Question

If $l,m,n$ denote the sides of pedal triangle opposite to the vertices $A,B,C$ respectively of triangle$ABC.$ Then the value of $\frac{l}{a^2}+\frac{m}{b^2}+\frac{n}{c^2}$ is :

• A. $\frac{a^2+b^2+c^2}{a^3+b^3+c^3}$
• B. $\frac{a^2+b^2+c^2}{2abc}$
• C. $\frac{a^3+b^3+c^3}{abc(a+b+c)}$
• D. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Answer 1

The correct option is B $\frac{a^2+b^2+c^2}{2abc}$

We know, $l=R\sin 2A=a\cos A,m=b\cos B,n=c\cos C$
So, $\frac{l}{a^2}+\frac{m}{b^2}+\frac{n}{c^2}=\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}\cdots \left(i\right)$
Also, we know, $\cos A=\frac{b^2+c^2-a^2}{2bc}\Rightarrow \frac{\cos A}{a}=\frac{b^2+c^2-a^2}{2abc}$
Similarly, we have $\frac{\cos B}{b}=\frac{a^2+c^2-b^2}{2abc}$ and $\frac{\cos C}{c}=\frac{a^2+b^2-c^2}{2abc}$
Now substituting these values in $\left(i\right)$ we have:
$\frac{l}{a^2}+\frac{m}{b^2}+\frac{n}{c^2}=\frac{a^2+b^2+c^2}{2abc}$