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

(→‎Solution 1)
(→‎Solution 4: format and simplify)
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== Solution 4==
 
== Solution 4==
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One only needs the angle bisector theorem, the segment addition postulate, and some simple algebra to solve this question.
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One only needs the angle bisector theorem to solve this question.
  
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The question asks for AF:DF. Apply the angle bisector theorem to <math>\triangle ABD</math> to get the ratio <math>\frac {AF}{DF}</math> :
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The question asks for <math>AF:FD = \frac{AF}{FD}</math>. Apply the angle bisector theorem to <math>\triangle BAD</math> to get
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<cmath>\frac{AF}{FD} = \frac{AB}{BD}.</cmath>
  
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<math>\frac {AF}{AB}</math> = <math>\frac {DF}{BD}</math> or, equivalently,
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<math>AB = 6</math> is given. To find <math>BD</math>, apply the angle bisector theorem to <math>\triangle ABC</math> to get
 +
<cmath>\frac{BD}{DC} = \frac{BA}{AC} = \frac{6}{8} = \frac{3}{4}.</cmath>
  
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<math>\frac {AF}{DF}</math> = <math>\frac {AB}{BD}</math>.  
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Since
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<cmath>BD + DC = BC = 7,</cmath>
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it is immediately obvious that <math>BD = 3</math>, <math>DC = 4</math> satisfies both equations.
  
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AB is given. To find BD apply the angle bisector theorem to <math>\triangle ABC</math> to get:
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Thus,
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<cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath>
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<math>\frac {BD}{AB}</math> = <math>\frac {CD}{AC}</math>
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~revision by [[User:emerald_block|emerald_block]]
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---> <math>\frac {BD}{AB}</math> = <math>\frac {BC - BD}{AC}</math> since, by the segment addition postulate, BD + CD = BC
 
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---> BD = <math>\frac {AB*BC}{AC + AB}</math>.
 
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Substituting this expression for BD into the proportion <math>\frac {AF}{DF}</math> = <math>\frac {AB}{BD}</math> yields:
 
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<math>\frac {AF}{DF}</math> = <math>\frac {AB}{BD}</math> = <math>\frac {AC + AB}{BC}</math> = <math>\frac {8 + 6}{7}</math> = 2.
 
  
 
==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}}

Revision as of 11:08, 3 November 2021

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, $CD = 4$.

There are two ways to solve from here. First way:

Note that $DB = 7 - 4 = 3.$ By the angle bisector theorem on $\triangle ADB,$ $\frac{AF}{FD} = \frac{AB}{DB} = \frac{6}{3}.$ Thus the answer is $\boxed{\textbf{(C)}\; 2 : 1}$

Second way:

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

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 $\boxed{\textbf{(C)}\; 2 : 1}$.

Solution 4

One only needs the angle bisector theorem to solve this question.

The question asks for $AF:FD = \frac{AF}{FD}$. Apply the angle bisector theorem to $\triangle BAD$ to get \[\frac{AF}{FD} = \frac{AB}{BD}.\]

$AB = 6$ is given. To find $BD$, apply the angle bisector theorem to $\triangle ABC$ to get \[\frac{BD}{DC} = \frac{BA}{AC} = \frac{6}{8} = \frac{3}{4}.\]

Since \[BD + DC = BC = 7,\] it is immediately obvious that $BD = 3$, $DC = 4$ satisfies both equations.

Thus, \[AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.\] ~revision by emerald_block

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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