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

Question

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!

Answer 1

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.

Answer 2

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

http://en.wikipedia.org/wiki/Reuleaux_triangle

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