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 a line) and are left with , so our triangle has area and so the answer is .
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 . You can now use Heron's Formula to finish. The answer is , or .
Let , and be the point of tangency (as stated in Solution 1). We can now let be . By using Power of a Point Theorem on A to the incircle, you get that . We can use it again on point D to the incircle to get the equation . Setting the two equations equal to each other gives , and it can be further simplified to be
Let lengths and be called and , respectively. We can write as and as . Plugging these into the equation, you get:
Additionally, by Median of a triangle formula, you get that
Refer back to the fact that . We can now plug in our variables.
Solving, you get that or , but the latter will result in a degenerate triangle, so . Finally, you can use Heron's Formula to get that the area is , giving an answer of
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