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2

$begingroup$

In an n-sided (n>4) regular polygon, label the vertices 0, 1, ..., n-1. For each vertex i, draw a pair of diagonals:

• from i to (i+2) mod n and


• from i to (i-2) mod n

Question: What is the ratio of the area of the internal polygon to the external polygon?

For n=5 and n=6, this process will result in the following diagrams. For the pentagon, the ratio would be area(ONRQP)/area(CBFED).

_{(source: uga.edu)}

Background: This page has a discussion about pentagons and notes that if CB=1 (unit length) then ON=1/$phi^2$ where $phi=(1+sqrt5)/2$ is the golden ratio. Since the area of a pentagon is proportional to the square of its side length (see here), area(ONRQP)/area(CBFED) = 1/$phi^4$ ~ 0.146.

But I don't know how to generalize it to a regular n-gon. It seems obvious that as n increases, the ratio approaches 1 but I wonder if there is a closed form expression as a function of n.

geometry trigonometry euclidean-geometry polygons golden-ratio

share|cite|improve this question

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

$endgroup$

add a comment | 

2

$begingroup$

In an n-sided (n>4) regular polygon, label the vertices 0, 1, ..., n-1. For each vertex i, draw a pair of diagonals:

• from i to (i+2) mod n and


• from i to (i-2) mod n

Question: What is the ratio of the area of the internal polygon to the external polygon?

For n=5 and n=6, this process will result in the following diagrams. For the pentagon, the ratio would be area(ONRQP)/area(CBFED).

_{(source: uga.edu)}

Background: This page has a discussion about pentagons and notes that if CB=1 (unit length) then ON=1/$phi^2$ where $phi=(1+sqrt5)/2$ is the golden ratio. Since the area of a pentagon is proportional to the square of its side length (see here), area(ONRQP)/area(CBFED) = 1/$phi^4$ ~ 0.146.

But I don't know how to generalize it to a regular n-gon. It seems obvious that as n increases, the ratio approaches 1 but I wonder if there is a closed form expression as a function of n.

geometry trigonometry euclidean-geometry polygons golden-ratio

share|cite|improve this question

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

$endgroup$

add a comment | 

$begingroup$

In an n-sided (n>4) regular polygon, label the vertices 0, 1, ..., n-1. For each vertex i, draw a pair of diagonals:

• from i to (i+2) mod n and


• from i to (i-2) mod n

Question: What is the ratio of the area of the internal polygon to the external polygon?

For n=5 and n=6, this process will result in the following diagrams. For the pentagon, the ratio would be area(ONRQP)/area(CBFED).

_{(source: uga.edu)}

Background: This page has a discussion about pentagons and notes that if CB=1 (unit length) then ON=1/$phi^2$ where $phi=(1+sqrt5)/2$ is the golden ratio. Since the area of a pentagon is proportional to the square of its side length (see here), area(ONRQP)/area(CBFED) = 1/$phi^4$ ~ 0.146.

But I don't know how to generalize it to a regular n-gon. It seems obvious that as n increases, the ratio approaches 1 but I wonder if there is a closed form expression as a function of n.

geometry trigonometry euclidean-geometry polygons golden-ratio

share|cite|improve this question

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

$endgroup$

share|cite|improve this question

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

share|cite|improve this question

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

edited Mar 11 at 8:29

Glorfindel

3,42981830

edited Mar 11 at 8:29

Glorfindel

3,42981830

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

asked Sep 17 '13 at 16:56

user2602740user2602740

15815

1 Answer
1

active

oldest

votes

1

$begingroup$

You mentioned that the area is proportional to the the square of the side length, but it could be difficult to calculate the side lengths that are involved.

Hint: The area is proportionate to the distance from the center.

Let $ omega$ be a $n$th root of unity, and $omega^i$ be the vertices of the polygon.

What is the distance of a side of the regular polygon from the origin?

$ | frac omega^i + omega^i+1 2 |$

What is the distance of a diagonal of the regular polygon from the origin?

$ | frac omega^i + omega^i+2 2| $

Hence, the ratio of areas is

$left| frac omega^1+omega^-1omega^frac12+omega^frac12 right|^2 = ...$

share|cite|improve this answer

edited Sep 17 '13 at 19:14

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given $omega^i$ as the vertices I can see that the length of a side of the polygon is |$omega^i$ - $omega^i+1$| and a similar expression for the diagonals. I don't know how to compute the distance of a side. Also, I just don't see how that will help me compute the areas.
  $endgroup$
  – user2602740
  Sep 17 '13 at 17:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user2602740 The distance from the side to the origin is the distance from the midpoint of the side to the origin, namely $fracomega^i + omega^i+1 2 $. What about the distance from the diagonal to the origin?
  $endgroup$
  – Calvin Lin
  Sep 17 '13 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As the number of sides increases and the ratio of inner area/outer area goes to 1, we are computing $pi.$
  $endgroup$
  – Fred Kline
  Dec 26 '14 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredKline Can you please show me how we are computing pi?
  $endgroup$
  – user2602740
  May 9 '15 at 18:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ![en.wikipedia.org/wiki/… should have a little info. Scroll down to Polygon approximation to a circle
  $endgroup$
  – Fred Kline
  May 9 '15 at 18:45
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

You mentioned that the area is proportional to the the square of the side length, but it could be difficult to calculate the side lengths that are involved.

Hint: The area is proportionate to the distance from the center.

Let $ omega$ be a $n$th root of unity, and $omega^i$ be the vertices of the polygon.

What is the distance of a side of the regular polygon from the origin?

$ | frac omega^i + omega^i+1 2 |$

What is the distance of a diagonal of the regular polygon from the origin?

$ | frac omega^i + omega^i+2 2| $

Hence, the ratio of areas is

$left| frac omega^1+omega^-1omega^frac12+omega^frac12 right|^2 = ...$

share|cite|improve this answer

edited Sep 17 '13 at 19:14

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given $omega^i$ as the vertices I can see that the length of a side of the polygon is |$omega^i$ - $omega^i+1$| and a similar expression for the diagonals. I don't know how to compute the distance of a side. Also, I just don't see how that will help me compute the areas.
  $endgroup$
  – user2602740
  Sep 17 '13 at 17:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user2602740 The distance from the side to the origin is the distance from the midpoint of the side to the origin, namely $fracomega^i + omega^i+1 2 $. What about the distance from the diagonal to the origin?
  $endgroup$
  – Calvin Lin
  Sep 17 '13 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As the number of sides increases and the ratio of inner area/outer area goes to 1, we are computing $pi.$
  $endgroup$
  – Fred Kline
  Dec 26 '14 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredKline Can you please show me how we are computing pi?
  $endgroup$
  – user2602740
  May 9 '15 at 18:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ![en.wikipedia.org/wiki/… should have a little info. Scroll down to Polygon approximation to a circle
  $endgroup$
  – Fred Kline
  May 9 '15 at 18:45
  
  

  
  
  
  

add a comment | 

1

$begingroup$

You mentioned that the area is proportional to the the square of the side length, but it could be difficult to calculate the side lengths that are involved.

Hint: The area is proportionate to the distance from the center.

Let $ omega$ be a $n$th root of unity, and $omega^i$ be the vertices of the polygon.

What is the distance of a side of the regular polygon from the origin?

$ | frac omega^i + omega^i+1 2 |$

What is the distance of a diagonal of the regular polygon from the origin?

$ | frac omega^i + omega^i+2 2| $

Hence, the ratio of areas is

$left| frac omega^1+omega^-1omega^frac12+omega^frac12 right|^2 = ...$

share|cite|improve this answer

edited Sep 17 '13 at 19:14

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given $omega^i$ as the vertices I can see that the length of a side of the polygon is |$omega^i$ - $omega^i+1$| and a similar expression for the diagonals. I don't know how to compute the distance of a side. Also, I just don't see how that will help me compute the areas.
  $endgroup$
  – user2602740
  Sep 17 '13 at 17:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user2602740 The distance from the side to the origin is the distance from the midpoint of the side to the origin, namely $fracomega^i + omega^i+1 2 $. What about the distance from the diagonal to the origin?
  $endgroup$
  – Calvin Lin
  Sep 17 '13 at 19:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As the number of sides increases and the ratio of inner area/outer area goes to 1, we are computing $pi.$
  $endgroup$
  – Fred Kline
  Dec 26 '14 at 11:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FredKline Can you please show me how we are computing pi?
  $endgroup$
  – user2602740
  May 9 '15 at 18:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ![en.wikipedia.org/wiki/… should have a little info. Scroll down to Polygon approximation to a circle
  $endgroup$
  – Fred Kline
  May 9 '15 at 18:45
  
  

  
  
  
  

add a comment | 

$begingroup$

You mentioned that the area is proportional to the the square of the side length, but it could be difficult to calculate the side lengths that are involved.

Hint: The area is proportionate to the distance from the center.

Let $ omega$ be a $n$th root of unity, and $omega^i$ be the vertices of the polygon.

What is the distance of a side of the regular polygon from the origin?

$ | frac omega^i + omega^i+1 2 |$

What is the distance of a diagonal of the regular polygon from the origin?

$ | frac omega^i + omega^i+2 2| $

Hence, the ratio of areas is

$left| frac omega^1+omega^-1omega^frac12+omega^frac12 right|^2 = ...$

share|cite|improve this answer

edited Sep 17 '13 at 19:14

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

$endgroup$

share|cite|improve this answer

edited Sep 17 '13 at 19:14

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114

answered Sep 17 '13 at 17:21

Calvin LinCalvin Lin

36.3k349114