The convex polyhedron is completely contained in the cube with the edge 1 Prove that the sum of squares of the area of the faces this polyhedron does not exceed 6. How to prove that?
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First, I claim that for any polygon $P$ contained in some plane in $3$-space, the square of the area of $P$ is equal to the sum of the squares of the areas of the projections of $P$ onto the three coordinate planes. It suffices to prove this in the case $P$ has area $1$. Let $\langle x,y,z\rangle$ be a vector normal to $P$. Then the projections of $P$ onto the $xy$-, $xz$-, and $yz$-planes have areas, respectively, $$ \frac{|z|}{\|\langle x,y,z\rangle\|}, \frac{|y|}{\|\langle x,y,z\rangle\|}, \frac{|x|}{\|\langle x,y,z\rangle\|}. $$ The sum of the squares of these numbers is $1$.
Now, suppose some convex polyhedron is contained in a unit cube with faces parallel to the coordinate planes. Consider the projection of the faces of that polyhedron onto the $xy$-axis. This gives us a bunch of polygons contained in a unit square. The area of each of the polygons is at most $1$, and the sum of their areas is at most $2$ (the polygons can overlap in pairs, but never in triples). The sum of the squares of a bunch of nonnegative numbers is bounded by the largest number times the sum of the numbers, so the sum of the squares of the areas of the projections is at most $2$. The same argument holds for the projections onto the other two coordinate planes. Adding these up, we find that the sum over faces of the polyhedron and coordinate planes of the square of the area of the projections is at most $6$. The claim now implies the sum of the squares of the areas of the faces of our polyhedron is at most $6$.
Julian Rosen
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