## A little circle problem

I’ve just seen this post from Colin Beveridge (@icecolbeveridge) answering this question:

Naturally I had a go at it before reading Colin’s solution. When I read his I found a lovely concise solution that we slightly different to mine, so I thought I’d share mine.

I started by just drawing a right angled triangle from the centre of the circle like so.

I seem to have cut off the denominator of 6 on the angle there. I know that the hypotenuse is 12-r (where r is radius) and the side opposite the known angle is r.

This means I can use the sine ratio of pi/6 to get

r/(12-r) = 1/2

Which leads to:

2r = 12 – r

Then

3r = 12

r = 4

Which is the same as Colin got.

I’ve seen questions like this on A level papers before and I know they often throw students, so I make sure I explore lots of geometry based problems and puzzles to combat this. I’d be interested to know which way you would approach this. Colin and I used a very similar approach, just differing in the point at which we introduced the 12. Which way did you do it?

## Proving Products

Just now one of the great maths based pages I follow shared this:

So naturally I figured I would have a go. I thoughts just get stuck in with the algebra and see what happens, normally a good approach to these things.

My first thought was that if I use 2n – 3, 2n -1, 2n + 1 and 2n +3 then tgere would be less to simplify later. I know that (2n + 1)(2n – 1) = 4n^2 – 1 and (2n – 3)(2n + 3) = 4n^2 – 9 so I multiples these together.

(4n^2 – 1)(4n^2 -9) = 16n^4 – 40n^2 + 9

I thought the best next move would be to complete the square:

(4n^2 – 5)^2 – 16

This shows me that the product of 4 consecutive odd numbers is always 16 less than a perfect square and as such that the product of 4 consecutive odd numbers plus 16 is always a square.

(4n^2 – 5)^2 – 16 + 16 = (4n^2 – 5)^2

A nice little proof to try next time you teach it to your year 11s.