## 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.

I would definitely avoid trig and aim to find the coordinates of D, A and P. D can be seen immediately as the equation is in the form y=mx+c, so it’s (0,6).

I always try to get Ss to think what gradient means, so -2 means you go right one, you go down 2. So to go down 6, you must go across 3 do A is (3,0). Then the perpendicular is the negative reciprocal so to go left 3 you must go down 1.5, so we get P (0,-1,5). So the length is 1.5 + 6 = 7.5

Nice question and a good one to demonstrate that Trig is actually often best avoided if possible.

Here is another one, although a bit more challenging!

Yes, I remember this question. Head of KS5 asked lots of us and I was the 1st person to answer in just about 2.5 minutes đź™‚

PS: I again used similar triangles and ratio to work it out.

Your third method is (essentially) the reasoning under the proof of the Theorem of Right Triangle Altitude (http://proofsfromthebook.com/2013/08/11/right-triangle-altitude-theorem/). I don’t know whether your curriculum includes that theorem and thus,if your students could use it as a shortcut.

It is indeed that proof. That theorem isn’t on the Curriculum though, which is a shame as it is a lovely theorem.

Triangle OAD and triangle OPA are similar. So tan(ODA)=OA:OD=OP:OA, gives OP=1.5. It follows that PD=7.5 (unit length).

Really no need to over think. I remember one of the students asked me to check this question. But she used the equation of a straight line and worked on it for about 5 minutes.

I had a glance and explained how I would do it in just about 30 seconds. Come on, it’s a non-cal paper. Do not over complicate it!

I’m inclined to agree in principle, although I feel for the student who had spent 5 minutes working on a perfectly valid solution then had her work belittled in such a manner.

I went straight to method 2 and worked out the answer in my head in about a minute or less

Stumbled across this…I’m an A level student and did it the perpendicular gradients way (method 2). I think that’s because at A2 I’m more used to working with algebra than using right angled triangle trig.

Interesting. I may try it on my A level students tomorrow and.she what they try.

I did try this with some of.y A level classes, they seemed to immediately go for Pythagoras’s Theorem but ealise perpendicular gradients is easier before getting too far on pythag

Hi Cav, As suggested, I did this before reading the rest of the blog. I did exactly the same as method 2, perpendicular gradient. I wonder if the fact that it was on my phone and couldn’t use a pencil meaning I had to store everything in my head had a bearing. Trig didn’t occur to me and despite the framing of the question in your blog I still tend to go for a solution straight away rather than mull over different possible methods. I think that this is what I’d encourage the pupils to do. However, if I was teaching this at GCSE I am sure I would use the similar triangles method as this would be straightforward once you start marking known lengths and angles on your diagram which is what I always teach pupils to do.

Hi Mark, thanks for your insights. It is an interesting question I think. I tried it with a few A level classes, they tended to immediately think Pythagoras’s Theorem but then realise perpendicular gradients were simpler.