## Catchy Mnemonics

I find most memory aids a little silly. Why learn a rhyme about horses when you can just learn the trig ratios? Why learn a rhyme about the duke of York when you can just remember the order the colours come in?

However, I find that music is a good way of remembering things. For some reason music is good for us to remember words. I can, for instance, remember the words to a great deal of 90s pop songs even though I didn’t like them and never chose to listen to them because I heard them out places and on TV so often that they got lodged in my brain forever.

This is something I have seen work well in learning maths facts. Year on year I hear pupils sing “mean is average, mean is average…” etc in lessons to remember the averages. And I also hear a great many variations on the circle song.

Last year when I was teaching kinematics one of the students said “Sir,play the SUVAT song.” I’d not heard of the SUVAT song and he found it on you tube and we listened to it. It’s simple and it’s catchy and it really helped him and his class remember those equations. So on Tuesday I played it to one of my mechanics classes. By the end of the leson I’d heard three people sing it and it has been stuck in my head all week.

*What do you think about mnemonics? Do they have a place? Have you any songs or rhymes that you use to remember things or that you encourage students to use? And do they help?*

## Circles and Triangles

Regular readers will know that I love a good puzzle. I love all maths problems, but ones which make me think and get me stuck a bit are by far my favourite. The other day Ed Southall (@solvemymaths) shared this little beauty that did just that:

I thought “Circles and a 3 4 5 triangle – what an awesome puzzle”, I reached for a pen an paper and drew out the puzzle.

I was at a bit of a loss to start with. I did some pythag to work some things out:

Eliminated y and did some algebra:

Wrote out what I knew:

And drew a diagram that didn’t help much:

I then added some additional lines to my original diagram:

Which made me see what I needed to do!

I redrew the important bits (using the knowledge that radii meet tangents at 90 degrees and that the line was 3.2 away from c but the center of the large circle was 2.5 away):

Then considered the left bit first:

Used Pythagoras’s theorem:

Then solved for x:

Then briefly git annoyed at myself because I’d already used x for something else.

I did the same with the other side to find the final radius.

A lovely puzzle using mainly Pythagoras’s theorem, circle theorems and algebra so one that is, in theory at least, accessible to GCSE students.

I hope you enjoyed this one as much as I did!

## A lovely old problem

Recently Ed Southall shared this problem from 1976:

I’m not entirely sure if it is from an A level or and O level paper. It covers topics that currently sit on the A level, but I think calculus was on the O level at some point. *Edit: it’s O level* I saw the question and couldn’t help but have a try at it.

First, I drew the diagram – of course:

I have the coordinates of P, and hence N so I needed to work out the coordinates of Q. To do this I differentiated to get the gradient of a tangent and followed to get the gradient of a tangent at P.

Next I found the equation, and hence the X intercept.

And then, because I’m am idiot, I decided to work out the Y coordinate I already knew and had used!

The word in brackets is duh…..

Now I had all three point.

It was a simple division to find the tangent ratio of the angle.

The next 2 parts were trivial:

And then I misread the question and assumed I’d been asked to find the shaded region (actually part d).

Because I decided calculators were probably not widely available in 1976 I did it without one:

I thought it was quite a lot of complicated simplifying, but then I saw part c and the nice answer it gives:

Which makes the simplifying in part d simpler:

*I thought this was a lovely question and I found it enjoyable to do. It tests a number of skills together and although it is scaffolded it still requires a little bit of thinking. I hope to see some nice big questions like this on the new specification.*

*Edit: The front cover of the paper:*

## Radical Exponents

Recently I saw this from brilliant.org on Facebook and it struck me as an interesting problem:

the first solution is trivial and obvious:

But the Facebook post said there was two, so I set out in search of the next one. As there were exponents I thought I’d take logs of both sides:

Then realised I could take logs to base X and make things a whole lot simpler….

So x = 9/4

As you can see it reduces to an easily solvable problem, and all that was left was to check the answer:

A lovely little problem that gives a good work out to algebra and log skills.

## Is one solution more elegant?

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:

I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

**Method 1**

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule, here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right. I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

**Method 2**

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

**Method 3**

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

**Method 4**

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

*(6 + x)^2 = 45 + 9 + x^2 *

x^2 + 12x + 36 = 54 + x^2

12x = 18

*x = 1.5*

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:

All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

*Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!*

Cross-posted to Betterqs here.

## Mathematical Style

Yesterday I read this post from Tom Bennison (@DrBennison). The post was written to start a conversation for a twitter chat that I unfortunately couldn’t make. It did, however, make me think.

He was questioning Wetherby mathematical elegance and style should be assessed at A level. Suggesting that solutions with more elegance should be awarded more marks.

Bizarrely, the example he used was almost exactly the same as a discussion if had with a year 13 class not long before I read his post. His example was finding the midpoint of a quadratic. He looked at two methods – completing the square and differentiation – and suggested that as CTS is more elegant that should be worth more.

I agree immensely that CTS is a preferable method with far more elegance, but I don’t think the marks should be different depending on the method you choose. I feel that we should be encouraging mathematical thought, trying to create young mathematicians who can apply themselves to a problem and find their own way through. I feel if we start assigning marks for elegance and style them we would be moving towards the “guess what’s in my head” style of assessment that I feel we need to be moving away from. The way to do well would be to spot from a question what the examiner wants, rather than to apply the mathematical tools at ones disposal and find a solution.

Back to that Y13 lesson I mentioned, we were looking back over some C3 functions work and one of the questions involved finding the range of a quadratic function – so obviously it was necessary to.find the minimum. A discussion ensued as to how to do this with students coming up with 3 valid methods. The two mentioned above, both of which I find quite elegant, although I do much prefer CTS. The third method was suggested by one student who said “it’s -b/a – you just do -b/a” I knew what he meant – he was saying that this was the x value where the minimum occurred and that you put that in to find y, but he didn’t really understand what it was or why. He’d come across the method online and has learned it as a trick. When I showed him it came from completing the square and looking at it as a graph transformation, I saw the light bulb come on.

It is an interesting discussion. Some methods are far more elegan, and some are just algorithmic tricks. I think that the lack of understanding with these tricks will lead to marks being lost. So perhaps this will self regulate.

*I’d love to hear your views in this, which way would you tackle finding the minimum of a quadratic? And do you think we should assign marks to elegance and style?*

## Infuriatingly impossible exam questions

Today I was working on some Vectors exam questions with my Y13 mechanics class and I came across this question:

A student had answered it and had gotten part d wrong. What he had done was this:

I have recreated is incorrect working.Obviously he had found out when the ship was at the lighthouse, instead of 10km away. I explained this to him and started to explain how he should have tackled this when a sudden realisation angered me.

Now for those if you that didn’t work through the question, here is the actual answer:

Can you see what had me infuriated?

This is an impossible answer! If the lighthouse is on the trajectory of the ship and it will hit said lighthouse at t=3 then that would stop the ship! At the very least it would slow it down!!!! In reality it would have to avoid the lighthouse and change trajectory. Meaning the second answer, T=5, would not happen under any circumstances!

My initial thought was:

“are they expecting students to spot this and discount the second answer? That’s a bit harsh.”So I checked the markscheme:

Nope, they are looking for both answers. Argh! I can understand using a real life context in mechanics, I really can. But why not check for this sort of thing!

What do you guys think? Is this infuriating or am I just getting get up over nothing? I’d love to hear your views in the comments or via social media.## Share this via:

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