Maximum surface area of polygons with sequential side lengths. [closed]
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What is the maximum area of a polygon with sides of lengths $1,2,3,ldots,N$?
Intuition tells me the polygon must be inscribed in a regular polygon with $1+2+3+ cdots +N$ sides.
What would be the limit of the surface area of a polygon with $N to infty$ if the series converge?
geometry limits polygons
edited Jan 2 at 1:34
David G. Stork
11k41432
asked Jan 2 at 1:24
BartusBartus
33
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closed as off-topic by Saad, KReiser, Leucippus, clathratus, Paul Frost Jan 2 at 11:16
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." – Saad, KReiser, Leucippus, clathratus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 2 more comments
$begingroup$
What is the maximum area of a polygon with sides of lengths $1,2,3,ldots,N$?
Intuition tells me the polygon must be inscribed in a regular polygon with $1+2+3+ cdots +N$ sides.
What would be the limit of the surface area of a polygon with $N to infty$ if the series converge?
geometry limits polygons
edited Jan 2 at 1:34
David G. Stork
11k41432
asked Jan 2 at 1:24
BartusBartus
33
$endgroup$
closed as off-topic by Saad, KReiser, Leucippus, clathratus, Paul Frost Jan 2 at 11:16
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." – Saad, KReiser, Leucippus, clathratus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
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– SmileyCraft
Jan 2 at 1:30
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
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@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
1
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43
|
show 2 more comments
$begingroup$
What is the maximum area of a polygon with sides of lengths $1,2,3,ldots,N$?
Intuition tells me the polygon must be inscribed in a regular polygon with $1+2+3+ cdots +N$ sides.
What would be the limit of the surface area of a polygon with $N to infty$ if the series converge?
geometry limits polygons
edited Jan 2 at 1:34
David G. Stork
11k41432
asked Jan 2 at 1:24
BartusBartus
33
$endgroup$
What is the maximum area of a polygon with sides of lengths $1,2,3,ldots,N$?
Intuition tells me the polygon must be inscribed in a regular polygon with $1+2+3+ cdots +N$ sides.
What would be the limit of the surface area of a polygon with $N to infty$ if the series converge?
geometry limits polygons
geometry limits polygons
edited Jan 2 at 1:34
David G. Stork
11k41432
asked Jan 2 at 1:24
BartusBartus
33
edited Jan 2 at 1:34
David G. Stork
11k41432
asked Jan 2 at 1:24
BartusBartus
33
edited Jan 2 at 1:34
David G. Stork
11k41432
edited Jan 2 at 1:34
David G. Stork
11k41432
edited Jan 2 at 1:34
David G. Stork
11k41432
11k41432
asked Jan 2 at 1:24
BartusBartus
33
asked Jan 2 at 1:24
BartusBartus
33
asked Jan 2 at 1:24
BartusBartus
33
33
closed as off-topic by Saad, KReiser, Leucippus, clathratus, Paul Frost Jan 2 at 11:16
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." – Saad, KReiser, Leucippus, clathratus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, KReiser, Leucippus, clathratus, Paul Frost Jan 2 at 11:16
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." – Saad, KReiser, Leucippus, clathratus, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:30
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
$begingroup$
@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
1
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43
|
show 2 more comments
1
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:30
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
$begingroup$
@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
1
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43
1
1
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:30
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:30
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
$begingroup$
@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
$begingroup$
@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
1
1
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43
|
show 2 more comments
1 Answer
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$begingroup$
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $sqrt{r^2-1/2},sqrt{r^2-2/2},...,sqrt{r^2-n/2}$. So the total area is $1sqrt{r^2-1/2}+2sqrt{r^2-2/2}+...+nsqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2pi$. So we have $sin^{-1}(frac{1/2}{r})+sin^{-1}(frac{2/2}{r})+...+sin^{-1}(frac{n/2}{r})=2pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $sqrt{4!}=2sqrt{6}approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $sqrt{r^2-1/2},sqrt{r^2-2/2},...,sqrt{r^2-n/2}$. So the total area is $1sqrt{r^2-1/2}+2sqrt{r^2-2/2}+...+nsqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2pi$. So we have $sin^{-1}(frac{1/2}{r})+sin^{-1}(frac{2/2}{r})+...+sin^{-1}(frac{n/2}{r})=2pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $sqrt{4!}=2sqrt{6}approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
$endgroup$
add a comment |
$begingroup$
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $sqrt{r^2-1/2},sqrt{r^2-2/2},...,sqrt{r^2-n/2}$. So the total area is $1sqrt{r^2-1/2}+2sqrt{r^2-2/2}+...+nsqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2pi$. So we have $sin^{-1}(frac{1/2}{r})+sin^{-1}(frac{2/2}{r})+...+sin^{-1}(frac{n/2}{r})=2pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $sqrt{4!}=2sqrt{6}approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
$endgroup$
add a comment |
$begingroup$
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $sqrt{r^2-1/2},sqrt{r^2-2/2},...,sqrt{r^2-n/2}$. So the total area is $1sqrt{r^2-1/2}+2sqrt{r^2-2/2}+...+nsqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2pi$. So we have $sin^{-1}(frac{1/2}{r})+sin^{-1}(frac{2/2}{r})+...+sin^{-1}(frac{n/2}{r})=2pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $sqrt{4!}=2sqrt{6}approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
$endgroup$
As I mentioned in the comments, you just need to place the vertices on a circle. So if the radius $r$ is known, then the area is the sum of the areas of a bunch of isosceles triangles with base length $1,2,3,...,n$ and slanted length $r$. So they have height $sqrt{r^2-1/2},sqrt{r^2-2/2},...,sqrt{r^2-n/2}$. So the total area is $1sqrt{r^2-1/2}+2sqrt{r^2-2/2}+...+nsqrt{r^2-n/2}$.
To find the radius $r$ note that the sum of the angles of the isosceles triangles must be $2pi$. So we have $sin^{-1}(frac{1/2}{r})+sin^{-1}(frac{2/2}{r})+...+sin^{-1}(frac{n/2}{r})=2pi$. I don't know if there is an analytic or geometric way to solve for $r$, but computationally you can simply do a binary search.
EDIT: By Brahmagupta's formula at least for $n=4$ we can calculate the maximal area is $sqrt{(s-1)(s-2)(s-3)(s-4)}$ where $s=(1+2+3+4)/2=5$, so the area is $sqrt{4!}=2sqrt{6}approx4.9$.
EDIT2: Apparently there are formulas like Brahmagupta's for up to $8$ sides. However, more than $8$ sides appears to be a currently unsolved problem. Hence, I expect you will not get a better answer than binary search.
However, as I mentioned in the comments $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$, so you can use this to approximate $A(n)$. However, binary search definitely allows more precise approximations.
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
edited Jan 2 at 2:19
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
answered Jan 2 at 2:02
SmileyCraftSmileyCraft
3,491518
3,491518
add a comment |
add a comment |
1
$begingroup$
I remember that solving for marimum area polygon given the side lengths simply means the polygon vertices lie on a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:30
$begingroup$
@SmileyCraft You are correct. For example, Maximum Polygon Area states this, along with 2 proofs.
$endgroup$
– John Omielan
Jan 2 at 1:32
$begingroup$
Also note this question is not apparently a duplicate, but it is quite closely related to, Area of irregular polygon using side edges.
$endgroup$
– John Omielan
Jan 2 at 1:38
$begingroup$
@SmileyCraft: Perhaps... but in what sequential order? Ahh... I see that it doesn't matter as the area of the external "caps" for each segment are unchanged.
$endgroup$
– David G. Stork
Jan 2 at 1:41
1
$begingroup$
Let $A(n)$ denote the maximum are of a polygon with sides of lengths $1,2,3,...,n$ and let $S(n)=1+2+3+...+n=n(n+1)/2$. Then at least we can say $lim_{ntoinfty}A(n)/(S(n))^2=1/pi$. This because we approximate area and circumference of a circle.
$endgroup$
– SmileyCraft
Jan 2 at 1:43