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Let $\Delta$ be the set of all triangles with two equal edges and be inscribed in a circle of radius $R$.

So, how do I show that:

Equilateral triangle in $\Delta$ is maximizing the area?

and

this equilateral triangle in $\Delta$ is also maximizing the circumference?

Help appreciated!

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$d$=distance from circumcenter to the other side.$$$$ by $AM\geq GM$ $S=\sqrt{R^{2}-d^{2}}(R+d)=\sqrt{(R-d)(R+d)^{3}}\leq\sqrt{27\left(\dfrac{(R-d)+\dfrac{R+d}{3}+\dfrac{R+d}{3}+\dfrac{R+d}{3}}{4}\right)^{4}}$ $$$$ the equality holds when $R-d=\dfrac{R+d}{3}$ $\Longrightarrow$ $R=2d$ $$$$ That is, it's equilateral

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Let one of the equal angles be $x$

So, the angles will be $x,x,\pi-2x$

So, using Law of Sines, $$\frac a{\sin x}=\frac b{\sin x}=\frac c{\sin(\pi-2x)}=2R$$ where $R$ circum-radius which is constant

So, the sides are $2R\sin x, 2R\sin x, 2R\sin(\pi-2x)=2R\sin2x$

So, the area will be $$\frac{2R\sin x \cdot 2R\sin x \cdot 2R\sin2x}{4R}=R^22\sin^2x\sin 2x$$

Now, $$2\sin^2x\sin 2x=\cos x\cdot4\sin^3x=\cos x(3\sin x-\sin3x)=\frac{3\cdot 2\sin x\cos x-2\sin3x\cos x}2=\frac{3\sin2x-(\sin4x+\sin2x)}2=\sin2x-\frac{\sin4x}2$$

Using Second Derivative Test prove this will be maximum if $x=\frac\pi3$ as $0<x<\frac\pi2$ as $\pi-2x>0$

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If the top angle of such an isosceles triangle is $\alpha$, then the equal sides have lengths $2R\sin\frac{\pi-\alpha}2=2R\cos\frac\alpha2$, the other side has length $4R\cos\frac\alpha2\cdot \sin\frac\alpha2$. Thus the area is $$ A = 4R^2\cos^2\frac\alpha2\cdot \sin \alpha$$ and the perimeter is $$ p = 4R\cos\frac\alpha2\cdot\left(1+ \sin\frac\alpha2\right).$$ Use your favorite technique to find the maxima of these expressions (trigonometric addition formulae may be helpful).

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