Question

In $△ABC$, circumradius, $R=3$. Let $O$ be the circumcentre and $H$ be the orthocentre, then the value of $\frac{1}{64}(A{H}^2+B{C}^2)(B{H}^2+A{C}^2)(C{H}^2+A{B}^2)$ is $3^k$, where $k$ is

Answer 1

In
$△AEH,\cos (90-C)=\frac{AE}{AH}\Rightarrow \sin C=\frac{AE}{AH}\Rightarrow AH=\frac{AE}{\sin C}$
In
$△ABE\cos A=\frac{AE}{c}\Rightarrow AE=c\cos A\therefore AH=\frac{c\cos A}{\sin C}\Rightarrow AH=\frac{2R\sin C\cos A}{\sin C}=2R\cos A$
Similarily $BH=2R\cos B$ and $CH=2R\cos C$
Now,
$\frac{1}{64}(A{H}^2+B{C}^2)(B{H}^2+A{C}^2)(C{H}^2+A{B}^2)=3^k\frac{1}{64}\left(\right(2R\cos A{)}^2+\left(2R\sin A{)}^2\right)\left(\right(2R\cos B{)}^2+\left(2R\sin B{)}^2\right)\left(\right(2R\cos C{)}^2+\left(2R\sin C{)}^2\right)=3^k$
$\Rightarrow \frac{1}{64}\left(64R^6\right)=3^k{R}^6=3^k$
since $R=3\Rightarrow k=6$