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

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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 $AF$ : $FD$ ?

$[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); [/asy]$

$\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2$

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

INDUCTIVE REASONING IS STILL REASONING

Tear a piece of your scrap paper and mark the length of $AF$ on it with your pencil. Do the same for $FD$ . Clearly $AF$ is twice $FD$ . Thus $\frac{AF}{FD} = \boxed{\textbf{(C)}\; 2 : 1}$

@@ Line 74: / Line 74: @@
 INDUCTIVE REASONING IS STILL REASONING
-Tear a piece of your scrap paper and mark the length of <math>AF</math> on it with your pencil. Do the same for <math>FD</math>. Clearly <math>AF</math> is twice <math>FD = \boxed{\textbf{(C)}\; 2 : 1}</math>
+Tear a piece of your scrap paper and mark the length of <math>AF</math> on it with your pencil. Do the same for <math>FD</math>. Clearly <math>AF</math> is twice <math>FD</math>. Thus <math>\frac{AF}{FD} = \boxed{\textbf{(C)}\; 2 : 1}</math>
 ==See Also==
 {{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}
 {{MAA Notice}}

2016 AMC 12A (Problems • Answer Key • Resources)
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