# 2021 AIME I Problems/Problem 12

## Problem

Let $A_1A_2A_3...A_{12}$ be a dodecagon (12-gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

The expected number of steps depends on the distance between the frogs, not on the order in which these distances appear. Let $E(a,b,c)$ where $a+b+c=9$ denote the expected number of steps that it takes for two frogs to meet if traversing in clockwise or counterclockwise order, the frogs are $a$, $b$ and $c$ vertices apart. Then

$E(3,3,3)=1+\frac{1}{4} E(3,3,3)+\frac{3}{4} E(1,3,5)$, giving $E(3,3,3)=\frac{4}{3}+E(1,3,5)$; (1)

$E(1,3,5)=1+\frac{1}{8}E(1,1,7)+\frac{1}{2}E(1,3,5)+\frac{1}{8}E(3,3,3)$, giving $E(1,3,5)=2+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(3,3,3)$; (2)

$E(1,1,7)=1+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(1,3,5)$, giving $E(1,1,7)=\frac{4}{3}+\frac{1}{3}E(1,3,5)$; (3)

Plug in (1) and (3) into (2), we see that $E(1,3,5)=4$. $E(3,3,3)=\frac{4}{3}+4=\frac{16}{3}$. Each step is one minute. The answer is $16+3=\boxed{19}$.

-Ross Gao