Prove that he distance between the circum-centre and the ortho-centre of a triangle ABC is `Rsqrt(1-8cosAcos BcosC.`

Explanation

Let O and H be the circumcenter and the orthocenter, respectively.
If OF is the perpendicular to AB, we have
`angleOAF = 90^(@) - angle AOF = 90^(@) -C`

Also, `angleHAL = 90^(@) -C`
Hence, `angle OAH = A - angle OAF - angle HAL`
`= A -2 (90^(@) -C)`
`= A + 2C -180^(@)`
`= A + 2C -(A + B +C) = C -B`
Also, `OA = R and HA = 2R cos A`
Now in `Delta AOH`
`OH^(2) = OA^(2) + HA^(2) - 2OA HA cos (angle OAH)`
`=R^(2) + 4R^(2) cos^(2) A - 4R^(2) cos A cos (C -B)`
`= R^(2) + 4R^(2) cos A [cos A - cos (C -B)]`
`=R^(2) -4R^(2) cos A [cos (B + C) + cos (C - B)]`
`= R^(2) -8R^(2) cos A cos B cos C`
Hence, `OH = R sqrt(1-8 cos A cos B cos C)`