## Saturday puzzle

One of the first things I saw this morning when I awoke was this post from solve my maths on facebook:

That’s interesting I thought, so I thought it have a go.

The radius is given to be 2. So we have an equilateral triangle side length 2. Using my knowledge of triangle and exact trig ratios I know the height of such a triangle is root 3 and as such so is the area.

Similarly, as the diagonal of the rectangle is 2 and the short side is 1 we can work out from Pythagoras’s Theorem that the longer side is root 3. And again it follows that so is the area.

Lastly we have the square, the diagonal is 2 and as such each side must be root 2, again this is evident from Pythagoras’s Theorem this gives us an area of 2.

Which leaves us a nice product of the areas as 6.

*I think* *that is correct, I’ve justvwoken up nd this post has been my working, so do about up if you spot an error. And I’d love to hear if youbsolved it a different way.*

## 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.

## An interesting area problem

Here’s an interesting little question for you:

*Have you worked it out? How long did it take you to see it?*

It took me a few seconds at least, I had screenshotted the picture and was reaching for the pencil when the penny dropped, and that’s why I thought it was an interesting question.

The answer is, of course, 100pi. This follows easily from the information you have as the diagonal of the rectangle is clearly a radius – the top left is on the circumference and the bottom right is on the centre.

*So why didn’t I spot it immediately? *

I think the reason for me not spotting it instantly might be the misdirection in the question, the needless info that the height of the rectangle had me thinking about 6, 8, 10 triangles before I had even discovered what the question was.

I see this in students quite often at exam time, they can get confused about what they’re doing and it links to this piece I wrote earlier about analogy mistakes. The difference is I wasn’t constrained by my first instinct but all too often students can be and they can worry that it must be solved in the manner they first thought of.

Earlier today a student was working on an FP1 paper and he was struggling with a parabola question, he had done exactly this, he had assumed one thing which wasn’t the right way and got hung up in it. When he showed me the problem my instinct was the same as his, but when I hit the same dead end he had I stepped back and said “what else do we know”, then saw the right answer. I’m hoping that by seeing me do this he will realise that first instincts aren’t always correct.

I’m going to try this puzzle on all my classes tomorrow and Friday and see if they can manage it!

*How quickly did you see the answer? Do you experience this sort of thinking from your students? I’d love to hear any similar experiences.*

Cross-posted to Betterqs here.

## Problem solving triangles

Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this one:

It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t, it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

*This post was cross posted to the BetterQs blog here.*

## Exact Trigonometric Ratios

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).

Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.

Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.

## Concentric Circles Area Puzzle

This morning I saw this post from Ed Southall (@solvemymaths):

And thought, that looks an interesting puzzle. I’ll have a little go. I think you should too, before reading any further…

Ok, so this is how I approached it. First I drew a sketch:

I assigned the arbitrary variables r and x to the radii of the larger and smaller circles respectively and used the fact that tangents are perpendicular to right angles, and the symmetry of isosceles triangles, to construct two right angled triangles.

I wrote an expression for the required area in r and x. Used Pythagoras’s Theorem to find an expression for x in terms or r, subbed it in and got the lovely answer of 25pi.

*An interesting little puzzle, did you solve it the same way? I’d love to hear alternative solutions.*

## 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!

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