2005 AIME I Problems/Problem 15
Problem
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find
Solution 1
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 .
Now, by Stewart's Theorem in triangle with cevian , we have
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 .
Solution 2
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 .
Solution 3
WLOG let E be be between C & D (as in solution 1). Assume . We use power of a point to get that and
Since now we have , in triangle and cevian . Now, we can apply Stewart's Theorem.
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 .
-Alexlikemath
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
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