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Difference between revisions of "1994 AIME Problems/Problem 2"

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== Solution ==
 
== Solution ==
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Call the center of the larger circle <math>O</math>. Extend the diameter <math>\overline{PQ}</math> to the other side of the square (at point <math>E</math>), and draw <math>\overline{AO}</math>. We now have a [[right triangle]], with [[hypotenuse]] of length <math>20</math>. Since <math>\displaystyle OQ = OP - PQ = 20 - 10 = 10</math>, we know that <math>OE = AB - OQ = AB - 10</math>. The other leg, <math>AE</math>, is just <math>\frac 12 AB</math>.
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Apply the [[Pythagorean Theorem]]:
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:<math>(AB - 10)^2 + (\frac 12 AB)^2 = 20^2</math>
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:<math>AB^2 - 20 AB + 100 + \frac 14 AB^2 - 400 = 0</math>
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:<math>\displaystyle AB^2 - 16 AB - 240 = 0</math>
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The [[quadratic formula]] shows that the answer is <math>\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}</math>. Discard the negative root, so our answer is <math>8 + 304 = 312</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=1|num-a=3}}
 
{{AIME box|year=1994|num-b=1|num-a=3}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 15:41, 10 April 2007

Problem

A circle with diameter $\overline{PQ}\,$ of length 10 is internally tangent at $P^{}_{}$ to a circle of radius 20. Square $ABCD\,$ is constructed with $A\,$ and $B\,$ on the larger circle, $\overline{CD}\,$ tangent at $Q\,$ to the smaller circle, and the smaller circle outside $ABCD\,$. The length of $\overline{AB}\,$ can be written in the form $m + \sqrt{n}\,$, where $m\,$ and $n\,$ are integers. Find $m + n\,$.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Call the center of the larger circle $O$. Extend the diameter $\overline{PQ}$ to the other side of the square (at point $E$), and draw $\overline{AO}$. We now have a right triangle, with hypotenuse of length $20$. Since $\displaystyle OQ = OP - PQ = 20 - 10 = 10$, we know that $OE = AB - OQ = AB - 10$. The other leg, $AE$, is just $\frac 12 AB$.

Apply the Pythagorean Theorem:

$(AB - 10)^2 + (\frac 12 AB)^2 = 20^2$
$AB^2 - 20 AB + 100 + \frac 14 AB^2 - 400 = 0$
$\displaystyle AB^2 - 16 AB - 240 = 0$

The quadratic formula shows that the answer is $\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}$. Discard the negative root, so our answer is $8 + 304 = 312$.


See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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