A geometry problem by Jitesh Mittal

(using radian measure throughout this solution). Let's consider the simplest situation where we know there is at least one diameter that can be drawn to divide the circle into equal parts: 2 equally spaced points.

In this case the circle is divided into two equal parts by a single diameter and the radian measure for each part is $\pi$ . To create another diameter, it will require 4 equally spaced points. Since the added points must be equally spaced, the second diameter will divide the (previously divided) 2 equal parts resulting in 4 equally divided sections of a circle, or in other words each part will be \(\frac{\pi}{2}.

We can see that if we want to keep adding diameters such that the points are evenly spaced, then they will always be a fraction of \(\pi\) radians.

Thus if we have 8 points evenly spaced it is equivalent to $\frac{2\pi}{8}$ , if we reduce this we get $\frac{\pi}{4}$ . We find that this evenly divides $\pi$ radians: $\frac{(\frac{\pi}{4})}{\pi} = \frac{\pi}{1}*\frac{4}{\pi} = 4$ . So we can see that this can be reduced down to the initial condition which must be satisfied, namely that we begin by dividing the circle into equal halves by a single diameter.

Therefore it can be seen that $\frac{2*\pi}{2009}$ does not divide $\pi$ radians evenly, This shows that the circle cannot be divided in half even once with 2009 equally spaced points and not even a single diameter can be determined.