2011 AMC 12A Problems/Problem 13
Contents
Problem
Triangle has side-lengths and The line through the incenter of parallel to intersects at and at What is the perimeter of
Solution 1
Let be the incenter of . Because and is the angle bisector of , we have
It then follows due to alternate interior angles and base angles of isosceles triangles that . Similarly, . The perimeter of then becomes
Solution 2
Let be the incenter. is the angle bisector of . Let the angle bisector of meets at and the angle bisector of meets at . By applying both angle bisector theorem and Menelaus' theorem,
Perimeter of
Solution 3
Like in other solutions, let be the incenter of . Let intersect at . By the angle bisector theorem, . Since , we have , so , so . By the angle bisector theorem on , we have , so , so . Because , the perimeter of must be , so our answer is .
Another way to find is to use mass points. Assign a mass of 24 to , a mass of 18 to , and a mass of 12 to . Then has mass 30, so .
Solution 4
We know that the ratio of the perimeter of and is the ratio of their heights, and finding the two heights is pretty easy. Note that the height from to is from Herons and then and also that the height from to is simply the height from to minus the inradius. We know the area and the semiperimeter so which gives us . Now we know that the altitude from to is so the ratios of the heights from for and is . Thus the perimeter of is so our answer is
-srisainandan6
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
Video Solution by OmegaLearn
https://youtu.be/5jwD5UViZO8?t=1013
~ pi_is_3.14
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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