Assume a convex figure $F\subset \mathbb R^2$ satisfies the following property: if $f:F\to \mathbb R^2$ is a distance-non-increasing map then its image $f(F)$ is congruent to a subset of $F$.
Is it true that $F$ is a round disk?
Comments:
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It is easy to see that the round disk has this property.
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One can reformulate the property: if for some set $G\subset\mathbb R^2$ there is a distance-non-contracting map $G\to F$ then there is a distance-preserving map $G\to F$. (The equivalence follows from Kirszbraun theorem)
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No bad map is known for the following figure: intersection of two discs say unit disc with center at (0,0) and a disc with radius 1.99 and center at (0,1) — see comments of Martin M. W. below. (That might be a counterexample.)
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Some figures as Reuleaux triangle are bad (see the comments below)
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The construction with two folds along parallel lines (see below) gives the following: If $F$ is good then for any point $x\in \partial F$ the restriction of $dist_x$ to $\partial F$ does not have local minima except $x$. (This property holds for any shape $C^2$-close to a round disc.)
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This problem was meant to be an exercise for school students, but I was not able to solve it :). It appears in print in 2008 (in Russian), see problem #5 in Плоское оригами и длинный рубль.
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One answer is accepted, BUT it only provides a solution for unbounded figures.
Best Answer
Here's a few thoughts on the question:
First, although this is probably obvious to everyone who's posted already, the region F must be bounded (if it is not $\mathbb{R}^2$). If not, then since it's convex, it must contain an infinite ray, and F must be contained in a half-space since it is convex. Take the projection onto the ray, then map the ray by arclength onto a spiral that can't live in any half-space, for example. This map then cannot be contained inside of F and is clearly length-decreasing.
So assume F is bounded. Then I think we may assume F is compact, by taking its closure. Any 1-Lipschitz map from F to $\mathbb{R}^2$ will extend to a 1-Lipschitz map from of the closure, and if the image lies in F, then its closure will lie in the closure. This probably doesn't help at all.
Edit: as Anton Petrunin has pointed out, the following argument is bogus: Now, the space of 1-Lipschitz maps to $\mathbb{R}^2$ is convex (and I think it is complete in the sup topology). Also, it's an easy exercise to see that any convex combination of 1-Lipschitz maps lying in isometric copies of F also lie in isometric copies of F (convex combinations of isometries give conformal affine maps with dilatation $\leq 1$). So to prove the claim for a given region F, we need "only" prove that extremal maps, i.e. ones which are not convex combinations of other maps, are contained in an isometric copy of F. I'm not sure if this helps, but there might be some literature on the convex structure of 1-Lipschitz maps which one could possibly exploit.