2004 AIME II Problems/Problem 12
Problem
Let be an isosceles trapezoid, whose dimensions are and Draw circles of radius 3 centered at and and circles of radius 2 centered at and A circle contained within the trapezoid is tangent to all four of these circles. Its radius is where and are positive integers, is not divisible by the square of any prime, and and are relatively prime. Find
Solution
Let the radius of the center circle be and its center be denoted as .
Clearly line passes through the point of tangency of circle and circle . Let be the height from the base of the trapezoid to . From the Pythagorean Theorem,
We use a similar argument with the line , and find the height from the top of the trapezoid to , , to be .
Now is simply the height of the trapezoid. Let be the foot of the perpendicular from to ; then . By the Pythagorean Theorem, so we need to solve the equation . We can solve this by moving one radical to the other side, and squaring the equation twice to end with a quadratic equation.
Solving this, we get , and the answer is .
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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