# 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.

EDIT : My question is really close of this one but I am unfamiliar with complex analysis.

• There is probably a nice proof using complex analysis. For instance, it is always true that there is a point $M$ on the unit circle such that $\prod_{i=1}^n MA_i > 1$. See math.stackexchange.com/questions/8142/…
– lhf
Aug 7, 2014 at 11:28

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 $1$.
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.