Subfield: Geometry


Question: Turbo the snail sits on a point on a circle with circumference 1. Given an infinite sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$. Turbo successively crawls distances $c_{1}, c_{2}, c_{3}, \ldots$ around the circle, each time choosing to crawl either clockwise or counterclockwise.

For example, if the sequence $c_{1}, c_{2}, c_{3}, \ldots$ is $0.4,0.6,0.3, \ldots$, then Turbo may start crawling as follows:
<img_3362>

Determine the largest constant $C>0$ with the following property: for every sequence of positive real numbers $c_{1}, c_{2}, c_{3}, \ldots$ with $c_{i}<C$ for all $i$, Turbo can (after studying the sequence) ensure that there is some point on the circle that it will never visit or crawl across.

Solution:
The largest possible $C$ is $C=\frac{1}{2}$.

For $0<C \leqslant \frac{1}{2}$, Turbo can simply choose an arbitrary point $P$ (different from its starting point) to avoid. When Turbo is at an arbitrary point $A$ different from $P$, the two arcs $A P$ have total length 1; therefore, the larger of the two the arcs (or either arc in case $A$ is diametrically opposite to $P$ ) must have length $\geqslant \frac{1}{2}$. By always choosing this larger arc (or either arc in case $A$ is diametrically opposite to $P$ ), Turbo will manage to avoid the point $P$ forever.

For $C>\frac{1}{2}$, we write $C=\frac{1}{2}+a$ with $a>0$, and we choose the sequence

$$
\frac{1}{2}, \quad \frac{1+a}{2}, \quad \frac{1}{2}, \quad \frac{1+a}{2}, \quad \frac{1}{2}, \ldots
$$

In other words, $c_{i}=\frac{1}{2}$ if $i$ is odd and $c_{i}=\frac{1+a}{2}<C$ when $i$ is even. We claim Turbo must eventually visit all points on the circle. This is clear when it crawls in the same direction two times in a row; after all, we have $c_{i}+c_{i+1}>1$ for all $i$. Therefore, we are left with the case that Turbo alternates crawling clockwise and crawling counterclockwise. If it, without loss of generality, starts by going clockwise, then it will always crawl a distance $\frac{1}{2}$ clockwise followed by a distance $\frac{1+a}{2}$ counterclockwise. The net effect is that it crawls a distance $\frac{a}{2}$ counterclockwise. Because $\frac{a}{2}$ is positive, there exists a positive integer $N$ such that $\frac{a}{2} \cdot N>1$. After $2 N$ crawls, Turbo will have crawled a distance $\frac{a}{2}$ counterclockwise $N$ times, therefore having covered a total distance of $\frac{a}{2} \cdot N>1$, meaning that it must have crawled over all points on the circle.

Note: Every sequence of the form $c_{i}=x$ if $i$ is odd, and $c_{i}=y$ if $i$ is even, where $0<x, y<C$, such that $x+y \geqslant 1$, and $x \neq y$ satisfies the conditions with the same argument. There might be even more possible examples.
To show that $C\le \frac12$

We consider the following related problem:

We assume instead that the snail Chet is moving left and right on the real line. Find the size $M$ of the smallest (closed) interval, that we cannot force Chet out of, using a sequence of real numbers $d_{i}$ with $0<d_{i}<1$ for all $i$.

Then $C=1 / M$. Indeed if for every sequence $c_{1}, c_{2}, \ldots$, with $c_{i}<C$ there exists a point that Turbo can avoid, then the circle can be cut open at the avoided point and mapped to an interval of size $M$ such that Chet can stay inside this interval for any sequence of the from $c_{1} / C, c_{2} / C, \ldots$, see Figure 5 . However, all sequences $d_{1}, d_{2}, \ldots$ with $d_{i}<1$ can be written in this form. Similarly if for every sequence $d_{1}, d_{2}, \ldots$, there exists an interval of length smaller or equal $M$ that we cannot force Chet out of, this projects to a subset of the circle, that we cannot force Turbo out of using any sequence of the form $d_{1} / M, d_{2} / M, \ldots$. These are again exactly all the sequences with elements in $[0, C)$.

<img_3381>

Figure 5: Chet and Turbo equivalence

Claim: $M \geqslant 2$.

Proof. Suppose not, so $M<2$. Say $M=2-2 \varepsilon$ for some $\varepsilon>0$ and let $[-1+\varepsilon, 1-\varepsilon]$ be a minimal interval, that Chet cannot be forced out of. Then we can force Chet arbitrarily close to $\pm(1-\varepsilon)$. In partiular, we can force Chet out of $\left[-1+\frac{4}{3} \varepsilon, 1-\frac{4}{3} \varepsilon\right]$ by minimality of $M$. This means that there exists a sequence $d_{1}, d_{2}, \ldots$ for which Chet has to leave $\left[-1+\frac{4}{3} \varepsilon, 1-\frac{4}{3} \varepsilon\right]$, which means he ends up either in the interval $\left[-1+\varepsilon,-1+\frac{4}{3} \varepsilon\right)$ or in the interval $\left(1-\frac{4}{3} \varepsilon, 1-\varepsilon\right]$.

Now consider the sequence,

$$
d_{1}, 1-\frac{7}{6} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{7}{6} \varepsilon, d_{2}, 1-\frac{7}{6} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{7}{6} \varepsilon, d_{3}, \ldots
$$

obtained by adding the sequence $1-\frac{7}{6} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{7}{6} \varepsilon$ in between every two steps. We claim that this sequence forces Chet to leave the larger interval $[-1+\varepsilon, 1-\varepsilon]$. Indeed no two consecutive elements in the sequence $1-\frac{7}{6} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{2}{3} \varepsilon, 1-\frac{7}{6} \varepsilon$ can have the same sign, because the sum of any two consecutive terms is larger than $2-2 \varepsilon$ and Chet would leave the interval $[-1+\varepsilon, 1-\varepsilon]$. It follows that the $\left(1-\frac{7}{6} \varepsilon\right)$ 's and the $\left(1-\frac{2}{3} \varepsilon\right)$ 's cancel out, so the position after $d_{k}$ is the same as before $d_{k+1}$. Hence, the positions after each $d_{k}$ remain the same as in the original sequence. Thus, Chet is also forced to the boundary in the new sequence.

If Chet is outside the interval $\left[-1+\frac{4}{3} \varepsilon, 1-\frac{4}{3} \varepsilon\right]$, then Chet has to move $1-\frac{7}{6} \varepsilon$ towards 0 , and ends in $\left[-\frac{1}{6} \varepsilon, \frac{1}{6} \varepsilon\right]$. Chet then has to move by $1-\frac{2}{3} \varepsilon$, which means that he has to leave the interval $[-1+\varepsilon, 1-\varepsilon]$. Indeed the absolute value of the final position is at least $1-\frac{5}{6} \varepsilon$. This contradicts the assumption, that we cannot force Chet out of $[-1+\varepsilon, 1-\varepsilon]$. Hence $M \geqslant 2$ as needed.

Final answer: $\frac{1}{2}$