EGMO 2023 | Problem 4 | Solution - 17 April 2023

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Problem 4.
Turbo the snail sits on a point on a circle with circumference $1$ . Given an infinite sequence of positive real numbers $c_1, c_2, c_3, \dots$ , Turbo successively crawls distances $c_1, c_2, c_3, \dots$ around the circle, each time choosing to crawl either clockwise or counterclockwise.
Determine the largest constant $C > 0$ with the following property: for every sequence of positive real numbers $c_1, c_2, c_3, \dots$ 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 answer is $C= \frac{1}{2}$

For $C= \frac{1}{2}$ , consider the following strategy for Turbo. Choose an arbitrary point $X$ (which is not Turbo's starting point), and always jump in the direction of $Y$ which is the antipode of $X$ . Indeed, this strategy works by induction.

Now, assume $C > \frac{1}{2}$ . Consider the following construction to guarantee every point is covered. First, start off the sequence with $\frac{1}{2}, \frac{1}{4}, \frac{1}{2}$ This will guarantee that turbo will have crawled across every point, except some interval $T_0$ of length $\frac{1}{4}$ , and that turbo will be standing right on an endpoint of $T_0$ . Now, let the next number in the sequence be the distance from any endpoint of $T_0$ to the antipode of the midpoint of $T_0$ . This forces Turbo to the antipode of the midpoint of $T_0$ , otherwise, he must cover $T_0$ and, by extension, the whole circle. From here, choose some $\frac{1}{2} < \alpha < C$ and make turbo move a distance of $\alpha$ . This reduces the size of the interval of uncovered points by exactly $\alpha-\frac{1}{2}$ . Let $T_1$ be the new set of uncovered points. Then, do the exact same process we did on $T_0$ to $T_1$ . In general we have $\text{length of }T_k = \frac{1}{4} - k \left ( \alpha - \frac{1}{2} \right )$ which is negative for sufficiently large $k$ .