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Can the Minkowski sum of two convex sets in the plane which are not polygons be a polygon? Explicitly my convex set is of the form

$ C= \{(x,y) \in \mathbb{R}^2 : x,y \geq 0 \text{ , } \sqrt{y}+\sqrt{x} \geq 1 \text{ and } y \leq 1-x\} $

I am interested in knowing if there is a convex set C' such that the Minkowski sum

$C + C' = P$

where

$ P = \{(x,y) \in \mathbb{R}^2: x,y \geq 0 \text{ and } 1/2 \leq x+y \leq 1 \}. $ Any help would be great!

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1 Answer 1

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The Minkowsky sum of two convex sets is a compact polygon only if the two summands are compact polygons.

Consider first the case that $C,C'$ are compact. A compact convex set is a polygon if and only if there are only finitely many points that can be obtained as intersection of a line with $C$. Let $c\in C$ and $\ell$ a line such that $\ell\cap C=\{c\}$. A line parallel to $\ell$ then intersects $C'$ in a segment $[c_1',c_2']$ of its boundary (where possibly $c_1'=c_2'$). Then $[c+c_1',c+c_2']$ is part of the boundary of $P$ parallel to $\ell$. Let the preceeding edge of $P$ be $[a,c+c_1']$ and the next edge be $[c+c_2',b]$ Then the lines parallel to these through $c$ bound $C$. The exterior angle at $c$ is at least as big as the smallest exterior angle of $P$. We conclude that there can be at most finitely many such points $c$. Thus $C$ is a polygon. By the same argument, $C'$ is a polygon.

The details of what happens when $C,C'$ are not assume closed (they must of course still be bounded) are left as an exercise. At least it is immediately clear that $\overline C$ and $\overline{C'}$ are polygons.

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