Question

Let ${}^{\prime }{P}^{\prime }$ is any arbitrary interior point of the triangle $ABC.H_a,H_b$ and $H_c$ are the length of altitudes drawn from the vertices $A,B$ and $C$ respectively.If $x_a,x_b$ and $x_c$ represent the distance of ${}^{\prime }{P}^{\prime }$ from sides $BC,CA$ and $AB$ respectively, then the minimum value of $191(\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c})$ must be

• A. $1719$
• B. $7179$
• C. $7119$
• D. $9917$

Answer 1

The correct option is A $1719$

We have $△={△}_{BPC}+{△}_{CPA}+{△}_{APB}$
$\Rightarrow △=\frac{1}{2}.a{x}_a+\frac{1}{2}b.x_b+\frac{1}{2}.c.x_c$
$\Rightarrow △=\frac{1}{2}(a{x}_a+b{x}_b+c{x}_c)$ .............$\left(1\right)$
Also,$△=\frac{1}{2}a{H}_a=\frac{1}{2}b{H}_b=\frac{1}{2}c{H}_c$ ..............$\left(2\right)$
From eqns$\left(1\right)$ and $\left(2\right)$, we get
$\frac{x_a}{H_a}+\frac{x_b}{H_b}+\frac{x_c}{H_c}=1$
Now, A.M$\ge$H.M
$\therefore \frac{\frac{x_a}{H_a}+\frac{x_b}{H_b}+\frac{x_c}{H_c}}{3}\ge \frac{3}{(\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c})}$
$\Rightarrow \frac{1}{9}\ge \frac{1}{{\left(\frac{1}{(\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c})}\right)}^2}$
$\Rightarrow \frac{1}{3}\ge \frac{1}{\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c}}$
or $\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c}\ge 9$
$\therefore 191(\frac{H_a}{x_a}+\frac{H_b}{x_b}+\frac{H_c}{x_c})\ge 191\times 9=1719$