Source: USAMO-2021-P2
Category: Graph Theory

Problem:
The Planar National Park is an undirected 3‑regular planar graph (i.e., all vertices have degree $3$). A visitor walks through the park as follows: she begins at a vertex and starts walking along an edge. When she reaches the other endpoint, she turns left. On the next vertex she turns right, and so on, alternating left and right turns at each vertex. She does this until she gets back to the vertex where she started. What is the largest possible number of times she could have entered any vertex during her walk, over all possible layouts of the park?

Solution:
The answer is $3$. 

We consider the trajectory of the visitor as an ordered sequence of turns. A turn is defined by specifying a vertex, the incoming edge, and the outgoing edge. Hence there are six possible turns for each vertex. Given one turn in the sequence, one can reconstruct the entire sequence of turns. This implies already that the trajectory of the visitor, when extended to an infinite sequence, is totally periodic (not just eventually periodic), because there are finitely many possible turns, so some turn must be repeated. So, any turn appears at most once in the period of the sequence, giving a naive bound of $6$ visits to any vertex for the original problem. 

However, it is impossible for both of the turns $a \to b \to c$ and $c \to b \to a$ to occur in the same trajectory. If so, then extending the path, we get a forward path $a \to b \to c \to d \to e \to \dotsb$ and a backward path $\dotsb \to e \to d \to c \to b \to a$. Because the walk's rules are deterministic and the alternating left-right turns are symmetric with respect to reversing direction, the two paths traverse the exact same sequence of vertices and edges in opposite directions. We assumed for contradiction that these two paths were part of the same trajectory, yet a single directed trajectory cannot traverse the same cycle in opposite directions; thus, they clearly never meet. This halves the maximum number of possible visits to any vertex, yielding an upper bound of $3$.

It remains to give a construction showing $3$ can be achieved. There are many valid constructions. One such construction yields the path $C \to A \to H \to I \to F \to G \to D \to B \to A \to H \to E \to F \to G \to J \to B \to A \to C$, which visits vertex $A$ exactly three times.

Answer: 3

Grading guidelines:
(Partial) 1. Formalized the walk as a deterministic process on “turn states” (vertex + incoming edge + outgoing edge, 6 per vertex), and deduced the walk is periodic (hence any fixed turn state cannot occur twice before returning). 2. Derived the naive bound that any vertex can be entered at most 6 times, by counting the 6 possible turn states at that vertex. 3. Proved the key restriction that a trajectory cannot contain both turns a→b→c and c→b→a at the same middle vertex b (i.e., the walk cannot use both orientations of the same local passage), and used it to improve the bound (even if not all the way to 3). (Almost) 1. Proved the sharp upper bound 3 by combining (i) periodicity/determinism on turn states with (ii) the impossibility of having both a→b→c and c→b→a in one trajectory, but left a small logical gap in justifying why the “forward” and “backward” extensions cannot intersect/merge into one closed walk. 2. Gave a construction of a 3-regular planar graph and a valid alternating left-right walk that revisits some vertex three times, but did not fully verify one local condition (3-regularity, planarity, or the left/right choices at one vertex). 3. Completed the proof of the upper bound 3, but omitted the existence example (or gave an example that achieves only 2 visits) while the rest of the argument is correct.