1985 AIME Problems/Problem 12
- 1 Problem
- 2 Solution 1 (Single Variable Recursion)
- 3 Solution 2 (Multivariable Recursion by Algebra)
- 4 Solution 3 (Multivariable Recursion by Table)
- 5 Solution 4 (Single Variable Version of Solution 2)
- 6 Solution 5 (Dynamic Programming)
- 7 Solution 6 (Generating Functions and Roots of Unity Filter / Casework)
- 8 Solution 7 (Partitions)
- 9 Solution 8 (Sequences)
- 10 Remark
- 11 See also
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)
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.
For all nonnegative integers the explicit formula for is and the explicit formula for is Finally, the requested probability is from which
Solution 2 (Multivariable Recursion by Algebra)
There are ways for the bug to make independent crawls without restrictions.
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
Solution 3 (Multivariable Recursion by Table)
Define notation as Solution 2 does.
In fact, we can generalize the following relationships for all nonnegative integers Using these equations, we recursively fill out the table below: Note that the paths from to and the paths from to have one-to-one correspondence. So, we must get for all values of
The requested probability is from which
Solution 4 (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, .
Solution 5 (Dynamic Programming)
Let be the probability the bug lands on vertex after crawling meters, be the probability the bug lands on vertex after crawling meters, and etc.
Note that and For the probability that the bug land on each vertex after meters is the sum of the probability the bug land on other vertices after crawling meters. So, we have It follows that
We construct the following table: Therefore, the answer is
Solution 6 (Generating Functions and Roots of Unity Filter / Casework)
The generating function for a problem with this general form ( states, steps) is , so the generating function of interest for this problem is . Our goal is to find the coefficients of every and add them up before dividing by . Here we have two ways to proceed:
1. Roots of Unity Filter
Let . We have that if , then From here, the desired probability is . Therefore, the answer is .
We can factor as The coefficients of will be the same as the coefficients of The possible values for would then be and Because the coefficients of and are equal and so are the coefficients of and To make an term, we need "" terms and one "" term multiplied together. There are ways to arrange these numbers. The coefficient of the term will be the sum of the number of the possible arrangements for these numbers: Thus, the coefficient of the term is From here, the desired probability is Thus, our answer is
Solution 7 (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 8 (Sequences)
Note that this problem is basically equivalent to the following: How many distinct sequences of integers are there such that for all and for all ?
Now consider the integers modulo Let be a new sequence of integers such that for all
Note that the condition is equivalent to that for all and since must be a multiple of
Thus, our desired number of paths is equivalent to the number of ordered septuples of positive integers such that for all and is congruent to
We will now proceed with casework on the number of s in the septuple.
One : There are ways to arrange the , and valid ways (the proof of this combinatorial identity is left as an exercise to the reader) to arrange the s and s.
Three s: There are ways to arrange the s, and valid ways to arrange the s and s.
Five s: There are ways to arrange the s, and valid ways to arrange the s and s.
Adding up our cases, we obtain valid sequences. Dividing by the total number of paths without restrictions, our desired probability is giving an answer of
Here is a similar problem from another AIME test: 2003 AIME II Problem 13, in which we have an equilateral triangle instead.
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