Can the Minkowski sum of two convex sets in the plane which are not polygons be a polygon?

Minkowski sum of convex sets in the plane which are not polygons
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
I am interested in knowing if there is a convex set C' such that the Minkowski sum
$C+{C}^{\prime }=P$
where
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Step 1
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 $\ell \cap C=\left\{c\right\}$. A line parallel to $\ell$ then intersects C′ in a segment $\left[{c}_{1}^{\prime },{c}_{2}^{\prime }\right]$ of its boundary (where possibly ${c}_{1}^{\prime }={c}_{2}^{\prime }$). Then $\left[c+{c}_{1}^{\prime },c+{c}_{2}^{\prime }\right]$ is part of the boundary of P parallel to $\ell$. Let the preceeding edge of P be $\left[a,c+{c}_{1}^{\prime }\right]$ and the next edge be $\left[c+{c}_{2}^{\prime },b\right]$. 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.
Step 2
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 $\overline{{C}^{\prime }}$ are polygons.