This is the reason why a half step was take,
a old number of steps is required to reach length \(1\) at \(\cfrac{\pi}{2}\) further down. Without the half steps the formulation of \(L\) will scream for an even factor.
\(M=4\cfrac{N}{n}\) is an even partition such that integer steps can be taken to \(N\). \(a\) is odd because \(N\) is odd due to \(n\) being odd. \(b\) is also odd for the same reason. So \(c\) is even such that \(a+b+c\) fills up \(M\) an even number. But full steps in \(c\) cannot reach \(\cfrac{\pi}{2}\). \(L\) is in odd steps only due to \(N\) and \(n\) being odd. So, a half step is taken to find the step count and \(c\) is recovered by doubling the half step count ( times \(2\)).
It is possible instead to step all the way to \(\cfrac{3\pi}{2}\) and then divide the count obtained by \(3\)? In this case the expression for length \(\bar{c}\) is\(\cfrac{3}{L}\) which is the reciprocal of an integer only if \(L\) has a factor \(3\). This is not the case for prime numbers other then \(3\), and the prove will fail.
We start not with \(n\) dividing a circle, we start with \(4\) dividing the circle.
Goodnight.
