Question

If $\frac{a^2}{r_1^2}+\frac{b^2}{r_2^2}+\frac{c^2}{r_3^2}=k$, then

• A. $k<144$
• B. $k$ $\ge$ $144$
• C. $k$ $=$ $144$
• D. $k$ $=$ $16$

Answer 1

Answer: (B)

$k$ $\ge$ $144$

We know that $\Delta =abc/4R$
so ${\Delta }_1=\frac{OB\times OC\times BC}{4R_1}=\frac{a{R}^2}{4R_1}$
$\Rightarrow \frac{a}{R_1}=\frac{4\Delta }{R^2}$ similarly
$\frac{b}{R^2}=\frac{4{\Delta }_2}{R^2},\frac{c}{R_3}=\frac{4{\Delta }_3}{R^2}$

Hence $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=\frac{4}{R^2}({\Delta }_1+{\Delta }_2+{\Delta }_3)=\frac{4\Delta }{R^2}$

$BOD=\frac{1}{2}BOC=A$

From $\Delta OBD,\frac{a/2}{2r_1}=tanA$
$\Rightarrow \frac{a}{r_1}=4tanA$, similarly $\frac{b}{r_2}=4tanB,\frac{c}{r_3}=4tanC$
Hence $\frac{a}{r_1}+\frac{b}{r_2}+\frac{c}{r_3}=4(tanA+tanB+tanC)$

$=4tanAtanBtanC[\because A+B+C={180}^o]$

$a=4r_{1}tanA=\frac{4{\Delta }_1{R}_1}{R^2}\Rightarrow \frac{R_1}{r_1}=\frac{R^{2}tanA}{{\Delta }_1}$

similarly $\frac{R_2}{r_2}=\frac{R^{2}tanB}{{\Delta }_2},\frac{R_3}{r_3}=\frac{R^{2}tanC}{{\Delta }_3}$
Hence
$\frac{R_1}{r_1}+\frac{R_2}{r_2}+\frac{R_3}{r_3}=R^2[\frac{tanA}{{\Delta }_1}+\frac{tanB}{{\Delta }_2}+\frac{tanC}{{\Delta }_3}]$

$k=\frac{a^2}{r_1^2}+\frac{b^2}{r_2^2}+\frac{c^2}{r_3^2}=16(ta{n}^{2}a+ta{n}^{2}B+ta{n}^{2}C)$

$ta{n}^{2}A+ta{n}^{2}B+ta{n}^{2}C\ge 3(ta{n}^{2}Ata{n}^{2}Bta{n}^{2}C{)}^{1/3}$ $(A.M.\ge G.M.)$ (1)
Similarly $tanA+tanB+tanC\ge 3(tanAtanBtanC{)}^{\frac{1}{3}}$

$\Rightarrow tanAtanBtanC\ge 3(tanAtanBtanC{)}^{\frac{1}{3}}$

$\Rightarrow (tanAtanBtanC{)}^{\frac{2}{3}}\ge 3$ (2)

From (1) and (2) we get
$k=16(ta{n}^{2}A+ta{n}^{2}B+ta{n}^{2}C)\ge 16\times 3\times 3=144$