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

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== Solution 1==
 
== 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>
 
 
and 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>
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Similarly, <math>CD = 4</math>.
 
Similarly, <math>CD = 4</math>.
  
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.
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Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>.
  
 
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math>
 
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math>
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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 3==
+
== Solution 2==
  
 
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>. There are two ways to continue from here:
 
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>. There are two ways to continue from here:
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<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math>
 
<math>2.</math> Apply the angle bisector theorem on <math>\triangle{ABD}</math> to get <math>\frac{AF}{FD} = \frac{AB}{BD} = \frac{6}{3} = \boxed{\textbf{(C)}\; 2 : 1}</math>
  
== Solution 4==
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==Solution 3==
 
+
Draw the third angle bisector, and denote the point where this bisector intersects <math>AB</math> as <math>P</math>. Using angle bisector theorem, we see <math>AE=48/13 , EC=56/13, AP=16/5, PB=14/5</math>. Applying [https://artofproblemsolving.com/wiki/index.php/Van_Aubel%27s_Theorem Van Aubel's Theorem], <math>AF/FD=(48/13)/(56/13) + (16/5)/(14/5)=(6/7)+(8/7)=14/7=2/1</math>, and so the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math>.
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>
 
 
 
 
 
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:44, 17 July 2020

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

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 2

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 3

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 $\boxed{\textbf{(C)}\; 2 : 1}$.

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions

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