Difference between revisions of "2014 AIME II Problems/Problem 8"
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Revision as of 23:26, 27 March 2014
Solution 1
Using the diagram above, let the radius of be , and the radius of be . Then, , and , so the Pythagorean theorem in gives . Also, , so Noting that , we can now use the Pythagorean theorem in to get
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives for a final answer of .