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

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==Solution 5 (Luck-Based)==
 
==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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Note that <math>AF</math> and <math>BD</math> look like medians. Assuming they are medians, we mark the answer <math>\boxed{\textbf{(C)}\ 2:1}</math> as we know that the centroid (the point where all medians in a triangle are concurrent) splits a median in a <math>2:1</math> ratio, with the shorter part being closer to the side it bisects.
 
~[[User:scthecool|scthecool]]
 
~[[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.
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Note: This is heavily luck based, and if the figure had not been drawn to scale, this answer would have been wrong. It is advised to not use this in a real competition unless absolutely necessary.
  
 
==Solution 6 (Cheese)==
 
==Solution 6 (Cheese)==

Revision as of 17:58, 18 January 2025

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 not been drawn to scale, this answer would have been wrong. It is advised to not use this in a real competition unless absolutely necessary.

Solution 6 (Cheese)

Assume the drawing is to-scale. Use your allotted ruler to measure out each side. Note that $AB:BC:AC$ is equal to $6:7:8$.

Measure out the length of $\overline{AF}$ in relation to $\overline{FD}$. This ratio is approximately $\boxed{\textbf{(C)}\ 2:1}$. Solution by juwushu.

Solution 7 (Mass Points Only)

Draw line $CF$, extending it to $AB$ at point $G$. Now, we have $3$ cevians that intersect at point $F$. Point $G$ has a mass of $6$, point $E$ has a mass of $8$, and point $F$ has a mass of $7$. Note that the masses of $E$, $D$, and $G$ are the sums of the masses of $A$ and $C$, $B$ and $C$, and $A$ and $B$ respectively. We see that the sum of the masses of $E$, $D$, and $G$ equals twice the sum of the masses of $A$, $B$, and $C$. The sum of the masses of $E$, $D$, and $G$ equals \[8 + 7 + 6 = 21\]. Therefore, the sum of the masses of $A$, $B$, and $C$ equals \[\frac{21}{2} = 10.5\]. We know that the masses of $B$ and $C$ sum to 7. Therefore, the mass of point $A$ is \[10.5 - 7 = 3.5\].

Hence, the ratio of $AF : FD = 7 : 3.5 = \boxed{\textbf{(C)}\ 2:1}$.

Solution by abirgh.

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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