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

Line 41: Line 41:
 
We can set <math>AD=0</math>, so fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>.
 
We can set <math>AD=0</math>, so fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>.
  
 +
==Solution 3==
 +
 +
Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get <math>\boxed{(c) 56}</math>
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/8ki_yMyE6no
 
https://youtu.be/8ki_yMyE6no

Revision as of 17:27, 25 August 2021

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 $56\implies \boxed{\textbf{(C)}}$.

Solution 3

Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get $\boxed{(c) 56}$

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

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