And another one….

Triangles and rectangles

 

I saw this just as I was settling down in bed last night. 1 1/2 hours later, and after double, treble and quadruple checking, my solution is as follows*. Another fab puzzle, putting into practice a variety of skills. And I even managed to find two routes to the answer!

IMG_0400

 

*Upon further inspection last night, I had made a mistake and posted a slightly different solution on Twitter. Still a bit rusty!

The rectangle puzzle

Today began with me reading through some tweets I’d favourited, including the latest from @mathsjem (who if, for some ridiculous reason, you’re not already following, you should follow immediately). In the section on problem solving, she remarked the following:

“It’s a good idea for maths teachers to try to solve unfamiliar problems every now and then (like the example below from ‏@dannytybrown) to remind ourselves that mathematical problem solving often requires patience, creativity and multiple attempts. We all experience frustration in problem solving, just like our students do, but we know that the satisfaction of eventually finding the solution is well worth it.”

I wholeheartedly endorse this view. It is something I have become acutely aware of recently and, as I will be teaching top set for the first time in my career from next week, I find this an excellent way of testing my rustier skills. I blogged previously about the looped polygons puzzle, and today I decided to tackle another of @dannytybrown’s puzzles, namely this one:

rectangle puzzle

I must admit I huffed and puffed for a while on this, but in actual fact I was only really able to solve it once I’d had a look at another of Danny’s puzzles which was mentioned in @mathsjem’s post:

rectangle problem

This had me confused for ages. I just couldn’t find a way into it. So I looked at solutions proposed by others and tried to understand them. Eventually, and after a lot of confusion on my part, I got there (although I am awaiting a reply from Danny based on what I regard as a key aspect of the solution with which I am not overly happy at present). In the end, the solution boils down to using trigonometric ratios to make finding the answer much easier.

So back I went to the paper folding puzzle. I had already jotted down what I knew and had deduced – the length is rt 3 x w, the width is w. This means that the base of the right angled triangle can be written as (rt 3 x w)/w. During my first attempt at the problem I had calculated the hypotenuse at being w + ((rt 3 x w)/3). When I went back to it and tried again from scratch, however, my curiosity was stirred, and I worked through it slightly differently.

We know that hyp^2 = w^2 + ((rt 3 x w)/3)^2. Working this out leaves hyp^2 = w^2 + 3w^2/9. This adds up to 4/3w^2, which when you root it gives a hypotenuse equal to 2w^2/rt 3. And if we rationalise the denominator? Well then the hypotenuse of the triangle equals (2 x rt 3 x w)/2 – twice the length of the base. This then means that the angle between the base and the hypotenuse must be 60 degrees.

On my first look at the problem, I really struggled with identifying how much of the folded section would fall outside the rest of the sheet. In fact, I couldn’t comprehend how I would even begin to calculate this. Having gone away and come back to it, however, I realised the thing I was trying to grasp earlier was that the fold line acts as a line of symmetry. So if I doubled the angle, that would allow me to calculate how far the sheet would fold across the existing section. Except that it was now quite straightforward. The angle being 60 degrees, added to the original 60 degrees, creates an equilateral triangle. The section that will fold over neatly onto the existing paper can be shown by drawing a diagonal from the bottom line (where the diagonal already ends) to the top left corner. The length of the diagonal is (2 x  rt 3 x w)/2 which, believe it or not, is the same as the length of the sheet from the left hand corner to the point the diagonal meets it. This means that there is half of a small rectangle (or 1/6 of the total shape) that will overhang when folded.

The original question asks for the ratio of the area of the new shape : the area of the original shape. The shape is 1/2 + 1/6 of the original (1/3 is folded over), leaving 2/3 of the original, meaning the ratio is 2:3.

I loved doing this problem, and I think it also helped having a break and focusing on another problem….particularly when that problem and this shared a key, common, but overlooked (in my eyes at least) bit of mathematics!

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!

Looped triangles – my solution

Up early on a Sunday for my long run (9 miles today, in case you wondered), I came across a tweet from @srcav:

As I had time to kill whilst my breakfast digested, I thought I’d have a go. Below is my solution. There has been some discussion as to whether this is right or not. I welcome any comments/corrections.

Initial thoughts – I looked at the site from which the puzzle was taken (http://fivetriangles.blogspot.co.uk/) and saw the following information:

“All Five Triangles problems share a characteristic: however opaque they may appear initially, none requires more than common mathematics skills learnt in classrooms throughout the world.”

I decided to solve the problem using angles and circle theorems. First, I worked out the angles I needed. I took the approach that there are 5 segments of 5cm in the diagram which run parallel to the line segments AC, CE, EB, BD, and AD. My belief (and this is where I may have gone wrong) is that this is effectively the tangent, where to string leaves the circumference of the circle. Because the tangent is always at 90 degrees to the radius, I thought we might be able to use this to find the sector of the circle the string does not touch, and use this work out the circumference touched by the string, and therefore the solution.

IMG_0002

Given that the interior angle of a pentagon is 108 degrees, the exterior must be 252 degrees. Because ABC is an isosceles triangle ( as AB = BC), angle BAC = 36 degrees. If we draw the line AD, we can see that angles CAD and DAE are also equal to 36 degrees. If we add in a point F so that segment AF bisects the reflex BAE, then FAE = 126 degrees, and FAD = 126 + 36 = 162 degrees. The same is true for FAC.

IMG_0003 (1)

I think a key part of the problem (and again, this may be a mistake in my thinking) is that AC is parallel to the segment of string running from the right hand side of the circle around point A to the right hand side of the angle around point C. The same is true for the other 4 5cm segments. I said above that FAC = 162 degrees. Drawing the tangent to the circle from point A gives a 90 degree angle, which leaves a remaining angle of 72 degrees. This means the difference between the angle from A to the tangent, and the angle created by drawing a perpendicular to FA, is 18 degrees. I therefore think that on each side of each circle, the string remains in contact for a further 18 degrees. This gives a total sector of 216 degrees.

(Interestingly enough, this means that the sector of the semi circle touched by the string is 108 degrees – I wonder if this is significant given the context of the problem?)

Taking this as a fraction of 360, and then multiplying by 5 (for there are 5 circles), gives an answer of 3. The circumference of each circle is 2 pi, so the total curved distance of the string is 6 pi. All that remains is to add the 25cm for each segment, giving a total of (25 + 6 pi)cm, or 43.85cm to 2 decimal places.

As I say, I may be wrong. Shoot away!