Prove that $\pi>3$ using geometry I was asked this question today in an interview.
Question: Prove that $\pi>3$ using geometry.
They gave me hints about drawing a unit circle and then inscribing an equilateral triangle and then proceeding. But I could not follow. Can anyone help?
 A: The inscribed $12$-gon in the unit circle has area $\frac{12}{2}\sin (2\pi/12)=3$. The area of the unit circle is $\pi$. Hence $\pi\ge 3$. 
A: The inscribed hexagon in the unit circle has perimeter $6$. The perimeter of the circle is $2\pi$, hence $\pi > 3$.
A: I am not sure if this is redundant, but:
If an equilateral triangle is inscribed in a unit circle, and if, on each side of the inscribed triangle, an isosceles triangle is further inscribed in the circle, then an equilateral hexagon with each side of length $=1$ results; but then $6 < 2\pi$ implies $3 < \pi$.
So is this something you are after?
A: With a little bit of cheating, you don't need the whole hexagon...  
Let $O$ be the centre of the unit circle with equilateral $\triangle ABC$ inscribed in it.  Extend $\vec {AO}$ to meet the circle at D.  
As $BC$ and $OD$ are perpendicular bisectors of each other,  $\triangle OBD$ is isosceles, and hence $|BD|=1$.  But this must be smaller than the minor arc subtended, which has length $\dfrac{\pi}3$. 
