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?

1. Let the square have side $x$ and let the brush have width $1$. Each of the four white corner triangles has legs $\frac{x-\sqrt2}{\sqrt2}$, so together they have area $(x-\sqrt2)^2$. If the painted fraction is $f$, the white area is $(1-f)x^2$, hence $(x-\sqrt2)^2=(1-f)x^2$.
2. Solving the quadratic for $x$: $x=\dfrac{\sqrt2\left(1+\sqrt{1-f}\right)}f$ (taking the larger root).
3. With $f=\frac89$: $\sqrt{1-f}=\frac13$, so $x=\sqrt2\cdot\frac{4/3}{8/9}=\sqrt2\cdot\frac32=\boxed{\dfrac{3\sqrt2}2}$.