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:**

It is easy to see that the

**round disk**has this property.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)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**.)Some figures as

**Reuleaux triangle**are bad (see the comments below)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.)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.

One answer is

**accepted, BUT**it only provides a solution for unbounded figures.