Difference between revisions of "2016 AMC 12A Problems/Problem 12"
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<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math> | <math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math> | ||
− | == Solution | + | == Solution 1== |
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> | By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math> |
Revision as of 21:56, 21 June 2020
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 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 as
. 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.