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

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== Solution ==
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==Problem 12==
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In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?
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<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>
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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>
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== Solution 1==
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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>
  
 
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
 
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
  
Similarly, <math>CD = 4</math>
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Similarly, <math>CD = 4</math>.
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There are two ways to solve from here.
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First way:
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Note that <math>DB = 7 - 4 = 3.</math> By the angle bisector theorem on <math>\triangle ADB,</math> <math>\frac{AF}{FD} = \frac{AB}{DB} = \frac{6}{3}.</math> Thus the answer is <math>\boxed{\textbf{(C)}\; 2 : 1}</math>
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Second way:
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Now, we use [[mass points]]. Assign point <math>C</math> a mass of <math>1</math>.
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<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math>
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Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math>
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<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
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So <math>\frac{AF}{FD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
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== Solution 2==
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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>1.</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>
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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>
  
Now, we use mass points.
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==Solution 3==
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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>.
  
Assign point <math>C</math> a mass of <math>1</math>.
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== Solution 4==
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One only needs the angle bisector theorem to solve this question.
  
Because <math>\frac{AE}{EC} = \frac{6}{7}, A</math> will have a mass of <math>\frac{7}{6}</math>
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The question asks for <math>AF:FD = \frac{AF}{FD}</math>. Apply the angle bisector theorem to <math>\triangle ABD</math> to get
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<cmath>\frac{AF}{FD} = \frac{AB}{BD}.</cmath>
  
Similarly, <math>B</math> will have a mass of <math>\frac{4}{3}</math>
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<math>AB = 6</math> is given. To find <math>BD</math>, apply the angle bisector theorem to <math>\triangle BAC</math> to get
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<cmath>\frac{BD}{DC} = \frac{BA}{AC} = \frac{6}{8} = \frac{3}{4}.</cmath>
  
<math>mE = mA + mC = \frac{13}{6}</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.
  
Similarly, <math>mD = mC + mB = \frac{7}{3}</math>
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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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~revision by [[User:emerald_block|emerald_block]]
  
The mass of <math>F</math> is the sum of the masses of <math>E</math> and <math>B</math>.
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==Solution 5 (Luck-Based)==
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Note that <cmath>AF</cmath> and <cmath>BD</cmath> look like medians. Assuming they are medians, we mark the answer <cmath>\boxed{\textbf{(C)}\ 2:1}</cmath> as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a <cmath>2:1</cmath> ratio, with the shorter part being closer to the side it bisects.
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~[[User:scthecool|scthecool]]
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Note: This is heavily luck based, and if the figure had been not drawn to scale, for example, this answer would have easily been wrong. It is thus advised to not use this in a real competition unless absolutely necessary.
  
<math>mF = mE + mB = \frac{7}{2}</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/Gjt25jRiFns?t=43
  
This can be checked with <math>mD + mA</math>, which is also <math>\frac{7}{2}</math>
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~ pi_is_3.14
  
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
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==See Also==
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{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 18:19, 20 March 2024

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 $\triangle ABD$ to get \[\frac{AF}{FD} = \frac{AB}{BD}.\]

$AB = 6$ is given. To find $BD$, apply the angle bisector theorem to $\triangle BAC$ 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

Solution 5 (Luck-Based)

Note that \[AF\] and \[BD\] look like medians. Assuming they are medians, we mark the answer \[\boxed{\textbf{(C)}\ 2:1}\] as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a \[2:1\] ratio, with the shorter part being closer to the side it bisects. ~scthecool Note: This is heavily luck based, and if the figure had been not drawn to scale, for example, this answer would have easily been wrong. It is thus advised to not use this in a real competition unless absolutely necessary.

Video Solution by OmegaLearn

https://youtu.be/Gjt25jRiFns?t=43

~ pi_is_3.14

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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