Difference between revisions of "2003 AIME II Problems/Problem 13"
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The transition matrix is { {0,1,1}, {1,0,1} , {1,1,0} } * (1/2) . Then use the exponentiation method of squaring ( A*A---(A^2)*(A^2)---(A^4*A^4)--(A^8*A^2) to get the transition value of 342. Divide by 2^10 for the probability, reduce fractions, for the result of 171+512 = 683. | The transition matrix is { {0,1,1}, {1,0,1} , {1,1,0} } * (1/2) . Then use the exponentiation method of squaring ( A*A---(A^2)*(A^2)---(A^4*A^4)--(A^8*A^2) to get the transition value of 342. Divide by 2^10 for the probability, reduce fractions, for the result of 171+512 = 683. | ||
− | + | ===Solution 5=== | |
+ | This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to <math>341.33/1024</math>. We can either round up or down. If we round down, we see <math>341/1024</math> cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, <math>342/1024</math> can be reduced to <math>171/512</math> where the sum 171+512= <math>\boxed{683}</math> is an accepted answer. | ||
== See also == | == See also == |
Revision as of 20:46, 12 August 2018
Contents
Problem
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects one of the two vertices where it is not currently located, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is where
and
are relatively prime positive integers, find
Solution
Solution 1 (Easiest)
Let represent the probability that the bug is at its starting vertex after
moves. If the bug is on its starting vertex after
moves, then it must be not on its starting vertex after
moves. At this point it has
chance of reaching the starting vertex in the next move. Thus
.
, so now we can build it up:
,
,
,
,
,
,
,
,
,
,
Thus the answer is
Solution 2
Consider there to be a clockwise and a counterclockwise direction around the triangle. Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., . Since
, it is only possible that
.
In the first case, we pick out of the ant's
steps to be clockwise, for a total of
paths. In the second case, we choose
of the steps to be clockwise, and in the third case we choose
to be clockwise. Hence the total number of ways to return to the original vertex is
. Since the ant has two possible steps at each point, there are
routes the ant can take, and the probability we seek is
, and the answer is
.
Solution 3
Label the vertices of the triangle with the ant starting at
. We will make a table of the number of ways to get to
in
moves
. The values of the table are calculated from the fact that the number of ways from a vertex say
in
steps equals the number of ways to get to
in
steps plus the number of ways to get to
in
steps.
Therefore, our answer is
Solution 4
Notice that this problem can be converted into a Markov Chain transition matrix.
The transition matrix is { {0,1,1}, {1,0,1} , {1,1,0} } * (1/2) . Then use the exponentiation method of squaring ( A*A---(A^2)*(A^2)---(A^4*A^4)--(A^8*A^2) to get the transition value of 342. Divide by 2^10 for the probability, reduce fractions, for the result of 171+512 = 683.
Solution 5
This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to . We can either round up or down. If we round down, we see
cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up,
can be reduced to
where the sum 171+512=
is an accepted answer.
See also
- 1985 AIME Problems/Problem 12 - very similar problem with a tetrahedron
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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