Can we find a point $M$ on the unit circle such that $\prod_{i=1}^n MA_i=1$? 
We are on $\Bbb{R}^2$. Let $A_1,\cdots,A_n$ be $n$ points on the unit circle.
Can we find a point $M$ on the unit circle such that $\prod_{i=1}^n MA_i=1$ ? ( of course I mean the distance here)

I think the answer is ye but I can't make any significant progress. I tried to parameterize the circle in polar coordinate.
Any ideas please ?
EDIT : My question is really close of this one but I am unfamiliar with complex analysis.
 A: As noted by lhf, it is proven in Lower bound on product of distances from points on a circle that there exists $M_0$ such that the product is strictly greater than one if $M = M_0$.
Trivially the product is $0$ if $M = A_1$. By continuity there is some $M$ between $M_0$ and $A_1$ for which the product is equal to $1$.
Yuval Filmus' answer in the question referenced above does not use complex analysis, I'll write the argument more in detail below:
Write the points in polar coordinates, i.e. $A_i = (\cos(\alpha_i), \sin(\alpha_i))$ and $M = (\cos(\theta), \sin(\theta))$. Consider
$$f(\theta) = \log \prod_{i=1}^n \|M - A_i\|.$$
We want to show that there exists $\theta$ such that $f(\theta) > 0$. Now we may write
$$\begin{eqnarray*}
\|M - A_i\| & = & \sqrt{(\cos(\theta) - \cos(\alpha_i))^2 + (\sin(\theta) - \sin(\alpha_i))^2} \\
& = & \sqrt{2 - 2 \cos(\theta) \cos(\alpha_i) - 2 \sin(\theta) \sin(\alpha_i)} \\
& = & \sqrt{2 - 2 \cos(\theta - \alpha_i)}.
\end{eqnarray*}$$
Thus
$$
\begin{eqnarray*}
f(\theta) & = & \sum_{i=1}^n \log\left(\sqrt{2 - 2 \cos(\theta - \alpha_i)}\right) \\
& = & \frac{1}{2} \sum_{i=1}^n \left(\log(2) + \log(1 - \cos(\theta - \alpha_i))\right) \\
& = & \frac{n}{2} \log(2) + \frac{1}{2} \sum_{i=1}^n \log(1 - \cos(\theta - \alpha_i)).
\end{eqnarray*}$$
Let us calculate the average of $f(\theta)$ on the unit circle.
$$\begin{eqnarray*}
\frac{1}{2\pi} \int_0^{2\pi} f(\theta) \, d\theta & = & \frac{n}{2} \log(2) + \frac{1}{2} \sum_{i=1}^n \frac{1}{2\pi} \int_0^{2\pi} \log(1 - \cos(\theta - \alpha_i)) \, d\theta \\
& = & \frac{n}{2} \log(2) + \frac{1}{2} \sum_{i=1}^n \frac{1}{2\pi} \int_0^{2\pi} \log(1 - \cos(\theta)) \, d\theta \\
& = & \frac{n}{2} \log(2) + \frac{n}{2} \cdot \frac{1}{2\pi} \int_0^{2\pi} \log(1 - \cos(\theta)) \, d\theta.
\end{eqnarray*}$$
Finally
$$\frac{1}{2\pi} \int_0^{2\pi} \log(1 - \cos(\theta)) \, d\theta = \frac{1}{2\pi} \int_0^{2\pi} \log(2 \sin^2(\theta/2)) \, d\theta = \log(2) + 4 \cdot \frac{1}{2\pi} \int_0^{\pi} \log(\sin(\theta)) \, d\theta,$$
and $\int_0^{\pi} \log(\sin(\theta)) \, d\theta = -\pi \log(2)$. See for example the question Computing the integral of $\log(\sin x)$.
Thus we have $\frac{1}{2\pi} \int_0^{2\pi} f(\theta) \, d\theta = 0$. In particular since $f(\theta)$ is negative close to $\theta = \alpha_i$, it must also be positive somewhere.
