Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 6 (Cheese))
 
(41 intermediate revisions by 21 users not shown)
Line 3: Line 3:
 
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>?
 
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>?
  
−
<pre style="color: gray">TODO: Diagram</pre>
+
<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>
  
 
<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>
 
<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>
  
−
== Solution ==
+
== Solution 1==
 +
 
 
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>
  
Line 14: Line 15:
 
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>.
+
There are two ways to solve from here.
 +
First way:
 +
 
 +
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>
 +
 
 +
Second way:
 +
 
 +
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>
Line 22: Line 30:
 
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
 
<math>mD = mC + mB = 1 + \frac{4}{3} = \frac{7}{3}</math>
  
−
So <math>\frac{AF}{AD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
+
So <math>\frac{AF}{FD} = \frac{mD}{mA} = \boxed{\textbf{(C)}\; 2 : 1}</math>
 +
 
 +
== 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:
 +
 
 +
<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>
 +
 
 +
<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 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>.
 +
 
 +
== Solution 4==
 +
One only needs the angle bisector theorem to solve this question.
 +
 
 +
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>
 +
 
 +
<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>
 +
 
 +
Since
 +
<cmath>BD + DC = BC = 7,</cmath>
 +
it is immediately obvious that <math>BD = 3</math>, <math>DC = 4</math> satisfies both equations.
 +
 
 +
Thus,
 +
<cmath>AF:FD = AB:BD = 6:3 = \boxed{\textbf{(C)}\ 2:1}.</cmath>
 +
~revision by [[User:emerald_block|emerald_block]]
 +
 
 +
==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.
 +
~[[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.
 +
 
 +
==Solution 6 (Cheese)==
 +
Assume the drawing is to-scale. Use your allotted ruler to measure out each side. Note that <math>AB:BC:AC</math> is equal to <math>6:7:8</math>.
 +
 
 +
Measure out the length of <math>\overline{AF}</math> in relation to <math>\overline{FD}</math>. This ratio is approximately <math>\boxed{\textbf{(C)}\ 2:1}</math>. Solution by [[User:juwushu|juwushu]].
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/Gjt25jRiFns?t=43
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2016|ab=A|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 16:13, 3 November 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.

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_12&oldid=231410"