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Difference between revisions of "2013 AMC 12A Problems/Problem 11"
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== Problem==
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Triangle <math>ABC</math> is equilateral with <math>AB=1</math>. Points <math>E</math> and <math>G</math> are on <math>\overline{AC}</math> and points <math>D</math> and <math>F</math> are on <math>\overline{AB}</math> such that both <math>\overline{DE}</math> and <math>\overline{FG}</math> are parallel to <math>\overline{BC}</math>. Furthermore, triangle <math>ADE</math> and trapezoids <math>DFGE</math> and <math>FBCG</math> all have the same perimeter. What is <math>DE+FG</math>?
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<asy>
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size(180);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
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real s=1/2,m=5/6,l=1;
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pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m;
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draw(A--B--C--cycle^^D--E^^F--G);
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dot(A^^B^^C^^D^^E^^F^^G);
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,N);
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label("$D$",D,S);
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label("$E$",E,NW);
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label("$F$",F,S);
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label("$G$",G,NW);
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</asy>
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<math>\textbf{(A) }1\qquad
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\textbf{(B) }\dfrac{3}{2}\qquad
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\textbf{(C) }\dfrac{21}{13}\qquad
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\textbf{(D) }\dfrac{13}{8}\qquad
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\textbf{(E) }\dfrac{5}{3}\qquad</math>
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==Solution==
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Let <math>AD = x</math>, and <math>AG = y</math>. We want to find <math>DE + FG</math>, which is nothing but <math>x+y</math>.
 
Let <math>AD = x</math>, and <math>AG = y</math>. We want to find <math>DE + FG</math>, which is nothing but <math>x+y</math>.
  
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Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math>
 
Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math>
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=10|num-a=12}}

Revision as of 17:38, 22 February 2013

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 $C$

See also

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