A remarkable fact about the unit circle; looking for a shape with an even more remarkable fact.

Question

You may have heard of the following remarkable fact about the unit circle:

If $n$ equally spaced points are drawn on a unit circle, and line segments are drawn from one of the points to each of the other points, then the product of the lengths of these $n-1$ line segments, equals $n$ (proof).

This made me wonder, does there exist a shape such that the same kind of product is not equal to $n$, but rather a constant? That is:

Does there exist a two dimensional shape (not necessarily convex) and a fixed point on its perimeter, such that, if $n$ equally spaced (in terms of distance along the perimeter) points, one of which is the fixed point, are drawn on the perimeter, and line segments are drawn from the fixed point to each of the other points, then the product $P(n)$ of the lengths of these $n-1$ line segments, is constant for all $n$?

Ellipse?

Suppose that the shape in question could be the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a$ and $b$ are positive, and the fixed point is on an axis of the ellipse, say $Q(0,-b)$.

We have $P(2)=P(4)\implies 2b=2b(a^2+b^2)\implies a^2=1-b^2$.

So the equation of the ellipse is $\frac{x^2}{1-b^2}+\frac{y^2}{b^2}=1$, or $y=\pm b\sqrt{1-\frac{x^2}{1-b^2}}$.

For $n=8$, suppose the point $A$ on the ellipse in the first quadrant, and the point $B$ on the ellipse in the fourth quadrant, have $x$-coordinate $s$. We have $P(8)=P(4)\implies AQ^2BQ^2=1$. That is,

$$\left(s^2+b^2\left(\sqrt{1-\frac{s^2}{1-b^2}}+1\right)^2\right)\left(s^2+b^2\left(\sqrt{1-\frac{s^2}{1-b^2}}-1\right)^2\right)=1$$

Here is the graph of $b$ against $s$.

According to the graph, $b\ge 1$, so $a^2=1-b^2\le 0$, so there is no ellipse.

Sine curves?

I also considered the graph $y=\pm a\sin\left(\frac{x}{a}\right), 0\le x\le a\pi$, with fixed point $(0,0)$.

Numerical investigation suggests that there exists a constant $C\approx 0.7571$ such that, as $n\to\infty$,

• If $a<C$ then $P(n)\to 0$.
• If $a=C$ then $\dfrac{P(n)}{n}\to L\approx1.07$.
• If $a>C$ then $P(n)\to\infty$.

So this kind of sine graph doesn't work.

Possibly related question

I asked the following question on MO: "Does there exist a continuous function $f(x)$ such that $f(0)=0$ and $0<\lim\limits_{n\to\infty}\prod\limits_{k=1}^n f\left(\frac{k}{n}\right)<\infty$?"

The answer was yes. However, my question now is not about limits. Also, the answer in the MO question involved spikes in the graph, which I don't think is possible in a graph of line segment length against distance along perimeter from the fixed point.

Answer 1

tl; dr: No such simple rectifiable closed curve of length $\ell < 2e$ exists.

I lean toward believing no such curve exists regardless of (finite) length, but don't see a promising approach.

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Fix a positive real number $\ell$. Let $\gamma$ be a simple rectifiable closed path of length $\ell$, parametrized by arc length on $[0, \ell]$. We'll view the choice of starting/ending point $\gamma(0)$ as the choice of point from which we draw chords. (For definiteness, assume the target space of $\gamma$ is an arbitrary-dimensional Euclidean space with distance function $d$; this can be generalized.)

Introduce the chordal distance function $$ d(s) = d\bigl(\gamma(0), \gamma(\ell s)\bigr),\quad 0 \leq s \leq 1. $$ This function is continuous, non-negative, $1$-periodic (if we extend $\gamma$ by periodicity), and zero precisely at integers.

The periodic extension of $d$ satisfies $$ d(s) \leq \ell|s|\quad\text{for $|s| \leq \tfrac{1}{2}$}, $$ since the chordal distance from $\gamma(0)$ to $\gamma(\ell s)$ does not exceed the "short" arc length $\ell|s|$ from $0$ to $\ell s$ along $\gamma$. This inequality is key to non-existence below.

(Generally, in fact, the triangle inequality guarantees $d$ is Lipschitz with constant $\ell$: For all real $s$ and $t$, $$ d(t) \leq d(s) + d\bigl(\gamma(\ell s), \gamma(\ell t)\bigr) \leq d(s) + \ell|s - t|, $$ and similarly with the roles of $s$ and $t$ reversed. That is, $$ |d(s) - d(t)| \leq \ell|s - t|. $$ Lipschitz continuity is used below only to make a geometric observation.)

For each integer $n \geq 2$, the points $\gamma(\frac{\ell j}{n})$ with $1 \leq j < n$ are "equally-spaced" along $\gamma$. The product of chord lengths is $$ C(n) := \prod_{j=1}^{n} d(\tfrac{j}{n}). $$ From now on, assume there exists a positive real number $C$ such that $C(n) = C$ for all $n \geq 2$. Our goal is to show no chordal distance function $d$ exists. This implies no such path $\gamma$ exists.

It will be convenient to (harmlessly) shift the domain of $d$ to be $[-\frac{1}{2}, \frac{1}{2}]$. If we divide $\gamma$ into $2n$ equal-length arcs, and "pair up" symmetrically-placed chord lengths, the resulting product is $$ C = d(\tfrac{1}{2}) \cdot \prod_{j = 1}^{n-1} d(-\tfrac{j}{2n}) \cdot d(\tfrac{j}{2n}), $$ independently of $n$. Our bound $|d(s)| \leq \ell|s|$ implies \begin{align*} C &\leq \tfrac{1}{2} \ell \cdot \prod_{j = 1}^{n-1} \biggl(\frac{\ell j}{2n}\biggr)^{2} \\ &= \frac{\ell^{2n-1}[(n-1)!]^{2}}{2^{2n-1} n^{2n-2}} = \frac{\ell^{2n-1}[n!]^{2}}{2^{2n-1} n^{2n}}. \end{align*} By the weak version of Stirling's formula, $n! < n^{n}\sqrt{n}/e^{n-1}$, we have $$ C < ne \biggl(\frac{\ell}{2e}\biggr)^{2n-1}\quad\text{for all $n \geq 2$.} $$ If $\ell < 2e$, the expression in parentheses is smaller than $1$, so the upper bound converges to $0$ as $n \to \infty$. Particularly, no path of length $\ell < 2e$ has constant products of chord lengths. This completes the proof of the assertion above.

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Since I'm unlikely to have time to work more on this in the near future, here are some additional observations that arose while thinking about this:

0. On one hand the existence of a naive length bound is striking, particularly since the bound involves $e$. But as the proof shows, there's something distinctive about taking products of arbitrarily-many evenly-spaced values from a fixed positive function and requiring those products to stay bounded and bounded away from zero.

1. The value $C = d(\frac{1}{2})$ is the length of the arclength-bisecting chord from $\gamma(0)$. Consider the products of chord lengths obtained by dividing the arclength into quarters, eighths, sixteenths, and so on. Since $$ C = \prod_{i=1}^{3} d(\tfrac{i}{4}) = d(\tfrac{1}{4}) \cdot d(\tfrac{1}{2}) \cdot d(\tfrac{3}{4}) $$ is positive, we have $$ 1 = \prod_{j=1}^{2} d(\tfrac{2j-1}{4}) = d(\tfrac{1}{4}) \cdot d(\tfrac{3}{4}). $$ Similarly, since $$ C = \prod_{i=1}^{7} d(\tfrac{i}{8}) = d(\tfrac{1}{8}) \cdot d(\tfrac{1}{4}) \cdot d(\tfrac{3}{8}) \cdot d(\tfrac{1}{2}) \cdot d(\tfrac{5}{8}) \cdot d(\tfrac{3}{4}) \cdot d(\tfrac{7}{8}), $$ we have $$ 1 = d(\tfrac{1}{8}) \cdot d(\tfrac{3}{8}) \cdot d(\tfrac{5}{8}) \cdot d(\tfrac{7}{8}) = \prod_{j=1}^{4} d(\tfrac{2j-1}{8}). $$ Generally, $$ 1 = \prod_{j=1}^{2^{n-1}} d(\tfrac{2j-1}{2^{n}}) $$ for every $n \geq 2$. In words, each time we bisect subintervals of $[-\frac{1}{2}, \frac{1}{2}]$ (or of $[0, 1]$), the product of the values of $d$ at the "newly-added" midpoints is unity.

The same is true, naturally, if we recursively subdivide $[0, 1]$ in an arbitrary way: The chord lengths of the initial subdivision multiply to $C$, and the product of chord lengths is preserved under subdivision, so the product of the "newly-added" chord lengths at each stage is unity.

In a related vein, if $\gamma$ has constant products of chord lengths, then for every $n \geq 2$ we have \begin{align*} C &= \prod_{j=1}^{n} d(-\tfrac{j}{2n+1}) \cdot d(\tfrac{j}{2n+1}) && \text{$(2n + 1)$-gon;} \\ C^{2} &= \prod_{j=1}^{n} d(-\tfrac{j}{2n}) \cdot d(\tfrac{j}{2n}) && \text{$2n$-gon, diameter double-counted;} \\[12pt] 1 &= \prod_{j=1}^{n-1} d(-\tfrac{j}{2n}) \cdot d(\tfrac{j}{2n}) && \text{$2n$-gon, diameter omitted;} \\ C &= \prod_{j=1}^{n-1} d(-\tfrac{j}{2n-1}) \cdot d(\tfrac{j}{2n-1}) && \text{$(2n -1)$-gon.} \end{align*} The first two products have $n$ factors; the last two have $(n - 1)$. Again, since $d$ is a fixed non-constant continuous function, these identities seem unlikely to be satisfiable. But I do not have a proof.

2. Here is some prospectively-useful notation. We'll work in the set of rationals in $(0, 1)$. If $S$ is a finite set of rationals in $(0, 1)$, define $\Pi(S)$ to be the product of chord lengths $d(s)$ taken over $s$ in $S$: $$ \Pi(S) = \prod_{s \in S} d(s). $$

For example, let $n \geq 2$ be an integer, and let $Z_{n}^{*}$ denote the set of rationals $j/n$ such that $1 \leq j < n$. The "regular $n$-gon" with $\gamma(0)$ as a vertex has vertices $\gamma(\frac{\ell j}{n})$ with $1 \leq j < n$. By hypothesis, the product of chord lengths is $$ C = \prod_{j=1}^{n-1} d(\tfrac{j}{n}) = \Pi(Z_{n}^{*}). $$

For each integer $m \geq 2$, let $U(m) \subseteq Z_{m}^{*}$ denote the set of rational numbers $j/m$ such that $1 \leq j < m$ and $\gcd(j, m) = 1$; namely, such that $j/m$ is in lowest terms. The notation is motivated by abstract algebra: $j/m \in U(m)$ if and only if the residue class $[j]$ is invertible mod $m$.

If $p$ is prime, then $Z_{p}^{*} = U(p)$ (i.e., every non-zero residue class is invertible), so $$ \Pi\bigl(U(p)\bigr) = \Pi\bigl(Z_{p}^{*}\bigr) = C. $$ Generally, $Z_{n}^{*}$ is the disjoint union of $U(m)$ as $m$ runs through the non-trivial divisors of $n$. For example, \begin{align*} Z_{6}^{*} &= \{\tfrac{1}{6}, \tfrac{1}{3}, \tfrac{1}{2}, \tfrac{2}{3}, \tfrac{5}{6}\} \\ &= \{\tfrac{1}{2}\} \cup \{\tfrac{1}{3}, \tfrac{2}{3}\} \cup \{\tfrac{1}{6}, \tfrac{5}{6}\} \\ &= U(2) \cup U(3) \cup U(6). \end{align*} Since $\Pi$ is multiplicative over disjoint unions, we have $$ C = \Pi(Z_{6}^{*}) = \Pi\bigl(U(2)\bigr) \cdot \Pi\bigl(U(3)\bigr) \cdot \Pi\bigl(U(6)\bigr) = C \cdot C \cdot \Pi\bigl(U(6)\bigr). $$ We conclude that $$ d(\tfrac{1}{6}) \cdot d(\tfrac{5}{6}) = \Pi\bigl(U(6)\bigr) = 1/C. $$ In words, the product of the first and fifth chords in a "regular hexagon" is the reciprocal of the product of all five chord lengths. Arguing similarly, we can calculate $\Pi\bigl(U(m)\bigr)$ recursively for arbitrary $m \geq 2$.

In this notation, the calculations under item 1. may be written $\Pi\bigl(U(2^{k})\bigr) = 1$ if $k \geq 2$: We have $\Pi\bigl(U(2)\bigr) = C$, and $$ Z_{2^{k}}^{*} = U(2) \cup U(4) \cup U(8) \cup \cdots \cup U(2^{k}). $$

These lines of thought arose in trying to force a discontinuity in $d$, either at $0$ or at $\frac{1}{2}$. The difficulty appears to be that although $d$ must be uniformly continuous, the number of factors $n$ in a product is inversely proportional to the "sample spacing" $1/n$. When the number of factors is large, small changes in individual factors can have a large effect on the product.

3. Inversely, if there exists a continuously-differentiable function $d$ that is positive except at integers and satisfies the "constant product of sample values" condition above, then the Lipschitz condition guarantees $|d'| \leq \ell$, and we can solve the differential equation $$ d'(t)^{2} + d(t)^{2} \theta'(t)^{2} = \ell^{2} $$ for $\theta$ on $[0, 1]$. The parametric curve $$ \gamma(t) = \bigl(d(t) \cos\theta(t), d(t) \sin\theta(t)\bigr) $$ satisfies the constant product of chord lengths condition relative to the origin. (I see no reason such a curve, if it exists, should be simple.)