Star Polygons

Home , Dodecagram, Heptagram, Octagram, Star (glyph), Star polygon

IB EXTENDED ESSAY

Candidate Name: Šimon Rovder Supervisor: RNDr. František Kosper Candidate No.: 000771-032 Word count: 3931

Šimon Rovder – 000771-032

Abstract

This essay presents the steps of research of star polygons with the goal of proving the existence of star polygons of any amount of vertexes. The only information this research began with was the knowledge of two star polygons (pentagram and heptagram) and the knowledge of the fact that they are called star polygons. This research, which started from nothing, discovered basic rules behind the existence of star polygons, presents a clear way of how to find them and successfully proves the existence of an infinite amount of star polygons. This essay presents a proof of the existence of all star polygons with the amount of vertexes greater than 5 and different from 6. Exploration of this behavior of the number 6 is later explained and a valid (yet unusual) way of drawing it was eventually found, demonstrated and explained.

In the further sections of this essay, the range of the research was expanded to include research and a proof of the fact that any amount of star polygons of any amounts of vertexes can all be drawn together within one larger star polygon.

The appendix of this essay contains other information and formulas (related to star polygons) discovered during this research. Some sections of this essay use that information, yet it is not directly linked with the research question. It is included in this essay mainly in reference to any future research based on this work, which is of course encouraged.

Šimon Rovder – 000771-032

Contents

Abstract ...... 2 Introduction ...... 4 Star Polygons ...... 5 Star Polygon Definition ...... 5 Included Star Polygons (the ones this research started with): ...... 6 Finding the star polygons ...... 8 Empirical search ...... 8 Finding the Vertex Jump Variable k ...... 9 Can any star polygon be drawn? ...... 14 Polygon projections within Star Polygons ...... 20 Inward Projections ...... 20 Outward Projections ...... 25 Can a combination of many different regular or star polygons be projected into a single star polygon? ...... 30 Conclusion ...... 36 Bibliography ...... 37 Appendix ...... 38 Angles at the vertexes of any order ...... 38 Length of any line segment within a star polygon of radius 1 ...... 39 Area of any star polygon of radius 1 ...... 41

Šimon Rovder – 000771-032

Introduction

“Can you draw a five point star with one move?” A popular puzzle among younger children and an interesting problem for a mathematician, if of course, slightly expanded.

“Can you draw a seven point star with one move” sounds a bit better, but “Can you draw an n point star with one move” sounds just about right for the initiation of actual research.

Seven years ago, I came across a seven vertex star drawn in a very similar fashion as the highly famous pentagram. I found it fascinating, yet I did not continue to pursue this beautiful area of geometry. Now I decided to come back to this problem and dedicate my extended essay to its expansion.

I decided to explore this problem by myself and with no research of anything already discovered. Thus what this essay presents, are all my own findings, except for the Prime number Theorem used once on page 11 in a proof related to prime numbers.

So, Can we draw an n vertex star polygon?

Šimon Rovder – 000771-032

Star Polygons

Star Polygon Definition

In order to even start working with the star polygons this puzzle was based on, there had to be a definition for them first.

The puzzle is:

Drawing a star polygon by one move (one continuous line), while only changing the direction of drawing at the vertexes of the star polygon.

As the whole idea of studying these star polygons originated from the existence of the pentagram and heptagram, the main definition of what will be considered a “star polygon” was derived from them as well. By this definition, a polygon has to fulfill the following criteria in order to be my star polygon:

1. The polygon is a single self-intersecting polygon.

2. The vertexes of a star polygon are located on the perimeter of one circumference

3. Adjacent points are equally far from one another.

4. All vertexes are connected to two other vertexes, skipping a constant amount of

vertexes with this connection.

By this definition, the range of this topic is clearly defined to the star polygons we want.

The second and third criteria apply to all regular polygons and the first and fourth are

Šimon Rovder – 000771-032 the main defining characteristics of star polygons. The absence of the circumference

(criterion 2) would eliminate any ability to order the vertexes and check the fourth criterion, while the absence of the third criterion would deform the star polygon and ruin equality in lengths of sides. So we can only check if a polygon is a star polygon if we know the second and third criteria are met.

Examples

So from the definition of a star polygon, we can take a look at a few valid examples of star polygons as well as at some other shapes excluded by this definition.

Included Star Polygons (the ones this research started with):

Figure 1 : Pentagram Figure 2 : Heptagram

Šimon Rovder – 000771-032

Excluded Polygons:

Star of David (in violation of the first Tilted Pentagram (In violation of Criteria 2 criterion – made of more than one and 3, and undeterminable for criterion 4): polygon):

Figure 3 : Star of David Figure 4 : Tilted Pentagram

Investigation of star polygons of an amount of vertexes of 1 and 2 can be excluded as well, since a polygon cannot be formed out of less than three vertexes. Also since a star polygon has all its vertexes connected to different two vertexes than it would if the polygon was regular, star polygons of less than 5 vertexes can be excluded from this research, as they have no way of fulfilling this criterion (not enough vertexes to change all connections). Thus this research will be focused on star polygons of 5 and more vertexes (the ones that correspond to the puzzle), making the pentagram the simplest star polygon there is.

Šimon Rovder – 000771-032

Finding the star polygons

Empirical search

The initial hypothesis was that for every regular polygon should be at least one corresponding star polygon. This was a valid assumption as after excluding the star polygons of 1, 2, 3 and 4 vertexes, it seems fairly plausible. To make orientation easier I decided to define a number n for each star polygon. This number would be the number of vertexes the star polygon had and should give some structure to any data I was to collect. So from all we know by now:

At the very beginning, the search for star polygons was empirical. It was a matter of drawing the vertexes and attempting to connect them by definition. Also the differences between the pentagram and the heptagram suggested the existence of yet another specific number for each star polygon. A number that defines the number of vertexes that are to be skipped when connecting vertexes to form the star polygon. This number will from now on be referred to as k. These two numbers n and k should come in specific combinations for each star polygon and should tell us that: “To draw a star polygon of n vertexes, you have to connect each vertex to the next kth vertex in both directions of the circumference on which they are”. A graphical representation of this in a heptagram

(n=7 and k=3) would be as follows:

Šimon Rovder – 000771-032

Figure 5 : Drawing a Star Polygon

So the rules seemed to be simple. For every n, the number k has to be found in order to draw the star polygon. So if a star polygon of every n is to exist, a k must exist for every n. The question is what is k for every n?

Finding the Vertex Jump Variable k

The initial assumption for the variable k was that whatever properties it has, they will have something to do with divisibility of n. This assumption was quickly followed by a range definition for k. In order for a star polygon to fulfill the first star polygon criterion, k has to be more than 1. Connecting each vertex to the two very closest vertexes, results only in a non-self-intersecting, regular polygon.

Šimon Rovder – 000771-032

Also k should be less than n/2, as having k greater than this value would result in the same star polygon as having k less than n/2 by the same amount as it was more. So the star polygon that could result in a k larger than the integer part of n/2 would already have been found by trying the ks smaller than it.

An immediate finding after the search began was that there were multiple ways of drawing some star polygons. The heptagram itself is an example of this. There were two different ways of drawing it, both of which fulfill all the criteria:

Figure 6 : Different forms of a Heptagram

Thus we see that k is not going to be a number. It is going to be a set of numbers. A set that will be unique for each n. We can see that a heptagram (n=7) can be drawn if k=2 and if k=3.

Šimon Rovder – 000771-032

This finding however is not in contradiction with anything found so far, so the search for a connection between n and k continued. These were the star polygons found for all ns from 5 to 12:

Legend:

Ø – Not looking for it – k is out of the range of ks for that n.

☓– Did not fulfill the criteria of a star polygon or did not result in the correct star polygon.

k n 2 3 4 5

5 Ø Ø Ø

6 ☓ Ø Ø Ø

7 Ø Ø

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8 ☓ Ø Ø

9 ☓ Ø

10 ☓ ☓ Ø

12 ☓ ☓ ☓

Table 1 : Empirical Search

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The relationship between n and k was clear from these star polygons. k has to fulfill the following criteria for a star polygon to exist:

1. k has to be greater than 1

2. k has to be smaller than n/2

3. k cannot have any common divisors with n (k and n must be relatively prime

(Weisstein, 2005))

Whenever the numbers n and k had common divisors, the star polygon collapsed into a polygon that had these numbers in a form where these common divisors were cancelled out. (They act like a fraction) An n=12 k=3 star polygon resulted in an n=4 k=1 polygon, which was actually a square, not a dodecagram.

Figure 7 : n=12 k=3 Square

Šimon Rovder – 000771-032

Knowing this we can now definitely say that a star polygon of n=6 do not exist, as it has no corresponding k. Knowing this concludes the search for star polygons and now we have all the knowledge needed to find all star polygons. Now we can look for the answer to our question. Can any star polygon be drawn?

Can any star polygon be drawn?

To answer this question, we need to essentially answer a different question. Do all ns have at least one corresponding k? We know that star polygons of n=1, n=2, n=3, n=4 and n=6 do not exist. Do all other exist? They do. And it can be proven like this:

We know that if n is greater than or equal to 5 there is at least one number that could possibly be k according to the first two criteria. The problem we need to solve is with the third criterion.

A number k exists for all odd numbers n if the integer part of n/2 is greater than or equal to 2. This is because odd numbers cannot be divided by two, so if n is odd, the number 2 fulfills the third criterion of k and all star polygons of an odd number of vertexes exists.

The question is for what n higher is the integer part of n/2 greater than or equal to 2?

We can find it using a very simple inequality:

Šimon Rovder – 000771-032

So we see that for all odd ns greater than or equal to 5, exists a star polygon and our research question is halfway answered.

If n is an even number, the question is more difficult, however, it can still be answered.

If a star polygon does not exist, it means its n has a common divisor with all numbers between 1 and n/2. As all prime numbers can only be divided by themselves and 1, an n that has no corresponding k has to be greater than or equal to the product of all prime numbers between 1 and n/2. Only that way can it have a common divisor with all the prime numbers in that interval (and no prime number can hence be k).

The amount of prime numbers between 1 and n/2 can be approximated by the Prime number Theorem which states that:

( )

Using this we can calculate a minimum amount of prime numbers between 1 and n/2.

Now we need to determine their values. This however cannot be done exactly. The only approximation we can make is using the following argument:

Šimon Rovder – 000771-032

The product of the first x prime numbers is always going to be more than the smallest of these prime numbers raised to the power of x.

That is:

Using this information, we can determine a certain value of n above which all star polygons will exist. This boundary will not tell us anything about the star polygons below this boundary, only above it, as we used approximations. If 2 to the power of the amount of prime numbers between 1 and n/2 is greater than n, the n has at least one corresponding k and a star polygon hence must exists:

( )

( ) ( )

( )

We know that n ˃ 4

( ) ( ) ( )

Šimon Rovder – 000771-032

This inequality gives us two resulting intervals: n ≥ 19,5802 and 0,761084 ≤ n ≤ 6,89479

As n must be a whole number, we can adjust them to these: n > 19

0 < n < 7

So we can see that the upper boundary beyond which a star polygon will always exist is

19. Existence of star polygons of an even n smaller than 19 are not proven mathematically in this essay, however, I provide for them the following empirical evidence of their existence:

Šimon Rovder – 000771-032 n Existing k Image

8 3

10 3

12 5

14 5

16 5

18 7

Table 2 : Empirical Proof

Šimon Rovder – 000771-032

So we can conclude that a star polygon can be drawn for every n other than 6, effectively answering our research question as: “NO”, since n=6 This case was “More- less” closed, but I noticed there was more to the star polygons that meets the eye at once. Something deeper within them that I expanded my research to. Something that would imply that even the star polygon of n=6 could be drawn by a single move. So I decided to expand my research question to another question: “Can a star polygon formed of multiple polygons be drawn by one move?”

Šimon Rovder – 000771-032

Polygon projections within Star Polygons

Inward Projections

An interesting discovery made during this research was projections of polygons within star polygons. What it is is best shown on the star polygon n=12 k=5.

It is visible that apart from the vertexes on the outmost circumference, there are more

“orders” of vertexes formed by the intersections of the sides.

Šimon Rovder – 000771-032

Figure 8 : Orders of vertexes

The amount of these orders is k-1 (-1 because we do not count the outmost vertexes)

Each order of these vertexes has the vertexes connected in a different way. One order closer to the center, they are connected in a way such that they form four separate triangles:

The vertexes a further order towards the center connect in a way such that they form three squares:

Šimon Rovder – 000771-032

Another order down they form two hexagons:

Šimon Rovder – 000771-032

And the last layer forms the regular polygon which can be found at the center of every star polygon

These were regular polygons projected into star polygons. But the cases get stranger. Within this n=16 k=7 star polygon, we can find two projections of the octagram:

Šimon Rovder – 000771-032

I decided to call this phenomenon projections, as it actually “projects” different polygons without intentionally drawing them. These projections proved to have a pattern. Back in trying to find the rules behind star polygons, we figured out that when attempting to draw a star polygon with a k that has a common divisor with n, the polygon collapsed into a different polygon that fulfilled this criterion, by dividing both n and k by its greatest common divisor automatically (n and k act like the numerator and denominator of a fraction). So for example attempting to draw a star polygon n=21 k=6 results in a star polygon n=7 k=2 (both divided by 3).

In finding projections of polygons in star polygons, we continuously decrease the number k by one to go one order towards the center, we find the common divisor of the new combination of n and k, and divide them both by it. This way we find the type of the polygon that will be projected. The amount of these projections is the common divisor.

Further description of why this phenomenon occurs can be seen on the linear relationship between angles in the different orders, described briefly in the appendix.

So in our n=12 k=5 star polygon, we can decrease k by one. We get n=12 k=4. Their common divisor is four, which means the amount of projected polygons will be four.

And the resulting star polygon is n=3 k=1, so the projected polygons will be triangles.

And we get four triangles.

Šimon Rovder – 000771-032

So our complete analysis of the aforementioned entire dodecagram:

Initial Star polygon’s n and k Projected polygon’s n and k Common divisor - amount in a fraction in the form of in a fraction in the form of: of projected polygons , gradually adjusting

In essence, star polygons not only can be formed of multiple polygons, they already are.

The real question is, whether this phenomenon can be controlled. It turns out it can. By reversing the analysis process.

Outward Projections

This process can also be reversed in the same way. Suppose we want to find a star polygon that projects somewhere within it three pentagrams of n=5 k=2. So we multiply n and k by 3 to get n=15 k>6. k has to be more than 6, because the pentagrams exist on the 6th order of vertexes. Now we gradually increase k until it fulfills all three criteria.

Šimon Rovder – 000771-032

We can stop at n=15 k=7, which should be a star polygon that projects three pentagrams into itself. And here is our result:

If a case arises when the new n and k do not arrive at a correct k while adjusting upwards before reaching n/2, both n and k must be multiplied by the required amount of projections again, to increase the difference between the new n and k, increasing the amount of integers between it and thus increasing the probability of finding a correct k.

In essence, what is done is that the structure that cannot be drawn is projected more times into a different star polygon, preserving the structure. As this interval can be

Šimon Rovder – 000771-032 stretched infinitely by further multiplication without adding a new common divisor into the system, one is bound to find a correct value for k eventually, as we have proven that all star polygons other than n=6 exist. In an infinite interval, one is bound to find a new prime number (Caldwell, 2012), hence a projecting polygon always exists, even if it is not the very first one that the number n suggests.

For example, two pentagrams (n=5 k=2) cannot be projected into a star polygon of n=10 k>4, as n=10 k>4 cannot be adjusted correctly. However, n=10 k>4 can be further projected twice into n=20 k>8 and n=20 k>8 can be adjusted to n=20 k=9 and hence drawn. It projects four pentagrams, but two can be ignored as the intended structure is achieved anyway. Simply it is achieved twice at once. Here is the result:

Šimon Rovder – 000771-032

This however proves one unusual fact. At the beginning of this research, star polygons made of more than one polygon were excluded, as they could not be drawn by one move. Now we see that they can. As a projection within a larger polygon. The n=12 k=5 is a demonstration of it. It projects the Star of David into it:

Šimon Rovder – 000771-032

Star of David: Star of David projected into a n=12 k=5 star polygon:

So it is possible to draw any star polygon formed of more identical regular or star polygons by one move, by projecting them into a larger one, answering the second question as YES. It was not the solution the initial puzzle was looking for in star polygons, yet it fulfills the rules the puzzle establishes. It draws a star polygon by one move and only changes the direction of drawing at the outmost vertexes. It is a less direct way of solving it, yet it should be considered correct.

It however lead to yet another question, which takes this topic to the very extreme.

“Can we draw any amount of any different polygons (star or regular) into a single star polygon as projections at different orders?”

Šimon Rovder – 000771-032

Can a combination of many different regular or star polygons be projected into a single star polygon?

This part is the final extension of the whole essay. First let’s see if it is possible to draw two different polygons (regular or star) by one move as projections in a third star polygon.

Let’s take two star polygons and attempt to do so. One of n=5 k=2 and the second one of n=7 k=3.

We have to imagine the two numbers as fractions of the form n/k (as n and k act like fractions):

and

Now we shall set the number of vertexes (numerators) to the same amount by multiplying both numbers in the first fraction by the numerator of the second fraction and both numbers in the second fraction by the numerator of the first fraction. We get:

and

So whatever star polygon will be the result, it will have n=35 vertexes and on its 15th order of vertexes from the center, the heptagram should be projected and on the 14th order of vertexes from the center, the pentagram should be projected. This means the

Šimon Rovder – 000771-032 amount of orders must be greater than 15 (k must be greater than 15). So we adjust the first fraction (the one with the greatest denominator and hence greatest needed k) upwards, until we reach a number that fulfills the criteria of k for n=35. That is 16. So the result is a polygon of n=35 k=16 and both the pentagram and heptagram can be found in it:

To prove that this process can be done for any two star polygons, we perform a universal calculation for two hypothetical star polygons of n=A k=a and n=B k=b. The fractions are as follows:

Šimon Rovder – 000771-032

and

Multiplying by numerators we get:

and

We will assume that aB is greater than bA, as this proof works the same for both ways anyway. Thus we have to adjust a star polygon of n=AB and k>aB to one k of which fulfills all the requirements of a k for that n. The amount of possible ks is however restricted to:

That is:

( )

There is no guarantee we will always find a corresponding k in this amount of possible values. We know that a star polygon of n=AB definitely exists, but we do not know whether any of its ks is greater than aB. We know however, that whether this star polygon structure exists or not, we can definitely project this structure more times into a larger, definitely existing star polygon, just like we projected the essentially non- existing n=6 k=2 Star of David into the n=12 k=5 dodecagram. All we have to do is multiply both its n=AB and k>aB by a common factor of both (as to not introduce any

Šimon Rovder – 000771-032 new common factor, yet expanding the amount of possible ks). A clear common factor is

B. Thus we get that a star polygon of n=AB and k>aB is projected B times into the polygon of n=ABB k>aBB. This as a fraction is:

And the range of possible ks is now:

( )

This amount of possible ks is B times as large as the initial amount of possible ks and the amount of different factors of n and k remained constant. Thus the amount of possible ks can be enlarged exponentially and can grow to infinity without increasing the amount of different factors. And if this is true, one is bound to come across a prime number that satisfies the criteria of k in this amount of growing possibilities at some point, as there is an infinite amount of prime numbers in an infinite interval. (Caldwell, 2012)

This proves that two different polygons (regular or star) can be projected together in one larger star polygon at its different orders of vertexes from the center. This can also be done recursively. Two star polygons into each of which one already projects two smaller star polygons, can be projected into yet another larger star polygon. Here you can see a demonstration of such recursive projections:

Šimon Rovder – 000771-032

n=7 k=2 n=5 k=2 n=8 k=3 n=9 k=4 (Heptagram) (Pentagram) (Octagram) (Eneagram)

} }

n=63 k=29 n=40 k=17 (Hexacontakaitrigram) (Tetracontagram)

(The heptagram is too small for me to even find. It is somewhere near the center)

Šimon Rovder – 000771-032

Together They form a: n=2520 k=1163 (Henchiliapentaheptoicosagram)

With the current resolution capabilities it is not really possible to see all four Table 3 : Recursive Projection

So yes, any two (or more) star polygons can be projected into a single polygon.

Šimon Rovder – 000771-032

Conclusion

This essay was aimed at answering the question whether it is possible to draw a star of any amount of vertexes with only one move (in the manner of drawing a pentagram).

This question was effectively answered as NO, but for all star polygons of the amount of vertexes different from 6 as YES. The main idea behind the problem however automatically excludes polygons of the amount of vertexes 1, 2,3 or 4, as they cannot all be connected in a different way anyway. The only problematic polygon was of 6 vertexes.

Extending the topic of this essay lead to further discoveries. One of them was the effective proof that even star polygons that could normally not be drawn by one move

(formed of more than one polygon), such as the problematic star polygon of six vertexes can be drawn by one move as a projection within a larger star polygon. This was not the solution the initial puzzle was looking for in star polygons, yet it fully fulfills the rules the puzzle establishes. It draws a star polygon by one move and only changes the direction of drawing at the outmost vertexes. It is a less direct way of solving it, yet it should be considered correct.

So yes, a star polygon of any amount of vertexes can be drawn, one way or the other.

Šimon Rovder – 000771-032

Bibliography

Selkirk, K., 1992. Longman Mathematics Handbook. Harlow: Longman.

Weisstein, E. W., 2005. Prime Number Theorem. [online] Available at:

[Accessed 19 November 2012].

Caldwell, C.K., 2012. Euclid's Proof of the Infinitude of Primes. [online] Available at:

[Accessed 22 November 2012].

Šimon Rovder – 000771-032

Appendix

This section contains all other information I discovered during my research. They were not directly connected to the research question, they are the product of my own curiosity, but I include them here as an appendix as it would be a waste not to pass it further.

Angles at the vertexes of any order

α = Angle t = Order from the center. (t ≤ k)

Linear relation pattern found by gradually deriving one angle from the previous ones and simplifying:

Šimon Rovder – 000771-032

The formula counts the RED angles!

Length of any line segment within a star polygon of radius 1

First order (side of the regular polygon at the center):

( ) ( )

Any other order:

[ ( ) ( )] ( )

l = length t = Order from the center. (t ≤ k)

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Found by the following assumption and the tangent function (using the knowledge of the angles formula):

Adjusted for a radius of one by the following ratio and the cosine function:

Šimon Rovder – 000771-032

Area of any star polygon of radius 1

{ [ ( )] [ ]}

( ( ) )

[ ( )]

Found by subtracting areas of excessive triangles from the complete regular polygon

(using the knowledge of the angles formula):

Adjusted for the radius of one in the same way as the lengths of the line segments.