2012 AMC 12A Problems/Problem 18
Problem
Triangle has , , and . Let be 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 . 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
Solution 3
We can use POP(Power of a point) to solve this problem. First, notice that the area of is . Therefore, using the formula that , where is the semi-perimeter and is the length of the inradius, we find that .
Draw radii to the three tangents, and let the tangent hitting be , the tangent hitting be , and the tangent hitting be . Let . By the pythagorean theorem, we know that . By POP, we also know that is also . Because we know that , we find that . We can rinse and repeat and find that . We can find by essentially coming in from the other way. Since , we also know that . By POP, we know that , so .
Let , for simplicity. We can change the equation into , which we find to be . Therefore, , which further implies that . After simplifying, we find , so
~EricShi1685
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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