2013 AMC 12A Problems/Problem 11

Revision as of 23:55, 6 February 2013 by Indianninja707 (talk | contribs)

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 $C$

Invalid username
Login to AoPS