Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== ==Solution==")
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
 +
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>?
  
 +
<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>
 +
 +
<math>\textbf{(A) }48\qquad
 +
\textbf{(B) }52\qquad
 +
\textbf{(C) }56\qquad
 +
\textbf{(D) }60\qquad
 +
\textbf{(E) }72\qquad</math>
  
 
==Solution==
 
==Solution==
 +
 +
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.
 +
 +
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>.

Revision as of 21:10, 7 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 $2 * (AD + DE) = 56$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_12&oldid=50890"