2013 AMC 12A Problems/Problem 11

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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 9-9x-x = 3x, and x = 9/13.

Again substituting, we find y = 12/13.

Therefore, x+y = 21/13, which is C

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