# Difference between revisions of "2013 AMC 12A Problems/Problem 11"

## Problem

Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$?

$[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real s=1/2,m=5/6,l=1; pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m; draw(A--B--C--cycle^^D--E^^F--G); dot(A^^B^^C^^D^^E^^F^^G); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NW); label("F",F,S); label("G",G,NW); [/asy]$

$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$

## Solution

Let $AD = x$, and $AG = y$. We want to find $DE + FG$, which is nothing but $x+y$.

Based on the fact that $ADE$, $DEFG$, and $BCFG$ have the same perimeters, we can say the following:

$3x = x + 2(y-x) + y = y + 2(1-y) + 1$

Simplifying, we can find that

$3x = 3y-x = 3-y$

Since $3-y = 3x$, $y = 3-3x$.

After substitution, we find that $9-10x = 3x$, and $x$ = $\frac{9}{13}$.

Again substituting, we find $y$ = $\frac{12}{13}$.

Therefore, $x+y$ = $\frac{21}{13}$, which is $C$

~sugar_rush

## See also

 2013 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 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.

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