# Difference between revisions of "2016 AMC 12A Problems/Problem 12"

## Problem 12

In $\triangle ABC$, $AB = 6$, $BC = 7$, and $CA = 8$. Point $D$ lies on $\overline{BC}$, and $\overline{AD}$ bisects $\angle BAC$. Point $E$ lies on $\overline{AC}$, and $\overline{BE}$ bisects $\angle ABC$. The bisectors intersect at $F$. What is the ratio $AF$ : $FD$? $[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7)), F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,NW); label("F",F,1.5*N); [/asy]$ $\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2$

## Solution 1

Applying the angle bisector theorem to $\triangle ABC$ with $\angle CAB$ being bisected by $AD$, we have $$\frac{CD}{AC}=\frac{BD}{AB}.$$

Thus, we have $$\frac{CD}{8}=\frac{BD}{6},$$

and cross multiplying and dividing by $2$ gives us $$3\cdot CD=4\cdot BD.$$

Since $CD+BD=BC=7$, we can substitute $CD=7-BD$ into the former equation. Therefore, we get $3(7-BD)=4BD$, so $BD=3$.

Apply the angle bisector theorem again to $\triangle ABD$ with $\angle ABC$ being bisected. This gives us $$\frac{AB}{AF}=\frac{BD}{FD},$$

and since $AB=6$ and $BD=3$, we have $$\frac{6}{AF}=\frac{3}{FD}.$$

Cross multiplying and dividing by $3$ gives us $$AF=2\cdot FD,$$

and dividing by $FD$ gives us $$\frac{AF}{FD}=\frac{2}{1}.$$

Therefore, $$AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.$$

## Solution 2

By the angle bisector theorem, $\frac{AB}{AE} = \frac{CB}{CE}$ $\frac{6}{AE} = \frac{7}{8 - AE}$ so $AE = \frac{48}{13}$

Similarly, $CD = 4$.

Now, we use mass points. Assign point $C$ a mass of $1$. $mC \cdot CD = mB \cdot DB$ , so $mB = \frac{4}{3}$

Similarly, $A$ will have a mass of $\frac{7}{6}$ $mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}$

So $\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}$

## Solution 3

Denote $[\triangle{ABC}]$ as the area of triangle ABC and let $r$ be the inradius. Also, as above, use the angle bisector theorem to find that $BD = 3$. There are two ways to continue from here: $1.$ Note that $F$ is the incenter. Then, $\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}$ $2.$ Apply the angle bisector theorem on $\triangle{ABD}$ to get $\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}$

## Solution 4

Draw the third angle bisector, and denote the point where this bisector intersects $AB$ as $P$. Using angle bisector theorem, we see $AE=48/13 , EC=56/13, AP=16/5, PB=14/5$. Applying Van Aubel's Theorem, $AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1$, and so the answer is $\boxed{\textbf{(C)}\; 2 : 1}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 