1 / 23

Activity 1-9: Siders

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.

simon-bowman
Download Presentation

Activity 1-9: Siders

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. www.carom-maths.co.uk Activity 1-9: Siders

  2. My local café features this pleasing motif in its flooring. I made a few copies and cut out the shapes…

  3. 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?

  4. 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.

  5. Will they meet up to make a polygon?

  6. 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.

  7. We need (if a, b, c are the angles of the triangle, and n1, n2, n3are the numbers of sides for the polygons): Adding the last three of these: Can we find whole numbers n1, n2, n3that satisfy this? Yes, we can:

  8. If we substitute these values in for n1, n2, n3, what are a, band c? a= 72, b = 18, c = 90.

  9. What if we look at quadrilaterals? Note: a quadrilateral could be an outsider. We need:

  10. Adding the first and third equations gives: a + b + c + d = 360 = 360 ± 360/n1 ± 360/n3. This can only mean n1 = n3while one sign is + and the other -. We also have n2 = n4while 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.

  11. 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...

  12. Which means we can make this shape too... So what happens if we consider the general m-sided polygon?

  13. 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.

  14. So the maximum number of sides our siders can have here is 6. Let’s try m = 5, with the equation: which leads to

  15. Let’s call an n-sided shape producing n different regular polygonseither 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?

  16. Adding equations 1, 3 and 5 here gives: Adding equations 2, 4 and 6 gives:

  17. Let’s pick the following values for the ni... This gives: which solve to give the six angles (in degrees):

  18. 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, 210 - a to add to 360, then this wonderful thing happens... We arrive at the angles 165o, 135o, 135o, 90o, 150oand 45o.

  19. A totally perfect hexagonal sider giving 6 regular polygons, and also tessellating the plane.

  20. Task: cut out some of these tiles and have a play... Hexagon Sheet pdf http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-9-1.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 Sheet pdf http://www.s253053503.websitehome.co.uk/carom/carom-files/carom-1-9-2.pdf Task: cut out some of these tiles and have a play...

  21. Or we might consider this tile...

  22. a = 165, b = 135, c = 135, d = 117, e = 123, f = 108.4, g = 116.6 The areas of the polygons are equal... Fiddlehead Tiles pdf http://www.s253053503.websitehome.co.uk/articles/mydirr/fiddlehead-sheet.pdf

  23. With thanks to:Mathematics in School, for publishing my original Siders article. Carom is written by Jonny Griffiths, hello@jonny-griffiths.net

More Related