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 a line) and are left with , so our triangle has area and so the answer is .
Solution 2
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 . You can now use Heron's Formula to finish. The answer is , or .
-Alexlikemath
Solution 3
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
~sky2025
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
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