2021 AIME I Problems/Problem 12
Problem
Let 
be a dodecagon (
-gon). Three frogs initially sit at 
and 
. 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 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
The expected number of minutes depends on the numbers of vertices between the frogs, in either clockwise or counterclockwise order.
Let 
denote the expected number of minutes for two frogs to arrive at the same vertex, such that the frogs are 
and 
vertices apart, in either clockwise or counterclockwise order. We wish to find 
Note that:
always holds.
if and only if two frogs arrive at the same vertex.- At the end of each minute, each frog has
possibilities. So, there are 
possibilities in total.
always holds.
if and only if two frogs arrive at the same vertex.
possibilities. So, there are 
possibilities in total.
We have the following system of equations:
Rearranging and simplifying each equation, we get
Substituting 
and 
into 
we obtain ![\[E(1,3,5)=2+\frac{1}{4}\left[\frac{4}{3}+E(1,3,5)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(1,3,5)\right],\]](http://latex.artofproblemsolving.com/4/9/7/49744701f40dc3926288f8a745188620222db049.png)
from which 
Substituting this into gives 
Therefore, the answer is 
~Ross Gao (Solution)
~MRENTHUSIASM (Reformatting and Minor Revisions)
See Also
| 2021 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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