There is an attractive theorem that says that if two plane figures are directly similar, then so is any convex combination of them. Below, $P_1$ and $P_2$ are directly similar polygons: they have the same angles in the same sequence; They are (positively) scaled, rotated, translated versions of one another. The convex combination illustrated $P_{12} = \frac{1}{3} P_1 + \frac{2}{3} P_2$ is also directly similar.
(Let me henceforth abbreviate "direcly similar" with "similar.") My question is:

Q1. Is there a natural extension to $\mathbb{R}^3$ and to higher dimensions?

Any extension cannot be an exact extension—some aspect has to give. For example, here are two cubes $P_1$ and $P_2$, the latter a $\frac{3}{4}$-scale rotated version of $P_1$. Shown is $P_{12} = \frac{1}{2} P_1 + \frac{1}{2} P_2$, and it is clearly not a cube. I would define similarity in $\mathbb{R}^3$ to require all faces to be similar with the same scale factor as well preserving all dihedral angles—in other words, $B$ is similar to $A$ if $B$ is a rotated, scaled, translated version of $A$.
But perhaps this is true?

Q2. Is the convex combination of two cubes (scaled, rotated, translated) always a parallelopiped? A parallelotope in $\mathbb{R}^d$?

Another approach would be to restrict the transformations:

Q3. Is there some condition on the transformations applied to the shapes that permits the similarity conclusion? Or: What is the widest class of transformations that leads to the similarity conclusion?

Certainly if $P_2$ is just a translated, scaled copy of $P_1$, then any convex combination is similar, in any dimension. Rotations are the culprit. But perhaps some rotations still lead to similarity.

It seems likely this has all been well-explored. If so, thanks for pointers!


First of all, we may forget about translations, since they just shift the convex combinations. Thus we may regard the direct similarity as a linear transform given by an orthogonal matrix $A$ multiplied by some constant $\alpha$. Then the transform mapping the otiginal polytope to the convex combination is given by $$ B=\lambda I+(1-\lambda)\alpha A. $$ This is also a linear transform, so the image of a parallelotope is also a parallelotope (if the transform is non-degenerate; notice that it may be degenerate only if $\lambda=\pm(1-\lambda)\alpha$ since the complex eigenvalues of $A$ have unital absolute values; in this case $A$ should have eigenvalue $\pm 1$, and this may happen even in the planar case). This answers Q2.

As for Q1 (and Q3), we need to check whether $B$ is an orthogonal matrix multiplied by a real number $c$ (provided that the polytope is solid). This is the case iff all eigenvalues of $B$ have the same absolute value, which reduces to the following two options:

1) All eigenvalues of $A$ coincide (thus $A$ is scalar, and the transform is a scaling); or

2) All the eigenvalues of $A$ are $e^{\pm i\theta}$ for some fixed $\theta$ (surely these two eigenvalues coinciding multiplicities). Thus $A$ acts as a composition of rotations in $d/2$ mutually orthogonal planes. This case cannot happen in odd-dimensional space (which answers Q3), but it sometimes happens in all even-dimensional spaces.

The phenomenon of the planar case is that for the plane there are no more options.

  • $\begingroup$ If object $P_1$ lies in a $k$-flat, and $P_2$ is obtained by rotation about an axis orthogonal to that $k$-flat (and then scaled), then I believe the similarity holds for convex combinations. For example, polygons in parallel planes in $\mathbb{R}^3$. $\endgroup$ – Joseph O'Rourke Jan 9 '14 at 1:59
  • 1
    $\begingroup$ @Joseph: that's why I speak on the case when the polytope is solid (i.e. contains an interior point). But it seems also that you need not the rotation in this axis, but again some $k/2$ rotations at the same angle in orthogonal 2-flats. $\endgroup$ – Ilya Bogdanov Jan 9 '14 at 4:09

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