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

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Note that because <math>\bar{DE}</math> and <math>\bar{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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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>.
 
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>.
 
Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>.

Revision as of 04:26, 7 February 2013

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 $2 * (AD + DE) = 56$.

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