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 $m + n$ .

Error creating thumbnail: Unable to save thumbnail to destination

Note: The diagram was not given during the actual contest.

Solution

Error creating thumbnail: Unable to save thumbnail to destination

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 $8 + 304 = \boxed{312}$ .

Video Solution by OmegaLearn

https://youtu.be/nPVDavMoG9M?t=32

~ pi_is_3.14

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

Page

Toolbox

Search

1994 AIME Problems/Problem 2

Contents

Problem

Solution

Video Solution by OmegaLearn

See also