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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!

I J
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    Two notes (1) I suspect that by "sphere", you mean a solid sphere (one that contains its interior). Most mathematicians call that a "ball", and use "sphere" to mean just the boundary. – David E Speyer Jun 02 '11 at 22:00
  • (2) You should specify that you are talking about convex closed sets (unless that's not what you want). I'm sure there are some stupid counter-examples otherwise by taking $S_1$ and $S_2$ to be dense subsets of balls, or deleting a small piece deep in the interior of a ball. – David E Speyer Jun 02 '11 at 22:01
  • Subject to those notes: Nice question! Looking forward to the answer. – David E Speyer Jun 02 '11 at 22:02
  • Oh right. I meant euclidean ball and convex closed sets. Edited – I J Jun 02 '11 at 22:13

3 Answers3

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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

Jim Belk
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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.

enter image description here

GEdgar
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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.

Guillaume Aubrun
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