## How to solve a maths problem, for ‘beginners’

On Sunday 14th December, @Srcav published a maths problem that I couldn’t help attempting; and I got very stuck, very quickly.  Yesterday, 9 days later, and with maybe 90 minutes of concerted thinking given over to the problem, I finally cracked it.  But I want to talk through that process in a way that people who aren’t proficient in maths might be able to follow. Because, this is the holy grail of maths teaching, this is what every maths teacher hopes their students will achieve: to be able to tackle complex, unfamiliar problems.  Maths problems don’t have obvious solutions, though the more experienced and more proficient a mathematician is, the faster they will adapt their experience to solve the novel problem.  And so, for students with so little knowledge and experience, helping them to tackle complex problems is extremely challenging for teachers; I think we’re far from having institutionally cracked this, despite many, many disparate thoughts existing out there.

I admit, I might not be able to explain this well enough in the time I have for everyone to follow… the more maths you know, the easier it will be.

So here’s the problem, along with my meandering, scattered notes and working:

Ahem… I’m not sure I’d approve if my students set out their work like that!  But then on the other hand, perhaps more complex problems, where you have to try things out, hit dead ends, go back, try again, persevere when the end is never in sight; perhaps they don’t lend themselves to neatly structured working, at least, not on first pass?

Alright, let’s break this down.

Step 1

Though I didn’t read @Srcav’s post in full, I did read the first few lines where he mentioned using some Trig.  I figured I’d give that a go and see what it got me (top right).  Not much.

I *was* able to look at the numbers for the angles though and conclude that they confirmed my gut instinct, that the diagonal through the rectangle would meet the square half way along the right hand side.  What next?

Step 2

I’ve seen geometric problems like this before, and I’ve solved them by drawing the shapes onto Cartesian Planes (graphs/axes) figuring out equations for the lines, and then solving them simultaneously to find the coordinates of points on the shape, so I gave that a go.

When you do this you can put the Origin (point (0,0)) wherever you like.  The centre of the circle is an obvious place.  If the origin is at the centre of a circle, then the equation of the circle is:

x^2 + y^2 = r^2

Where r is the circle’s radius, the thing we’re trying to find!  If I knew just *one* point on the circle, one value for x and y, I could find the value of r.  But I put the origin in the centre of the circle, I don’t know how far from the origin that rectangle is, so can’t use it to find a point on the circle so substitute in.  Had to abandon this route.

Step 3 – Sticking to the Graph

We’re going to use the graph idea for a while now, but I’ve moved the origin so that it’s at the top left of the rectangle.  By putting it here, I know that the point where the rectangle and the circle touch/intersect is (2, -1).  But… the equation of the circle has now changed.  Since its centre is no longer on the origin, the equation is more complicated:

(x – x1)^2 + (y – y1)^2 = r^2

x1 and y1 are numbers we don’t know, that tell us how far from the origin the centre of the circle is.  To create the circle’s equation before, we only needed 1 piece of information, the coordinates of just one point.  By moving the origin I now *have* the coordinates of one point… but I actually now need *three* pieces of information, or three coordinates, to figure out the equation of the circle, and therefore the value of r!

Step 4 – Finding another coordinate

Because the square and circle are rotationally symmetrical, I realised I could rotate that rectangle by 90 degrees, and it would still touch the circle.  This was how I found the second coordinate on the diagram above, (1, -2).  But I still need one more somehow…

Step 5 – Trying to find a third coordinate

I tried all manner of silly things that tested my understanding of how simultaneous equations work.  Bit more Trig to see if that would help… in short, it failed, and there was another option that I’d been toying with that I hadn’t really tried out properly yet…

Step 6 – Similar Shapes

This is another problem that me and a bunch of friends spent somewhere well over an hour trying to solve.

Some people solved this using Pythagoras (probably the easiest solution), one Cambridge maths graduate solved it using the mathematics of infinite series… arguably unnecessarily complex!  I solved it using something thinking about similar shapes that I find tricky to communicate… but which worked.  So, maybe I could use a similar idea here…

Basically, because the rectangle and circle are inside a square, I reasoned that if a second, larger rectangle had its top left corner attached to the bottom right of our original rectangle, then it could be scaled up in proportion to the original rectangle, and my gut told me, it would meet the right hand side of the square at its mid-point.

This is what the middle-right diagram is all about.

Problem is, I’ve no idea how to communicate my thinking here clearly… In short, what I’m trying next only works if that rectangle meets the square at its mid-point; I need to prove that that’s the case!

I used the earlier Trig and my knowledge of various angles I gained from it to reason this to be true.  I drew the angles onto the diagram.  There’s now a strange kind of symmetry going on that I can see… but I find very difficult to explain.  Let’s put it like this:

Magic happens, then the diagram I drew is true.

Oh no wait!  …I’d already worked out the equation of the diagonal as y = -x/2,and proved that it met the square at its midpoint on the right hand side – I got that from all the early Trig.  So if my two rectangles were joined the way I described, and were similar shapes, then that same diagonal must also be a diagonal of the bigger rectangle, meaning that its bottom right corner must touch the square half way along its right hand side.

I was then able to look at the angles I took from the Trig to see that the second green dotted line must have the equation y = -2x, and must intersect with the rectangle at its bottom left corner.

Step 7 – Solving equations simultaneously to work out coordinate points

Given that, (thanks magic!), I could create equations for the vertical line that forms the left side of the big rectangle, and for the second green dotted line.  Solving them simultaneously gives me the coordinates of the bottom left corner of the rectangle (2, -4).  The coordinates of the top left corner were always known (2, -1), which means I can now work out the height of the big rectangle.

Step 8 – Demonstrating Kahneman System 1 thinking, or what my chemistry teacher called: “Intelligent idiocy”

Looking at this now, it’s painfully obvious to me that I’ve already solved the problem.  The big rectangle has a height of 4, the small one a height of 1, the total height is 5, I’ve reasoned that this length is equal to half the length of the square, therefore the radius of the circle, 5, done.

That’s not what I did.

At the very bottom right, the idea of some kind of proportional reasoning still firmly glued to the forefront of my mind, I opted instead to calculate the length of the of the big rectangle as well.  8cm, turns out to be.  I then added that to the 2cm length of the small rectangle, then divided by 2.  Like an idiot.

Conclusion

So there we go… that’s loosely what it looked like for me trying to solve this problem; perhaps it gives a hint as to why we struggle to much to ‘teach’ this way of thinking to students.

After this I read @Srcav’s post, only to find that he had two much more elegant ways of solving it.  One problem is that when I had the origin at the centre of the circle, the real units we were given at the start meant I was thinking all in terms of real numbers.  If I’d done what he did and build a diagram with a few unknowns and worked through with them, I might have solved all this much more quickly, and with a method that’s much simpler.

Turns out there’s another way of solving this, once again using Pythagoras’ Theorem, which is yet simpler still.  Interestingly, @Srcav hit upon it when he decided he wasn’t happy enough with his original method.

Problem Solving

Teaching problem solving to kids in schools is the holy grail of mathematics teaching.  It is insanely complex, difficult, and despite many wider and disparate thoughts on the issue, we’re nowhere close to solving it.

But while I haven’t solved it myself, I am convinced that it is a problem with a solution, and that it will be solved in the not-so-distant future.  It’s pretty exciting when you consider what might be achieved by an entire population able to reason so powerfully…