Say that a collection of $n$ convex solids in $\mathbb{R}^3$ are mutually touching if, for every pair of them, they share some positive-measure part of their boundaries. (That is, we're not permitting the solids to touch only via an edge or a corner.)
As an example, if we "thicken up" an $n=4$ configuration in 2 dimensions (the maximum, since $K_5$ is nonplanar), then placing a large cube on top of the thickened configuration yields $n=5$ mutually touching convex regions. (By thickening it into a sort of wedge shape, we can put convex solids on both the top and bottom, for $n=6$.)
Do such configurations exist for arbitrarily high $n$? For infinitely many solids at once? What if we require the solids be congruent to each other? (Note that if we relax the convexity requirement, the answer to all of these is pretty easily seen to be yes.)
I vaguely recall reading a paper which provided an affirmative answer to the arbitrarily high $n$ case with congruent solids; I think it used the Voronoi cells given by equally spaced points on a helix, though it's been several years since I read it. Any pointers to this paper, or an equivalent result, would be appreciated (if I'm not just imagining it)!
Best Answer
Based on your description, I think I've found an argument that works. However many points there are, space them on a segment of helix defined by the following equation in cylindrical coordinates:
$$ (r, \theta, z) = (1, \theta, \frac{\theta}{2}) \hspace{1cm}\text{for } 0\leq\theta\leq\frac{\pi}{2} $$
The relevant solids will be the Voronoi cells of these points, cut off at some large distance. We want to check the existence of a boundary between every chosen pair of points on our helix segment. It will suffice to show that for every pair of points on the helix (not just the ones we happened to choose), there will be some point $Q$ in space that is equidistant from the pair and is closer to that pair of points than to any other point on the helix. (The existence of such a point shows that there must be a boundary between the two corresponding cells, with $Q$ lying on that boundary.)
The general idea of the proof is that if you're standing a few steps behind the center of a circle and looking straight through the center and across to the opposite side, the points with a larger angle to your line of sight will be closer to you. On the other hand, if you're looking at a line, the points with a larger angle will be farther. In the helix, which is kind of a cross between a circle and a line, these effects can be balanced, and the nearest points to you will be at an intermediate angle to you line of sight.
Let $P_1 = (1, \theta_1, \theta_1/2)$ and $P_2 = (1, \theta_2, \theta_2/2)$. We'll want to find a point $Q$ in space that is closer to those two points than to any other point on the helix segment. We'll choose $Q$ to be the following, for some $a$, which we'll specify later:
$$(-a, \frac{\theta_1+\theta_2}{2}, \frac{\theta_1+\theta_2}{4})$$
Define $R(\phi)=(1, \phi+\frac{\theta_1+\theta_2}{2}, \frac{\phi}{2}+\frac{\theta_1+\theta_2}{4})$. (This is always a point on the helix.) Note that since the helix segment only spans an angle of $\pi/2$, we only need to worry about $\phi$ between $-\pi/2$ and $\pi/2$. Also, note that $R(\phi^*)=P_1$ and $R(-\phi^*)=P_2$ for $\phi^*=\frac{\theta_1-\theta_2}{2}$.
We can now compute the squared distance between $R(\phi)$ and $Q$:
$$ s^2 = \frac{\phi^2}{4} + (\sin\phi)^2 + (a+\cos\phi)^2 = \frac{\phi^2}{4} + 1 + a^2 + 2a\cos\phi $$
Setting the derivative to 0, so we can find possible locations of minimal distance:
$$ \frac{ds^2}{d\phi} = \frac{\phi}{2} - 2a\sin\phi = 0 $$
This yields the solution $\phi=0$ for all $a$, and an additional pair of non-trivial solutions, $\phi = -\phi_a, \phi_a$ for $\frac{1}{4}<a<\frac{\pi}{4}$. (We're only counting solutions in $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is why there's an upper limit on $a$ too.)
We have $\lim_{a\to \frac{1}{4}}\phi_a = 0$ and $\lim_{a\to \frac{\pi}{4}}\phi_a = \frac{\pi}{2}$, and $\phi_a$ varies continuously with $a$. So we can choose an $a$ to produce any desired value of $\phi_a$ between 0 and $\frac\pi2$. Further, we can compute the second derivative of $s^2$, which is:
$$ \frac{d^2s^2}{d\phi^2} = \frac{1}{2} - 2a\cos\phi $$
This is negative for $a>\frac{1}{4}$, indicating that the $\phi=0$ critical point is a local maxima. Therefore, the neighbouring $-\phi_a$ and $\phi_a$ solutions must be local minima, and since $s^2(\phi)$ is an even function, they must have the same value of $s^2$ for both. Also, since there are no other possible local minima for $\phi\in(-\frac{\pi}{2}, \frac{\pi}{2})$, $\phi_a$ and $-\phi_a$ must give the global minima on that interval. Therefore, for given values of $\theta_1$, $\theta_2$, we choose $a$ such that $\phi_a = \phi^* = \frac{\theta_1 - \theta_2}{2}$, ensuring that $P_1$ and $P_2$ will correspond to $\phi_a$ and $-\phi_a$ and will therefore be the closest points to $Q$.