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

Revision as of 02:53, 7 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 $C$

Revision as of 23:55, 6 February 2013 (view source) Indianninja707 (talk \| contribs) ← Older edit	Revision as of 02:53, 7 February 2013 (view source) Epicwisdom (talk \| contribs) m (moved 2013 AMC 12A Problems/Problems 11 to 2013 AMC 12A Problems/Problem 11: Wrong title ("Problems 11")) Newer edit →
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