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

(Created page with "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 ...")
 
 
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Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y.
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== Problem==
  
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Based on the fact that ∆ADE, DEFG, and BCFG have the same perimeters, we can say the following:
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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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3x = x + 2(y-x) + y = y + 2(1-y) + 1
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<asy>
 +
size(180);
 +
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
 +
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;
 +
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);
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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);
 +
</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==
 +
 
 +
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>.
 +
 
 +
Based on the fact that <math>ADE</math>, <math>DEFG</math>, and <math>BCFG</math> have the same perimeters, we can say the following:
 +
 
 +
<math>3x = x + 2(y-x) + y = y + 2(1-y) + 1</math>
  
 
Simplifying, we can find that
 
Simplifying, we can find that
  
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3x = 3y-x = 3-y
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<math>3x = 3y-x = 3-y</math>
 +
 
 +
Since <math>3-y = 3x</math>, <math>y = 3-3x</math>.
 +
 
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After substitution, we find that <math>9-10x = 3x</math>, and <math>x</math> = <math>\frac{9}{13}</math>.
 +
 
 +
Again substituting, we find <math>y</math> = <math>\frac{12}{13}</math>.
 +
 
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Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math>
  
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Since 3-y = 3x, y = 3-3x.
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==Video Solution==
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https://www.youtube.com/watch?v=XQpQaomC2tA
  
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After substitution, we find that 9-9x-x = 3x, and x = 9/13.
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~sugar_rush
  
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Again substituting, we find y = 12/13.
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== See also ==
 +
{{AMC12 box|year=2013|ab=A|num-b=10|num-a=12}}
  
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Therefore, x+y = 21/13, which is C
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[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:12, 24 November 2020

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 $C$

Video Solution

https://www.youtube.com/watch?v=XQpQaomC2tA

~sugar_rush

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

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

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