Ttdfjt

Why will we fail creating a self sustaining off world colony?

Would skyscrapers tip over if people fell sideways?

How do I tell my girlfriend she's been buying me books by the wrong author for the last nine months?

Hard for me to understand one tip written in "The as-if rule" of cppreference

Russian equivalents of 能骗就骗 (if you can cheat, then cheat)

Customs and immigration on a USA-UK-Sweden flight itinerary

Early 2000s movie about time travel, protagonist travels back to save girlfriend, then into multiple points in future

Two palindromes are not enough

What was the point of separating stdout and stderr?

The alcoholic village festival

Is it OK to say "The situation is pregnant with a crisis"?

Could you fall off a planet if it was being accelerated by engines?

Does it make sense to (partially) create a conlang that you don't intend to actually use in the story?

Installed software from source, how to say yum not to install it from package?

What verb for taking advantage fits in "I don't want to ________ on the friendship"?

Why is my 401k manager recommending me to save more?

Could all three Gorgons turn people to stone, or just Medusa?

Perform mirror symmetry transformation of 3D model (in OBJ)

What are the children of two Muggle-borns called?

Dynamic Sql Query - how to add an int to the code?

Position representation of spin states and spin operators

Why was Pan Am Flight 103 flying over Lockerbie?

How soon after takeoff can you recline your airplane seat?

How does mmorpg store data?

Do equal angles necessarily mean a polygon is regular?

Conditions on polygon to ensure equal interior angles or opposite sidesA formula for the exterior angles of an irregular polygon, given interior angles?Regular polygon Interior anglesIf all the 4 sides of a quadrilateral are equal but only 3 of its angles are equal. Is it a square?If I have an $n$-gon, for odd $ngeq5$ with all angles equal to the angles of the regular $n$-gon, is my $n$-gon necessarily regular?Interior angles of irregular polygonn-sided regular polygonIs it possible for a shape to have equal angles but not equal sides/ vice versa?Shape of the limiting polygonIf a quadrilateral has a pair of equal opposite sides, and a pair of equal opposite angles, then is it necessarily a parallelogram?

18

2

$begingroup$

In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.

geometry

share|cite|improve this question

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

$endgroup$

• 
  
  
  

  
  50
  

  
  
  
  
  
  

  
  $begingroup$
  What about a rectangle?
  $endgroup$
  – Peter Foreman
  Jun 21 at 15:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, what an obvious example I missed! Of course, now I must ask are there any others, an infinite number?
  $endgroup$
  – Jamminermit
  Jun 21 at 15:27
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Any shape with an even number of sides can be made regular and then 'stretched' parallel to one of the sides to create an irregular shape with each angle the same.
  $endgroup$
  – Peter Foreman
  Jun 21 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a regular polygon. Move one side so it still runs parallel to the original (then another side etc). The equilateral triangle is a special case.
  $endgroup$
  – Mark Bennet
  Jun 21 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is, in my opinion, another reason why the triangle is special. Of course in hindsight it's not very surprising, but at least it's something that beyond triangles we need more criteria to assure regularity. In this sense it would seem that there are more $n$-gons (with $n>3$) than triangles, for each $n.$
  $endgroup$
  – Allawonder
  Jun 21 at 16:32
  

  
  
  
  

 | 
show 3 more comments

18

2

$begingroup$

In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.

geometry

share|cite|improve this question

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

$endgroup$

• 
  
  
  

  
  50
  

  
  
  
  
  
  

  
  $begingroup$
  What about a rectangle?
  $endgroup$
  – Peter Foreman
  Jun 21 at 15:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, what an obvious example I missed! Of course, now I must ask are there any others, an infinite number?
  $endgroup$
  – Jamminermit
  Jun 21 at 15:27
  

  
  
  
  


• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Any shape with an even number of sides can be made regular and then 'stretched' parallel to one of the sides to create an irregular shape with each angle the same.
  $endgroup$
  – Peter Foreman
  Jun 21 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Draw a regular polygon. Move one side so it still runs parallel to the original (then another side etc). The equilateral triangle is a special case.
  $endgroup$
  – Mark Bennet
  Jun 21 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is, in my opinion, another reason why the triangle is special. Of course in hindsight it's not very surprising, but at least it's something that beyond triangles we need more criteria to assure regularity. In this sense it would seem that there are more $n$-gons (with $n>3$) than triangles, for each $n.$
  $endgroup$
  – Allawonder
  Jun 21 at 16:32
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.

geometry

share|cite|improve this question

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

$endgroup$

share|cite|improve this question

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

share|cite|improve this question

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

asked Jun 21 at 15:20

JamminermitJamminermit

3581313 bronze badges

3 Answers
3

active

oldest

votes

30

$begingroup$

Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.

The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.

share|cite|improve this answer

edited Jun 21 at 16:36

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NateEldredge Done. Note convexity not required.
  $endgroup$
  – Ethan Bolker
  Jun 21 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't need this CMS-1500 claim form any more...
  $endgroup$
  – Marco13
  Jun 22 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 I think you posted your comment on the wrong site ...
  $endgroup$
  – Ethan Bolker
  Jun 22 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It referred to the reverse of the page that your image was scanned from, but true: It certainly wasn't "on topic", in this regard...
  $endgroup$
  – Marco13
  Jun 22 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 Now I understand. Recycled paper.
  $endgroup$
  – Ethan Bolker
  Jun 22 at 19:40
  

  
  
  
  

add a comment | 

43

$begingroup$

Here are four pentagons all with interior angles of $108^circ$. Only the largest is regular. The generalization to any regular polygon should be clear.

share|cite|improve this answer

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Only polygon where this not works is a regular triangle.
  $endgroup$
  – M. Winter
  Jun 22 at 12:34
  

  
  
  
  

add a comment | 

0

$begingroup$

The side lengths need not be equal. It is amenable to a vector representation.

If segments of length of an isotropic force set $F_i$ acts on a particle ( components $F_xi,F_yi$ of consecutive positive integers $i$ ) then the vectors sum up to zero:

$$ Sigma_i^n F_x=0 $$
$$ Sigma_i^n F_y=0 $$

Each vector joins tail to head of vector and the force polygon comes back to point of start after all $n$ rotations of $dfrac2 pin$.

So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.

share|cite|improve this answer

edited Jun 25 at 15:13

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

$endgroup$

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3269812%2fdo-equal-angles-necessarily-mean-a-polygon-is-regular%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

30

$begingroup$

Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.

The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.

share|cite|improve this answer

edited Jun 21 at 16:36

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NateEldredge Done. Note convexity not required.
  $endgroup$
  – Ethan Bolker
  Jun 21 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't need this CMS-1500 claim form any more...
  $endgroup$
  – Marco13
  Jun 22 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 I think you posted your comment on the wrong site ...
  $endgroup$
  – Ethan Bolker
  Jun 22 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It referred to the reverse of the page that your image was scanned from, but true: It certainly wasn't "on topic", in this regard...
  $endgroup$
  – Marco13
  Jun 22 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 Now I understand. Recycled paper.
  $endgroup$
  – Ethan Bolker
  Jun 22 at 19:40
  

  
  
  
  

add a comment | 

30

$begingroup$

Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.

The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.

share|cite|improve this answer

edited Jun 21 at 16:36

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @NateEldredge Done. Note convexity not required.
  $endgroup$
  – Ethan Bolker
  Jun 21 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I hope you don't need this CMS-1500 claim form any more...
  $endgroup$
  – Marco13
  Jun 22 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 I think you posted your comment on the wrong site ...
  $endgroup$
  – Ethan Bolker
  Jun 22 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It referred to the reverse of the page that your image was scanned from, but true: It certainly wasn't "on topic", in this regard...
  $endgroup$
  – Marco13
  Jun 22 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Marco13 Now I understand. Recycled paper.
  $endgroup$
  – Ethan Bolker
  Jun 22 at 19:40
  

  
  
  
  

add a comment | 

$begingroup$

Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.

The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.

share|cite|improve this answer

edited Jun 21 at 16:36

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 21 at 16:36

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

answered Jun 21 at 15:30

Ethan BolkerEthan Bolker

51.6k55 gold badges6161 silver badges130130 bronze badges

43

$begingroup$

Here are four pentagons all with interior angles of $108^circ$. Only the largest is regular. The generalization to any regular polygon should be clear.

share|cite|improve this answer

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Only polygon where this not works is a regular triangle.
  $endgroup$
  – M. Winter
  Jun 22 at 12:34
  

  
  
  
  

add a comment | 

43

$begingroup$

Here are four pentagons all with interior angles of $108^circ$. Only the largest is regular. The generalization to any regular polygon should be clear.

share|cite|improve this answer

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Only polygon where this not works is a regular triangle.
  $endgroup$
  – M. Winter
  Jun 22 at 12:34
  

  
  
  
  

add a comment | 

$begingroup$

Here are four pentagons all with interior angles of $108^circ$. Only the largest is regular. The generalization to any regular polygon should be clear.

share|cite|improve this answer

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

$endgroup$

share|cite|improve this answer

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

answered Jun 21 at 16:11

steven gregorysteven gregory

19.4k33 gold badges2828 silver badges6161 bronze badges

0

$begingroup$

The side lengths need not be equal. It is amenable to a vector representation.

If segments of length of an isotropic force set $F_i$ acts on a particle ( components $F_xi,F_yi$ of consecutive positive integers $i$ ) then the vectors sum up to zero:

$$ Sigma_i^n F_x=0 $$
$$ Sigma_i^n F_y=0 $$

Each vector joins tail to head of vector and the force polygon comes back to point of start after all $n$ rotations of $dfrac2 pin$.

So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.

share|cite|improve this answer

edited Jun 25 at 15:13

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

The side lengths need not be equal. It is amenable to a vector representation.

If segments of length of an isotropic force set $F_i$ acts on a particle ( components $F_xi,F_yi$ of consecutive positive integers $i$ ) then the vectors sum up to zero:

$$ Sigma_i^n F_x=0 $$
$$ Sigma_i^n F_y=0 $$

Each vector joins tail to head of vector and the force polygon comes back to point of start after all $n$ rotations of $dfrac2 pin$.

So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.

share|cite|improve this answer

edited Jun 25 at 15:13

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

$endgroup$

add a comment | 

$begingroup$

The side lengths need not be equal. It is amenable to a vector representation.

If segments of length of an isotropic force set $F_i$ acts on a particle ( components $F_xi,F_yi$ of consecutive positive integers $i$ ) then the vectors sum up to zero:

$$ Sigma_i^n F_x=0 $$
$$ Sigma_i^n F_y=0 $$

Each vector joins tail to head of vector and the force polygon comes back to point of start after all $n$ rotations of $dfrac2 pin$.

So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.

share|cite|improve this answer

edited Jun 25 at 15:13

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 25 at 15:13

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges

answered Jun 25 at 15:05

NarasimhamNarasimham

21.7k66 gold badges2222 silver badges5858 bronze badges