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Let $C$ be a circle of radius $r$ with $n$ points. Prove that there is a point on the circle such that the product of the distances from this point to the other $n$ points is greater than $r^n$. So we seem to be looking at chords that are of length $\leq 2r$. To show the existence of such a point, would you use a constructive argument? Or more of an indirect argument? Likewise, to show that the inequality holds, would we invoke certain "famous" inequalities such as AM-GM or Cauchy-Schwarz?

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Are the points on the circumference of the circle? I hope so, because otherwise with n=1 and the point at the center it fails. –  Ross Millikan Oct 28 '10 at 4:38
    
Yes (more characters). –  PEV Oct 28 '10 at 4:45
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Your questions are pretty non-questions... One would use whatever one would use! –  Mariano Suárez-Alvarez Oct 28 '10 at 4:49

3 Answers 3

up vote 11 down vote accepted

Let $r>0$, $D=\{z\in\mathbb C:|z|<r\}$, and $S=\{z\in\mathbb C:|z|=r\}$. Pick $n\in\mathbb N$ and distinct points $z_1,\dots,z_n\in S$, and consider the polynomial $f(z)=\prod_{i=1}^n(z_i-z)$.

Since $f$ is holomorphic on $\mathbb C$ and non-constant, its absolute value attains its maximum on the closed disc $D\cup S$ at some point $p$ of $S$ and not inside $D$. Since $|f(0)|=r^n$, it follows that $|f(p)|> r^n$.

This is precisely what you wanted.

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@Trevor: This argument depends only on the maximum modulus theorem, which is not on your list of candidates :) –  Mariano Suárez-Alvarez Oct 28 '10 at 5:03
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Nice! I was going to post a proof using only real-valued functions, but what it boiled down to was the fact that $\ln\lvert f\rvert$ is harmonic, and that's elegantly subsumed by your argument. A good reminder that 2D problems are often best viewed in terms of complex numbers! –  Rahul Oct 28 '10 at 5:06

We can take $r=1$ by scaling.

The distance between $e^{i\alpha}$ and $e^{i\beta}$ is $2|\sin\frac{\alpha-\beta}{2}|$. Denoting the given points by $e^{i\alpha_i}$ and the required point by $e^{i\beta}$, the total distance is

$ \prod_i 2|\sin\frac{\alpha_i-\beta}{2}| $

Now let's take a random angle $\beta$. The average multiplicative value of sine on $[0,\pi]$ is $1/2$, see Wikipedia. Therefore the multiplicative expected value of the product is $1$.

If you want to be more formal, replace "multiplicative" by taking of the logarithm.

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Using complex numbers, but a more elementary (and constructive) approach than the Maximum Modulus theorem.

Consider the unit circle. It is enough if we show that the product of distances is greater than $1$.

In fact, we will show that the product of distances of some point is at least $2$.

Let the points be $\displaystyle z_1, z_2, \dots z_n$. Rotate the circle so that $\displaystyle (-1)^n \prod_{j=1}^{n} z_j = 1$

Let $\displaystyle P(z) = \prod_{j=1}^{n}(z-z_j)$.

The product of distances from $\displaystyle z$ to $\displaystyle z_j$ is given by $\displaystyle |P(z)|$.

Now let $\displaystyle \omega_j$ be the $\displaystyle n$ $n^{th}$ roots of unity.

Since for $\displaystyle 1 \le k < n$, we have that $\displaystyle \sum_{j=1}^{n} (\omega_j)^k = 0$

we have that

$\displaystyle \sum_{j=1}^{n} P(w_j) = 2n$

Thus $\displaystyle \sum_{j=1}^{n} |P(w_j)| \ge |\sum_{j=1}^{n} P(w_j)| = 2n$

Hence there is some $j$ for which $|P(w_j)| \ge 2$.

So we have shown that there is a point whose product of distances from the $n$ points is at least $2r^n$.

In fact, I believe we can also show that if for all points, the product of distances is $\leq 2r^n$, then the $n$ points must be equally spaced! (I will leave that for you :-)).

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