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. 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 n 1, n 2, n 3 are the numbers of sides for the polygons): Adding the last three of these: Can we find whole numbers n 1, n 2, n 3 that satisfy this? Yes, we can:
If we substitute these values in for n 1, n 2, n 3, 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/n 1 ± 360/n 3. This can only mean n 1 = n 3 while one sign is + and the other -. In other words, we are free to choose n 1 and n 2, and the resulting quadrilateral will create a regular n 1 -agon both as an outsider and an insider, and a regular n 2 -agon both as an outsider and an insider. We also have n 2 = n 4 while one sign is + and the other -.
Here we have n 1 = 5, n 2 = 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 n 1 = 5 and n 2 = 7 for a, b, c and d, we get (in degrees): a, – a, a, 108 – a. So we could choose a so that the quadrilateral is cyclic, which gives a =
Which means we can make this shape too... So what happens if we consider the general m-sided polygon?
Let us now stipulate that n 1, n 2,... n m, 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 n i... 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 sider tessellates and thus becomes totally perfect? It turns out that if you choose the three angles a, a - 15, a to add to 360, then this wonderful thing happens... We arrive at the angles 165 o, 135 o, 135 o, 90 o, 150 o and 45 o.
A totally perfect hexagonal sider giving 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 = , g = ). There is a pleasing opening-out effect as the number of sides grows. Septagon Sheet pdf Task: cut out some of these tiles and have a play... carom/carom-files/carom pdf carom/carom-files/carom pdf
Or we might consider this tile...
Fiddlehead Tiles pdf a = 165 , b = 135 , c = 135 , d = 117 , e = 123 , f = , g = The areas of the polygons are equal... articles/mydirr/fiddlehead-sheet.pdf
With thanks to: Mathematics in School, for publishing my original Siders article. Carom is written by Jonny Griffiths,