2016 AMC 12A Problems/Problem 12
Contents
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 to solve this question.
The question asks for . Apply the angle bisector theorem to to get
is given. To find , apply the angle bisector theorem to to get
Since it is immediately obvious that , satisfies both equations.
Thus, ~revision by emerald_block
Solution 5 (Luck-Based)
Note that and look like medians. Assuming they are medians, we mark the answer as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a ratio, with the shorter part being closer to the side it bisects. ~scthecool Note: This is heavily luck based, and if the figure had been not drawn to scale, for example, this answer would have easily been wrong. It is thus advised to not use this in a real competition unless absolutely necessary.
Video Solution by OmegaLearn
https://youtu.be/Gjt25jRiFns?t=43
~ pi_is_3.14
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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