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 $C$

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2013 AMC 12A Problems/Problem 11

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