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Difference between revisions of "2004 AIME II Problems/Problem 12"

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== Solution ==
 
== Solution ==
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{{solution}}
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Let the radius of the center circle be <math>r</math> and its center be denoted as <math>O</math>. The center obviously has an x-coordinate of <math>3</math>.
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So line <math>AO</math> passes through the point of tangency of circle <math>A</math> and circle <math>O</math>. Now let <math>y</math> be height from the base of trapezoid to O. Now, from Pythagoreum Theorem,
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<math>3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}</math>.
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We use a similar argument with the line <math>AD</math>, and find the height from the top of the trapezoid to <math>O</math>, <math>z</math>, to be <math>z = \sqrt {r^2 + 4r}</math>.
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Now <math>y + z = 5</math>, so we solve the equation <math>\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = 5</math>
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Solving this, we get <math>r = \frac { - 60 + 48\sqrt {3}}{23}</math>
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So <math>k + m + n + p = 60 + 48 + 3 + 23 = \fbox{134}</math>.[
 
== See also ==
 
== See also ==
 
* [[2004 AIME II Problems/Problem 11 | Previous problem]]
 
* [[2004 AIME II Problems/Problem 11 | Previous problem]]

Revision as of 20:06, 8 December 2007

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

See also

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