Difference between revisions of "2022 AMC 12A Problems/Problem 3"

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−	Let each side length of the square be <math>x.</math> Since four of the rectangles (excluding the middle one) contribute two side lengths (length and width) to the perimeter of the square, we can equate <math>4x,</math> the perimeter of the square, to the sum of the dimensions of four rectangles. By simply plugging in different combinations, we can see that the only possibility for the middle rectangle is <math>4x=1+6+5+6+2+7+2+3=32,</math> meaning the rectangle in the middle is <math>\boxed{B}</math>.	+
−	~~-lightningowl64~~

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