Question

Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6\sqrt{3}r$.

Answer 1

$AD$ is altitude of isosceles triangle,

$AB=ACOD=OE$ as it is isosceles triangle

$AE=AF,BD=BE$ and $CD=CE$ as tangents drawn from exterior point is equal

So, $AB+BC+CA=$ perimeter

$AF+BF+BD+CD+AE+EC=PAAF+BF+(BF+BD+CD+FC)=2AF+4BD$

In $△AFO||AF=\frac{OF}{tan\theta }=\frac{r}{tan\theta }BD=ADtan\theta =(AO+OD)tan\theta =(\frac{r}{sin\theta }+r)tan\theta$

So, $P\left(\theta \right)=\frac{2r}{tan\theta }+4\frac{(r+rsin\theta )}{sin\theta }\frac{sin\theta }{cos\theta }P\left(\theta \right)=2rcot\theta +4sec\theta +4rtan\theta P\left(\theta \right)=0$ for maximum/ minimum

$P\left(\theta \right)=r(-2cose{c}^2\theta +4sec\theta tan\theta +4se{c}^2\theta )=0$

$2(-\frac{1}{{\sin }^2\theta }+\frac{2\sin \theta }{{\cos }^2\theta }+\frac{2}{{\cos }^2\theta })=0or,-{\cos }^2\theta +2{\sin }^3\theta +2{\sin }^2\theta =0or,-1-{\sin }^2\theta +2{\sin }^3\theta +2{\sin }^2\theta =02{\sin }^3\theta +3{\sin }^2\theta -=0\left(\sin \theta +1\right)\left(2{\sin }^2\theta -\sin \theta -1\right)=0\theta \ne -1as\theta >90°2{\sin }^2\theta -2\sin \theta +\sin \theta -1=0(2\sin \theta -1)(\sin \theta +1)=0\sin \theta \ne 1$

So, $\sin \theta =\frac{1}{2}\theta =\frac{\pi }{6}$

If $f^{{}^{\prime }}\left(x\right)$ changes sign so $\theta =\frac{\pi }{6}$ is maxima

$r=[2\sqrt{3}+4\times \frac{2}{\sqrt{3}}+4\times \frac{1}{\sqrt{3}}]=\frac{18}{\sqrt{3}}r=6\sqrt{3}r$