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

Revision as of 15:25, 12 August 2023

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 $2 * (AD + DE) = \boxed{\textbf{(C) }{56}}$ .

Solution (Cheese)

We can set point $F$ to be on point $C$ , and point $D$ to be on point $A$ .

This makes a degenerate parallelogram with sides of length $28$ and $0$ , so it has a perimeter of $28 + 28 = \boxed{\textbf{(C) }{56}}$

Video Solution

https://youtu.be/CCjcMVtkVaQ

~sugar_rush

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

2013 AMC 10A (Problems • Answer Key • Resources)
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

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

@@ Line 37: / Line 37: @@
   \boxed{\textbf{(C) }{56}}</math>.
-==Solution(Cheese)==
+==Solution (Cheese)==
 We can set point <math>F</math> to be on point <math>C</math>, and point <math>D</math> to be on point <math>A</math>.
-This makes a degenerate parallelogram with sides of length <math>28</math> and <math>0</math>, so it has a perimeter of <math>28 + 28 = 56</math>.
+This makes a degenerate parallelogram with sides of length <math>28</math> and <math>0</math>, so it has a perimeter of <math>28 + 28 =
+\boxed{\textbf{(C) }{56}}</math>
 ==Video Solution==

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