# Pushing convex bodies together

Given two convex bodies $A$ and $B$, in $\mathbb R^3$ let's say. We define $A(t)$ and $B(t)$ as $A+xt$ and $B+yt$ where $x,y$ are two arbitrary points. (That is the Minkowski sum, so the two bodies are moving at constant velocity in the $x$ and $y$ directions, and $t$ is the time variable.) Can one show that the function $$f(t)=\operatorname{Vol}(A(t)\cap B(t))$$ is unimodal? That is nondecreasing up to some point, and then nonincreasing.

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Gjergji, I suspect that you ask for the 3D version because this is known in 2D. Can you give a link to the corresponding result? – Wadim Zudilin Jun 4 '10 at 10:28
I suspect the proof, even in the general case, not to be very hard, but I don't have any reference. I was discussing Kneser-Poulsen with a friend today, and thought that the question above is one of the simplest one could start asking in a series of related questions. – Gjergji Zaimi Jun 4 '10 at 11:13

The sets $\{ (A(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ and $\{ (B(t),t)|t\in \mathbb{R} \} \subset \mathbb{R}^4$ are convex, their intersection $K$ is a bounded convex set, and $f(t)$ is the volume of the slice of $K$ at height $t$. By Brunn-Minkowski inequality, this is log-concave, so definitely unimodal.
Thank you Thorny! This is nice. By the way, you mean $f(t)^{1/3}$ to be concave right? Unimodality follows in any case. – Gjergji Zaimi Jun 4 '10 at 12:30
If $f^{1/3}$ is concave then $f$ is log-concave. Thorny's kind of formulation is frequently preferred when it is possible, because it's dimension-free. – Mark Meckes Jun 4 '10 at 13:35
The intersection $K$ need not be bounded. Consider the case $A \cap B \ne 0$ and $x = y$. – Steve Huntsman Jun 4 '10 at 13:46