
. Let the common external tangent of
meet the center line
at a point H. Let
be the tangency points of the circles
with their common external tangent. From the similar right triangles
,
and by Pythagorean theorem for the right angle triangle ,
Let O be the center and the radius of the circle
. Since
,
Let the common external tangent of the circles
intersects the circle
at points A, B, so that the points H, A, B follow on the tangent in this order. Let T be the midpoint of the chord AB. Then
. Since the angle
is right, the right angle triangles
are similar,
and
The power of the point H to the circle is equal to
Substituting or
into the formula for HA + HB leads to the same quadratic equation for HA or HB:
Since HA, HB are the 2 roots of this quadratic equation, their difference is
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |