Track / Stage 5 / 264 of 400 #864 of 2000

Problem 864

AIME late

Geometry Difficulty 5.7 Prove it THE 68th ROMANIAN MATHEMATICAL OLYMPIAD · Romania

Let $ABCD A'B'C'D'$ be a cube with side length $1$ . An ant walks on the cube's faces, starting from $A$ and ending at $C'$ . The ant moves only on the cube's edges or on the diagonals of its faces. Knowing that the ant never passes through the same point twice, find the maximal length of such a walk.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

We claim that the maximal length of a walk is $3 + 4\sqrt{2}$ . An example of such a walk is $A \to A' \to B \to D \to D' \to C \to B' \to C'$ . Since the cube has $8$ vertices, the ant's walk consists of at most $7$ steps, each of length either $1$ or $\sqrt{2}$ .

The length of a $6$ steps walk is at most $6\sqrt{2} < 3 + 4\sqrt{2}$ , hence the maximal walk must have exactly $7$ steps.

We claim that the number of "diagonal" steps is at most $4$ . For this purpose, let us color black the vertices $A, C, B', D'$ , and white the other four. Observe that a "diagonal" step changes the color, while an "edge" step changes it. Since $A$ and $C'$ have different colors, we deduce that the walk must contain an odd number of "edge" steps. By inspection, we see that a walk with one "edge" step and $6$ "diagonal" ones self-intersects, thus proving the claim.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.