Minkowski Sum of Two Convex Closed Sets – Euclidean Ball

convex optimizationconvex-analysiseuclidean-geometry

If for two convex closed sets $S_1$ and $S_2$, the Minkowski sum is a Euclidean ball then can $S_1$ and $S_2$ be anything other than Euclidean balls themselves. I suspect they can be but I haven't found a counterexample. I don't have experience with Minkowski sums so any help will be appreciated.

Thanks!

Best Answer

This is almost certainly false. The following animation shows two convex shapes (with outlines shown in red and green) whose Minkowski sum is a disk of radius 3 (with outline shown in blue). The green shape is an ellipse with major and minor radii 1 and 1/2, which uniquely determines the red shape.

enter image description here

I do not have a proof that the red shape is convex, but it shouldn't be too hard to check.

Incidentally, here is the Mathematica code I used to produce this animation:

MyPlot = ParametricPlot[{3*{Cos[t], Sin[t]}, With[{u = ArcTan[-Sin[t], Cos[t]/2]}, 3*{Sin[u], -Cos[u]} - {Cos[t], Sin[t]/2}]}, {t, 0, 2 Pi}]; myframes = Table[With[{u = ArcTan[-Sin[t], Cos[t]/2]}, With[{pt = 3*{Sin[u], -Cos[u]} - {Cos[t], Sin[t]/2}}, Show[MyPlot, ParametricPlot[pt + {Cos[r], Sin[r]/2}, {r, 0, 2 Pi}, PlotStyle -> Darker[Green]], Graphics[{PointSize[Large], Point[pt]}]]]], {t, 0, 2 Pi - Pi/20, Pi/20}]; ListAnimate[myframes]

Edit: Here is a simpler solution using two congruent shapes. The boundary of each shape is the union of two circular arcs, each of which is congruent to 1/4 of the blue circle.

enter image description here

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