What is $C'$?

Assuming that $C'\equiv C,$ and setting $\angle BAC=\angle AOM=\theta,$ we have: $AC=R\sin(2\theta),$ and $AA'=R\tan\frac{\theta}{2},$ where $R$ is the radius of the circle. Then $f(\theta)=\frac{[ABC]}{[A'B'C]}=\frac{AC^2}{A'C^2}=\frac{\sin^2(2\theta)}{\left(\sin(2\theta)-\tan\frac{\theta}{2}\right)^2}.$ To simplify set $x=\tan\frac{\theta}{2},$ and rewrite $f(x)=\frac{16(1-x^2)^2}{(-x^4-6x^2+3)^2}.$ Since $x>0,$ it is easy to show that as long as $x<\sqrt{2\sqrt{3}-3}$ (or $\theta<68.53^\circ$), the quadratic in $x^2$ in the denominator is positive, and therefore we can write $\sqrt{f(x)}=\frac{4(1-x^2)}{3-6x^2-x^4}.$ We are asked to show that $\sqrt{f(x)}<\sqrt{2},$ which is obvious by solving the resulting quadratic in $x^2.$

I'm sorry this might be a stupid question. But I'm from the US and have never seen notation like this. What does the $S_{ABC}$ notation mean in this context.

Prasanthinalluri wrote: I'm sorry this might be a stupid question. But I'm from the US and have never seen notation like this. What does the $S_{ABC}$ notation mean in this context. Area of $ABC$.