A circle with diameter 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 large circle,CD¯tangent at Q to the smaller circle, and the smaller circle outside ABCD. The length of AB¯can be written in the form m+nwhere m and n are integers. Find m9m+4

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A circle with diameter 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 large circle, $\bar{CD}$ tangent at Q to the smaller circle, and the smaller circle outside ABCD. The length of $\bar{AB}$ can be written in the form $m + \sqrt{n}$ where m and n are integers. Find $\frac{m}{9 m + 4}$

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answer is 312.

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Detailed Solution

Call the center of the larger circle O. extend the diameter $\bar{PQ}$ to the other side of the square (at point E), and draw $\bar{AO}$ We now have right tirangle, with
hypotenuse of length 20.
Since OQ = OP - PQ = 20 -10 = 10 ,
we know that OE = ABH -OQ = AB -10 .
The other leg. AE, is just $\frac{1}{2}$ AB
Apply the Pythagorean theorem:
$\begin{array}{l} (AB - 10)^{2} + (\frac{1}{2} AB) = 20^{2} \\ {AB}^{2} - 20 AB + 100 + \frac{1}{4} {AB}^{2} - 400 = 0 \\ {AB}^{2} - 16 AB - 240 = 0 \end{array}$
The quadratic formula shows that the answer is
$\frac{16 \pm \sqrt{16^{2} + 4 .240}}{2} = 8 \pm \sqrt{304}$
Discard the negative root,
so out answer is 8 + 304 = 312