The 2012 STEP Results Discussion Thread

I have a little problem that I devised in S3 just now to keep you entertained until 4.30:

Consider a circle $\gamma$ with centre O and radius 1, and also condsider the line BAC which is tangent to $\gamma$ where C is on the circle. By considering the angle AOB, show that $\displaystyle\frac{ \pi}{4}= \sum_{n=1}^{ \infty} \frac{(-1)^{n-1}}{2n-1}$

I have a little problem that I devised in S3 just now to keep you entertained until 4.30:

Consider a circle $\gamma$ with centre O and radius 1, and also condsider the line BAC which is tangent to $\gamma$ where C is on the circle. By considering the angle AOB, show that $\displaystyle\frac{ \pi}{4}= \sum_{n=1}^{ \infty} \frac{(-1)^{n-1}}{2n-1}$

I have a little problem that I devised in S3 just now to keep you entertained until 4.30:

Consider a circle $\gamma$ with centre O and radius 1, and also consider the line BAC which is tangent to $\gamma$ where C is on the circle. By considering the angle AOB, show that $\displaystyle\frac{ \pi}{4}= \sum_{n=1}^{ \infty} \frac{(-1)^{n-1}}{2n-1}$

EDIT: I just used the word consider too many times

haven't had a proper go but is it something along the lines of over/under estimating the arc length of the part of the circle subtended on an angle pi/4 which converges to the exact value?

1 minute...

Well, how did you find it?

my thoughts on STEP II: It was an easy paper. 4&6 were relatively well known theorems and 11 was the easiest STEP II question I've very seen

Wont have made my offer... pretty bad tbh. Managed 1 complete solution, 2 90% complete and 3 partial =[