# If the Minkowski sum of two convex closed sets is a Euclidean ball, then can the two sets be anything other than Euclidean balls?

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!

#### Solutions Collecting From Web of "If the Minkowski sum of two convex closed sets is a Euclidean ball, then can the two sets be anything other than Euclidean balls?"

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.

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.

Here’s mine. Done before I saw Jim’s solution (honest). But after seeing his, I animated mine, too (using Maple).

Two copies of the Reuleaux triangle

same size, one rotated by 180 degrees from the other.

For $\mu$ a Borel positive measure on the sphere $S^{n-1}$, consider a continuous Minkowski sum of segments
$$K_\mu = \int_{S^{n-1}} [0,\theta] \, \mathrm{d}\mu (\theta) .$$
The set $K_\mu$ is convex (it could be defined by its support function, then the integral becomes a usual one). Now observe that (1) $K_{\mu+\nu} = K_\mu + K_\nu$ (2) by rotation invariance, $K_\mu$ is a Euclidean ball if $\mu$ is the uniform measure (3) there are many ways to write the uniform measure as a sum of positive measures.