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edited Sep 12 2010 at 16:44 |
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.
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edited Mar 9 2010 at 17:35 |
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edited Dec 12 2009 at 18:03 |
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.
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edited Dec 7 2009 at 3:42 |
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edited Dec 2 2009 at 21:48 |
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. 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.) One can reformulate the propertyas following: 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.
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edited Dec 1 2009 at 18:12 |
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. 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.) One can reformulate the property as following: if for some set $G\subset\mathbb R^2$ there is a distance-non-contracting map $G\to F$ then there is an isometry 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)below. (That might be a counterexample.) 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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edited Nov 30 2009 at 3:33 |
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?
Comments:
It is easy to see that the round disk has this property. 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.) One can reformulate the property as following: if for some set $G\subset\mathbb R^2$ there is a distance-non-contracting map $G\to F$ then there is an isometry $G\to F$. (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). 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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edited Nov 30 2009 at 3:23 |
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. 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.) One can reformulate the property as following: if for some set $G\subset\mathbb R^2$ there is a distance-non-contracting map $G\to F$ then there is an isometry $G\to F$. (The equivalence follows from Kirszbraun theorem) 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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edited Nov 30 2009 at 2:31 |
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?
Comments:
It is easy to see that the round disk has this property. 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.
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edited Nov 29 2009 at 18:17 |
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. Some figures as ellipse and Reuleaux triangle are bad (see the comments below) 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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edited Nov 28 2009 at 23:01 |
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. Some figures as ellipse and Reuleaux triangle are bad (see the comments below) 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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edited Nov 28 2009 at 17:56 |
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 discdisk?
Comments:
- It is easy to see that the round disc disk has this property.
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edited Nov 28 2009 at 17:19 |
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edited Nov 28 2009 at 17:12 |
Self-enclosed figureWhen 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 disc?
Comments:
- It is easy to see that the round disc has this property.
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edited Nov 28 2009 at 16:18 |
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edited Nov 28 2009 at 4:11 |
Assume a convex figure $F\subset \mathbb R^2$ satisfy satisfies the following property: if $f:F\to \mathbb R^2$ is a distance non-expanding distance-non-increasing map then its image $f(F)$ is congruent to a subset of $F$.
Is it true that $F$ is a round disc?
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asked Nov 28 2009 at 2:57 |
Self-enclosed figure
Assume a convex figure $F\subset \mathbb R^2$ satisfy the following property: if $f:F\to \mathbb R^2$ is a distance non-expanding map then its image $f(F)$ is congruent to a subset of $F$.
Is it true that $F$ is a round disc?
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