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Difference between revisions of "2013 AMC 12A Problems/Problem 9"

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== Problem==
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In <math>\triangle ABC</math>, <math>AB=AC=28</math> and <math>BC=20</math>.  Points <math>D,E,</math> and <math>F</math> are on sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to <math>\overline{AC}</math> and <math>\overline{AB}</math>, respectively.  What is the perimeter of parallelogram <math>ADEF</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 r=5/7;
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pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);
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pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));
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pair E=extension(D,bottom,B,C);
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pair top=(E.x+D.x,E.y+D.y);
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pair F=extension(E,top,A,C);
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draw(A--B--C--cycle^^D--E--F);
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dot(A^^B^^C^^D^^E^^F);
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label("$A$",A,NW);
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label("$B$",B,SW);
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label("$C$",C,SE);
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label("$D$",D,W);
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label("$E$",E,S);
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label("$F$",F,dir(0));
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</asy>
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<math>\textbf{(A) }48\qquad
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\textbf{(B) }52\qquad
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\textbf{(C) }56\qquad
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\textbf{(D) }60\qquad
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\textbf{(E) }72\qquad</math>
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==Solution==
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Note that because <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to the sides of <math>\triangle ABC</math>, the internal triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles.
 
Note that because <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to the sides of <math>\triangle ABC</math>, the internal triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles.
  
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Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>.
 
Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}}

Revision as of 17:37, 22 February 2013

Problem

In $\triangle ABC$, $AB=AC=28$ and $BC=20$. Points $D,E,$ and $F$ are on sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$, respectively. What is the perimeter of parallelogram $ADEF$?

[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]

$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$

Solution

Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$, the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$, and are therefore also isosceles triangles.

It follows that $BD = DE$. Thus, $AD + DE = AD + DB = AB = 28$.

Since opposite sides of parallelograms are equal, the perimeter is $2 * (AD + DE) = 56$.


See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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