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# Re-exploration of regular n-gon polygons.

## Question

A discrete function that takes the number of sides of a regular polygon and tells you the measure of one of its inner angles. A regular triangle has three sides and its inner angle is 60 degrees. A regular quadrilateral has four sides and its inner angle is 90 degrees. A regular pentagon has five sides and its inner angle is 108 degrees.

That's a recipe for a regular pentagon right there. Draw a 108 degree angle between two segments with the same length. Then draw another 108 degree angle on the last segment. And another, and another, until the segments reconnect and you have a regular polygon with five sides. We can write a table: We can graph those values: We can also write an equation: That equation perfectly describes the discrete values in that graph. But the equation is stupid. It doesn't know it's only supposed to describe those discrete values. We can put in other values and, like a sucker, it'll give us a number, even though it isn't supposed to and even though that number won't make any sense.

Like n = 3.5. A regular polygon with 3.5 sides? No such thing. But if we throw n = 3.5 into that function, it gives us the number 77.1 degrees.

Maybe that's just gibberish, the result of pushing this function machine beyond its warranty. But maybe it isn't.

What if we tried to draw a regular 3.5-gon in the same way we did the regular 5-gon up there?

When you make the shape, ask yourself the following question: But where is the 3.5 in that shape? Maybe you see how the number 3.5 turned into the number 77.1 and how the number 77.1 turned into that _____ shape. But where is the 3.5 in that shape?

It may be helpful to see 3.5 as the rational number 7/2)

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check out what happens when you connect the vertices of the 3.5-gon. Anything?

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Just picturing it in my head, I'm guessing that the resulting shape will be a

regular first order 7-point star.

n-point stars by extending the sides of a regular n-gon. Then, extend the sides of that star to create a star of the next order. The highest order n-point star achievable is found with the formula ceiling(n/2) - 2.

(I invented the term "first order star" some time ago when I was playing around with regular stars some time ago. You get first order

Edit:

The angle of 77.1

° is found when I divide 540° (the sum of the interior angles of a 7-point star) by 7. You can find a couple of cool explanations for getting 540° here if you can't figure out how I got that.

Edit2:

7 / 2 = 3.5 So does plugging in the formula with n where n is half of any integer yield a regular n-point star?
Edited by ParaLogic
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Eugh editing time ran out just when I noticed my mistake in Edit2. So ignore that one.

Is it possible that plugging

n/2 into the equation (where n is an odd integer greater than 4) yields an n-point star? Obviously you won't get a star if you plug in an even number...

Need to test more!

Edit: Yep, it seems to work with 5/2 and 9/2 as well!

Edit2: I think I have reached a conclusion!!. (post coming up)

Edited by ParaLogic
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All right. I've cluttered up this thread very much, but this particular topic intrigued me very much. I believe my conclusion is correct, although I didn't actually prove it. (Someone else can handle that. ) I'm starting with this equation and ignoring the process outlined for creating a regular polygon right now. If n is an integer greater than 2, then y gives the measure of an interior angle of a regular n-gon, as BMAD showed. But if n can be any rational number of form a/b (where a and b are integers, of course), then it doesn't always work out.

The following statement should be true in all cases: If ceiling(a/2) - 2 ≥ b and a ≥ 5, then plugging a/b into the formula yields the measure of an interior angle of a regular (b-1)th order a-point star.

Now you may ask, what about 6/2? It's clearly going to give you a regular polygon with 3 sides, right? True, if you take that angle and apply BMAD's process for creating a regular polygon, that is exactly what will happen. That's why I'm ignoring it. But if you look at a regular 6-point star, you will see that it's just two triangles stacked on top of each other. Therefore, my statement still holds true.

So a regular pentagon has interior angles of 108°. But so does a 30-point star of order 5. Just picture the 30-point star as six pentagons stacked on top of each other (with a slight offset between each one).

You can only apply BMAD's technique to certain numbers (like 7/3 or 7/2) because those resulting stars cannot be separated into distinct shapes.

So we've got that behind us.

What if

b-1 is greater than the highest possible order of an a-point star? (See: 7/4, 11/6, etc.)

Edit: Did I take the question in the OP too far? It was pretty vague...

Edited by ParaLogic
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Looking at bonanova's post, maybe I did take the question too far...

Either way, I just realized that for numbers like 7/4 and 11/6:

They are less than or equal to 2, which produces negative or zero answers using the equation, so they don't matter.

Edited by ParaLogic
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Looking at bonanova's post, maybe I did take the question too far...

Either way, I just realized that for numbers like 7/4 and 11/6:

They are less than or equal to 2, which produces negative or zero answers using the equation, so they don't matter.

I disagree paralogic. i believe having a negative angle would just mean that the angle is found in the "opposite direction" as the convention I used to make my pentagon.

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• 0   The three preceding attachments were made with a simple geometric program using the angle measures until it completes a shape. It is there if you look for it. You can find where the n-sides occurs. In other words if i made a n=4 regular shape, i will create a square with n=4 sides. If i made a n=6 regular shape, i would make a hexagon with n=6 sides. Where is the n in these pictures? The first picture x=1.75, the second picture x=3.5, and in the last picture x=4.25.

note: i used x to denote that i am talking about any rational number since n is typically reserved for natural numbers.

second note: in my own exploration of the problem i noticed that there are asymtopes when x is an irrational number. can anyone confirm this?

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I don't like the implications of negative angles, but if you use them, then

Used in the equation:

7/1 is the same as 7/6 (except negative)
7/2 is the same as 7/5 (negative)
7/3 is the same as 7/4 (negative)

In general, using a rational number a/b gives the same answer as using a/(a-b), except negative. It's easy to prove that algebraically.

1.75 = 7/4

3.5 = 7/2

4.25 = 17/4

Using that form, you can predict the shapes.

As for irrational numbers...well I don't (want to) know.

Edited by ParaLogic
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I don't like the implications of negative angles, but if you use them, then

Used in the equation:

7/1 is the same as 7/6 (except negative)

7/2 is the same as 7/5 (negative)

7/3 is the same as 7/4 (negative)

In general, using a rational number a/b gives the same answer as using a/(a-b), except negative. It's easy to prove that algebraically.

1.75 = 7/4

3.5 = 7/2

4.25 = 17/4

Using that form, you can predict the shapes.

As for irrational numbers...well I don't (want to) know.

I fully agree with your analysis Paralogic but now how do you identify the 'sides' of the shapes. In the case of 1.75-a-gon for example, where is the 1.75 sides? Also if a pentagon makes the same shape that a 1.25-a-gon makes (which i can confirm that it does -except the shape is flipped if created in the way described in the op) then is the shape the same? Does the shape have 5 sides or 1.25 sides or both?

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