Question

If $p_1,p_2,p_3$ are the altitudes of a triangle which circumscribes a circle of diameter $\frac{16}{3}$unit, then the least value of $p_1+p_2+p_3$ must be:

• A. $=24$
• B. $\ge 24$
• C. $<24$
• D. $\le 24$

Answer 1

The correct option is B $\ge 24$

We have $p_1=\frac{2△}{a},p_2=\frac{2△}{b},p_3=\frac{2△}{c}$
$\therefore \frac{p_1+p_2+p_3}{3}\ge {\left(p_1{p}_2{p}_3\right)}^{\frac{1}{3}}$
$={\left[\frac{{\left(2△\right)}^3}{abc}\right]}^{\frac{1}{3}}$
$=2△{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$
$\Rightarrow p_1+p_2+p_3\ge 6△{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$
$=6rs{\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$
$=6r\left(\frac{a+b+c}{2}\right){\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$
$=9r\left(\frac{a+b+c}{3}\right){\left(\frac{1}{abc}\right)}^{\frac{1}{3}}$
Since A.M$\ge$G.M we have
$\ge 9r$
$9\times \frac{8}{3}=24$
$\Rightarrow p_1+p_2+p_3\ge 24$