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

Revision as of 02:52, 23 January 2023

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 $m + n$ .

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

Video Solution by OmegaLearn

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

~ pi_is_3.14

Solution

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

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

Revision as of 02:52, 23 January 2023

Contents

Problem

Video Solution by OmegaLearn

Solution

See also

@@ Line 5: / Line 5: @@
 Note: The diagram was not given during the actual contest.
+== Video Solution by OmegaLearn ==
+https://youtu.be/nPVDavMoG9M?t=32
+~ pi_is_3.14
 == Solution ==