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
Define the distance between two frogs as the number of sides between them that do not contain the third frog.
Let 
denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are 
and 
(in either clockwise or counterclockwise order). Without the loss of generality, assume that 
We wish to find 
Note that:
- At any moment before the frogs stop jumping, the only possibilities for
are 
and 
does not depend on the actual positions of the frogs, but depends on the distances between the frogs.- At the end of each minute, each frog has
outcomes. So, there are 
outcomes in total.
are 
and 
outcomes. So, there are 
outcomes in total.
We have the following system of equations:
Rearranging and simplifying each equation, we get
Substituting 
and 
into 
we obtain ![\[E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],\]](http://latex.artofproblemsolving.com/5/3/8/538a8689508403bf582d64c79a6443797483ab3d.png)
from which 
Substituting this into gives 
Therefore, the answer is 
~Ross Gao (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
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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