Do the given perimeter and area corresponds to many shapes? [on hold]
I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?
geometry area plane-geometry
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
put on hold as off-topic by José Carlos Santos, jgon, KReiser, Cesareo, user91500 Jan 5 at 8:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, jgon, KReiser, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?
geometry area plane-geometry
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
put on hold as off-topic by José Carlos Santos, jgon, KReiser, Cesareo, user91500 Jan 5 at 8:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, jgon, KReiser, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
1
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57
add a comment |
I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?
geometry area plane-geometry
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
I have a perimeter P and area A of a planar shape. How to prove that there are many shapes that corresponds to those perimeter and area values?
geometry area plane-geometry
geometry area plane-geometry
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
asked Jan 4 at 15:18
Ivan BuninIvan Bunin
174111
174111
put on hold as off-topic by José Carlos Santos, jgon, KReiser, Cesareo, user91500 Jan 5 at 8:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, jgon, KReiser, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by José Carlos Santos, jgon, KReiser, Cesareo, user91500 Jan 5 at 8:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, jgon, KReiser, Cesareo, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
1
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57
add a comment |
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
1
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
1
1
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57
add a comment |
2 Answers
2
active
oldest
votes
If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $frac1{2pi}P$.
If $P^2<4A$, there is no such shape.
If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.
edited Jan 4 at 16:11
amWhy
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
add a comment |
If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)
Perhaps this proof works for what you have in mind.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $frac1{2pi}P$.
If $P^2<4A$, there is no such shape.
If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.
edited Jan 4 at 16:11
amWhy
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
add a comment |
If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $frac1{2pi}P$.
If $P^2<4A$, there is no such shape.
If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.
edited Jan 4 at 16:11
amWhy
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
add a comment |
If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $frac1{2pi}P$.
If $P^2<4A$, there is no such shape.
If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.
edited Jan 4 at 16:11
amWhy
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
If $P^2=4A$, there is only one planar shape (up to translation) that matches: the disc of radius $frac1{2pi}P$.
If $P^2<4A$, there is no such shape.
If $P^2>4A$, you may start with a disc of area $A$, which has too short perimeter. If you deform a tiny part of the perimeter outside and another part inside, you can achieve that the aea remains unchanged, while the perimeter takes on as large a value as you want. As there is a lot of freedom of choice involved in this process, the claim follows.
edited Jan 4 at 16:11
amWhy
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
edited Jan 4 at 16:11
amWhy
192k28225439
edited Jan 4 at 16:11
amWhy
192k28225439
edited Jan 4 at 16:11
amWhy
192k28225439
192k28225439
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
answered Jan 4 at 15:35
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
add a comment |
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
How it is possible to deform a tiny part of disk perimeter outside? Can you draw this?
– Ivan Bunin
Jan 4 at 15:46
1
1
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
there should be a $pi$ in there... $4pi Ale P^2$.
– robjohn♦
Jan 4 at 15:58
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@robjohn So, when $P^2>4 pi A$ there are infinity set of possible shapes?
– Ivan Bunin
Jan 4 at 16:07
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@IvanBunin: yes. If more context is added to the question, I can post an answer.
– robjohn♦
Jan 4 at 20:50
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
@robjohn The Magnus' answer looks enough for me. But if you have a more strong or powerful prove, you are welcome to present it.
– Ivan Bunin
Jan 4 at 22:04
add a comment |
If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)
Perhaps this proof works for what you have in mind.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)
Perhaps this proof works for what you have in mind.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)
Perhaps this proof works for what you have in mind.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you start with a rectangle and have an indent on it, like in shape A, then you can always move that indent to any other place and the perimeter and the area will remain the same, as in shape B (starting with an "outdent" would work too of course.)
Perhaps this proof works for what you have in mind.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 20:54
MagnusMagnus
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 4 at 20:54
MagnusMagnus
1113
answered Jan 4 at 20:54
MagnusMagnus
1113
1113
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Magnus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Is that the entire information given? Or something more specific? (Eg polygon with fixed number of sides or whatever).
– timtfj
Jan 4 at 15:34
Yes, I need a common prove.
– Ivan Bunin
Jan 4 at 15:44
1
That depends on the particular perimeter and area. Note that for all shapes, $4pitext{Area}letext{Perimeter}^2$, and if $4pitext{Area}=text{Perimeter}^2$, there is only one shape, a circular disk.
– robjohn♦
Jan 4 at 15:57