Source: USATST-2023-P1
Category: Extremal Problems

Problem:
There are $2022$ equally spaced points on a circular track $\gamma$ of circumference $2022$. The points are labeled $A_1$, $A_2$, \dots, $A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\arc{PQ}$ of $\gamma$; if $\overline{PQ}$ is a diameter of $\gamma$, she moves along either semicircle. Determine the maximal possible sum of the lengths of the $2022$ arcs which Bunbun traveled, over all possible labellings of the $2022$ points.

Solution:
Replacing $2022$ with $2n$, the answer is $2n^2 - 2n + 2$. (When $n=1011$, the number is $2042222$.) 

\paragraph{Construction.} Let the $2n$ points along the track be labeled $0, 1, \dots, 2n-1$ in order. The maximum distance is achieved by the sequence of points $A_{2i-1} = i-1$ and $A_{2i} = n+i-1$ for $i=1, 2, \dots, n$. (For example, when $n=5$, the sequence of points is $0, 5, 1, 6, 2, 7, 3, 8, 4, 9$.) This corresponds to alternating between points in the two halves of the circle.

\paragraph{First proof of bound.} Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$. 

\begin{claim*}
The distance of the leg of the journey $A_{2i-1} \to A_{2i} \to A_{2i+1}$ is at most $2n - d_i$.
\end{claim*}

\begin{proof}
The point $A_{2i}$ lies on one of the two arcs from $A_{2i-1}$ to $A_{2i+1}$, which have lengths $d_i$ and $2n-d_i$. The sum of the shorter arc distances $A_{2i-1} \to A_{2i}$ and $A_{2i} \to A_{2i+1}$ is at most the length of the arc containing $A_{2i}$. Thus, we get a bound of $\max(d_i, 2n-d_i) = 2n - d_i$.
\end{proof}

That means the total distance is at most 
\[
\sum_{i=1}^n \left( 2n - d_i \right) = 2n^2 - (d_1 + d_2 + \dots + d_n).
\]

\begin{claim*}
We have 
\[
d_1 + d_2 + \dots + d_n \ge 2n-2.
\]
\end{claim*}

\begin{proof}
The left-hand side is the total length of the closed walk $A_1 \to A_3 \to \dots \to A_{2n-1} \to A_1$. Among the $n$ distinct points visited, two of them must be at a distance of at least $n-1$ apart. The closed walk is divided into two paths connecting these two points. If one path consists of a single step, its length is at least $n-1$, and the other path consists of $n-1$ steps which contribute at least $1$ each, giving a sum of at least $(n-1) + (n-1) \cdot 1 = 2n-2$. More generally, by the triangle inequality, each of the two paths connecting the two points has length at least $n-1$, so the total length of the walk is always at least $2(n-1) = 2n-2$.
\end{proof}

\paragraph{Second proof of bound.} Draw the $n$ diameters through the midpoints of the $2n$ arcs between adjacent points on the track.

\begin{claim*}[Interpretation of distances]
The distance between any two points equals the number of diameters crossed to travel between the points.
\end{claim*}

\begin{proof}
Clear, as each step of length $1$ along the shorter arc between two points crosses exactly one such diameter.
\end{proof}

With this in mind, call a diameter \emph{critical} if it is crossed by all $2n$ arcs. 

\begin{claim*}
At most one diameter is critical.
\end{claim*}

\begin{proof}
Suppose there were two critical diameters; these divide the circle into four sectors, each containing at least one of the $2n$ points. Then all $2n$ arcs cross both diameters, and so every hop must travel between opposite sectors. But this means that the sequence of hops alternates between two opposite sectors, so the points in the other two sectors are never accessed --- contradiction.
\end{proof}

\begin{claim*}
Every diameter is crossed an even number of times.
\end{claim*}

\begin{proof}
Clear: crossing a diameter moves Bunbun from one semicircle to the opposite one, so the diameter needs to be crossed an even number of times for the closed loop to return to its origin.
\end{proof}

This immediately implies that the maximum possible total distance is achieved when one diameter is crossed all $2n$ times, and every other diameter is crossed $2n-2$ times, for a total distance of at most 
\[
n \cdot (2n-2) + 2 = 2n^2 - 2n + 2.
\]

Answer: 2042222

Grading guidelines:
(Partial) 1. Gave (and correctly evaluated) a concrete labeling achieving total distance $2n^2-2n+2$ (e.g. alternating between opposite halves: $0,n,1,n+1,\dots,n-1,2n-1$). 2. Proved the key local bound for a 2-step leg: for $d_i$ the shorter distance from $A_{2i-1}$ to $A_{2i+1}$, showed \(\operatorname{dist}(A_{2i-1},A_{2i})+\operatorname{dist}(A_{2i},A_{2i+1})\le 2n-d_i\). 3. Introduced the diameter-crossing interpretation (distance equals number of crossed mid-diameters) and proved at least one of: “every diameter is crossed an even number of times” or “at most one diameter can be crossed by all $2n$ hops (critical diameter)”. (Almost) 1. Obtained the global upper bound $\sum (2n-d_i)=2n^2-\sum d_i$ and argued $\sum d_i\ge 2n-2$, but left a gap in justifying $\sum d_i\ge 2n-2$ (e.g. the choice of two odd-index points at distance $\ge n-1$ or the triangle-inequality argument for the two connecting paths). 2. Used the diameter-crossing method to derive the bound $2n^2-2n+2$, but did not fully justify one of the structural claims (e.g. why at most one diameter is critical, or why each diameter is crossed an even number of times in the closed tour). 3. Solution gives the correct maximum value and a matching construction, but has a minor indexing/boundary omission (e.g. handling the final hop $A_{2n}\to A_1$ or the diameter case) while the main method is correct.