Must a curve $\eta \colon [a, b] \to \mathbb{R}^2$ intersect the curves $\eta + \frac{\eta(b) - \eta(a)}{n}$ ($n \geqslant 1$)?

This is the fourth - and with any luck, the last! - in a series of distinct but related questions. I believe that the three previous questions have all been answered satisfactorily. At any rate, this is indisputably true of the most recent one (which was also asked by me). It summarises the history of the problem so far, and it makes a general conjecture, which was found to be too general: If $\gamma\colon[a,b]\to\mathbb{C}$ is continuous and $\gamma(b)=-\gamma(a)$, must the curves $\gamma$ and $e^{ic}\gamma$ intersect for all real $c$?.

The present question is trivial for $n = 1$. That its answer is positive for $n = 2$ is almost a consequence of the two detailed answers [also, for all I know, of the third, less detailed answer] to the second question in the series: A continuous curve intersects its 90 degrees rotated copy?.

The similar proposition for $\mathbb{R}^k$ ($k \ne 2$) is trivially true if $k = 1$, and (almost as trivially) false if $k > 2$ and $n > 1$. For example, if $k = 3$, any polygonal path whose vertices are $(0, 0, 0)$, $(1, 0, 0)$, $(1, 1, 0)$ and $(1, 1, 1)$, in that order, does not intersect its image under translation via the vector $\left(\frac1n, \frac1n, \frac1n\right)$.

For $\mathbb{R}^2$, if there is any counterexample [it seems entirely possible to me that there is one - but my judgement has been almost infallibly wrong, so far!], then there exists a counterexample that is piecewise linear, i.e. a polygonal path.

Proof For all $z \in \mathbb{R}^2$ [I'll use $\mathbb{R}^2$ and $\mathbb{C}$ almost interchangeably, whenever the use of the complex number system seems advantageous, but no serious complex analysis is ever involved - only the basic properties of the complex exponential function], and $\delta > 0$, let $N(z, \delta)$ denote the open disc with centre $z$ and radius $\delta$. For all $S \subset \mathbb{R}^2$, write $N(S, \delta) = \bigcup_{z \in S} N(z, \delta)$.

$N(S, \delta)$ is open, for all $S \subset \mathbb{R}^2$. If $S$ is connected, then so is $N(S, \delta)$. A connected, open subset of $\mathbb{R}^2$ is path-connected. Indeed (as is well-known, easily proved, and important for the argument here), any two of its points can be joined by a polygonal path. (The edges of this path can even be chosen to be all parallel to one or other of the axes, but we'll not need this result.)

For any curve [by which term of course I mean a continuous function] $\eta \colon [a, b] \to \mathbb{R}^2$, denote its image $\eta([a, b])$ by $[\eta]$: this is a connected, compact subset of $\mathbb{R}^2$. For $v \in \mathbb{R}^2$, define the curve $\eta + v \colon [a, b] \to \mathbb{R}^2$ by $(\eta + v)(t) = \eta(t) + v$ ($a \leqslant t \leqslant b$). Clearly $[\eta + v] = [\eta] + v$, where $S + v = \{z + v : z \in S\}$ ($S \subseteq \mathbb{R}^2$).

If $[\eta] \cap [\eta + v] = \emptyset$, then, by compactness, there exist $z^* \in [\eta]$, $w^* \in [\eta + v]$ such that $|z - w| \geqslant |z^* - w^*|$ for all $z \in [\eta]$, $w \in [\eta + v]$. Let $|z^* - w^*| = 2\delta$. Then $\delta > 0$, and $N([\eta], \delta) \cap N([\eta + v], \delta) = \emptyset$. Let $\eta^*$ be a polygonal path joining $\eta(a)$ to $\eta(b)$ in $N([\eta], \delta)$. Then $\eta^* + v$ is a polygonal path joining $\eta(a) + v$ to $\eta(b) + v$ in $N([\eta + v], \delta)$, whence $[\eta^*] \cap [\eta^* + v] = \emptyset$.

If $\eta$ is a counterexample to the conjecture, then $[\eta] \cap [\eta + v] = \emptyset$, where $$ v = \frac{\eta(b) - \eta(a)}{n} = \frac{\eta^*(b) - \eta^*(a)}{n}, $$ for any polygonal path $\eta^*$ joining $\eta(a)$ to $\eta(b)$. It has just been shown that there exists such an $\eta^*$ satisfying $[\eta^*] \cap [\eta^* + v] = \emptyset$, i.e. $\eta^*$ is also a counterexample. $\square$

A conjecture only slightly weaker than the present one is:

If $\gamma \colon [a, b] \to \mathbb{C}$ is continuous and $\gamma(b) = -\gamma(a)$, then the curves $\gamma$ and $e^{i\pi/n}\gamma$ intersect for every positive integer $n$.

Proof The proposition concerning $\gamma$ is trivial if $0 \in [\gamma]$, so we assume $0 \notin [\gamma]$. In that case, there exists a branch $\eta$ of $\operatorname{Log}\gamma$ on $[a, b]$.

[My answer, to the second question in this series of four, quotes passages from an elementary treatment of the existence of branches of $\operatorname{Arg}\gamma$, of which this is a simple consequence.]

Conversely, given any curve $\eta \colon [a, b] \to \mathbb{C}$, we can define $\gamma \colon [a, b] \to \mathbb{C}$ by $\gamma(t) = \exp(\eta(t)) \ne 0$ ($a \leqslant t \leqslant b$); and then $\eta$ is a branch of $\operatorname{Log}\gamma$ on $[a, b]$.

For any two curves $\gamma, \eta$ that are related in the way just defined, we have: $$ \gamma(b) = -\gamma(a) \text{ if and only if, for some } m \in \mathbb{Z}, \, \eta(b) - \eta(a) = (2m + 1){\pi}i. $$

Suppose that the main conjecture of the present question is true; and that we are given a curve $\gamma \colon [a, b] \to \mathbb{C}$. Let $\eta$ be as just defined in terms of $\gamma$. Then, for every positive integer $n$, $[\eta]$ intersects $[\eta + v]$, where $$ v = \frac{\eta(b) - \eta(a)}{(2m + 1)n} = \frac{i\pi}{n}. $$ But $[\gamma] = \exp([\eta])$, and $[e^{i\pi/n}\gamma\big] = \exp([\eta] + v) = \exp([\eta + v])$; therefore $[\gamma]$ intersects $[e^{i\pi/n}\gamma\big]$, as required.

There is also a partial converse. Suppose now that the "slightly weaker" conjecture is true; and that we are given a curve $\eta \colon [a, b] \to \mathbb{R}^2$. If $\eta(b) = \eta(a)$, there is nothing to prove, so from now on we suppose that $\eta(b) \ne \eta(a)$.

The truth-value of the proposition concerning $\eta$ is unaffected by translation, rotation, and scaling, so we can assume without loss of generality that $\eta(a) = 0$ and $\eta(b) = i\pi$. Now let $\gamma$ be defined in terms of $\eta$ as above. Because $\gamma(a) = 1$ and $\gamma(b) = -1$, we can apply the given conjecture, concluding that $[\gamma] \cap \big[e^{i\pi/n}\gamma\big] \ne 0$.

Unfortunately, all we can deduce from this is that for every positive integer $n$, there exists an integer $m$ such that $\eta$ intersects $\eta + \left(2m + \frac1n\right)(\eta(b) - \eta(a))$. If there is an argument that enables us to recover the full conjecture from this unsatisfactory result, I haven't thought of it yet. This explains, I hope, what I meant by saying that the second conjecture is "slightly weaker" than the main one; also, why I said at the outset that the answers to the second question in the series "almost" prove this question's main conjecture in the case $n = 2$. $\square$

Experiments by several users drawing polygonal paths in GeoGebra [other programs are available!] have so far failed to produce a counterexample to the conjecture (expressed in the "weaker" form). Here is a failed attempt for $n = 3$ (illustrating the main form of the conjecture in this question):

Failing to construct a counterexample with n = 3.

It's a racing certainty that someone will now produce a counterexample with embarrassing ease. :)


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