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2016 AMC 12A Problems/Problem 12: Difference between revisions

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Much easier solution (solution 1)
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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==
 
Applying the angle bisector theorem to <math>\triangle ABC</math> with <math>\angle CAB</math> being bisected by <math>AD</math>, we have
 
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath>
 
Thus, we have
 
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath>
 
and cross multiplying and dividing by <math>2</math> gives us
 
<cmath>3\cdot CD=4\cdot BD.</cmath>
 
Since <math>CD+BD=BC=7</math>, we can substitute <math>CD=7-BD</math> into the former equation. Therefore, we get <math>3(7-BD)=4BD</math>, so <math>BD=3</math>.
 
 
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected.  This gives us
 
<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath>
 
and since <math>AB=6</math> and <math>BD=3</math>, we have
 
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath>
 
Cross multiplying and dividing by <math>3</math> gives us
 
<cmath>AF=2\cdot FD.</cmath>
 
Dividing by <math>FD</math> gives us
 
<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath>
 
Therefore,
 
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath>
 
== Solution 2==
 
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>


Line 24: Line 62:
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>


== Solution 2==
== Solution 3==
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>.  Note that <math>F</math> is the incenter. Then, <math>\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}</math>
Denote <math>[\triangle{ABC}]</math> as the area of triangle ABC and let <math>r</math> be the inradius. Also, as above, use the angle bisector theorem to find that <math>BD = 3</math>.  Note that <math>F</math> is the incenter. Then, <math>\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}</math>


== Solution 3 ==
== Solution 4==
We denote <math>CD</math> by <math>y</math> and <math>DB</math> by <math>x</math>. Then, with the Angle Bisector Theorem in triangle <math>ACB</math> with angle bisector <math>AD</math>, we have  
We denote <math>CD</math> by <math>y</math> and <math>DB</math> by <math>x</math>. Then, with the Angle Bisector Theorem in triangle <math>ACB</math> with angle bisector <math>AD</math>, we have  
<math>\frac{x}{6}=\frac{y}{8}</math> or <math>y=\frac{4x}{3}.</math> However, <math>x+y=7,</math> so <math>x+\frac{4x}{3}=7</math> or <math>x=3.</math>  
<math>\frac{x}{6}=\frac{y}{8}</math> or <math>y=\frac{4x}{3}.</math> However, <math>x+y=7,</math> so <math>x+\frac{4x}{3}=7</math> or <math>x=3.</math>  

Revision as of 17:45, 17 February 2016

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

\[AF=2\cdot FD.\]

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 $BD = 3$. Note that $F$ is the incenter. Then, $\frac{AF}{FD} = \frac{[\triangle{AFB}]}{[\triangle{BFD}]} = \frac{AB * \frac{r}{2}}{BD * \frac{r}{2}} = \frac{AB}{BD} = \boxed{\textbf{(C)}\; 2 : 1}$

Solution 4

We denote $CD$ by $y$ and $DB$ by $x$. Then, with the Angle Bisector Theorem in triangle $ACB$ with angle bisector $AD$, we have $\frac{x}{6}=\frac{y}{8}$ or $y=\frac{4x}{3}.$ However, $x+y=7,$ so $x+\frac{4x}{3}=7$ or $x=3.$ Now, we use the Angle Bisector Theorem again in triangle $ADB$ with angle bisector $BF.$ We get $\frac{AF}{6}=\frac{FD}{3}$ or $\frac{AF}{FD}=\frac{2}{1},$ which gives us the answer $\frac{AF}{AD} =\boxed{\textbf{(C)}\; 2 : 1}$

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

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