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I have two solid two-dimensional shapes on a plane, $S_1$ and $S_2$, with boundaries $B_1$ and $B_2$. If the boundaries of the two shapes are allowed to touch, but no points internal to the boundaries are allowed to intersect, is it true that the boundaries of $S_1$ and $S_2$ can touch at only a single point on a plane (allowing for rotation, translation, reflection, etc. of the shapes) if and only if $S_1$ and $S_2$ are strictly convex? How does this generalize to higher dimensions?

Update in response to WimC's comment: Can we say that $S_1$ and/or $S_2$ are strictly convex if and only if their boundaries can be touched together at most at a single point?

Update 2: Perhaps we can rescue the original statement of this problem, requiring $S_1$ and $S_2$ to both be strictly convex, if we require that there should only ever be a single point of intersection for arbitrary dilations of either shape?

Update 3: In response to Rahul's comment, the converse statement of want to show here is that, provided neither $S_1$ or $S_2$ are strictly convex, then there MUST exist some configuration of the shapes that allows for multiple simultaneous contacts along their boundaries. We know that if both shapes are strictly convex then there can only be one point of contact, and we've discussed conditions where one point of contact can exist when only one shape is strictly convex, but it seems more challenging to show that one shape being strictly convex is both necessary and sufficient for forbidding multiple points of simultaneous contact.

For weakly convex functions, i.e. convex hulls etc. with straight lines in their boundaries, can we make a similar statement about the number of edges that can touch?

Please note that this is a (hopefully) more focused version of an earlier question, now removed, which asked for the properties of $B_1$ and $B_2$ that guaranteed some strict upperbound for the number of points $k$ where $B_1$ and $B_2$ can touch.

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If $B_1$ is a circle with radius $r$ then $B_2$ need not be convex as long as its "outward" radius of curvature is greater than $r$ everywhere. –  WimC Dec 30 '12 at 17:14
    
@WimC Is it true that at least one of the shapes must be strictly convex? –  WhiteAutumn Dec 30 '12 at 17:42
    
Is the question that two shapes can touch at multiple points if and only if neither is strictly convex? Because that seems trivial: take two distinct points where the shapes meet, and consider whether points on the line segment joining the points lie in either set. –  Rahul Dec 30 '12 at 20:47
    
@Rahul Narain Yes, but how can we prove the existence of those two points? –  WhiteAutumn Dec 30 '12 at 21:19
    
Sorry, my previous comment is incorrect. If two shapes touch at multiple points, then we can only conclude that at least one of them is nonconvex. But I guess what you really want to know is whether, if both shapes are nonconvex, there always exists a configuration in which they touch at multiple points. Perhaps you should emphasize that in the question. –  Rahul Dec 30 '12 at 21:48
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1 Answer

No.

If you take $S_1 = \{(x,y) | y \leq 0\}$ and $S_2 = \overline{B}((0,1),1)$, then $S_1 \cap S_2 = \{(0,0)\}$, but $S_1$ is not strictly convex by any definition.

One set may be non-convex, and still have only one touching point. Assuming that touching means $S_1 \cap S_2 \neq \emptyset$ and $S_1^\circ \cap S_2^\circ = \emptyset$, then define $S_1 = B((0,0),2)\setminus B((2,0), 1)$ and $S_2(x,y)=B((x,y),\frac{1}{2})$. If you draw a picture, then it should be clear that if $(x,y)$ are such that $S_1$ and $S_2(x,y)$ are touching, then the intersection contains at most one point. (It is not hard to prove, just messy.)

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$S_2$ is strictly convex no? So at least one of $S_1$ and $S_2$ is, as claimed (in my interpretation of the question). –  WimC Dec 30 '12 at 18:26
    
It says 'iff $S_1$ and $S_2$ are strictly convex'? –  copper.hat Dec 30 '12 at 18:30
    
There's an update below that... –  WimC Dec 30 '12 at 18:31
    
The update seems to be defining 'strictly convex' of a pair $S_1,S_2$ as being able to touch only at one point. In that case, the question becomes slightly vacuous? –  copper.hat Dec 30 '12 at 18:41
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I think the sets need to be simply connected; otherwise two annuli form a counterexample. Maybe you might as well assume that they are homeomorphic to a closed disc. –  Niels Diepeveen Jan 1 '13 at 2:31
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