Let $P$ be the $1$-skeleton of a convex polyhedron fixed in $\mathbb{R}^3$, and $|P|$ the sum of the Euclidean lengths of the edges of $P$. Let $P_1, P_2, P_3$ be the perpendicular projections of $P$ onto the Cartesian coordinate planes, and $|P_i|$ the sum of the lengths of the segments of $P_i$.

For example, for the particular placement of $P$ a unit edge-length regular tetrahedron shown below, $|P_1|+|P_2|+|P_3|$ is nearly double $|P|=6$:

          $|P_1|$ (red) $=1+\sqrt{\frac{2}{3}}+\sqrt{\frac{11}{3}}$. $|P_2|$ (green) $=1+\sqrt{3}$. $|P_3|$ (blue) $=3+\sqrt{3}$. $\Sigma \approx 11.2$.

Conjecture. For any placement of any convex polyhedron $P$, $|P_1|+|P_2|+|P_3| \ge |P|$, with equality uniquely achieved by the cube.

For a unit edge-length cube $P$, $|P|=12$ and $|P_i|=4$ when oriented so that each projection is a square. So I'm conjecturing that the cube hides its edges in projection more effectively than any other convex polyhedron. Can anyone see a proof or a counterexample?

I would also be interested in which orientations of the regular tetrahedron minimize $\Sigma |P_i|$.

The higher-dimensional analog could be the subject of a future post.

  • $\begingroup$ As @GuillaumeAubrun correctly remarks, the claim should be for a rectangular box, not just a cube. $\endgroup$ – Joseph O'Rourke Sep 13 '18 at 23:31

Edit: this does not answer the OP question, as it considers $1$-dimensional projections instead of $2$-dimensional projections.

For every vector $(x_1,x_2,x_3) \in \mathbf{R}^3$, we have $$ |x_1| + |x_2| + |x_3| \geq \sqrt{x_1^2+ x_2^2+x_3^2} $$ with inequality only for multiples of basis vectors.

Summing this inequality over all edges of $P$ yields that $|P_1|+|P_2|+|P_3| \geq |P|$, with inequality for polytopes all whose edges are parallel to basis vectors, which are essentially cubes (up to dilatation in each of the three directions).

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    $\begingroup$ Upon reflection, I am not sure how your analysis takes into account one edge hiding another, as occurs with a cube...? $\endgroup$ – Joseph O'Rourke Sep 13 '18 at 15:46
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    $\begingroup$ It doesn't, and actually doesn't answer the question at all $\endgroup$ – Guillaume Aubrun Sep 13 '18 at 16:11
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    $\begingroup$ The inequality should rather be $\sqrt{x^2+y^2}+\sqrt{x^2+z^2}+\sqrt{y^2+z^2}\ge 2\sqrt{x^2+y^2+z^2}$ and the hiding effect is offset by the $2$ factor since each interval can be covered at most twice. $\endgroup$ – fedja Nov 12 '18 at 22:27
  • $\begingroup$ @fedja great obversation ! $\endgroup$ – Guillaume Aubrun Nov 13 '18 at 9:22

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