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www.carom-maths.co.uk. Activity 1-9: Siders. 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. - PowerPoint PPT Presentation
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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, [email protected]