Activity 1-9: Siders

www.carom-maths.co.uk

My local café features this pleasing motif in its flooring.

I made a few copies and cut out the shapes…

Playing with the kite shapes, I found myself doing this:

‘Would these tilings meet up to form a polygon?’ I wondered.

Task: are exact polygons created here?

In the left-hand case, I said the tile was acting as an INSIDER,while in the right-hand case, it acted as an OUTSIDER.

Tiles like this I called SIDERS.

Will they meet up to make a polygon?

The triangle - the simplest shape we could start with...

A triangle can only ever work as an insider (why?).

Could it give three different regular polygons in this way?

Task: prove the inside angle of a regular n-agon is(180 – 360/n)o.

We need (if a, b, c are the angles of the triangle, and n1, n2, n3 are the numbers of sides for the polygons):

Adding the last three of these:

Can we find whole numbers n1, n2, n3 that satisfy this?

Yes, we can:

If we substitute these values in for n1, n2, n3,what are a, b and c?

a = 72, b = 18, c = 90.

What if we look at quadrilaterals?Note: a quadrilateral could be

an outsider.

We need:

Adding the first and third equations gives:

a + b + c + d = 360 = 360 ± 360/n1 ± 360/n3.

This can only mean n1 = n3 while one sign is + and the other -.

In other words, we are free to choose n1 and n2,and the resulting quadrilateral will create

a regular n1-agon both as an outsider and an insider, and a regular n2-agon both as an outsider and an insider.

We also have n2 = n4 while one sign is + and the other -.

Here we have n1 = 5, n2 = 7.

It turns out that we get some freedom in choosing the angles for our shape here.

If we solve the above four equations with n1 = 5 and n2 = 7 for a, b, c and d, we get (in degrees):

a, 231.4... – a, 20.6... + a, 108 – a.

So we could choose a so that the quadrilateral is cyclic, which gives a = 79.7...

Which means we can make this shape too...

So what happens if we consider the general m-sided polygon?

Let us now stipulate that n1, n2, ... nm, must all be different.

If m = 6, note 1/3+1/4+1/5+1/6+1/7+1/8 = 341/280 > 1 = LHS,so m = 6 is possible.

If m = 7, note 1/3+1/4+1/5+1/6+1/7+1/8+1/9 = 3349/2520 < 1.5 = LHS,so m = 7 is impossible.

Let’s try m = 5, with the equation:

So the maximum number of sides our siders can have here is 6.

which leads to

Let’s call an n-sided shape producing n different regular polygons

either as an insider or an outsider ‘a perfect sider’.

When I showed my perfect 5-sided sider to a colleague; her comment was, ‘Does it tile on its own?’

It does not!

Such a shape would surely deserve the name, ‘a totally perfect sider’.

So far we have a sider (the quadrilateral),a perfect sider (the pentagon)

and a totally perfect sider (the triangle).

Can we find a totally perfect hexagonal sider?

Adding equations 1, 3 and 5 here gives:

Adding equations 2, 4 and 6 gives:

Let’s pick the following values for the ni...

This gives:

which solve to give the six angles (in degrees):

Once again, we get some freedom here.

Can we choose a so that our perfect sidertessellates and thus becomes totally perfect?

It turns out that if you choose the three angles a, a - 15, 210 - a to add to 360, then

this wonderful thing happens...

We arrive at the angles 165o, 135o, 135o, 90o, 150o and 45o.

A totally perfect hexagonal sidergiving 6 regular polygons, and also tessellating the plane.

Hexagon Sheet pdf

Notice also that since

it is possible to create a seven-sided sider that generates the regular polygons from 3 through to 8 sides

(a = 165, b = 135, c = 135, d = 117, e = 123, f = 108.4, g = 116.6). There is a pleasing opening-out effect as the number of sides grows.

Septagon Sheetpdf

Task: cut out some of these tiles and have a play...

Task: cut out some of these tiles and have a play...

http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-9-1.pdf

http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-9-2.pdf

Or we might consider this tile...

Fiddlehead Tiles pdf

a = 165, b = 135, c = 135, d = 117, e = 123, f = 108.4, g = 116.6The areas of the polygons are equal...

http://www.s253053503.websitehome.co.uk/articles/mydirr/fiddlehead-sheet.pdf

With thanks to:Mathematics in School, for publishing my original Siders article.

Carom is written by Jonny Griffiths, hello@jonny-griffiths.net