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 4)
Line 39: Line 39:
 
One only needs the angle bisector theorem and some simple algebra to solve this question.
 
One only needs the angle bisector theorem and some simple algebra to solve this question.
  
−
The question asks for AF:DF. Applying the angle bisector theorem to <math>\triangle ABD</math> yields the ratio <math>\frac {AF}{DF}</math> :
+
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> :
  
 
<math>\frac {AF}{AB}</math> = <math>\frac {DF}{BD}</math> or, equivalently,  
 
<math>\frac {AF}{AB}</math> = <math>\frac {DF}{BD}</math> or, equivalently,  
Line 56: Line 56:
  
 
<math>\frac {AF}{DF}</math> = <math>\frac {AB}{BD}</math> = <math>\frac {AC + AB}{BC}</math> = <math>\frac {8 + 6}{7}</math> = 2.
 
<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 13:42, 31 December 2020

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, $CD = 4$.

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

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 and some simple algebra to solve this question.

The question asks for AF:DF. Apply the angle bisector theorem to $\triangle ABD$ to get the ratio $\frac {AF}{DF}$ :

$\frac {AF}{AB}$ = $\frac {DF}{BD}$ or, equivalently,

$\frac {AF}{DF}$ = $\frac {AB}{BD}$.

AB is given. To find BD apply the angle bisector theorem again to $\triangle ABC$ to get:

$\frac {BD}{AB}$ = $\frac {CD}{AC}$

---> $\frac {BD}{AB}$ = $\frac {BC - BD}{AC}$, since BD + CD = BC

---> BD = $\frac {AB*BC}{AC + AB}$.

Substituting this expression for BD into the proportion $\frac {AF}{DF}$ = $\frac {AB}{BD}$ yields:

$\frac {AF}{DF}$ = $\frac {AB}{BD}$ = $\frac {AC + AB}{BC}$ = $\frac {8 + 6}{7}$ = 2.

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=141156"