Question

Circumradius of a $△ABC$ is $2$, O is the circumcentre, H is the orthocentre then $\frac{1}{64}(A{H}^2+B{C}^2)(B{H}^2+A{C}^2)(C{H}^2+A{B}^2)$ is equal to

• A. $64$
• B. $16$
• C. $\frac{1}{64}$
• D. $1$

Answer 1

The correct option is A $64$

$LetatriangleABC,Histheorthocenter$
$Then$
$AH=2R\cos A=\left(\frac{a}{\sin A}\right)\cos A=a\cot A$
$similarly$
$BH=2R\cos B=\left(\frac{b}{\sin B}\right)\cos B=b\cot B$
$CH=c\cot C$
$\frac{1}{64}(A{H}^2+B{C}^2)(B{H}^2+A{C}^2)(C{H}^2+A{B}^2)$
$\frac{1}{64}(a^2{\cot }^{2}A+a^2)(b^2{\cot }^{2}B+b^2)(c^2{\cot }^{2}C+c^2)$
$\frac{1}{64}{a}^2{b}^2{c}^2({\cot }^{2}A+1)({\cot }^{2}B+1)({\cot }^{2}C+1)$
$\frac{a^2{b}^2{c}^2}{64}\left(\cos e{c}^{2}A\right)\left(\cos e{c}^{2}B\right)\left(\cos e{c}^{2}C\right)$
$\frac{a^2{b}^2{c}^2}{64}\left(\frac{1}{{\sin }^{2}A{\sin }^{2}B{\sin }^{2}C}\right)$
$\frac{a^2{b}^2{c}^2}{64}\left(\frac{1}{\frac{a^2}{4R^2}\frac{b^2}{4R^2}\frac{c^2}{4R^2}}\right)$
$\frac{R^2{R}^2{R}^2}{1}$
$\mathrm{4.4.4}$
$64$