differential geometry - Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral?

Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral? I gather it doesn't because most of the proofs I've seen use derivatives etc. If so why doesn't it work?

Consider a triangle $ABC$ inscribed inside a circle $\Gamma$. Assume WLOG one of the sided is fixed then one can easily see that the other two sides being equal maximizes the area of the triangle (it has the greatest height). Now consider one of the two equal sides - using that side as a base we can see from the same argument as before that the triangle must in fact be equilateral to maximize the area (this triangle has the greatest height as before).

Thanks in advance.

Answer

Let us some analytic geometry using your idea: suppose our circle is $\,x^2+y^2=R^2\;$ , and we're going to fix, as you did, one of our sides as being parallel to the $\,x-$axis for simplicity, so that its end points are $\,A=(a,b)\;,\;\;B=(-a,b)\;,\;\;a>0\;,\;a^2+b^2=R^2\;,\;\;b\le 0\;$ .

Let the other vertex be $\,C=(0,R)\;$ since, as you remarked, for the above two points of the triangle, the maximal height is obtained when the other two sides are equal, which means the third vertex is on the sides perpendicular bisector, which is easy to see is the $\,y-$axis.

Thus, the area of the triangle depends on the distance $\,R-b=R-\sqrt{R^2-a^2}\,$ and on the horizonal side length's $\,2a\;$ :

$$f(a):=a\left(R-\sqrt{R^2-a^2}\right)\implies f'(a)=R-\sqrt{R^2-a^2}+\frac{a^2}{\sqrt{R^2-a^2}}\stackrel ?=0\iff$$

$$R\sqrt{R^2-a^2}-R^2+a^2+a^2=0\implies(2a^2-R^2)^2=R^2(R^2-a^2)\implies$$

$$R^4-4R^2a^2+4a^4=R^4-R^2a^2\implies a^2\left(4a^2-3R^2\right)=0\implies$$

$$\implies a=\frac{\sqrt3}2R$$

so that the slope of the line

$$CA\,\;,\;\;A=\left(\frac{\sqrt3}2R\,,\,-\frac12R\right)\;,\;C=(0,R)\;\,\;\;\text{is}$$

$$-\frac{\frac32R}{\frac{\sqrt3}2R}=-\sqrt3=\tan\frac{-\pi}3$$

and we're done...