Difference between revisions of "2021 AIME I Problems/Problem 12"
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Plug in (1) and (3) into (2), we see that <math>E(1,3,5)=4</math>. <math>E(3,3,3)=\frac{4}{3}+4=\frac{16}{3}</math>. Each step is one minute. The answer is <math>16+3=\boxed{19}</math>. | Plug in (1) and (3) into (2), we see that <math>E(1,3,5)=4</math>. <math>E(3,3,3)=\frac{4}{3}+4=\frac{16}{3}</math>. Each step is one minute. The answer is <math>16+3=\boxed{19}</math>. | ||
| + | |||
| + | -Ross Gao | ||
==See also== | ==See also== | ||
Revision as of 09:55, 12 March 2021
Problem
Let 
be a dodecagon (12-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 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
, giving 
; (1)
, giving 
; (2)
, giving 
; (3)
Plug in (1) and (3) into (2), we see that 
. 
. Each step is one minute. The answer is 
.
-Ross Gao
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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
