2004 AIME II Problems/Problem 12

Problem

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$ and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p.$

Solution

Let the radius of the center circle be $r$ and its center be denoted as $O$ . The center obviously has an x-coordinate of $3$ .

So line $AO$ passes through the point of tangency of circle $A$ and circle $O$ . Now let $y$ be height from the base of trapezoid to O. Now, from Pythagorean Theorem, $3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}$ .

We use a similar argument with the line $AD$ , and find the height from the top of the trapezoid to $O$ , $z$ , to be $z = \sqrt {r^2 + 4r}$ .

Now $y + z = 5$ , so we solve the equation $\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = 5$

Solving this, we get $r = \frac { - 60 + 48\sqrt {3}}{23}$

So $k + m + n + p = 60 + 48 + 3 + 23 = \fbox{134}$ .

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
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2004 AIME II Problems/Problem 12

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