Difference between revisions of "2021 AIME I Problems/Problem 12"

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==Problem==
 
==Problem==
These problems will not be available until the 2021 AIME I is released on Wednesday, March 10, 2021.
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Let <math>A_1A_2A_3...A_{12}</math> be a dodecagon (12-gon). Three frogs initially sit at <math>A_4,A_8,</math> and <math>A_{12}</math>. 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 <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==

Revision as of 15:49, 11 March 2021

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

See also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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