# Geometric proof that twice the square of the diagonal of a rectangular cuboid equals the surface area if and only if it is a cube

In the answers to this question it is shown that the surface area of a cube equals twice the square of the length of its diagonal.

This is very much related to the following algebraic claim:

Three numbers $$a,b,c\in \mathbb{R}$$ satisfy $$a=b=c$$ if and only if $$a^2 + b^2 + c^2 = ab + bc + ac$$.

Given a rectangular cuboid with sides $$a,b,c$$, the LHS of the equality is the square of the length of the diagonal. The RHS of the equation is half of the surface area. Therefore, A geometric formulation of the algebraic claim would be:

The surface area of a rectangular cuboid equals twice the square of the length of its diagonal if and only if it's a cube.

I would like to know if anyone can come up with a geometric proof of this claim. To start with, I don't even have any geometric intuition for why this equality is correct for a cube.

For a rectangle with diagonal $$d$$ you have (see figure): $$2\cdot\text{area of rectangle}=\text{area of rhombus}\le d^2,$$ where equality holds only if the rhombus is a square, i.e. if the rectangle is a square.
Let now $$A_1$$, $$A_2$$, $$A_3$$ be the areas of three concurring faces of a cuboid, with diagonals $$d_1$$, $$d_2$$, $$d_3$$. By the above remark we have: $$\text{surface area of cuboid}=2A_1+2A_2+2A_3\le d_1^2+d_2^2+d_3^2=2(\text{diagonal of cuboid})^2,$$ where equality holds only if all faces are squares.
• Thank you for the answer, I like it very much. I'd say that the last step: $d_1^2+d_2^2+d_3^2 = 2D^2$ is still somewhat algebraic, and doesn't use the main diagonal in a geometric way. Nevertheless I'll accept the answer in a day or two if no one would come up with something better. Apr 7, 2023 at 7:37
In cuboid $$AG$$ join $$AG$$ and $$AC$$. By Pythagorean theorem, $$AG^2=CG^2+AC^2$$, and $$AC^2=CD^2+AD^2=CD^2+CB^2$$. Therefore$$AG^2=CG^2+CD^2+CB^2$$and$$2AG^2=2CG^2+2CD^2+2CB^2$$But the surface area of the cuboid is$$2CG\times CD+2CG\times CB+2CD\times CB$$and this equals $$2AG^2$$ only if $$CG=CD=CB$$, i.e. only if the faces are squares and the cuboid is a cube.