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Which polygon is the one?
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There is a unit radius circle and you are trying to form a polygon where its all edges located on the circle, such as below:
What is the biggest value of sum of squares of sides of such a polygon possible?
mathematics geometry
asked 1 hour ago
Oray
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up vote
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There is a unit radius circle and you are trying to form a polygon where its all edges located on the circle, such as below:
What is the biggest value of sum of squares of sides of such a polygon possible?
mathematics geometry
asked 1 hour ago
Oray
15.1k435145
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There is a unit radius circle and you are trying to form a polygon where its all edges located on the circle, such as below:
What is the biggest value of sum of squares of sides of such a polygon possible?
mathematics geometry
asked 1 hour ago
Oray
15.1k435145
There is a unit radius circle and you are trying to form a polygon where its all edges located on the circle, such as below:
What is the biggest value of sum of squares of sides of such a polygon possible?
mathematics geometry
mathematics geometry
asked 1 hour ago
Oray
15.1k435145
asked 1 hour ago
Oray
15.1k435145
asked 1 hour ago
Oray
15.1k435145
asked 1 hour ago
Oray
15.1k435145
asked 1 hour ago
Oray
15.1k435145
15.1k435145
add a comment |Â
add a comment |Â
2 Answers
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up vote
4
down vote
If you have an $n$-sided polygon then each edge has length $2sinfracpi n$ so we are asking what $n$ maximizes $2nsin^2!fracpi n$. Writing $theta=fracpi n$, this is the same as maximizing $S=fracsin^2thetatheta$, where $theta$ is allowed to take a certain set of values the largest of which is
$fracpi3$.
For the moment, allow $theta$ to vary continuously between $0$ and $fracpi3$. The derivative of $S$ is $frac2thetasinthetacostheta-sin^2!thetatheta^2$ whose numerator is $sinthetacostheta,(2theta-tantheta)$; on $0<theta<fracpi3$ it's not hard to see that both factors are positive. Hence the biggest $S$ can be on this range is when
$theta=fracpi3$.
In reality $theta$ can't take every possible value between $0$ and $fracpi3$. But
it can be $fracpi3$, and we have just found that no other value in that range makes $S$ larger. So $fracpi3$ is best possible; our polygon should be a triangle.
I can't help thinking that there should be a more immediate proof. For instance, observe that
when you go from $n$ to $2n$ sides, you do it by replacing each side with two, and the new triangles you've added on are all obtuse, from which it's easy to see that the sum of squares goes down. Maybe something in this spirit can be made more general, but right now I'm not seeing how.
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
add a comment |Â
up vote
1
down vote
The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
answered 56 mins ago
tmpearce
1,528510
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
If you have an $n$-sided polygon then each edge has length $2sinfracpi n$ so we are asking what $n$ maximizes $2nsin^2!fracpi n$. Writing $theta=fracpi n$, this is the same as maximizing $S=fracsin^2thetatheta$, where $theta$ is allowed to take a certain set of values the largest of which is
$fracpi3$.
For the moment, allow $theta$ to vary continuously between $0$ and $fracpi3$. The derivative of $S$ is $frac2thetasinthetacostheta-sin^2!thetatheta^2$ whose numerator is $sinthetacostheta,(2theta-tantheta)$; on $0<theta<fracpi3$ it's not hard to see that both factors are positive. Hence the biggest $S$ can be on this range is when
$theta=fracpi3$.
In reality $theta$ can't take every possible value between $0$ and $fracpi3$. But
it can be $fracpi3$, and we have just found that no other value in that range makes $S$ larger. So $fracpi3$ is best possible; our polygon should be a triangle.
I can't help thinking that there should be a more immediate proof. For instance, observe that
when you go from $n$ to $2n$ sides, you do it by replacing each side with two, and the new triangles you've added on are all obtuse, from which it's easy to see that the sum of squares goes down. Maybe something in this spirit can be made more general, but right now I'm not seeing how.
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
add a comment |Â
up vote
4
down vote
If you have an $n$-sided polygon then each edge has length $2sinfracpi n$ so we are asking what $n$ maximizes $2nsin^2!fracpi n$. Writing $theta=fracpi n$, this is the same as maximizing $S=fracsin^2thetatheta$, where $theta$ is allowed to take a certain set of values the largest of which is
$fracpi3$.
For the moment, allow $theta$ to vary continuously between $0$ and $fracpi3$. The derivative of $S$ is $frac2thetasinthetacostheta-sin^2!thetatheta^2$ whose numerator is $sinthetacostheta,(2theta-tantheta)$; on $0<theta<fracpi3$ it's not hard to see that both factors are positive. Hence the biggest $S$ can be on this range is when
$theta=fracpi3$.
In reality $theta$ can't take every possible value between $0$ and $fracpi3$. But
it can be $fracpi3$, and we have just found that no other value in that range makes $S$ larger. So $fracpi3$ is best possible; our polygon should be a triangle.
I can't help thinking that there should be a more immediate proof. For instance, observe that
when you go from $n$ to $2n$ sides, you do it by replacing each side with two, and the new triangles you've added on are all obtuse, from which it's easy to see that the sum of squares goes down. Maybe something in this spirit can be made more general, but right now I'm not seeing how.
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
add a comment |Â
up vote
4
down vote
up vote
4
down vote
If you have an $n$-sided polygon then each edge has length $2sinfracpi n$ so we are asking what $n$ maximizes $2nsin^2!fracpi n$. Writing $theta=fracpi n$, this is the same as maximizing $S=fracsin^2thetatheta$, where $theta$ is allowed to take a certain set of values the largest of which is
$fracpi3$.
For the moment, allow $theta$ to vary continuously between $0$ and $fracpi3$. The derivative of $S$ is $frac2thetasinthetacostheta-sin^2!thetatheta^2$ whose numerator is $sinthetacostheta,(2theta-tantheta)$; on $0<theta<fracpi3$ it's not hard to see that both factors are positive. Hence the biggest $S$ can be on this range is when
$theta=fracpi3$.
In reality $theta$ can't take every possible value between $0$ and $fracpi3$. But
it can be $fracpi3$, and we have just found that no other value in that range makes $S$ larger. So $fracpi3$ is best possible; our polygon should be a triangle.
I can't help thinking that there should be a more immediate proof. For instance, observe that
when you go from $n$ to $2n$ sides, you do it by replacing each side with two, and the new triangles you've added on are all obtuse, from which it's easy to see that the sum of squares goes down. Maybe something in this spirit can be made more general, but right now I'm not seeing how.
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
If you have an $n$-sided polygon then each edge has length $2sinfracpi n$ so we are asking what $n$ maximizes $2nsin^2!fracpi n$. Writing $theta=fracpi n$, this is the same as maximizing $S=fracsin^2thetatheta$, where $theta$ is allowed to take a certain set of values the largest of which is
$fracpi3$.
For the moment, allow $theta$ to vary continuously between $0$ and $fracpi3$. The derivative of $S$ is $frac2thetasinthetacostheta-sin^2!thetatheta^2$ whose numerator is $sinthetacostheta,(2theta-tantheta)$; on $0<theta<fracpi3$ it's not hard to see that both factors are positive. Hence the biggest $S$ can be on this range is when
$theta=fracpi3$.
In reality $theta$ can't take every possible value between $0$ and $fracpi3$. But
it can be $fracpi3$, and we have just found that no other value in that range makes $S$ larger. So $fracpi3$ is best possible; our polygon should be a triangle.
I can't help thinking that there should be a more immediate proof. For instance, observe that
when you go from $n$ to $2n$ sides, you do it by replacing each side with two, and the new triangles you've added on are all obtuse, from which it's easy to see that the sum of squares goes down. Maybe something in this spirit can be made more general, but right now I'm not seeing how.
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
answered 28 mins ago
Gareth McCaughan♦
56.3k3140218
56.3k3140218
add a comment |Â
add a comment |Â
up vote
1
down vote
The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
answered 56 mins ago
tmpearce
1,528510
add a comment |Â
up vote
1
down vote
The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
answered 56 mins ago
tmpearce
1,528510
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
answered 56 mins ago
tmpearce
1,528510
The maximum value you can get is
9, from a triangle
Because
As you increase the number of sides, the square of the sides gets smaller faster than the increase in the number
answered 56 mins ago
tmpearce
1,528510
answered 56 mins ago
tmpearce
1,528510
answered 56 mins ago
tmpearce
1,528510
answered 56 mins ago
tmpearce
1,528510
1,528510
add a comment |Â
add a comment |Â
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