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Does a circle give the largest smallest distance between two points that bisecting the perimeter
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The title may seem a bit confusing, let's use math notation.
Let $c:mathbbRto mathbbR^2$ be a simple closed curve parametrized by length. A pair of points on the curve that bisect the perimeter means any pair in the form $(c(t),c(t+fracL2))$ where $L$ is the total length of $c$. Now I am interested in the smallest distance between those pairs, i.e. $$inf_tin[0,L/2)|c(t)-c(t+L/2)|.$$
Let's call this smallest bisection length.
Now it seems that fixing $L$, the shape that maximizes the smallest bisection length is the circle with radius $L/(2pi)$. So, is this true?
Any help will be appreciated! Another question may be related to this is what shape will give the maximum area given a fixed perimeter, whose answer is a circle.
geometry geometric-topology combinatorial-geometry
asked Aug 1 at 1:34
Alton Wang
62
 |Â
show 1 more comment
up vote
1
down vote
favorite
The title may seem a bit confusing, let's use math notation.
Let $c:mathbbRto mathbbR^2$ be a simple closed curve parametrized by length. A pair of points on the curve that bisect the perimeter means any pair in the form $(c(t),c(t+fracL2))$ where $L$ is the total length of $c$. Now I am interested in the smallest distance between those pairs, i.e. $$inf_tin[0,L/2)|c(t)-c(t+L/2)|.$$
Let's call this smallest bisection length.
Now it seems that fixing $L$, the shape that maximizes the smallest bisection length is the circle with radius $L/(2pi)$. So, is this true?
Any help will be appreciated! Another question may be related to this is what shape will give the maximum area given a fixed perimeter, whose answer is a circle.
geometry geometric-topology combinatorial-geometry
asked Aug 1 at 1:34
Alton Wang
62
1
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The title may seem a bit confusing, let's use math notation.
Let $c:mathbbRto mathbbR^2$ be a simple closed curve parametrized by length. A pair of points on the curve that bisect the perimeter means any pair in the form $(c(t),c(t+fracL2))$ where $L$ is the total length of $c$. Now I am interested in the smallest distance between those pairs, i.e. $$inf_tin[0,L/2)|c(t)-c(t+L/2)|.$$
Let's call this smallest bisection length.
Now it seems that fixing $L$, the shape that maximizes the smallest bisection length is the circle with radius $L/(2pi)$. So, is this true?
Any help will be appreciated! Another question may be related to this is what shape will give the maximum area given a fixed perimeter, whose answer is a circle.
geometry geometric-topology combinatorial-geometry
asked Aug 1 at 1:34
Alton Wang
62
The title may seem a bit confusing, let's use math notation.
Let $c:mathbbRto mathbbR^2$ be a simple closed curve parametrized by length. A pair of points on the curve that bisect the perimeter means any pair in the form $(c(t),c(t+fracL2))$ where $L$ is the total length of $c$. Now I am interested in the smallest distance between those pairs, i.e. $$inf_tin[0,L/2)|c(t)-c(t+L/2)|.$$
Let's call this smallest bisection length.
Now it seems that fixing $L$, the shape that maximizes the smallest bisection length is the circle with radius $L/(2pi)$. So, is this true?
Any help will be appreciated! Another question may be related to this is what shape will give the maximum area given a fixed perimeter, whose answer is a circle.
geometry geometric-topology combinatorial-geometry
asked Aug 1 at 1:34
Alton Wang
62
edited Aug 1 at 2:40
asked Aug 1 at 1:34
Alton Wang
62
asked Aug 1 at 1:34
Alton Wang
62
asked Aug 1 at 1:34
Alton Wang
62
62
1
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50
 |Â
show 1 more comment
1
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50
1
1
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50
 |Â
show 1 more comment
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1
That can't be. You can take the rope of length $L$ and cram it inside a disc of radius $epsilon$ still being a simple curve, for every $epsilon >0$.
– JessicaMcRae
Aug 1 at 1:56
Maybe it's a circle if the curve is required to enclose a convex region?
– Carl Schildkraut
Aug 1 at 2:15
@JessicaMcRae, Thank you. You are totally right about this question. I confused myself before and wrote "find the shape of smallest largest bisection length". In fact the correct version should be "find the shape of largest smallest bisection length". I am really sorry about that. Do you have any thought on the problem now?
– Alton Wang
Aug 1 at 2:43
Perhaps use "the smallest distance $D$ between those pairs, i.e. $$D = inf_tin[0,L/2)|c(t)-c(t+L/2)|$$", and "the shape that maximizes $D/L$"? In the case of a circle, $D$ is a constant (diameter of the circle), and $D/L = 1/pi$.
– Nominal Animal
Aug 1 at 9:20
@NominalAnimal Yes. You could also consider the shape that maximize the ratio D/L, but here I choose to fix L and maximize D, which should be equivalent.
– Alton Wang
Aug 1 at 16:50