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 
: 
?
![[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); [/asy]](http://latex.artofproblemsolving.com/5/0/b/50b0f8f69ee6677a5d5e89b5d864839232ff3faa.png)
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 ![$[\triangle{ABC}]$](http://latex.artofproblemsolving.com/e/a/1/ea10fad601af0fb0660213a0e841d49ba3dfcad0.png)
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, ![$\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}$](http://latex.artofproblemsolving.com/b/0/4/b044d5ddc28420135d6b984fec10d0fc2f79f655.png)
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. 
