The Arc Midpoint Find
To continue on the topic of circular arc midpoint computation, consider the following proposition which is valid in .
Let AMB be an arc (see figure) with center C, midpoint M, and angular measure 2\theta \ne \pi. If points M, C, A and B have x-coordinates \mu, c, 2a, and 2b respectively, then
\mu = c + (a + b - c)\sec\theta

The same holds for the other n-1 coordinates y, z, ... w.
Below we present two different approaches to the proof of the proposition.
First approach:
As discussed in Dot Product Finds Arc Midpoint, M obeys the vector equality \vec{OM} = \vec{OC} \pm \left( \vec{CA} + \vec{CB} \right)/(2\lambda)^{1/2}, where \lambda = 1+\left( \vec{CA} \cdot \vec{CB}\right)/r^2 = 1 + \cos2\theta, r is a radius of the arc. Hence, without loss of generatlity, x-coordinates satisfy \large \mu = c \pm \frac{(2a-c) + (2b-c)}{\sqrt{2(1+\cos2\theta)}} = c + (a + b - c)\sec\theta
Second approach:
Let T, with x=t, be the midpoint of the chord AB. Denote |CT|=h. Then, without loss of generaliry, |t-c|:|\mu - c|=h:r. Since h:r=|\cos\theta|, and t = a+b, we get |\mu - c|=|a+b-c||\sec\theta|, which obviously holds with no modulus as well. So, \mu = c + (a + b - c)\sec\theta
What if 2\theta = \pi? In that case infinitely many different values of \mu can be found.
Copyright © August, 2011 by Oleksandr G. Akulov
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