1985 AIME Problems/Problem 12
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[hide]Problem
Let , , and be the vertices of a regular tetrahedron, each of whose edges measures meter. A bug, starting from vertex , observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let be the probability that the bug is at vertex when it has crawled exactly meters. Find the value of .
Solution 1 (Single Variable Recursion)
For all nonnegative integers let be the probability that the bug is at vertex when it has crawled exactly meters. We wish to find
Clearly, we have For all note that after crawls:
- The probability that the bug is at vertex is and the probability that it crawls to vertex on the next move is
- The probability that the bug is not at vertex is and the probability that it crawls to vertex on the next move is
Together, the recursive formula for is Two solutions follow from here:
Solution 1.1 (Recursive Formula)
We evaluate recursively: Therefore, the answer is
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 1.2 (Explicit Formula)
Let for some function and constant For all the recursive formula for becomes Solving for we get For simplicity purposes, we set which gives Clearly, is a geometric sequence with the common ratio Substituting and into produces the first term of the geometric sequence.
So, the explicit formula for is and the explicit formula for is Finally, the requested probability is from which
~MRENTHUSIASM
Solution 2 (Multivariable Recursion)
Denominator
There are ways for the bug to make independent crawls without restrictions.
Numerator (Recursion)
Let denote the number of ways for the bug to crawl exactly meters starting from vertex and ending at vertex where and is a positive integer. We wish to find
Since the bug must crawl to vertex or on the first move, we have where
More generally, we get For note that
- Base Case:
- Recursive Case:
Clearly, is a geometric sequence with the first term and the common ratio Therefore, its explicit formula is Recall that By and we rewrite recursively: Answer
The requested probability is from which
~MRENTHUSIASM
Solution 3 (Single Variable Version of Solution 2)
Let denotes the number of ways that the bug arrives at after crawling meters, then we have .
Notice that there is respectively way to arrive at for each of the different routes after the previous crawls, excluding the possibility that the bug ends up at after the th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation Quick calculations yield Thus, .
~Nafer
Solution 4 (Partitions)
We can find the number of different times the bug reaches vertex before the th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at
Define to be the number of paths of length which start and end at but do not pass through otherwise. Obviously In general for the bug has three initial edges to pick from. From there, since the bug cannot return to by definition, the bug has exactly two choices. This continues from the nd move up to the th move. The last move must be a return to so this move is determined. So
Now we need to find the number of cycles by which the bug can reach at the end. Since we know that cannot be used, as on the th move the bug cannot move from to So we need to find the number of partitions of using only and These are and We can calculate these and sum them up using our formula. Also, order matters, so we need to find the number of ways to arrange each partition: Finally, this is a probability question, so we divide by or The answer is
Solution 5 (Educated Guess)
There exists a simple heuristic method to arrive at the answer to this question, due to Simon Rubinstein-Salzedo, as follows: after a couple of moves, the randomness of movement of the bug and smallness of the system ensures that we should expect its probability distribution to be very close to uniform. In particular, we would expect to be very close to for decently-sized , for example . (In fact, from looking at the previous solutions we can see that it is already close when , and in fact, the earlier values are also the best possible approximations given the restraints on where the bug can be.) Since we know the answer is of the form , we realize that must be very close to , as indeed it is. The answer is .
Solution 6 (Educated Guess)
Only do this if you don't know how to solve the problem and want to make a good guess. Since there are vertices of a tetrahedron, there is approximately an probability of coming back to after moves. Dividing by gives a number between and If the bug continuously alternates its location from to some other vertex, in the end, it will not be at Therefore we choose the smaller number,
~Rocket123
See also
- 2003 AIME II Problems/Problem 13 - very similar problem with an equilateral triangle
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