2016 AMC 12A Problems/Problem 12
Problem 12
In , , , and . Point lies on , and bisects . Point lies on , and bisects . The bisectors intersect at . What is the ratio : ?
Solution 1
By the angle bisector theorem,
so
Similarly, .
There are two ways to solve from here. First way:
Note that By the angle bisector theorem on Thus the answer is
Second way:
Now, we use mass points. Assign point a mass of .
, so
Similarly, will have a mass of
So
Solution 2
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 3
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 .
Solution 4
One only needs the angle bisector theorem, the segment addition postulate, and some simple algebra to solve this question.
The question asks for AF:DF. Apply the angle bisector theorem to to get the ratio :
= or, equivalently,
= .
AB is given. To find BD apply the angle bisector theorem to to get:
=
---> = since, by the segment addition postulate, BD + CD = BC
---> BD = .
Substituting this expression for BD into the proportion = yields:
= = = = 2.
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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