Mixing
Internal polygon formed by drawing diagonals in a regular polygon
$begingroup$
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).
http://intermath.coe.uga.edu/tweb/gwin1-01/luce/Actvty3/7030_activity_3_files/image005.gif
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
edited Jan 22 at 8:15
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).
http://intermath.coe.uga.edu/tweb/gwin1-01/luce/Actvty3/7030_activity_3_files/image005.gif
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
edited Jan 22 at 8:15
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).
http://intermath.coe.uga.edu/tweb/gwin1-01/luce/Actvty3/7030_activity_3_files/image005.gif
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
edited Jan 22 at 8:15
Glorfindel
3,42981830
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
$endgroup$
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).
http://intermath.coe.uga.edu/tweb/gwin1-01/luce/Actvty3/7030_activity_3_files/image005.gif
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
geometry trigonometry euclidean-geometry polygons golden-ratio
edited Jan 22 at 8:15
Glorfindel
3,42981830
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
edited Jan 22 at 8:15
Glorfindel
3,42981830
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
edited Jan 22 at 8:15
Glorfindel
3,42981830
edited Jan 22 at 8:15
Glorfindel
3,42981830
edited Jan 22 at 8:15
Glorfindel
3,42981830
3,42981830
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
asked Sep 17 '13 at 16:56
user2602740user2602740
15815
15815
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$left| frac{ omega^1+omega^{-1}}{omega^{frac{1}{2}}+omega^frac{1}{2}} right|^2 = ...$
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 $frac{omega^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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f496559%2finternal-polygon-formed-by-drawing-diagonals-in-a-regular-polygon%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$left| frac{ omega^1+omega^{-1}}{omega^{frac{1}{2}}+omega^frac{1}{2}} right|^2 = ...$
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 $frac{omega^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
$begingroup$
