1994 AIME Problems/Problem 2

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 $m + n$.

1994 AIME Problem 2.png

Solution

1994 AIME Problem 2 - Solution.png

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 $OQ = OP - PQ = 20 - 10 = 10$, we know that $OE = AB - OQ = AB - 10$. The other leg, $AE$, is just $\frac 12 AB$.

Apply the Pythagorean Theorem:

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

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 (ProblemsAnswer KeyResources)
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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