[Math] Point inside a square

Question

If a point is randomly chosen from a square, whose sides are equal to 1, find the probability that the point's distance from a side of the square is less than the distance of that point from the closest square diagonal.

I am not sure how to picture this at all, really confusing. Any help is appreciated.

Answer 1

A point in the shaded areas above is closer to a side than a diagonal, with $AO$ and $BO$ bisecting $\angle CAB$ and $\angle ABC$ respectively, and $\triangle BOA$ repeated around the square because of symmetry. $O$ happens to be the incentre of $\triangle ABC$, and the inradius is $$r=\frac{A(\triangle ABC)}s=\frac{1/4}{(\sqrt2+1)/2}=\frac{\sqrt2-1}2$$ The ratio $\frac{A(\triangle BOA)}{A(\triangle ABC)}=\frac{OD}{CD}$ then gives the desired probability: $$\frac{OD}{CD}=\frac r{1/2}=\sqrt2-1=0.414213\dots$$