Suppose $\triangle ABC$ is an equilateral triangle inscribed in the unit circle C(0,1). Suppose $\triangle ABC$ is an equilateral triangle inscribed in the unit circle C(0,1). Find the maximum value of $$\overline{PA}\cdot\overline{PB}\cdot\overline{PC}$$
where $P$ is a variable point in $\bar{D}(0,2).$
I am trying to figure a proof for this problem out. I have it all drawn out so I could see conceptually what is going on but I am having trouble finding the answer. I believe the sides of the equilateral triangle are going to be $\sqrt{3}$ but I am not sure about the rest. this question is actually blowing my mind trying to find an answer.
 A: An equilateral triangle inscribed in the unit circle must have $0$ as center. Hence, there is a fixed $t\in [0,2\pi)$ such that $$A = e^{it}, \quad B = e^{i(t+2\pi/3)} ,\quad C = e^{i(t+4\pi/3)} $$ A variable point $P \in \overline{D}(0,2)$ can be written $re^{ix}$ with variables $r \in [0,2]$ and $x\in [0,2\pi).$ Now, you have to find the maximum of $$|(e^{it}-re^{ix})(e^{i(t+2\pi/3)}-re^{ix})(e^{i(t+4\pi/3)}-re^{ix})|.$$ after some simplifications, this can be rewritten $$|e^{3it}-r^3e^{3ix}|=\sqrt{(\cos(3t)-r^3\cos(3x))^2+(\sin(3t)-r^3\sin(3x))^2}.$$ After expanding the squares, you have to find $$\max_{r,x}(1+r^6-2r^3\cos(3(t-x))).$$ But $r$ is positive and $\cos(3(t-x))$ has $-1$ as minimum value. Hence you have to find $$\max_{r\in [0,2]} r^6+2r^3+1 =\max_{r\in [0,2]}(r^3+1)^2=81.$$ Finally the desired maximum is $$\sqrt{81}=9.$$
A: Supposing
$$\triangle ABC =
\left((0,1) ,\,
\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2} \right) ,\,
\left(+\frac{\sqrt{3}}{2},-\frac{1}{2} \right) \right)$$
 and $P =(x,y) \in D(0,2) = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 2\}$ and $$\Pi^2 = \overline{PA}\cdot\overline{PB}\cdot\overline{PC},$$
we write
\begin{align}
\Pi^2 &= \sqrt{(x-0)^2+(y-1)^2} \times \sqrt{\left(x-\frac{-\sqrt{3}}{2}\right)^2+\left(y-\frac{-1}{2}\right)^2}\times \sqrt{\left(x-\frac{\sqrt{3}}{2}\right)^2+\left(y-\frac{-1}{2}\right)^2}\\
&=x^6+3 x^4 y^2+3 x^2 y^4+6 x^2 y+y^6-2 y^3+1.
\end{align}
Now, using polar coordinates, such $x\mapsto \rho \cos\theta$ and $x\mapsto \rho \sin\theta$, for $(\rho,\theta) \in [0,2] \times [0,2\pi)$, we have
\begin{align}
\Pi^2 &= 2 \rho ^3 \sin (3 \theta )+\rho ^6+1,
\end{align}
whose maximization is trivial, yielding $\rho=2$ and $\theta = \frac{\pi}{6}+ \frac{2\pi}{3} k, k=\{0,1,2\}$. This result is rather intuitive, being the set of opposite points the vertices of the triangle, where $\Pi^2=81$, implying $\Pi=9$.
