Difference between revisions of "2013 AMC 10A Problems/Problem 12"

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\textbf{(D) }60\qquad
 
\textbf{(D) }60\qquad
 
\textbf{(E) }72\qquad</math>
 
\textbf{(E) }72\qquad</math>
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[[Category: Introductory Geometry Problems]]
  
==Solution==
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==Solution 1==
  
 
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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It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>.
 
It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>.
  
Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>.
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The opposite sides of parallelograms are equal (you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is <math>2 \times (AD + DE) = 56\implies \boxed{\textbf{(C)}}</math>.
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==Solution 2==
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We can set <math>AD=0</math>, so fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>.
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==Video Solution==
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https://youtu.be/8ki_yMyE6no
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2013|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2013|ab=A|num-b=11|num-a=13}}
 
{{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2013|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 17:00, 26 December 2020

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 1

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, $AD + DE = AD + DB = AB = 28$.

The opposite sides of parallelograms are equal (you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is $2 \times (AD + DE) = 56\implies \boxed{\textbf{(C)}}$.

Solution 2

We can set $AD=0$, so fakesolving, we get $56\implies \boxed{\textbf{(C)}}$.

Video Solution

https://youtu.be/8ki_yMyE6no

~savannahsolver

See Also

2013 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 10 Problems and Solutions
2013 AMC 12A (ProblemsAnswer KeyResources)
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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