2003 AIME II Problems/Problem 13

Revision as of 16:37, 16 May 2021 by Th1nq3r (talk | contribs) (Solution 7 (trees))


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 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)

Note that since we can abuse symmetry, the number of solutions of $A,B$ and $C$, is (say) $3$ times the number of solutions to $A$. We can construct a graph as follows: Start off with vertex $A$. We now want to count how many walks from the vertex $A$ we started on ends in $A$ at the bottom. Now $A$ can either be $B$ or $C$. Continuing on in this fashion, we see that $a_n$, the number of times the walk of the graph ends in $A$, is $2^{n-1}-a_{n-1}$. So we now have to find $a_{10}$. (We also have that the total number of walks is $2^{n}$). Then noticing that $a_3=2$, we have that $a_{10}=342$. So the answer is $\frac{342}{2^{10-1}}=\frac{342}{1024}=\frac{171}{512}$. So we have $171+512=683$.

See also

2003 AIME II (ProblemsAnswer KeyResources)
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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