G C S E

[kuiperbelt]

Please see link below:

https://ibb.co/GskRDQx

[ClareTG0ld]

Something seems to be off in the diagram, because c is undefined, and y isn't used. But on the assumption that c is meant to be y, you can apply Pythagoras twice:

w^2 + x^2 = y^2
(w+1)^2 + (x+1)^2 = (y+1)^2

expanding out the second one using (w+1)^2 = w^2 + 2w + 1, etc., gives:

w^2 + 2w + 1 + x^2 + 2x + 1 = y^2 + 2y + 1
w^2 + x^2 + 2w + 1 + 2x + 1 = y^2 + 2y + 1

cancelling the w^2 + x^2 = y^2 from the first triangle, gives just

2w + 1 + 2x + 1 = 2y + 1

and now subtract one from both sides, et voila!

[ClareTG0ld]

Going back to one of my earlier long answers, you might wonder how I knew to do this. The answer is that I didn't. But:

1. I see a right-angled triangle, I do Pythagoras. What else is there?
2. Then I see algebra in brackets, I expand. Again, nothing else to do.
3. Then I notice that there's a lot of repetition in the two equations, so I exploit that to "cross out" everything in the long equation that appears in the shorter one.
4. By "chance" this happens to give the answer we were asked about, but I didn't necessarily need to know that it *would* in order to try the above.

I suppose there's a bit of a lie there: I didn't only realise this would work having finished it, and the whole thing sort of flashed through my head in half a second once I'd whispered the word "Pythagoras" to myself. But that's because I've had an entire extra lifetime to practise this as compared with GCSE students. The key thought process is still more about taking standard "tools" out, and playing around with them, knowing that at least one of these "tools" will work and having the patience to try them -- and occasionally fail, and need to pull out a different tool! But one will work. Don't panic!