Question

If $a>3$ and $A,B,C$ are variable angles of $△ABC,$ such that $\sqrt{2a^2-9}\cot A+2a\cot B+\sqrt{2a^2+9}\cot C=12a,$ then the minimum value of ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C=$

Answer 1

Let two vectors, $\overrightarrow{P}=\sqrt{2a^2-9}\hat{i}+2a\hat{j}+\sqrt{2a^2+9}\hat{k}$ and $\overrightarrow{Q}=\cot A\hat{i}+\cot B\hat{j}+\cot C\hat{k}$
If $\theta$ is the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$
then $\cos \theta =\frac{\overrightarrow{P}\cdot \overrightarrow{Q}}{|\overrightarrow{P}||\overrightarrow{Q}|}=\frac{\sqrt{2a^2-9}\cot A+2a\cot B+\sqrt{2a^2+9}\cot C}{\sqrt{8}a\sqrt{{\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C}}$
$\Rightarrow \cos \theta =\frac{12a}{\sqrt{8}a\sqrt{{\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C}}\Rightarrow \frac{144a^2}{8a^2({\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C)}\le 1[\because {\cos }^2\theta \le 1]\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C\ge 18$
$\therefore ({\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C{)}_{min}=18$