Prove that out of all inscribed triangles in a circle, an equilateral one has the biggest possible perimeter.
I am not sure if a non-geometrical approach would be valid but you might want to take a look at it.
Refer to the diagram i attached, there's a circle ABC with center O having a triangle inscribed in it. The line from the vertex, passing through the center cuts the base into two equal parts.This is also one of the properties of circle. Angle COB is 2 theta and angle CAB is theta(properties of circle). Also, draw lines connecting OC and OB respectively. Angle BOA = 180 - theta and OA and OB are the radius of the circle, call it r.
Using the cosine rule,
$\displaystyle AB^2=r^2+r^2+2(r)(r)cos (180-\theta)$
$\displaystyle AB=r\sqrt{2}\cdot \sqrt{1-\cos \theta}$
Note that AB=AC (congruent triangles)
The base, $\displaystyle CB^2=r^2+r^2+2(r)(r)\cos 2\theta$
$\displaystyle w=r\sqrt{2}\cdot \sqrt{1+\cos 2\theta}$
The perimeter of the triangle,
$\displaystyle P=2r\sqrt{2}\cdot \sqrt{1-\cos \theta}+r\sqrt{2}\cdot \sqrt{1+\cos 2\theta}$
$\displaystyle \frac{dP}{d\theta}=r\sqrt{2}(\frac{1}{\sqrt{1-\cos \theta}}-\frac{2\cos \theta}{\sqrt{1+\cos 2\theta}})
$
Set that to be 0, the first factor is not valid because $\displaystyle \theta\neq 0,180$
Solving the second factor,
$\displaystyle 4\cos^3 \theta-2\cos^2 \theta=0$
$\displaystyle \cos \theta=0.5$
Hence, $\displaystyle \theta=60$.