2014 AIME II Problems/Problem 8
Problem
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 .
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 .
- 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.
Solution 2
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 .
Solution 3
We use the notation of Solution 1 for triangle We use Cosine Law for and get: . vladimir.shelomovskii@gmail.com, vvsss
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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