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

(Much easier solution (solution 1))
(another solution (luck-based) and a warning about using it in a real competition)
 
(31 intermediate revisions by 15 users not shown)
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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
+
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math>
  
−
<cmath>\frac{CD}{AC}=\frac{BD}{AB}.</cmath>
+
<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
  
−
Thus, we have
+
Similarly, <math>CD = 4</math>.
  
−
<cmath>\frac{CD}{8}=\frac{BD}{6},</cmath>
+
There are two ways to solve from here.
 +
First way:
  
−
and cross multiplying and dividing by <math>2</math> gives us
+
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>
  
−
<cmath>3\cdot CD=4\cdot BD.</cmath>
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Second way:
  
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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>.
+
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>
  
−
Apply the angle bisector theorem again to <math>\triangle ABD</math> with <math>\angle ABC</math> being bisected.  This gives us
+
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math>
  
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<cmath>\frac{AB}{AF}=\frac{BD}{FD},</cmath>
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<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
  
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and since <math>AB=6</math> and <math>BD=3</math>, we have
+
So <math>\frac{AF}{FD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
  
−
<cmath>\frac{6}{AF}=\frac{3}{FD}.</cmath>
+
== Solution 2==
  
−
Cross multiplying and dividing by <math>3</math> gives us
+
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:
  
−
<cmath>AF=2\cdot FD.</cmath>
+
<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>
  
−
Dividing by <math>FD</math> gives us
+
<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>
  
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<cmath>\frac{AF}{FD}=\frac{2}{1}.</cmath>
+
==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>.
  
−
Therefore,
+
== Solution 4==
 +
One only needs the angle bisector theorem to solve this question.
  
−
<cmath>AF:FD=\frac{AF}{FD}=\frac{2}{1}=\boxed{\textbf{(C)}\; 2 : 1}.</cmath>
+
The question asks for <math>AF:FD = \frac{AF}{FD}</math>. Apply the angle bisector theorem to <math>\triangle ABD</math> to get
 +
<cmath>\frac{AF}{FD} = \frac{AB}{BD}.</cmath>
  
−
== Solution 2==
+
<math>AB = 6</math> is given. To find <math>BD</math>, apply the angle bisector theorem to <math>\triangle BAC</math> to get
 +
<cmath>\frac{BD}{DC} = \frac{BA}{AC} = \frac{6}{8} = \frac{3}{4}.</cmath>
  
−
By the angle bisector theorem, <math>\frac{AB}{AE} = \frac{CB}{CE}</math>
+
Since
 +
<cmath>BD + DC = BC = 7,</cmath>
 +
it is immediately obvious that <math>BD = 3</math>, <math>DC = 4</math> satisfies both equations.
  
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<math>\frac{6}{AE} = \frac{7}{8 - AE}</math> so <math>AE = \frac{48}{13}</math>
+
Thus,
 +
<cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath>
 +
~revision by [[User:emerald_block|emerald_block]]
  
−
Similarly, <math>CD = 4</math>.
+
==Solution 5 (Luck-Based)==
−
 
+
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.
−
Now, we use mass points. Assign point <math>C</math> a mass of <math>1</math>.
+
~[[User:scthecool|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.
−
<math>mC \cdot CD = mB \cdot DB</math> , so <math>mB = \frac{4}{3}</math>
 
−
 
 
−
Similarly, <math>A</math> will have a mass of <math>\frac{7}{6}</math>
 
  
−
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
+
== Video Solution by OmegaLearn ==
 +
https://youtu.be/Gjt25jRiFns?t=43
  
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So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
+
~ pi_is_3.14
−
 
 
−
== 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>
 
−
 
 
−
== 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
 
−
<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>
 
−
Now, we use the Angle Bisector Theorem again in triangle <math>ADB</math> with angle bisector <math>BF.</math> We get
 
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<math>\frac{AF}{6}=\frac{FD}{3}</math> or
 
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<math>\frac{AF}{FD}=\frac{2}{1},</math> which gives us the answer <math>\frac{AF}{AD} =\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 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 (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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