# Teachers are amazing.

For the last few days I’ve been thinking about the art of teaching – in part because it’s my job to train people to teach and I have a new set of trainees, but also because my recent(ish) trip to Japan was a bit of a revelation in so much as discovering how highly regarded teaching as an art actually is.

There’s a lot of talk about mastery of mathematics these days, but mastery of teaching is, currently, of greater interest to me.

It’s only when trainees have been teaching for a while that they start to fully appreciate just how many micro-decisions are involved in teaching. Behaviour management alone is a careful tightrope of reading people, anticipating their responses before (!), during and after they’ve made them, and subtly manipulating them to a place where their likely actions are the most predictable and controllable. It’s hugely complicated, but as it becomes part of daily working life, we probably don’t step back and notice how amazing it is that we’re even able to do it. All that before we even consider the difficulties of imparting knowledge and abstract concepts onto young impressionable minds. Take a moment to reflect on the complexity of the skills you possess. You’re fucking brilliant.

# Problem Solving #4

I stumbled across this question in a book yesterday:

I know the problem, and in fact have solved it a few times before, but upon looking at it I couldn’t quite remember how I solved it, so I thought I’d have another stab at it. It didn’t go well…!

I knew I needed something like this diagram, because there’s just nothing else I can think of doing with the tiny amount of information I’m given.

More specifically, I just need this half (I think). But this is where I got into a pickle. I stared at the above diagram for a little while, convinced it was what I needed, but completely lost at how it helps me. Sure I have one side, and what I need is part of the triangle, but surely I need *more* information??

Now as I said, I have solved this problem a few times, and in fact, the first time it was shown to me I solved it in about a minute or two, but for whatever reason I couldn’t for the life of me get past this point yesterday. The reason why I couldn’t progress was simple – I wasn’t following my own advice on solving problems. Specifically, I’d recommend to students just finding as many things as you can until you find a way in. So after staring at that little triangle for far too long, I decided to just use Pythagoras anyway, despite “knowing” that I didn’t have enough information for it to be useful.

$R^2 = r^2 + 25$

So what? I surely needed to replace R with some kind of expression using r and a number… but I don’t have that information, so I have a ‘useless’ two variable equation and nothing else to work with.

I re-read the question and realised I’d been hung up with the seemingly useless application of Pythag instead of focusing on the fact that the question concerns area. Finally, I had that moment of clarity.

$\pi R^2 = \pi r^2 + 25\pi$

The bit I want is the shaded bit, so I can effectively ignore $\pi r^2$

So

$Area=25\pi$

Weird how it took me so long to ‘see’ it this time, and I spotted it so quickly a couple of years ago. I probably shouldn’t overthink that…

# Opening the floodgates

Today universities received their bidding allocations for 2018-19 teacher training. The process goes a little like this: At the end of an academic year, universities and schools direct partners bid for the maximum number of trainees they are allowed to recruit per subject to NCTL. So if I want, say, 20 maths trainees, I put 20 in a little box, and cross my fingers that a) NCTL let me have 20, and b) I can get 20 trainees.

In theory, if I meet or roughly meet my allocations, I’m pretty likely to get the same numbers approved next year (I still have to bid, it doesn’t roll over) – or even an increase if I’m feeling lucky. If I want new subjects or a relatively large increase in a particular subject, I have to write about why I think I deserve them. You might wonder why NCTL caps numbers at all, particularly with the current teacher shortages. In essence, part of this is down to allowing for fair competition across providers, and ensuring that areas that do not have particular shortages do not take on more teachers than the locality needs. However, this year (again), NCTL has effectively opened the floodgates and allowed all providers unlimited recruitment in all subjects (secondary and primary) apart from for PE (max 4 places) and salaried places. This seems to be an attempt to get as many new teachers into schools in September 2019 as humanly possible to help ease the explosive combination of falling recruitment, falling retention, and rising student numbers. To some extent this decision is of course welcome (although small providers will once again be pitching against large providers who can effectively just sponge up all the applicants in the area). That aside, more teachers trained = more progress gained (i made that up, you can keep it).

But of course there’s a but…

Opening the floodgates assumes there’s water waiting on the other side.

Furthermore, is there another strategy to try and cling on to our existing teachers? To fix the leaky dam that is perilously close to bursting? It seems to me there are a lot of people out there with a PGCE who aren’t in schools anymore, and a lot with one foot out the door.

No doubt we can expect a glossy tv campaign at some point soon boasting golden fleece salaries, perfect children and young spritely teachers laughing in well lit classrooms as their bunsun burner flickers under a test tube of mysterious fluorescent gunk.

When the advert finishes we can return to our newspapers to read about how kids today are a mess and it’s the fault of schools – and teachers need to teach more stuff like how to be a balanced member of society, and how to have British values (the old ones not the new ones). Then we can go to work the next day and feign surprise when a colleague tells us they’re packing it all in because they can make more money tutoring.

This juxtaposition of dragging down the profession in the media (and society), whilst simultanously celebrating the glossy career of teaching and pretending that part of the problem is that we are restricted by how many teachers we can train in one year does no-one any favours.

We need honest dialogue. We need more voices fighting our corner and celebrating what we do. We need to have that difficult conversation about how the decline in the perception of youth behaviour is also about parenting and the media instead of just using teachers as a piñata. We need bigger school budgets, continual structured subject specific training, competitive salaries and an overhaul of the nonsense pressures of unnecessary workload.

Make the job appealing and people will come look around. Maybe they’ll even stay a while.

# Sorry we keep lying to you…

It’s well documented that many school students and adults alike are less than fond of mathematics. It tends to be a theme I discuss on here. I’ve singled out over-emphasis on speed of processes, misguided attempts at trying to convince students it’s entirely relevant to their daily lives, and of course, not explaining things,  as factors. These are all ways in which we, the teachers, are sometimes subconsciously influencing things – but it’s certainly not all our fault though. We live in a country where, typically, it’s a badge of honour to be crap at maths and still miraculously live a normal life. Our bloated curriculum and imbalanced subject hierarchy don’t exactly help either. I was reading a maths book for trainee teachers the other day and I came across a familiarly painful explanation of the column method of subtraction, stating that you ‘cannot subtract three from two, so you must borrow…” blah blah blah. It got me thinking – just how much do we lie to our poor little mushy brained students when they’re absorbing what they think is the truth? What are the consequences as the truth suddenly changes conveniently and we forget to tell them? Perhaps I should hashtag this as #fakemaths. I know the best hashtags. Nobody does hashtags better than me. I digress…

Below is a letter I have kindly drafted for my maths teachers over the years to send back to me. I’ll even pay for the postage.

Dear Ed,

By strange coincidence, we, your former maths teachers, have all ended up in a room together with only a pen and some paper and your name and address on a sticky label. We figured it meant something, so we started talking and realised we’ve all been guilty of a bad thing or three. We hereby apologise for the following truths we accidentally lied to you about (because reasons):

1. Really sorry we told you that you can’t subtract a big number from a little number. We just meant don’t do it or the algorithm is harder to interpret.
2. Whoops, you know the squares? They’re rectangles too. You could have had that mark.
3. My method isn’t the preferred method, it’s just one of many, and efficiency is subjective.
4. Showing all the steps really doesn’t mean you’re a better mathematician, it just means I can mark your work against the criteria I’ve been given.
5. The even numbers in the book weren’t harder than the odd numbers, I just didn’t have anything else to give you to do.
6. You don’t need to be good at maths to go to a supermarket.
7. You don’t tend to use algebra explicitly in restaurants.
8. We told you ‘because it works’ because we didnt know why it worked.
9. Opinions (before we’ve told you facts) aren’t really wrong as such.
10. Equals doesn’t mean ‘find the answer’
11. Algebra is not the same as 2 apples + 1 apple = 3 apples
12. We weren’t really adding a zero at all – but shhhh don’t tell anyone.
13. You don’t “just have to memorise it”.
14. There’s no such thing as ‘borrowing’ in maths – except for forgotten pencils and rubbers.
16. Mathematicians struggle too.

Hope we didn’t accidentally destroy the best subject in the world, but we totally understand if you think we’re lying to you about that.

Signed blah.

# Problem Solving #3

“A square is constructed as shown. Point A is the midpoint of a side. Prove that triangle BCD is isosceles (and not equilateral)”

If you don’t want to be swayed by my thinking, solve it now rather than reading ahead.

I like proof questions because they give you a little comfort blanket – you know the answer before you start. I stared at the diagram a little while, explaining internally how each line is constructed as a kind of mental warm up I suppose. Then I thought to myself “I’ll try and find the side lengths”. I think I was swayed to pursue sides rather than angles because I identified this triangle fairly quickly and figured it’d probably be useful:

I gave it side lengths of 2 and 1 for easy numbers, and so the hypotenuse must be

$\sqrt{5}$

Then I focused for a little while on this red triangle:

I hoped it’d be similar, but couldn’t convince myself that it was by using side lengths. At this point I only know the hypotenuse is 2.

So I thought maybe I could with angles.

I don’t know the values for the red and green angles. I could work them out as I know the side lengths, but I don’t want to yet. Instead, I noted that the green and red angles sum to 90 degrees. Therefore, this angle must also be ‘red’:

And as angles in a triangle sum to 180, and we have a 90 degree angle at point C, then the third angle in the red triangle must be ‘green’, so it’s a similar triangle to the one I started with. Phew!

Why is that useful? Well, now I can calculate side DC using similarity.

$\frac{DC}{2}=\frac{2}{\sqrt{5}}$

$DC = \frac{4\sqrt{5}}{5}$

Well, I’ve proved that triangle BCD isn’t equilateral at least.

Now to find side BC…

I know that angle BDC must also be ‘red’:

I could just use the Cosine rule and find side BC to be 2 or the surd for CD (I assume it would come out at 2 by inspection), but then I’d have to actually work out those angles, and it was late and I needed to get to work early because I’ve been locked out of my work emails all weekend (stupid password change requirements are so inconvenient). I had one last look before bed and thought ‘maybe there’s a way to show that angle DCB is also ‘red’…

The next morning I ate some toast (this is a theme, maths and toast. I highly recommend it). and stared at this diagram:

I stared for a good long while, thinking of that angle next to the right angle. My mind got a bit confused with itself as one of my unwritten approaches in my head was ‘assume it IS red, then what else’, and I found myself in a loop of using something else that would be true if it WAS red, to show that it was red… that doesn’t work in maths sadly.

I even sketched this to see if it would jolt any thoughts out of my brain:

It didn’t. Not that I’m consciously aware of anyway. As I brushed the crumbs from my face and sipped my obligatory morning mug of Yorkshire Tea, an idea finally struck.

I focused again on the red triangle:

It’s a right-angled triangle, so I can circumscribe it. Would that help?

It was a better idea than no idea. A few more details can be deduced now:

So the top line is a tangent, and the midpoint on the side of the square is the centre of the circle…

And the dotted line is a radius, so now I have another pair of red and green angles… i feel like i’m making progress now, but I couldn’t quite make what felt like a final move to unlock the proof. So i drew a second circle as I lost hope that the first would be sufficient on its own (perhaps it is and I missed something):

Notice I’ve ditched that tangent, it’s not important. At last, from this second circle I can happily convince myself that everything works, and the original triangle BCD is isosceles:

I know the two green angles are green from the previous diagram, and i know the dots A, B, C, D & F all lie on the new circle because …

Triangle BDF forms a right triangle for the new circle drawn.

Triangle DBA forms a right triangle with the same height, so the midpoint of its hypotenuse is the same as for BDF – so it’s the same circumscribed circle (so A lies on the circle from BDF)

We know ECD is 90 degrees, so DCA is 90 degrees, so that can be circumscribed.

DA is the hypotenuse of DBA and DCA so all three right triangles have the same circle circumscribing them all. I’m now convinced C lies on both circles.

That’s important because it tells us that BFC is also circumscribed by the same circle, and so FCB is 90 degrees and so DCB is… RED!

I have a feeling I might not have needed both circles for my solution. But I’ve moved on.

Here is the solution approach provided with the problem:

# Problem Solving #2

Whilst eating some toast I felt the need to have a go at another geometry problem this morning:

(not drawn to scale)

“If the radius of both circles is 2, and the area of both the equilateral triangle and rectangle are the same, find the dimensions of the rectangle”

I liked the diagram and the idea so I thought I’d have a go.

This question felt easier than the last one in that i have a direction to go in right from the start. I’m told the areas are the same, and I can use the diagonal of the rectangle to get a pythagorean formula, and a second formula from the triangle, then solve. So there’s less pondering at the initial stage:

Using the red triangle, we get

$x^2+y^2=4^2$

The second formula looked a bit trickier to determine. I decided to go for this triangle:

Which is a 30,60,90 triangle, and so has side ratios $1:2:\sqrt{3}$

I mixed up which side matched which part of the ratio initially, but it didnt look right so I went back and corrected it fairly quickly.

$\frac{a}{2}=\sqrt{3}\ so\ a=2\sqrt{3}$

Now I have the side length, I can derive the area:

$2\sqrt{3}*0.5*3=3\sqrt{3}$

And so I have my second formula for the rectangle:

$xy=3\sqrt{3}$

squaring this gives :

$x^2y^2=27, so\ x^2=\frac{27}{y^2}$

Substitute into the other equation:

$y^2+\frac{27}{y^2}=16$

$y^4-16y^2+27=0$

Now I have a quartic, which I’m not happy about. Fortunately as I only have a power of 4 and 2 to deal with, I can substitute with say, z:

$z = y^2$

$z^2-16z+27=0$

$z=8\pm\sqrt{37}$

$y^2=8\pm\sqrt{37}$

This is getting yucky. At this point I square rooted everything and started messing with

$xy=3\sqrt{3}$

Which got even messier to the point that I stopped and had a rethink. I went back to

$y^2=8\pm\sqrt{37}$

And it dawned on me that if I use

$x^2=16-(8\pm\sqrt{37})$

then I can see by observation that when:

$y^2=8+\sqrt{37}$

$x^2=8-\sqrt{37}$

and when

$y^2=8-\sqrt{37}$

$x^2=8+\sqrt{37}$

Hence the rectangle has dimensions $\sqrt{8+\sqrt{37}} ,\ \sqrt{8-\sqrt{37}}$

# Problem Solving

Someone tweeted me this puzzler a couple of days ago (I forget who it was, apologies):

“If the square has side 2cm, what is the radius of the circle?”

I had a stab at it today and thought I’d share my process, which may well be awful, but as Bob Hoskins famously didn’t quite say, “it’s good to share”

My first thought was “it’s a circle, I need a radius, where can I get one”. So I drew in a few lines:

I know that where the dotted diagonal lines cross is the centre of the circle, and I drew in the bold black line because I know I can calculate it, which might come in handy.

I thought I better have a visual reminder of what lengths are equal, in the hope it might prompt a new thought:

Then I just stared at it for a while.

I couldn’t visualise how calculating the black diagonal line would help me, even though I could do it. So I drew in another line as I thought about it:

Which helped me spot the blue arrow thing which kicked in a thought about circle theorems. At this point, note I didn’t have a direction I was going in, I was just trying stuff out. Testing the waters to see where it’s shallow. I thought about the circle theorem thing, and eventually decided against it, and ditched the idea. However, looking at the diagram with the two angles on it made me realise the direction I wanted to take:

I want to work with this triangle, because it has the radius (twice, bonus) as its side length, and I can use the length that I know I can calculate (the longest side of the green triangle, which was mentioned in the first diagram). So now all I need is the angles inside the green triangle and off I go. And to get those, I can use the big red bottom left triangle:

So now I have some ugly trig to work through:

[/insert elevator music whilst ugly trig is being worked through]

and out pops r = 1.25 cm.

The nice answer makes me suspect there’s a far nicer approach to this. I’ll ponder it some more.

*Update: A few people have sent me their approaches, which are predictably simpler than mine. Interestingly, some are asking why I ‘chose to solve it in this complicated way’. I didn’t choose to! This is where my first thoughts took me. Anyway, here are some alternative approaches: