Let $\Delta$ be the set of all triangles with two equal edges inscribed in a circle of radius $R$.
So, how do I show that:
1, The equilateral triangle in $\Delta$ is the one maximizing the area.
2, The equilateral triangle in $\Delta$ is the one maximizing the circumference?
Help greatly appreciated!
Say the angle between the two equal edges of the triangle is $\theta$. Can you work out the lengths of the edges as a function of $\theta$? Once you have the lengths of the edges can you find the area and perimeter (I'm assuming that's what you meant by circumference) with respect to $\theta$, and then maximize?
We can prove that Isosceles Triangles have Two Equal Angles
Now using Law of Sines, $$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R$$
Let $A=B=x\implies C=\pi-2x,\sin C=\sin(\pi-2x)=\sin 2x$
So, $x=A>0$ and $\pi-2x=C>0\implies x<\frac\pi2\implies 0<x<\frac\pi2$
So, $a=b=2R\sin x,c=2R\sin2x$
So, the circumference will be $=a+b+c=2R(2\sin x+\sin2x)$ where $R$ is constant
and the area will be $\frac{abc}{4R}=\frac{(2R\sin x)^2(2R\sin2x)}{4R}=4R^2\sin^3x\cos x$ where $R$ is constant
We can apply Second Derivative Test w.r.t. $x$, to find the required extreme values as follows
Let $f(x)=2\sin x+\sin2x\implies f'(x)=2(\cos x+\cos2x)$
For the extreme values of $f(x),f'(x)=0$
$\implies \cos2x+\cos x=0\implies 2\cos^2x+\cos x-1=0$ $\implies \cos x=\frac{-1\pm\sqrt{1^2-4\cdot2\cdot(-1)}}{2\cdot2}=-1$ or $\frac12$
As $0<x<\frac\pi2,\cos x>0\implies \cos x=\frac12\implies x=\frac\pi3$
$$\text{ Now, show that }f''(\frac\pi3)<0$$
Let $g(x)=8\sin^3x\cos x$ which can be simplified to $2\sin2x-\sin4x$ (See the comment below)
For $g'(x)=0,$ $4(\cos2x-\cos4x)=0\implies 2\cos^22x-\cos2x-1=0$
$$\implies \cos2x=1\text{ or }-\frac12$$
As $0<x<\frac\pi2\implies 0<2x<\pi \implies \cos x\ne1\implies \cos 2x=-\frac12=\cos\frac{2\pi}3$
$\implies 2x=2n\pi\pm \frac{2\pi}3\implies 2x=\frac{2\pi}3\iff x=\frac\pi3$ as $0<2x<\pi$
Now, show that $g''(\frac\pi3)<0$
We shall prove that the inscribed equilateral triangle has maximal area and maximal circumference among all inscribed triangles; a fortiori it then has maximal area, resp. circumference, among inscribed isosceles triangles.
For both questions the optimal triangle is acute: If the angle at $C$ is $>{\pi\over2}$ reflect $C$ at the diameter parallel to $c$ and obtain a triangle that obviously has larger area as well as larger circumference. Therefore we may assume that the midpoint $M$ of the circle is in the interior of our triangle $\Delta$. Denote by $\phi_i$ $\ (1\leq i\leq 3)$ the angles under which the three sides of $\Delta$ are seen from $M$, and let $r$ be the radius of the circle. Then $${\rm area}(\Delta)={r^2\over2}\sum_{i=1}^3\sin \phi_i\ .$$ Now $0<\phi_i<\pi$ $\ (1\leq i\leq 3)$. Since the function $\sin$ is concave in the interval $[0,\pi]$ Jensen's inequality tells us that $${1\over3}\sum_{i=1}^3\sin \phi_i\leq \sin\left({1\over3}\sum_{i=1}^3 \phi_i\right)=\sin{2\pi\over3}={\sqrt{3}\over2}\ ;$$ and furthermore that we have strict inequality unless all $\phi_i={2\pi\over3}$. It follows that $${\rm area}(\Delta)\leq{3\sqrt{3}\over4}\ r^2\ ,$$ and that we have equality sign only for equilateral inscribed triangles.
Similarly $${\rm circumference}(\Delta)=2r\sum_{i=1}^3\sin{\phi_i\over2}\ ,$$ and the same argument as above proves that $${\rm circumference}(\Delta)\leq3\sqrt{3}\ ,$$ with equality only for equilateral triangles.
