Prove that the least perimeter of an isoceles triangle in which a circle of radius $r$ can be inscribed is $6r\sqrt3$.
I have seen answer online on two sites. One is on meritnation but the problem is that answer is difficult and bad formatting. Other answer on topperlearning but that answer make uses of trigonometric functions. And I want to solve it without trignometric function.
So please can someone provide easy method.
Let $\Delta ABC$ be our triangle, $AB=BC=ax$ and $AC=a$.
Hence, since $ax+ax>a$, we get $x>\frac{1}{2}$ and $$r=\frac{2S}{ax+ax+a}=\frac{\frac{2a\sqrt{a^2x^2-\frac{a^2}{4}}}{2}}{2(2ax+a)}=\frac{a}{2}\sqrt{\frac{2x-1}{2x+1}}.$$ Thus, we need to prove that $$2ax+a\geq3\sqrt3a\sqrt{\frac{2x-1}{2x+1}}$$ or $$(2x+1)^3\geq27(2x-1).$$ Let $f(x)=(2x+1)^3-27(2x-1)$
Thus, $f'(x)=6(2x+1)^2-54=24(x-1)(x+2)$, which says that $x_{min}=1$
and we are done!
Let $\Delta ABC$ be our triangle, where $AB=BC$ and $\measuredangle ABC=2x$.
Hence, $P_{\Delta ABC}=2r\left(\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)$ and we need to prove that $\min\limits_{\left(0,\frac{\pi}{2}\right)}f=3\sqrt3$, where
$$f(x)=\cot{x}+2\tan\left(\frac{\pi}{4}+\frac{x}{2}\right).$$
Indeed,
$$f'(x)=\frac{1}{\cos^2\left(\frac{\pi}{4}+\frac{x}{2}\right)}-\frac{1}{\sin^2x}=\frac{\left(\sin{x}-\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)\left(\sin{x}+\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)}{\cos^2\left(\frac{\pi}{4}+\frac{x}{2}\right)\sin^2x}.$$
Since $\sin{x}+\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)>0$ for $0 we see that $f'(x)=0$ on $\left(0,\frac{\pi}{2}\right)$ for $x=\frac{\pi}{6}$ only and since $f'(x)<0$ on $\left(0,\frac{\pi}{6}\right)$ and $f'(x)>0$ on $\left(\frac{\pi}{6},\frac{\pi}{2}\right)$, we see that $f$ gets on $\left(0,\frac{\pi}{2}\right)$ a minimal value for $x=\frac{\pi}{6}$. Since $f\left(\frac{\pi}{6}\right)=3\sqrt3$, we are done!