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?










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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.












  • 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
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$



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.
    $endgroup$
    – 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














0












0








0


1



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 1:34









David G. Stork

11k41432




11k41432










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














  • 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










1 Answer
1






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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.






share|cite|improve this answer











$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 2:19

























        answered Jan 2 at 2:02









        SmileyCraftSmileyCraft

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        3,491518















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