A Problem from Gaokao 2018

Set up Cartesian coordinates with the origin at one of the vertices and the axes aligned with the edges.

Write the plane as $ax+by+cz=d$ . To make the angles equal, the dot products of $(a,b,c)$ with $\vec{i},\vec{j},\vec{k}$ must be equal , so we can write $x+y+z=1+p$ . The area of the cross section $S$ will be $\sqrt{3}A$ , where $A$ is the area of the orthogonal projection $D$ of $S$ onto the $xy-$ plane. For $0\leq p \leq 1$ , the region $D$ is the unit square minus two isocleles right triangles with sides $p$ and $1-p$ , resp., so $A=\frac{1}{2}+p(1-p)$ . The maximum $A=\frac{3}{4}$ is attained when $p=\frac{1}{2}$ , and the answer is $\sqrt{3}A=\boxed{\frac{3\sqrt{3}}{4}}$ .