Not all circles are orthogonal

Denote the circumcentre by $O$ and the nine-point centre by $N$ . Because the circumcircle and the nine-point circle are orthogonal to each other and have radii $R$ and $R/2$ respectively, we have

$R^2 +( \frac {R}{2})^2 = ON ^2$

or $5 \times R^2 = (2 \times ON)^2 = OH^2 = R^2(1 - 8cosAcosBcosC)$ , where $H$ is the orthocentre of the triangle.

So $5 = 1- 8cosAcosBcosC$ and therefore $cosAcosBcosC = \frac {-1}{2}$