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1994 AIME Problems/Problem 2

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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.

Image:1994 AIME Problem 2.png

Solution

Image: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 (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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