2016 AMC 12A Problems/Problem 12
Contents
[hide]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 not been drawn to scale, this answer would have been wrong. It is advised to not use this in a real competition unless absolutely necessary.
Solution 6 (Cheese)
Assume the drawing is to-scale. Use your allotted ruler to measure out each side. Note that is equal to
.
Measure out the length of in relation to
. This ratio is approximately
. Solution by juwushu.
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 |
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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