Orthocenter, Circumcenter, and Circumradius In triangle $ABC,$ let $a = BC,$ $b = AC,$ and $c = AB$ be the sides of the triangle. Let $H$ be the orthocenter, and let $O$ and $R$ denote the circumcenter and circumradius, respectively. Express $HO^2$ in terms of $a,$ $b,$ $c,$ and $R.$
I know what orthocenter, circumcenter, and circumradius are, however, I am having trouble expressing it in terms of a, b, c, and R.

In the standard notation we obtain: $$OH^2=\vec{OH}\cdot\vec{OH}=\left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2=$$
$$=3R^2+2\sum_{cyc}\vec{OA}\cdot\vec{OB}=3R^2+2R^2\sum_{cyc}\cos\measuredangle AOB=$$
$$=3R^2+2R^2\sum_{cyc}\cos2\gamma=3R^2+2R^2\sum_{cyc}\left(2\left(\frac{a^2+b^2-c^2}{2ab}\right)^2-1\right)=$$
$$=R^2\sum_{cyc}\left(\left(\frac{a^2+b^2-c^2}{ab}\right)^2-1\right)=$$
$$=\frac{R^2}{a^2b^2c^2}\sum_{cyc}c^2(a^4+b^4+c^4+a^2b^2-2a^2c^2-2b^2c^2)=$$
$$=\frac{R^2}{a^2b^2c^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2).$$
Now, we can prove that $$\frac{R^2}{a^2b^2c^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)=9R^2-a^2-b^2-c^2.$$
Indeed, we need to probe that:
$$\frac{1}{16S^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)=\frac{9a^2b^2c^2}{16S^2}-a^2-b^2-c^2$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2-2a^2b^2c^2)+\sum_{cyc}(2a^2b^2-a^4)\sum_{cyc}a^2=0$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2-2a^2b^2c^2)+$$
$$+\sum_{cyc}(2a^4b^2+2a^4c^2+2a^2b^2c^2-a^6-a^4b^2-a^4c^2)=0,$$
which is obvious.
Id est, we got also the following known nice formula:
$$OH^2=9R^2-a^2-b^2-c^2.$$
