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:
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:
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!