Point nearer the centre of a square than its sides

Choose a point uniformly at random inside a square. What is the probability that the point is closer to the square's centre than to the nearest side of the square?

$\tfrac{4\sqrt2-5}{3}\approx0.219$. Take the square $[-1,1]^2$; in the quarter where the right side is nearest, the condition $\sqrt{x^2+y^2}<1-x$ is the parabola region $x<\tfrac{1-y^2}{2}$. Intersecting with $|y|\le x$ (where $|y|\le\sqrt2-1$) and integrating gives the result, confirmed by simulation.