Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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Similarly, <math>CD = 4</math>. | Similarly, <math>CD = 4</math>. | ||
− | Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>. | + | Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>. |
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math> | <math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math> |
Revision as of 21:21, 26 August 2018
Problem 12
In ,
,
, and
. Point
lies on
, and
bisects
. Point
lies on
, and
bisects
. The bisectors intersect at
. What is the ratio
:
?
Solution 1
Applying the angle bisector theorem to with
being bisected by
, we have
Thus, we have
and cross multiplying and dividing by gives us
Since , we can substitute
into the former equation. Therefore, we get
, so
.
Apply the angle bisector theorem again to with
being bisected. This gives us
and since and
, we have
Cross multiplying and dividing by gives us
and dividing by gives us
Therefore,
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 AB as P. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.