Difference between revisions of "2013 AMC 12A Problems/Problem 11"

(Created page with "Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y. Based on the fact that ∆ADE, DEFG, and BCFG have the same perimeters, we can say the following: 3x ...")
 
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Let AD = x, and AG = y. We want to find DE + FG, which is nothing but x+y.
+
Let <math>AD = x</math>, and <math>AG = y</math>. We want to find <math>DE + FG</math>, which is nothing but <math>x+y</math>.
  
Based on the fact that ∆ADE, DEFG, and BCFG have the same perimeters, we can say the following:
+
Based on the fact that <math>ADE</math>, <math>DEFG</math>, and <math>BCFG</math> have the same perimeters, we can say the following:
  
3x = x + 2(y-x) + y = y + 2(1-y) + 1
+
<math>3x = x + 2(y-x) + y = y + 2(1-y) + 1</math>
  
 
Simplifying, we can find that
 
Simplifying, we can find that
  
3x = 3y-x = 3-y
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<math>3x = 3y-x = 3-y</math>
  
Since 3-y = 3x, y = 3-3x.
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Since <math>3-y = 3x</math>, <math>y = 3-3x</math>.
  
After substitution, we find that 9-9x-x = 3x, and x = 9/13.
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After substitution, we find that <math>9-10x = 3x</math>, and <math>x</math> = <math>\frac{9}{13}</math>.
  
Again substituting, we find y = 12/13.
+
Again substituting, we find <math>y</math> = <math>\frac{12}{13}</math>.
  
Therefore, x+y = 21/13, which is C
+
Therefore, <math>x+y</math> = <math>\frac{21}{13}</math>, which is <math>C</math>

Revision as of 23:55, 6 February 2013

Let $AD = x$, and $AG = y$. We want to find $DE + FG$, which is nothing but $x+y$.

Based on the fact that $ADE$, $DEFG$, and $BCFG$ have the same perimeters, we can say the following:

$3x = x + 2(y-x) + y = y + 2(1-y) + 1$

Simplifying, we can find that

$3x = 3y-x = 3-y$

Since $3-y = 3x$, $y = 3-3x$.

After substitution, we find that $9-10x = 3x$, and $x$ = $\frac{9}{13}$.

Again substituting, we find $y$ = $\frac{12}{13}$.

Therefore, $x+y$ = $\frac{21}{13}$, which is $C$

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