2003 AIME II Problems/Problem 13

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 $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Solution

Solution 1 (Easiest)

Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\frac{1}{2}(1-P_{n-1})$ . $P_0=1$ , so now we can build it up:

$P_1=0$ , $P_2=\frac{1}{2}$ , $P_3=\frac{1}{4}$ , $P_4=\frac{3}{8}$ , $P_5=\frac{5}{16}$ , $P_6=\frac{11}{32}$ , $P_7=\frac{21}{64}$ , $P_8=\frac{43}{128}$ , $P_9=\frac{85}{256}$ , $P_{10}=\frac{171}{512}$ ,

Thus the answer is $171+512=$ $\boxed{683}$

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., $\#CW - \#CCW \equiv 0 \pmod{3}$ . Since $\#CW + \#CCW = 10$ , it is only possible that $(\#CW,\, \#CCW) = (5,5), (8,2), (2,8)$ .

In the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \choose 5}$ paths. In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise. Hence the total number of ways to return to the original vertex is ${10 \choose 5} + {10 \choose 8} + {10 \choose 2} = 252 + 45 + 45 = 342$ . Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\frac{342}{2^{10}} = \frac{171}{512}$ , and the answer is $\boxed{683}$ .

Solution 3

Label the vertices of the triangle $A,B,C$ with the ant starting at $A$ . We will make a table of the number of ways to get to $A,B,C$ in $n$ moves $n\leq10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ steps equals the number of ways to get to $B$ in $n-1$ steps plus the number of ways to get to $C$ in $n-1$ steps.

$\begin{array}{|l|ccc|} \multicolumn{4}{c}{\text{Table}}\\\hline \text{Step}&A&B&C \\\hline 1 &0 &1 &1 \\ 2 &2 &1 &1 \\ 3 &2 &3 &3\\ \vdots &\vdots&\vdots&\vdots \\ 10 &342 &341 &341 \\\hline \end{array}$ Therefore, our answer is $512+171=\boxed{683}.$

Notice the pattern that there are $\left\lceil\frac{2^n}{3}\right\rceil$ way to get to $A$ for even $n$ moves. Thus, there are $\left\lceil\frac{2^{10}}{3}\right\rceil=342$ ways.

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 (guess & check)

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 $341.33/1024$ . We can either round up or down. If we round down, we see $341/1024$ 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, $342/1024$ can be reduced to $171/512$ where the sum 171+512= $\boxed{683}$ is an accepted answer.

Solution 6 (generating functions)

The generating function for this is $(x+x^2)$ since an ant on any vertex of the equilateral triangle can go $120$ degrees or $240$ degrees to a side and simplifying $(x^{120}+x^{240})$ gets you $(x+x^2)$ . Since $360$ degrees brings you back to the original vertex then we must find the sum of the coefficients that share a variable with a power divisible by $3$ .

Since we take this rotation $10$ times, our function becomes $(x+x^2)^{10}$ which is the same as $x^{10}(x+1)^{10}$ . This completely simplified is $x(x+1)^{10}$ and since your maximum power is $11$ , we only have to find the coefficients for $3$ , $6$ , and $9$ ( $0$ would apply here but the $x$ is the lowest power there is).

For $x^9$ , the coefficient is ${10 \choose 2}$ , and the same goes for $x^3$ . For $x^6$ , the coefficient is ${10 \choose 5}$ and the final sum for the numerator is $2*{10 \choose 2} + {10 \choose 5}$ . The total sum is $90+252=342$ and for the denominator, it was simply $2^{10}$ and this simplified was $171/512$ . Therefore the sum is $683$ .

Solution 7 (trees)

Start of with any vertex. Denote $a_n$ the number of paths ending where it started. Then notice that for a path to end in the vertex you started on, you must have for the $(n-1)$ case that of the $2^{n-1}$ total paths, we must take away the paths that end in the $(n-1)$ -st term where it started off. (Because imagine on the $(n-1)$ move that it landed on vertex $A$ . Then if we wanted to find the number of length $n$ paths that end at $A$ , we would be able to find that path, because on the $(n-1)$ -st move it landed at $A$ . You can't go from $A$ to $A$ ). Hence we have the recursion $a_n=2^{n-1}-a_{n-1}$ , with $a_3 = 2$ . Now reiterating gives us $a_{10} = 342$ , so that the probability is $\frac{342}{2^{10}} = \frac{171}{512}$ . So we have $171 + 512 = \boxed{683}$ . ~th1nq3r

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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