2007 AMC 10A Problem 19

Let $s$ be the side, $w$ the brush width, and $x$ the leg of one unpainted isosceles right triangle. Each triangle has area $\tfrac18 s^2,$ so $\tfrac12 x^2 = \tfrac18 s^2$ and $x = \tfrac{s}{2}.$ The leg plus the brush width is half the diagonal: $x + w = \tfrac{\sqrt2}{2} s.$ Thus $w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.$ Therefore $$\dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2.$$ Thus, the correct answer is C.