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2016 AMC 12A Problem 9

Problem 9 of 25EasierAlgebraGeometry

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is a−2b,\dfrac{a-\sqrt{2}}{b}, where aa and bb are positive integers. What is a+b?a+b?

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Solution

Let xx be the common side length. The diagonal of the unit square has length 2\sqrt{2} and consists of two small-square diagonals (each x2x\sqrt2) plus one small-square side length x,x, so 2x2+x=2. 2x\sqrt2+x=\sqrt2. Solving, x=222+1=2 (22−1)(22+1)(22−1)=4−27. \begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered} Thus a=4,a=4, b=7,b=7, and a+b=11.a+b=11. Thus, the correct answer is E.
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