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, .
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
You can just guess. So it obviously has to be greater than 1. and 2:1 is greater than 1. Also. 2:1 = 50% if you do 1/2 right? and well you have 5 choices, divide it by 10 since 10 is 25% of 40. 40 comes from 2 + 16 in 2016 + 12 = 20, 50% = 10 thus the ratio of 10 : 20 is 5: 10 is 1 : 2 hence 2:1 is your answer.
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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