Given a quadrilateral with a given perimeter P, I need to find the one with the maximum area. I already know that the maximum area of a rectangle with a given perimeter P the one with the max area is a square. But i have no idea how to find the maximum area when it's not a triangle.
I tried to use Bretschneider's formula:
$$S=\sqrt{(s-a)(s-b(s-c)(s-d)-\frac{1}{2}abcd[1+\cos{(\angle A+\angle C)}}]$$
but there are angles in this formula.
I also tried to use this:
$$S=\frac{1}{2}ab \sin \angle A +\frac{1}{2}cd \sin \angle C $$
I know for a fact that I need to use inequality of arithmetic and geometric means so I tried to use the second formula by using this:
$\sqrt{xy} \leq \frac{1}{2} (x+y) \ \ \ \ \ /\uparrow^2$
$xy\leq \frac{1}{4}(x+y)^2$
but agian, I have no idea what do I do with the angles.
Let's break this down:
You have the formula
$$S=\sqrt{(s-a)(s-b)(s-c)(s-d)-\frac{1}{2}abcd[1+\cos{(\angle A+\angle C)}}]$$
which as a sum of opposing angles at $A$and $C$. The cosine of this sum is at its minimum of $-1$ when the opposite angles are s___________y.
So now you know that each pair of opposing angles has that property. Now consider the sides. With the formula reduced to
$S=\sqrt{(s-a)(s-b)(s-c)(s-d)}$
$\text{*Brahmagupta's Formula* for a cyclic quadrilateral}$
you now use AM-GM to render
$(s-a)(s-b)(s-c)(s-d)\le[(1/4)((s-a)+(s-b)+(s-c)+(s-d))]^4$
with equality holding when $a=b=c=d$ for any fixed $s$. So the sides must be e___l, and you can even work out how each side is then related to $s=(a+b+c+d)/2$.
You conclude that all sides are e___l and the opposing angles are s___________y. This should be enough to determine the shape of the quadrilateral and get you s____ed away.