Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


share|improve this question
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 Speyer Jun 2 '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 Speyer Jun 2 '11 at 22:01
Subject to those notes: Nice question! Looking forward to the answer. –  David Speyer Jun 2 '11 at 22:02
Oh right. I meant euclidean ball and convex closed sets. Edited –  I J Jun 2 '11 at 22:13

2 Answers 2

up vote 16 down vote accepted

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

share|improve this answer
Jim, what software did you use to make the animation? –  Arjang Jun 2 '11 at 23:20
@Arjang It's just Mathematica. If you make a list of frames (myframes in the code above), then you can use Export to save as an animated gif, e.g. Export(filename, myframes). –  Jim Belk Jun 2 '11 at 23:24
Nicely done. I wish I could give another +1. –  Ross Millikan Jun 3 '11 at 0:43
As everyone else has been saying, nice answer! –  David Speyer Jun 3 '11 at 13:27

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.

enter image description here

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.