Orthocenter, Circumcenter, and Circumradius

Question

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.

Answer 1

This is a quite famous formula actually.

$HO^2 = 9R^2 - (a^2 + b^2 + c^2)$

For more details see here:

https://www.cut-the-knot.org/arithmetic/algebra/DistanceOH.shtml

Depending on your level (say if you are in 7th or 8th grade e.g.),
it may be quite a difficult formula to prove.

Answer 2

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: $$OH^2=9R^2-a^2-b^2-c^2.$$