# Overturned polygons: shapes with less than two sides

Hugh Duncan explores polygons with a shortage of edges There is a continuum of polygons with two or more sides (see ‘Between a Square Rock and a Hard Pentagon’), including fractional-sided shapes. There is also a continuum of negative sided polygons with sides numbering below minus two (see ‘Thinking Outside the Box’), the ‘holes’ left by their positive counterparts. What happens between minus two and plus two sides? Do shapes with a shortage of edges exist, and if so, what do they look like? Well I think they do and I’m taking their side! Read on…

Back in 2012 I searched the Internet for polygons that lacked a bit of an edge, but found nothing, so I worked on them myself. Let us have a look at the thinking process behind these little-known polygons.

## Another twist of the knife: overturned polygons

We are looking below two sides. How about a nice round fractional example, such as 1½ sides? Let’s give it a name, let’s call it a triamisagon. Good! Another mysterious-sounding object. For $n = 1.5$ and using the equation for the external angle $\Phi$, for a polygon with $n$ sides:

$$\Phi=360/n$$

one gets:

$$\Phi=360/1.5=240°$$

This will make the internal angle at each vertex as:

$$180-240=-60°$$

Look! The internal angle has now become negative! A (regular) 3/2-sided polygon contains angles, each of which is -60°! What new monster have we created?! This is where we go down the rabbit hole, if we haven’t already gone there that is! Okay let’s draw it! Consider the first side as being drawn horizontally from left to right (see above). The second side would start with a dotted line extending from the right end of the first line to find the original direction (east) and hence, next, the external angle. To draw this angle, one can now measure anti-clockwise from the dotted line until 240° is reached. Note the second side has now crossed over the first side and ends 60° below  the first line. The second side can now be drawn 60° below the first. The crossing over explains why the internal angle is -60°. I have left a ‘loop’ at the vertex, to show that it has overturned the first side. The third side can be drawn as before, creating another angle of -60°, including the loop of course. Before we join the second and third, we first have to turn again through 240° anti-clockwise, creating a third loop. Three sides have been drawn and taken two turns (3 lots of 240° = 720° = 2 turns), to complete a closed polygon, hence that is 3/2 sides per turn or 1.5 as expected. This polygon looks like an equilateral triangle. It is like taking an iron bar and over bending it into three equal parts.

Compared to the normal +3 sided equilateral triangle, the triamisagon looks the same, and if one ignores the loops at each vertex, the +3/2-agon is the 3-sider but flipped upside down. A triangle of +3 sides has its arrows pointing anti-clockwise, the convention used in maths for a positive rotation. At first glance, the 3/2-agon has the arrows seemingly pointing clockwise, but this is because of the loops at each vertex. If one was a car following a route of this shape, then the car always turns anti-clockwise.

For our triamisagon, as there are three internal angles of -60° each, then that makes a total internal angle of:

$$3 \times (-60) = -180°,$$

and that occurs in two turns so -180/2 = -90° per turn. Using the formula $180 \times (n-2)$, the theoretical internal angle for $n = -1.5$ is

$$180 \times (n-2) = 180 \times (1.5 – 2) = 180 \times (-0.5) = -90°,$$

as expected! Triamisagons exist! Let’s look at one with $n = 1⅓$ sides. This could be called a tesseratritagon. It would have an external angle $\Phi = 360/1.33333 = 270°$. The internal angle $\theta$ can now be found: $\theta =180 – 270 = -90°$. This shape can be constructed in the same way as the 3/2-agon. It will take four sides to complete the shape, while it has taken only three turns. Four sides in three turns is 4/3 or 1.33333 sides per turn. The shape looks like a square, but with the loops at the vertices, showing it has been constructed in a different way. Between 1 and 2 sides there would be an infinite number of these overturned polygons and a selection of them is shown below):

Due to the symmetry of polygons and anti-polygons there would be an equivalent set of shapes in the negative half of the polygonic continuum, shown below:

Does anything like these shapes exist out there in the real world? Well yes it does! Who remembers the old Knitting Nancy? See figure below of me in action! Slip roads off motorways and most model car racing kits allow one to include a few loops. Clothes pegs, the Adobe logo and hand exercisers contain them too.

## The one-sided argument: the monagon or henagon

Hopefully you have already accepted the two-sided polygon-the digon or biangle: an edge that goes from one point to a second, then comes back along the same line. Are we now ready for the one sided shape?! It sounds impossible doesn’t it, but that hasn’t stopped us before has it! Let us set the number of sides, n to one and calculate the internal angle:

$$(n-2) \times 180 = (1-2) \times 180 = -180°!$$

The external angle, the one we must turn through at the end of the line in order to complete our shape, is:

$$\Phi = 360/n = 360/1 = 360°.$$

A complete turn. We start drawing with one straight, horizontal edge going right (east). At the end, now we turn through 360° anticlockwise. This is a complete revolution, so we are actually facing the original direction. Not only that, we have created a little loop at this end (see diagram below). As we have not (yet) met up with the end of any edge, we draw a second edge from this point. This ‘second’ side continues in the same direction as the first, to the right (east). Now we turn through 360° again and once more we are still facing east, with a second loop. If we repeat this, the process will continue, forever adding another side, doing a little pirouette then edge, loop, edge loop, edge loop. We find that we made one side after one complete turn, two sides after two turns and so on, so $n$ sides in $n$ revolutions.

In other words, this is one side per turn, 1/1, hence it is a one-sided polygon by our chosen definition! We could reduce the loops to their infinitely small size and our one side per turn shape will indeed look like a one long side that goes off into the distance in each direction. As we look at it, the area below the looping line would be the ‘inside’ of the shape while the area above it would be the ‘outside’. See diagram below: I have tried to picture this in some other real world way. The closest I have come is to imagine that our two dimensional surface is perhaps a cylindrical surface. Imagine taking a sheet of A4 paper and we draw a horizontal line across the middle of the paper. We then take the sheet and roll it into a cylinder, such that the right hand edge of the paper joins onto the left hand edge. Our 2D universe is the outside surface of this paper. The line we drew now makes a simple ‘belt’ or equator round the middle. Voila, a one-sided polygon. Above the line is the outside, while below this line is the inside. See below left.

## Would the real one-sided polygon stand up?

According to Wolfram, a polygon with one side is called a henagon. There is no shape that exists in Euclidean space with one (straight) edge. It is typically drawn as a circular line that joins back up with itself (see right). The circumference of the henagon is just one edge and when drawn like this, it clearly has an inside and an outside. I would like to suggest that my equator around a cylindrical 2D world version and my infinite line with equally spaced loops should also be an acceptable alternative.

## Turn again, polygon (shapes with between 1 and 2/3 sides)

What if we now go below one side-that is, less than a side per turn? Shall we try a polygon with 3/4 sides?! Yes we shall! We are clearly entering ‘one’-derland! Sorry. With $n = ¾$ the external angle will be given by:

$$\Phi = 360/ ¾ = 480°.$$

This is turning through more than one revolution between sides. Crazy! Draw a horizontal line pointing east. At the end, turn anti-clockwise through 480° and draw a second line. Note we covered a complete revolution and then managed another 120°. See below, second shape from the left. Notice also that the little loop we drew to distinguish these low-sided shapes is now on the inside! Now turn through another 480° anti-clockwise and draw a third side. Like the bending of the iron bar, we have made a full loop plus a bit more and this loop is also inside the shape. All we have to do is make our final spin of 480 and join onto the start of the first side (see below, second shape from the left). We have made 3 sides but we have turned through $3 \times 480 = 1440°$. This is four compete turns (1440/360 = 4), so 3 sides in four turns would be written as ¾ or ¾ of a side per turn. This looks like the equilateral triangle (and our 3/2 sided polygon, which was also looking like an equilateral triangle), but this time with the little loops inside the vertices.

If we were to repeat the same process for other simple fractions between 2/3 and 1, we get the next section of polygons that look like our original ones with more than two sides, but with the identifying internal loops. See a selection of them above. Not surprisingly, there will be a similar set of anti-polygons between -1 and -2/3 on the other side of zero that will look like the set above, but as ‘holes’ with the equivalent shapes.

## And turn again (2/3 down to ½) and so on…..

Let’s keep going before the maths police stop us. Take $n = 3/5$ and the external angle is $360 \times 5/3 = 600°$. So we draw our starting line and turn 600° anti-clockwise, which is one complete turn of 360° and then a further 240°. We do this two more times and we get a shape with 3 sides which has turned $3 \times 600 = 1800°$ or five revolutions, i.e. three sides in five turns or 3/5 sides per turn. See below. Note we have what looks like a triangle again but the loops are on the outside again and they are double loops. These double outside loopers are found between 2/3 sides and ½ sides. A selection of shapes from this range is shown below. This system ends at ½ with a string of one siders with double loops between them.

## From a half to zero

There’s a pattern to these fractional polygons below two sides and all the way to zero. The next batch would have double loops inside and the one after that triple outside loops, then triple inside loops and so on. This continues all the way down to zero sides, each time adding an outer or inner loop and each stage taking up less and less of the polygonic number line. A few of those shapes closer to zero are shown below. These systems flip flop between outside and inside loopers, increasing an additional loop each time and the switching happens more frequently as we approach zero, but hopefully the trend is clear. It is now time to bring the topic to a close.

## The zeragon: the zero sided polygon

And so we arrive at the singularity. A polygon with no edges. Our integer $n$ for the number of sides of our polygon is set equal to zero: $n = 0$. With that, we can find the total internal angle:

$$\Phi = (n-2)180 = (0-2)180 = -2 \times 180 = -360°$$

The internal angle is -360°, so, being negative, this indicates the outside angle. As it has no sides/edges, would that not just make a no-sided polygon a point? One could then picture that there is an angle of 360° all around the outside the point, all the way round the point? This would constitute the -360° given by the formula. See below:

A zeragon is a point, a vertex, and has no sides, so technically it can have no angle ‘inside’ as there is no inside. As with the monagon and those negative sided polygons, the outside angle is taken to be negative. As seen in the diagram above, there is a complete 360º all the way round the outside of the point, so being outside it is -360º rather than +360º and this too obeys the general equation.

A zeragon as we have described it and tried to define it, is just a point, though in diagrams I have represented it as a very small circle (that is for the benefit of those with poor eyesight like me!). Now look at the diagram to the right. There it is, floating like a tiny speck above the same two-dimensional flat world that we have placed behind it (see the blue dot). It is casting its own tiny shadow onto the plane below. As it contains one vertex, it has a vertex number of +1 and hence must be positive no? So is it a positive zeragon? Would that mean, that the absence of a point (i.e. no vertex) would be a negative zeragon, as it would have a vertex number of -1? It could be drawn as a tiny pinhole in the plane of our 2D world. Being a hole, it would cast a shadow ‘inside’ itself. See the yellow dot in the right hand diagram. I have called it a negative zeragon, or possibly an anti-zeragon? Does that mean there are two types of zeragons?! Take a deep breath and let it all sink in. Is this now the complete polygonic continuum?! THE END (OR IS IT?!) Hugh Duncan graduated from UCL having studied astronomy. He taught physics and maths at the International School of Nice for over thirty years. Hugh was a punk when he started UCL in 1976 and guess what, he still is! Check out his band Old Age Spies on YouTube. He recently retired and has just had his first science fantasy book published ‘Life On Mars – The Vikings Are Coming’ inspired by his time at UCL. Check it out too!