Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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== Solution 1== | == Solution 1== | ||
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== Solution 2== | == Solution 2== |
Revision as of 19:54, 14 May 2020
Problem 12
In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?
Solution 1
Simply notice that AD and BE intersect at the triangle's centroid, as they bisect angle CAB and angle ABC respectively. Therefore, due to the properties of a centroid, we know that the ratio of AF to FD is 2:1. Answer: C ~the_noob_moment
Solution 2
By the angle bisector theorem,
so
Similarly, .
Now, we use mass points. Assign point a mass of .
, so
Similarly, will have a mass of
So
Solution 3
Denote as the area of triangle ABC and let be the inradius. Also, as above, use the angle bisector theorem to find that . There are two ways to continue from here:
Note that is the incenter. Then,
Apply the angle bisector theorem on to get
Solution 4
Draw the third angle bisector, and denote the point where this bisector intersects as . Using angle bisector theorem, we see . Applying Van Aubel's Theorem, , and so the answer is .
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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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