Looped polygons (odd number sides)

So much of my Sunday was occupied with the looped pentagon problem. Having solved that, I was intrigued to see how the problem would play out for other polygons. I have spent most of my time since focusing on odd-sided polygons (because I think even-sided ones are relatively straightforward, although I need to check) and thought I might share what I had come up with so far.

Firstly, I have to say this has been a bit of an eye-opening experience for me. Not having a solution, or any real help, has added to the intrigue. I’ve extended the problem, I’ve made connections and generalisations, and I’ve refined my approach all the time until now, where I feel I have a method to share. It’s the kind of thing I’d love to spend time on with students….if only!

The general problem is like this: a polygon is marked, with circles of 1cm radius placed around its vertices. A piece of string in looped around each vertex and pulled tight so there is no sag. The challenge is to calculate the length of string needed. In the original pentagon problem, the distance from vertex a (at the top) to vertex c (bottom left) is given as 5cm. I will refer to it simply as l.

The main issue I have found so far, and which I may well go back and revisit at some point, is deciding the order in which the string is looped. Once you move to a heptagon there are options. In all cases I present below, I have opted to go for the furthest vertex from the top vertex, which I have called the bottom left. The string then loops back up to the vertex immediately to the right of the top vertex, and so on.

I had, you will recall, noticed that l was parallel to the string – it must be so, as extending the string would create a tangent to the centre of the circle. My approach therefore involved a lot of angle theory. Beginning with the equilateral triangle, I split the interior angle in half to give 30 degrees, as this allowed me, by deduction, to calculate the angle from due North to the string. This done, and the two added together to give 150 degrees, I subtracted the tangential right angle to leave an angle of 60 degrees from due North to the tangent. Drawing a right angled triangle showed me that the angle from the centre of the circle to the tangent was 30 degrees from East, meaning that on both sides of the circle, the string was wrapped around an extra 30 degrees before moving towards the next triangle.

This gives a total contact with the circle of 180 + 30 + 30 = 240 degrees, or 2/3 of the circle’s circumference. The LENGTH of string needed can be calculated by 2/3 x 2 x pi for one circle, and then multiplied by 3 for all 3 circles. This gives a total of 4 pi for the circles, plus 3l for the straight bits, for a total of 3l + 4 pi. Knowing that the pentagon gave a solution of 5l + 6 pi for the length of string, I began to conjecture that for any s sided polygon, the length of string needed was sl + (s +1) pi.

I next tried a nonagon, being a shape with the number of sides being a factor of 360. And this was where I got stuck for a while. I guessed the answer was 9l + 10 pi, but this was based on me guessing the angle from due North to the bottom left circle. I eventually realised that the answer may lie inside the polygon – in the centre, in fact. By taking a line segment from the centre of the polygon to each vertex, I had the angles there, ready to go. So for the nonagon, each vertex was a movement of 40 degrees (360/9). Using angles on parallel lines, I could see that the angle created by drawing an isosceles triangle from the points due North, in the centre of the polygon and the bottom left vertex, gave me the angle I needed to subtract from 180 degrees. In this case that angle is 10 degrees, meaning the tangent/due North angle was 170. This, to cut things a little short, means that 10 degrees was the angle beyond 180 where the string was touching the circle on each side. Here, this resulted in 180 + (2 x 10) = 200 degrees, a total of 5/9 of the circle’s circumference. Again, the total curved contact can be found by calculating 5/9 x 2 x pi x 9, which equals 10 pi. So the total length is 9l + 10 pi.

Having found a shortcut, I then tried the heptagon. This was particularly brave given my lack of a scientific calculator, but I got there in the end. The large angle in the isosceles triangle I needed was 3/7 x 360, so the small angle worked out at 12 6/7 degrees. This was the angle beyond 180…..etc etc….and I eventually arrived at a length of 7l + 8 pi.

I had spotted another pattern here, and stopped to investigate. The proportion of the circle which was touching the string was 2/3, 3/5, 4/7, and 5/9, and can be expressed as (s-1)/s for a s-sided polygon. What I also found fascinating was angle from due North vertex to bottom left vertex as a proportion of 360: 1/3, 2/5, 3/7, 4/9…clearly this will gradually get closer to 1/2 without ever reaching it, and can be expressed as 1/2(s – 1) / s for an s sided shape. And if you add this angle to the sector touched by the string, you get 1 every time. Pretty neat, huh?

I realised now that this allows me to quickly work out the information I need without having to worry too about the angles – so for a 15-sided shape, we take 7/15 (168 degrees) as the largest angle in the isosceles triangle, meaning the angle beyond 180 degrees will be 4/15 (6 degrees) on each side, or 8/15 (12 degrees) in total. This will give total string contact of 15l + 16 pi.

With odd sided polygons, I think the next step is to investigate whether a different (but still repetitive) looping arrangement changes the total  amount of string needed. I’m not sure at the moment, but I do know I have some strategies which will hopefully allow me to find the answer pretty quickly!


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