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

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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
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==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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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>.
 
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>, by fakesolving, we get <math>56\implies \boxed{\textbf{(C)}}</math>.
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==Solution 3==
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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 56, or <math>\boxed{(C)}</math>.
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==Solution 4==
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Similar to Solution 1, we find that 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 and similar to each other. We know that the perimeter of <math>ADEF</math> can be found by <math>2(l + w)</math>, and <math>DE</math> will be the length and the width will be <math>EF</math>. These can be found by looking at the long sides of <math>\triangle DBE</math> and <math>\triangle FEC</math> respectively. We then find the ratio of long sides to the short side of <math>\triangle ABC</math> to be <math>7/5</math> by <math>28/20</math>, which applies to the other triangles since they are similar. We set up an expression, calling <math>BE</math> as <math>x</math> and <math>EC</math> as <math>20 - x</math>, and substitute it into the perimeter equation knowing our long sides from the ratio:
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<math> 2(7x/5 + 7/5(20-x)) </math>
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<math> 2(7x/5 + 28 - 7x/5) </math>
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<math> 2(28) </math>
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<math> 56\implies \boxed{\textbf{(C)}}</math>
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~neeyakkid23
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==Video Solution==
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https://youtu.be/8ki_yMyE6no
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~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 19:59, 10 September 2024

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 $AD=0$, by 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 56, or $\boxed{(C)}$.

Solution 4

Similar to Solution 1, we find that the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$, and are therefore also isosceles triangles and similar to each other. We know that the perimeter of $ADEF$ can be found by $2(l + w)$, and $DE$ will be the length and the width will be $EF$. These can be found by looking at the long sides of $\triangle DBE$ and $\triangle FEC$ respectively. We then find the ratio of long sides to the short side of $\triangle ABC$ to be $7/5$ by $28/20$, which applies to the other triangles since they are similar. We set up an expression, calling $BE$ as $x$ and $EC$ as $20 - x$, and substitute it into the perimeter equation knowing our long sides from the ratio:

$2(7x/5 + 7/5(20-x))$

$2(7x/5 + 28 - 7x/5)$

$2(28)$

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

~neeyakkid23

Video Solution

https://youtu.be/8ki_yMyE6no

~savannahsolver

See Also

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
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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