Difference between revisions of "2021 AIME I Problems/Problem 12"
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Let <math>E(a,b,c)</math> denote the expected number of minutes for two frogs to meet, such that the frogs are <math>a,b,</math> and <math>c</math> vertices apart, in either clockwise or counterclockwise order. Note that | Let <math>E(a,b,c)</math> denote the expected number of minutes for two frogs to meet, such that the frogs are <math>a,b,</math> and <math>c</math> vertices apart, in either clockwise or counterclockwise order. Note that | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>a+b+c=9</math> always holds.</li><p> | ||
+ | <li>If at least one of <math>a,b,</math> or <math>c</math> is <math>0,</math> then <math>E(a,b,c)=0.</math></li><p> | ||
+ | <li>At the end of each minute, each frog has <math>2</math> possibilities. So, there are <math>2^3=8</math> possibilities in total.</li><p> | ||
+ | </ol> | ||
+ | We have the following system of equations: | ||
+ | <cmath>\begin{align*} | ||
+ | E(3,3,3)&=1+\frac{2}{8}E(3,3,3)+\frac{6}{8}E(1,3,5), \\ | ||
+ | E(1,3,5)&=1+\frac{1}{8}E(1,1,7)+\frac{4}{8}E(1,3,5)+\frac{1}{8}E(3,3,3), \\ | ||
+ | E(1,1,7)&=1+\frac{2}{8}E(1,1,7)+\frac{2}{8}E(1,3,5). | ||
+ | \end{align*}</cmath> | ||
+ | Rearranging and simplifying each equation, we get | ||
+ | <cmath>\begin{align*} | ||
+ | E(3,3,3)&=\frac{4}{3}+E(1,3,5), &(1) \\ | ||
+ | E(1,3,5)&=2+\frac{1}{4}E(1,1,7)+\frac{1}{4}E(3,3,3), &\hspace{13mm}(2) \\ | ||
+ | E(1,1,7)&=\frac{4}{3}+\frac{1}{3}E(1,3,5). &(3) | ||
+ | \end{align*}</cmath> | ||
+ | Substituting <math>(1)</math> and <math>(3)</math> into <math>(2),</math> we obtain <cmath>E(1,3,5)=4,</cmath> from which <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>. | |
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~Ross Gao (Solution) | ~Ross Gao (Solution) |
Revision as of 02:34, 7 July 2021
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 distances between the frogs, in either clockwise or counterclockwise order.
Let denote the expected number of minutes for two frogs to meet, such that the frogs are
and
vertices apart, in either clockwise or counterclockwise order. Note that
always holds.
- If at least one of
or
is
then
- At the end of each minute, each frog has
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
from which
. Each step is one minute. 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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.