Question

Let $f$, $g$, $h$ be the lengths of the perpendiculars from the circumcentre of the $\Delta ABC$ on the sides BC, CA and AB respectively.

If $\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\lambda \frac{abc}{fgh},$ then the value of $"\lambda "$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{1}{4}$

Given f,g,h be length of $\perp$er from circumcentre of $\triangle ABC on sides , BC,CA,AB.

Let $△$ ABC be equilateral triangle

$f=\frac{1}{3}AD$

$a=b=c$

$AD=\frac{\sqrt{3}}{2}a$

$f=\frac{\sqrt{3}}{2}\times \frac{1}{3}\times a=\frac{a}{2\sqrt{3}}\to \left(1\right)$

$g=\frac{a}{2\sqrt{3}}\to \left(2\right)h=\frac{a}{2\sqrt{3}}\to \left(3\right)$

$\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=x\frac{abc}{fgh}\to \left(4\right)$

subtitute $\left(1\right),\left(2\right),\left(3\right)in\left(4\right)$

$\frac{a}{\frac{a}{2\sqrt{3}}}+\frac{a}{\frac{a}{2\sqrt{3}}}+\frac{a}{\frac{a}{2\sqrt{3}}}=\lambda \frac{aa\times a}{\frac{a}{2\sqrt{3}}\times \frac{a}{2\sqrt{3}}\times \frac{a}{2\sqrt{3}}}$
$2\sqrt{3}+2\sqrt{3}+2\sqrt{3}=\lambda \times 8\times 3\sqrt{3}6\sqrt{3}=\lambda \times 8\times 3\sqrt{3}\lambda =\frac{1}{4}\therefore \lambda =\frac{1}{4}$