Question

The least perimeter of an isoceles triangle circumscribed about a circle of radius ${}^{\prime }{r}^{\prime }$ is $kr\sqrt{3}$.Then ${}^{\prime }{k}^{\prime }$ is

Answer 1

Let $\angle QAO=\theta$
$\Rightarrow AQ=r\cot \theta$
$\Rightarrow AO=r\text{cosec}\theta$
$\therefore \tan \theta =\frac{x}{AO+ON}$
$\Rightarrow x=(r\text{cosec}\theta +r)\tan \theta \cdots \left(i\right)$
As, $BQ=BN,CN=CP$ and $AP=AQ\left(\text{tangents to a circle from same point}\right)$
So, Perimeter $=P=4x+2AQ$
$\Rightarrow P=4\left(r\tan \theta \right)(\text{cosec}\theta +1)+2r\cot \theta$
$\Rightarrow P=4r(\sec \theta +\tan \theta )+2r\cot \theta \cdots \left(ii\right)$
$\Rightarrow \frac{dP}{d\theta }=4r(\sec \theta \cdot \tan \theta +{\sec }^2\theta )-2r{\text{cosec}}^2\theta$
$\Rightarrow \frac{dP}{d\theta }=2r\cdot \frac{2{\sin }^3\theta +3{\sin }^2\theta -1}{{\sin }^2\theta \cdot {\cos }^2\theta }$
$\{\because \theta \text{is a variable}\}$
$\Rightarrow \frac{dP}{d\theta }=0$
$\Rightarrow 2{\sin }^3\theta +3{\sin }^2\theta -1=0$
$\Rightarrow (\sin \theta +1{)}^2(2\sin \theta -1)=0$
$\Rightarrow \sin \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{6}$
For $\theta =\frac{\pi }{6},f^{\prime }\left(\theta \right)$ changes sign from negative to positive, so a local minima.
$\Rightarrow \text{at}\theta =\frac{\pi }{6}$ Perimeter is minimum
$\Rightarrow P_{Minimum}=6\sqrt{3}r$ from $\left(ii\right)$