When shorter means smaller?

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?

• 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.

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If you marked it as "open-problem", can you give a reference? Is it known for ellipsoids? –  Boris Bukh Nov 28 '09 at 11:32
If an ellipse is "skinny," then it seems that you can fold the ellipse along a line through the center that meets the minor axis at a small angle to get an $f$ that doesn't have this property. I'm not sure about when the ellipse is close to being circular. –  Richard Kent Nov 28 '09 at 16:31
Do sets of constant width have this property? –  Gil Kalai Nov 28 '09 at 17:19
How does one map a Reuleaux triangle into something that's not congruent to a subset? –  David Eppstein Nov 28 '09 at 17:21
Reuleaux triangle is bad; one can fold Reuleaux triangle along one line so that one vertex will stick (a bit) from the opposite side. –  Anton Petrunin Nov 28 '09 at 18:22

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.

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(1) Nice observation on boundness --- (for me figure is bounded, but anyway it is nice). (2) Your statement "convex combination of 1-Lipschitz maps lying in isometric copies of F also lie in isometric copies of F" is wrong. –  Anton Petrunin Dec 10 '09 at 5:17
(1) I guess this wasn't mentioned in the statement of the problem, but it is clearly assumed. (2) ok, maybe I made a mistake. The idea I had was that if one takes isometric copies of F, we may normalize so that we have F and $A(F) +b, A \in SO(2), b\in \mathbb{R}^2$. Then for $t\in [0,1]$, $t F + (1-t)(A F+b) = (t+(1-t)A)F + (1-t)b$ gives a convex combination of the figures. As I mentioned, this transformation is a conformal affine map, so lies in an isometric copy of F. But maybe I made a mistake in the computation. What's your counterexample? –  Ian Agol Dec 10 '09 at 17:35
Counterexample: Take convex combination of two projections of disc to different lines. –  Anton Petrunin Dec 10 '09 at 17:57
ah right, I was interpolating same points to same points, which is stupid. Can I downvote my answer? –  Ian Agol Dec 10 '09 at 18:24

I will break tradition of the accumulation of interesting comments and post a suggested partial solution (of something). The structure of mathoverflow is somehow not perfect for an incremental discussion (even though it is great in many ways).

The main idea so far for relevant contractive maps are folding the plane in half. So you could change the question and ask if any convex shape other than a circle has the property that it always fits inside of itself if you fold it once across a chord. [Edit 1: If you do ask, the answer is that one fold is not enough for a dented circle, as Martin points out in the comment.] A Rouleaux triangle does not have this property, but maybe it is interesting to check other regular Rouleaux polygons.

I think that no ellipse (other than a circle) has this property. You have to be a little careful because if you take an ellipse that is not round and not too thin, then if you fold it across its short semiaxis, the half-ellipse can fit inside of the original ellipse in non-standard ways.

If you fold the ellipse across its long semiaxis instead, then trivially the half ellipse only fits in the original ellipse in one way (up to symmetry). Suppose that you tilt this chord slightly, but keep it passing through the center, and then cut the ellipse in half. Then I think that this kind of half ellipse also fits in the original ellipse in one way. If that is correct, then if you fold the ellipse along this slightly tilted chord, then the folded shape does not fit in the original ellipse. [Edit 2: Anton says that it is not true, and that this half-ellipse which is cut at a slight diagonal can be movable within the original ellipse. I do not know whether it can be moved far enough, but I will refrain from speculating.]

A similar trick works for any regular odd-sided polygon $P$. $P$ has a longest diagonal. Make a chord which is close to this diagonal and parallel to it, so that the region on one side that has the majority of vertices has less than half of the area. This subregion only fits in $P$ in one way, so if you fold along this chord the folded shape does not fit. [Edit 2: At least this case of the argument actually works.]

I conjecture that if $K$ is any convex shape whose longest chords are isolated, then either by tilting or offsetting a longest chord, you can make a fold that does not fit in $K$. [Edit 2: A foolish conjecture as long as the ellipsoid case is in doubt.]

On the other hand, a constant-width body has the opposite property. It has an entire circle of longest chords, in a natural sense a maximal family of them.

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To your question about a convex shape that can always be folded once on a chord: how about a disk with a little "chip" taken out? Say, the points of a disk of radius 1 centered at the origin, for which $x \le .99$. (Folding on a chord bigger than the chip gives you something that fits in the shape, since a full disk folded this way would fit; and folding on small chords is harmless too since they're close to tangent to the boundary.) I imagine Anton's parallel line trick kills this example with two folds, though. –  Martin M. W. Nov 29 '09 at 22:11
This is a shape whose longest chords are far from unique, which later on I suggested is the case that needs new ideas. But I agree that it is a clear counterexample to the modified question that I offered at the beginning: One fold is not always enough. –  Greg Kuperberg Nov 29 '09 at 22:27
"Then I think that this kind of half ellipse also fits in the original ellipse in one way." --- that is not true. If ellipse is almost round one can rotate such half moving it inside –  Anton Petrunin Nov 30 '09 at 2:17
It is not clear if "parallel line trick" kills this example (at first I made a wrong statement --- now it is corrected). If one cuts "chip" by a circle of radius >1 with the center inside of the unit disc then I do not see suitable map... –  Anton Petrunin Nov 30 '09 at 2:38
OK, interesting! And I agree, something like the intersection of the unit disk at (0,0) and the disk of radius 1.99 at (-1,0) does seem to rule out some other tricks involving folding on not-quite-parallel lines. –  Martin M. W. Nov 30 '09 at 3:06

An incomplete answer; but perhaps it helps to rephrase the problem as below. The reason the round circle does have this property is that without loss of generality, the map $f$ fixes the origin; and then since $f$ makes all distances shorter, QED.

To simplify things, I will assume that $F$ is bounded. Define $&rho;_&theta;$ to be a euclidean rotation (about the origin) through angle $&theta;$.

For $z\in F$, consider the function $d_z(x,y):=\sup\{&lambda;: x\not\in y+&lambda; (F-z)\}$; If $F$ were symmetric under $-1$, this would be a classical distance function; it does satisfy the triangle inequality by reasoning about Minkowski sums: if $x\in a F$ and $y\in x+b F$, then $y\in (a+b) F$. It is also translation-invariant: $d_z(x+v,y+v)=d_z(x,y)$.

We want to understand for which $F$ does $f$ being euclidean-short imply that for some $&theta;$ and some $z\in F$, $&rho;_&theta;\circ f$ is $d_z$-short.

Some easy observations, if $F$ is not a round circle, then (by relative compactness) assume that neither $d_z < |\cdot|$ nor $d_z > |\cdot|$; and then I think building a folding-type counterexample should be easy? But as I said before, it's an incomplete answer.

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I do not see how it might help... –  Anton Petrunin Nov 29 '09 at 2:02
not to be snarky, but that suggests to me you're not trying; although perhaps my notation/jargon are too opaque? –  some guy on the street Dec 3 '09 at 23:15

Inspired by the origami question, the only new suggestion that I can make is to mention the origamist Robert J. Lang. According to the biography on his web site, he is also a physicist. There was an open problem called the napkin-folding problem: Does there exist a contractive map from the square to the plane that increases its perimeter? Lang found a counterexample using his experience in origami.

Actually, I just discovered that you (Anton) already know this story, following a link on the Wikipedia page. However, my remaining comment is that maybe Lang is a good reference or expert for this new question.

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I think it simply time to prove that Martin's example is indeed a counterexample... BTW, do you read Russian? –  Anton Petrunin Dec 10 '09 at 2:54
With a little practice I can read Cyrillic letters at a rate of about 10 letters per minute. For instance, I eventually parsed "Петрунин". –  Greg Kuperberg Dec 10 '09 at 3:08
Maybe you are right that it's just time to establish the counterexample. For one thing, if you assume that the center of the shape M (the point that was the center before the circle was dented) is a fixed point, then already many crumpled images of M fit inside M. That was the proof for a circle; it still has a lot of strength to it. –  Greg Kuperberg Dec 10 '09 at 6:55

Edit: I could not make the second step work :-(

Here goes my attempt: assume $F$ is a bounded connected closed convex set which is not a round disc. We will like to show that there is a distance non-increasing map $f:\mathbb{R}^2 \to \mathbb{R}^2$ such that $f(F)$ is not congruent to any subset of $F$.

Step 1: There is a closed disc $D$ which satisfies the following properties:
(1) $D \not \supseteq F$, and
(2) for all subset $S$ of $D$, if $S \supsetneqq D \cap F$, then $S$ is not congruent to any subset of $F$.
To see it: start with any disc $D_0$ with diameter $d$ smaller than the diameter of $F$. If there exists a subset $S_0$ of $D_0$ such that $S_0 \supsetneqq D_0 \cap F$ but $S_0$ is congruent to a subset $S_1$ of $F$, then moving $D_0$ (i.e. translating the center of $D_0$) by appropriate amount, we obtain a disc $D_1$ of diameter $d$ which contains $S_1$. If $D_1$ does not satisfy (2), then repeat this process. Because $F$ is bounded and area of $D \cap F$ is an increasing sequence, there can be only two possibilities: either
(a) there exists a disc of diameter $d$ which satisfies (2), or
(b) $F$ contains a disc of diameter $d$.
Since $F$ is not a disc, there exists $d <$ diameter($F$) such that (b) does not hold for $d$. Then by the above argument, (a) is true, which completes step 1.

Step 2: Let $D$ be as in step 1. WLOG assume $D$ is centered at the origin. Construct an area non-increasing map $f: \mathbb{R}^2 \to \mathbb{R}^2$ such that $f$ is identity on $D$ and $f$ maps a point of $F \setminus D$ to $D \setminus F$. There has to be a better way to see it than what I propose now, but here we go. Let me first describe the idea: let $P$ be one of the "corner" points of $D \cap F$ such that there are points of $F$ near $P$ and outside $D$. Pick such a point $Q$ and construct $f$ in a way that $f$ rotates $Q$ a bit and then retracts into $D$. The point is that you would want to use retraction to offset the increase of distance caused by the (non-uniform) rotation.

On a second thought, I will stop now and go get some food, and try to have the construction typed in the meantime. But most probably someone will have a better idea by the time I get back.

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Another tact; consider the set of foldings of our convex shape. Since the fold axis partitions the perimeter into two parts, and by convexity both have finite length, so one of them is at least half the length of the total perimeter.

edit: this needs fixing/to be made more precise --- see comments below. end edit

I claim * (without proof) that the longer part determines the congruence of the image with a subset of $F$, up to a Euclidean symmetry of $F$; whence w.l.o.g., we may assume the folding map fixes the longer part. Then we clearly require that the shorter part be mapped into $F$ --- in fact, it must be mapped to the intersection of $F$ with the half-plane containing the longer part.

Since we are interested in all such foldings, we may consider in particular the foldings that exactly bisect the perimeter; but then by oposite containments, such a folding must be a congruence of the two parts, and hence the folding axis is in fact a symmetry axis of $F$. An intermediate value theorem argument shows that there is such a folding map with axis in any direction, so $F$ has at least a circular group of symmetries; that is, $F$ is a circle.

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so, obviously, the unproved claim is the now-missing ingredient; but I rather think it's obvious. –  some guy on the street Nov 30 '09 at 1:10
Your claim is false. Take an isosceles right triangle and fold it along a segment passing through the hypotenuse that is parallel to one side and nearly bisects the other. –  Richard Kent Nov 30 '09 at 1:46
this suggests that I should talk about turning angles instead of arclength. –  some guy on the street Nov 30 '09 at 1:53
Oh, nevermind. You're clearly assuming that F has the desired property. Sorry. –  Richard Kent Nov 30 '09 at 1:54
OK, strictly speaking that doesn't resolve the "up to symmetry" part of the claim ... }:( –  some guy on the street Nov 30 '09 at 1:56