2005 AIME I Problems/Problem 15
Let , and be the points of tangency of the incircle with , and , respectively. Without loss of generality, let , so that is between and . Let the length of the median be . Then by two applications of the Power of a Point Theorem, , so . Now, and are two tangents to a circle from the same point, so by the Two Tangent Theorem and thus . Then so and thus .
Our earlier result from Power of a Point was that , so we combine these two results to solve for and we get
Thus or . We discard the value as extraneous (it gives us an equilateral triangle) and are left with , so our triangle has area and so the answer is .
Use Power of a point similar to the first solution to find that and that the side , where is one third of the median's length. Then use systems of law of cosines, creating two triangles, with with angle , and with the same angle. Solving the system yields . Solving using Heron's Formula gets the answer , or .
WLOG let E be be between C & D (as in solution 1). Assume . We use power of a point to get that and
or if , we get a degenerate triangle, so , and thus AB = 26. You can now use Heron's Formula to finish. The answer is , or .
|2005 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|