# Generalization of the non-existence of a monostatic planar body

Domokos, Papadopulos, and Ruina showed that there does not exist a convex planar rigid body of uniform density which has only one orientation of stable equilibrium and one orientation of unstable equilibrium when placed on a level surface under the influence of gravity. In three dimensions, such a body does exist. Therefore, let us stay in two dimensions, but consider different ways of trying to generalize this result (which I will refer to as Problem 0). Consider the following two problems.

Problem 1: Is there a convex body of uniform density $0<\rho<1/2$ which has only one stable equilibrium when floating in an incompressible liquid of unit density? This problem has been considered by Varkonyi, without any definite results, but with some results that suggest at least that for $\rho\simeq0$ and $\rho\simeq1/2$ no such body exists. Note that the limit $\rho\to0$ corresponds to the original problem and the symmetry of the case $\rho=1/2$ makes the existence of at least two stable orientations trivial.

Problem 2: Consider two congruent convex rigid bodies, one fixed and the other moving under the influence of a potential energy function given by the negative of the overlap area between the bodies. Can the shape of such a body be chosen so that there is only one stable equilibrium (namely the configuration of perfect overlap)? I first heard this problem from Paul Chaikin and then later saw some numerical results, unpublished at the moment, by Etienne Marcotte, a physics student at Princeton University, which suggest that no such shape exists.

It is clear that these problems have a common flavor. What is the best way to unify them? Here is my attempt: Let $K$ be a convex two-dimensional body, and denote by $K_{\mathbf{x},\theta}$ the congruent body obtained from $K$ by rotating about the origin by an angle $\theta$ and then translating by a vector $\mathbf{x}$. Let $\mu$ be a log-concave measure, and define a potential function $U(\mathbf{x},\theta)=-\mu(K_{\mathbf{x},\theta})$. For any fixed $\theta$, it is easy to show that $U(\mathbf{x},\theta)$ is convex as a function of $\mathbf{x}$ and therefore achieves its minimum in a convex region. Let $u(\theta)$ denote this minimum. Conjecture: if there are angles $\theta_1$ and $\theta_2$ such that $u(\theta)$ is monotonic non-decreasing for $\theta_1\le\theta\le\theta_2$ and monotonic non-increasing for $\theta_2\le\theta\le\theta_1+2\pi$, then $u(\theta)$ is constant.

For Problem 1, the density of $\mu$ is given by $C-\rho y$ for $0<y<\operatorname{diam} K$ and $C+(1-\rho)y$ for $-\operatorname{diam} K<y<0$, where $C$ is large enough to make sure the density is non-negative. For Problem 2, $\mu$ is simply the Lebesgue measure restricted to $K$.

My question is whether anyone can point out a counterexample? I could not think of one, but it is possible I am missing something obvious. Other comments would be welcome as well.

Edit: since I did not receive any suggestions for counterexamples of my conjecture, I would like to ask a different question to go with the same exposition. The solution of Problem 0 is often identified as a special case of the four-vertex theorem (4VT). However, the conventional proof does not use the 4VT in any way. In the proof that does use the 4VT, it is not the boundary of the body that is the curve on which the theorem is invoked, and it is really unclear, at least to me, what the significance is of the curve on which it is invoked (see Appendix B of this paper). Can anybody clear this up? What is the connection, really, of the 4VT to Problem 0? Is there any hope of its extending to Problems 1 and 2 and the conjecture? What is the significance of the curve on which the 4VT is invoked in Problem 0?

• @Yoav: Do you know if Problems 1 and 2 have positive solutions in $\mathbb{R}^3$? – Joseph O'Rourke Aug 28 '12 at 11:05
• @Joseph: not that I know of. – Yoav Kallus Aug 28 '12 at 13:55
• @Joseph: The gomboc should be a solution for 3D Problem 1 for small enough nonzero $\rho$. – Yoav Kallus Sep 17 '12 at 18:10