A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width? $\text{(A)}\ 2\sqrt {2} + 1 \qquad$ $ \text{(B)}\ 3\sqrt {2}\qquad$ $ \text{(C)}\ 2\sqrt {2} + 2 \qquad$ $ \text{(D)}\ 3\sqrt {2} + 1 \qquad$ $ \text{(E)}\ 3\sqrt {2} + 2$
Solution
Let the brush width be $1$ and the side length be $x$. A strip of width $1$ along a diagonal meets each side within $\frac12\cdot\sqrt2=\frac{\sqrt2}2$ of the corner.
So the unpainted part consists of $4$ isosceles right triangles, one on each side, each with hypotenuse $x-2\cdot\frac{\sqrt2}2=x-\sqrt2$ and area $\frac14(x-\sqrt2)^2$.
Half the square is unpainted, so each triangle has area $\frac18x^2$: $(x-\sqrt2)^2=\frac{x^2}2$, $x-\sqrt2=\frac{x}{\sqrt2}$, and $x=\frac{\sqrt2}{1-\frac1{\sqrt2}}=2\sqrt2+2$.
A paint brush is swept along both diagonals of a square to produce a symmetric painted area. If $\frac89$ of the area of the square is painted, what is the ratio of the side length of the square to the brush width?
$\text{(A)}\ 2\sqrt {2} + 1 \qquad$ $ \text{(B)}\ 3\sqrt {2}\qquad$ $ \text{(C)}\ 2\sqrt {2} + 2 \qquad$ $ \text{(D)}\ 3\sqrt {2} + 1 \qquad$ $ \text{(E)}\ 3\sqrt {2} + 2$ 
