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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!

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1  
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 17 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

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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
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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

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

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

enter image description here

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