2014 AIME II Problems/Problem 8
Circle with radius 2 has diameter . Circle D is internally tangent to circle at . Circle is internally tangent to circle , externally tangent to circle , and tangent to . The radius of circle is three times the radius of circle , and can be written in the form , where and are positive integers. Find .
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 .
- Notice that C, E and the point of tangency to circle C for circle E will be collinear because C and E intersect the tangent line at a right angle, implying they must be on the same line.
Consider a reflection of circle over diameter . By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii , , and , and the big circle has radius .
Descartes' Circle Theorem gives
Note that the big circle has curvature because it is internally tangent. Solving gives for a final answer of .
We use the notation of Solution 1 for triangle We use Cosine Law for and get: . email@example.com, vvsss
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