2012 AMC 12A Problems/Problem 18
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[hide]Problem
Triangle has , , and . Let denote the intersection of the internal angle bisectors of . What is ?
Solution 1
Inscribe circle of radius inside triangle so that it meets at , at , and at . Note that angle bisectors of triangle are concurrent at the center (also ) of circle . Let , and . Note that , and . Hence , , and . Subtracting the last 2 equations we have and adding this to the first equation we have .
By Heron's formula for the area of a triangle we have that the area of triangle is . On the other hand the area is given by . Then so that .
Since the radius of circle is perpendicular to at , we have by the pythagorean theorem so that .
Solution 2
We can use mass points and Stewart's to solve this problem. Because we are looking at the Incenter we then label with a mass of , with , and with . We also label where the angle bisectors intersect the opposite side , , and correspondingly. It follows then that point has mass 52. Which means that is split into a ratio. We can then use Stewart's to find . So we have . Solving we get . Plugging it in we get . Therefore the answer is
-Solution by arowaaron
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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