trianglestrigonometry
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.
The formula for length of $AG = \frac{2}{3}\cdot\frac{1}{2}\sqrt{AC^2-BC^2+AB^2}$. Similarly, the length of $BG=\frac{2}{3}\cdot\frac{1}{2}\sqrt{AB^2+BC^2-AC^2}$. Now, we can find the circumradius of the triangle is $$\frac{AB\cdot BG\cdot AG}{4\sqrt{s(s-AB)(s-BG)(s-AG)}}$$, where $s=\frac{AG+BG+AB}{2}$.
If you are given $AB,BC,AC$, surely you cannot get the wrong answer unless there is a calculation error.
ABC is any triangle
H is the orthocenter of the ABC triangle
J is the orthocenter of the BCH triangle
BC=common side
altitudes:
r is a line passing through B, perpendicular to AC.$( H \; ∈ \; r)$
s is a line passing through C, perpendicular to AB.$( H \; ∈ \; s)$
${ H }=r ∩ s$
BCH triangle:
t is a line passing through B, perpendicular to CH.$( J \; ∈ \; t)(t≡\overleftrightarrow{AB})$
u is a line passing through C, perpendicular to BH.$( J \; ∈ \; u)(u≡\overleftrightarrow{AC})$
$t ∩ u=J=\overleftrightarrow{AB}∩\overleftrightarrow{AC}=A$
Best Answer
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 $$(2\cos^2A-1)+(2\cos^2B-1)+(2\cos^2C-1)=-4\cos A\cos B\cos C-1$$