Find area of the largest possible triangle inscribed in the cardioid $r=1-\cos\theta$. The best I can do is the triangle with vertices (in rectangular coordinates) $(0,1)$, $(0,-1)$ and $(-2,0)$. Can you do better or prove this yields maximal area?
I can show this is maximal for all triangles with either one vertex at $(0,0)$ or with one side that contains the point $(0,0)$.
 A: Perhaps you could use herons formula, which gives the area of a triangle as:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where:
$$s=\frac{a+b+c}{2}$$
If we represent the vertices as vectors $\alpha,\beta,\gamma$ then the lengths are just:
$$a=|\alpha-\beta|$$
$$b=|\beta-\gamma|$$
$$c=|\gamma-\alpha|$$
A: The accepted solution is perfectly correct, but I'm still bothered by the "algebraic intensity" of it.
If (and this may be a big "if") one can first argue that the triangle of maximum area must be isosceles and symmetric about the $x$-axis, then there are two (easy-ish) cases:
Case 1: One vertex is at the origin.
Then we need to maximize the function $A(x)=-\frac{1}{2}(1-\cos x)^2\sin(2x)$ on the interval $(\pi/2,\pi)$.  This function is maximized at $x=\frac{4\pi}{5}$, with maximum value of about 1.556. (This case can be done by hand.)
Case 2: One vertex is at the point $(-2,0)$.
Then we need to maximize the function $A(x)=(2-\cos x(\cos x-1))(1-\cos x)\sin x$ on the interval $[\pi/2,\pi)$.  This function is maximized (with help from Wolfram|Alpha) at $x\approx1.723$ (roughly $98.7^\circ$), with maximum value of about 2.078.
In both cases, $x$ is the angle between the positive $x$-axis and the ray with initial point $(0,0)$ passing through a vertex of the triangle (other than $(0,0)$ or $(-2,0)$, as the case may be).
