Difference between revisions of "2021 AIME I Problems/Problem 12"
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==Solution== | ==Solution== | ||
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+ | 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 | ||
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+ | E(3,3,3)=1+1/4E(3,3,3)+3/4E(1,3,5), giving E(3,3,3)=4/3+E(1,3,5); (1) | ||
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+ | E(1,3,5)=1+1/8E(1,1,1)+1/2E(1,3,5)+1/8E(3,3,3), giving E(1,3,5)=2+1/4E(1,1,1)+1/4E(3,3,3); (2) | ||
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+ | E(1,1,1)=1+1/4E(1,1,1)+1/4E(1,3,5), giving E(1,1,1)=4/3+1/3E(1,3,5); (3) | ||
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+ | Plug in (1) and (3) into (2), we see that E(1,3,5)=4. E(3,3,3)=4/3+4=16/3. | ||
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==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=11|num-a=13}} | {{AIME box|year=2021|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:34, 11 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
E(3,3,3)=1+1/4E(3,3,3)+3/4E(1,3,5), giving E(3,3,3)=4/3+E(1,3,5); (1)
E(1,3,5)=1+1/8E(1,1,1)+1/2E(1,3,5)+1/8E(3,3,3), giving E(1,3,5)=2+1/4E(1,1,1)+1/4E(3,3,3); (2)
E(1,1,1)=1+1/4E(1,1,1)+1/4E(1,3,5), giving E(1,1,1)=4/3+1/3E(1,3,5); (3)
Plug in (1) and (3) into (2), we see that E(1,3,5)=4. E(3,3,3)=4/3+4=16/3.
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 |
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