2016 AMC 12B Problems/Problem 17
Contents
Problem
In shown in the figure, , , , and is an altitude. Points and lie on sides and , respectively, so that and are angle bisectors, intersecting at and , respectively. What is ?
Solution 1
Get the area of the triangle by Heron's Formula: Use the area to find the height with known base : Apply the Angle Bisector Theorem on and , we get and , respectively. To find , , , and , apply variables, such that is and is . Solving them out, you will get , , , and . Then, since according to the Segment Addition Postulate, and thus manipulating, you get =
Solution 2
Let the intersection of and be the point . Then let the foot of the altitude from to be . Note that is an inradius and that , where is the semiperimeter of the triangle.
Using Heron's Formula, we see that , so .
Then since and are parallel, and .
Thus, and , so .
By the Dual Principle, and . With the same method as Solution 1, and . Then
Solution 3 (FAST)
lies on altitude , which we find to have a length of by Heron's Formula and dividing twice the area by . From H we can construct a segment with on such that is parallel to . A similar construction gives on such that is parallel to . We can hence generate a system of ratios that will allow us to find . Note that such a system will generate a rational number for the ratio . Thus, we choose the only answer that has a term in it, giving us .
Solution 4
Let and . Then, . By the Pythagorean Theorem on right triangles and , we have Subtracting the prior from the latter yields . So, , , and . Continue with Solution 1.
Video Solution
https://www.youtube.com/watch?v=ccB-z4_OHqw
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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