Circular Domain, Complex Range!

The image can be :

$H-Orthocentre\\S-Circumcentre$

We know $N$ , centre of nine point circle bisects $\overline{HS}$ .

Thus $\displaystyle\begin{aligned}ON &= \frac{1}{2} OS\\&=\dfrac{R}{2}\sqrt{1-8\cos A\cos B\cos C}\\ &=\dfrac{R}{2}\sqrt{3+2\sum\cos 2A}\end{aligned}$

And nine point circle is smaller than circumcircle, so that the maximum common chord passes through the centre $N$ of the nine point circle. Also, we know that radius of nine point circle is half of circumradius i.e. $\large r_{9pt \odot le} = \frac{R}{2}$

In the right angle triangle hence formed,

$\displaystyle \sqrt{R^2 - \dfrac{R^2}{4}} =\dfrac{\sqrt{3}}{2}R$

Using the aforesaid relation,

$\displaystyle\begin{aligned}\dfrac{R\sqrt{3}}{2} &= \dfrac{R}{2}\sqrt{3+2\sum\cos 2A}\\ 3 &= 3 + 2\sum\cos 2A \end{aligned}\\\\\Huge \therefore \boxed{\sum\cos 2A = 0}$