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Let $K$ and $L$ be two two-dimensional convex bodies, and consider their Cartesian product $K\times L\subseteq \mathbb R^4$. Now let $U_\theta\subseteq \mathbb R^4$ be the two-dimensional subspace spanned by $(1,0,\cos\theta,-\sin\theta)$ and $(0,1,\sin\theta,\cos\theta)$. Let $f(\theta) = \max_{\mathbf{x}\in\mathbb{R}^4} |(K\times L)\cap(U_\theta+\mathbf{x})|$, that is, the maximal cross-sectional area of $K\times L$ parallel to $U_\theta$. I have reason to believe, and would like to prove, that $f(\theta)$ cannot be unimodal. I call $f(\theta)$ unimodal if it is strictly increasing for $a\le\theta\le b$ and then strictly decreasing for $b\le\theta\le a+2\pi$. I will not say for now what my reasons are to believe that $f(\theta)$ cannot be unimodal, because I don't want to bias the reader's impression of what the best approach to this problem might be.

If anyone has any suggestions for approaches which might be beneficial to consider I would be very happy. Also, if you have some ideas for solving special cases, such as the case $K=L$, that would also be helpful. In the case $K=L$, we have $f(\theta)=f(-\theta)$, $f(0)$ is automatically a maxmimum, and $f(\pi)$ is either a local maximum or a local minimum. So, unimodality occurs if and only if $f(\pi)$ is a minimum and there are no other local extrema.

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The function can be a constant and constants seem to be unimodal by your definition. –  Sergei Ivanov Apr 7 '13 at 8:07
    
Yes, you're right. I actually should have written "increasing" instead of "weakly increasing". I will edit accordingly. –  Yoav Kallus Apr 7 '13 at 15:37

1 Answer 1

Here is a non-constant counter-example to the original version (with weak monotonicity).

First observe that $f(\theta) $ (up to a factor $\sqrt2$ or so) equals the maximum area of intersection of $K$ and a translate of $R_\theta(L)$ where $R_\theta$ is the rotation of the plane through the angle $\theta$.

Let $K$ be the triangle with vertices $(1,0)$, $(-1,0)$ and $(0,1)$. Let $L$ be a narrow isosceles triangle $ABC$ with $AB=AC=1$ and $BC=\varepsilon$. I assume that $BC$ is horizontal and $A$ is above $BC$.

For most values of $\theta$ (more precisely, for all but a small neighborhood of $\pi$) $R_\theta(L)$ can be translated so as to fit inside $K$, hence $f(\theta)=area(L)$. But for $\theta\approx\pi$ this is not possible and hence $f(\theta)<area(L)$. It seems that (although it is cumbersome to verify) $f(\pi)$ is the minimum and $f$ is monotone at both sides of $\pi$.
           Triangles
          (Image added by J.O'Rourke.)

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@Sergei: I took the liberty of adding a figure. Hope I illustrated your pretty idea correctly. –  Joseph O'Rourke Apr 7 '13 at 19:19
    
It seems from numerics that it is indeed the case that $f$ is weakly unimodal. So this is a valuable example to keep in mind. However, as I have pointed out, I am interested in the cases where $f(\theta)$ is not constant on any interval. Particularly, I want to eliminate the cases that $K$ or $L$ are circular disks and the cases where $R_\theta(L)\subset K$ for some $\theta$. –  Yoav Kallus Apr 7 '13 at 19:43

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