# Circumradius of triangle ABC where H is orthocentre and $(AH)(BH)(CH) = 3$ and $(AH)^2 + (BH)^2 + (CH)^2 = 7$

Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC. Given $(AH)(BH)(CH) = 3$

and $(AH)^2 + (BH)^2 + (CH)^2 = 7$

Then what is the circumradius of triangle?

I know that AH = 2RcosA and so but I get stuck after two or three steps.

As you wrote, we have $$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C$$ Now using that $$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1\tag1$$ (the proof for $(1)$ is written at the end of this answer)
we get $$\frac{7}{4R^2}+2\cdot\frac{3}{8R^3}=1,$$ i.e. $$(R+1)(2R-3)(2R+1)=0$$ to have $$\color{red}{R=\frac 32}$$
Let us prove $(1)$.
We have \begin{align}\cos 2A+\cos 2B+\cos 2C&=2\cos(A+B)\cos(A-B)+\cos 2C\\\\&=-2\cos C\cos(A-B)+2\cos^2C-1\\\\&=-2\cos C(\cos(A-B)-\cos C)-1\\\\&=-2\cos C\ (\cos(A-B)+\cos(A+B))-1\\\\&=-4\cos A\cos B\cos C-1\end{align} from which we have $$(2\cos^2A-1)+(2\cos^2B-1)+(2\cos^2C-1)=-4\cos A\cos B\cos C-1$$