2021 JMC 10 Problems/Problem 23

Problem

An invisible ant and an anteater, at the same constant speed of $1$ edge length per second, start at (not necessarily distinct) randomly chosen vertices of a cube. Each second, the ant first pings its location to the anteater, then randomly chooses one of the $3$ edges emerging from its vertex to traverse immediately. The anteater traverses the edge on the closest path to the ping at the same time the ant travels. If multiple optimal paths exist, one is randomly chosen. The anteater eats the ant if at some time they are both at the same point, not necessarily a vertex. What is the ant's expected lifespan in seconds?

$\textbf{(A) }\dfrac{41}{16}\qquad\textbf{(B) }\dfrac{11}{4}\qquad\textbf{(C) }\dfrac{49}{16}\qquad\textbf{(D) }\dfrac{27}{8}\qquad\textbf{(E) }\dfrac{55}{16}$

Solution

The crux of this problem is noting that the anteater could only make forward progress. Let $e_i$ be the life expectancy of the ant where the closest path from the anteater to the ant is a distance of $i$ edges, and $e_0 = 0$ . Note that $e_1 = \tfrac{1}{3} \cdot -\tfrac{1}{2} +\tfrac{2}{3} e_1 +1 \implies e_1 = \tfrac{5}{2}.$ Here, $-\tfrac{1}{2}$ comes from the death of the ant occurring before the step has gone through completely. Now, for $e_2$ , the anteater chooses one of two congruent paths. We have $e_2 =\tfrac{1}{3} \cdot 0 + \tfrac{2}{3} e_2 +1 \implies e_2 =3.$ Similarly, we obtain $e_3 = \tfrac{2}{3} e_1 + \tfrac{1}{3} e_3 + 1 \implies e_3 = 4.$ The life expectancy of the ant is $\tfrac{1}{8} e_0 +\tfrac{3}{8} e_1 +\tfrac{3}{8} e_2 +\tfrac{1}{8} e_3 = \tfrac{41}{16}$ .

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2021 JMC 10 Problems/Problem 23

Problem

Solution