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1985 AIME Problems/Problem 12

Revision as of 02:31, 25 July 2021 by MRENTHUSIASM (talk | contribs)

Problem

Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$, 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 $p = \frac{n}{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly $7$ meters. Find the value of $n$.

Solution 1 (Single Variable Recursion)

For all nonnegative integers $k,$ let $P(k)$ be the probability that the bug is at vertex $A$ when it has crawled exactly $k$ meters. We wish to find $p=P(7).$

Clearly, we have $P(0)=1.$ For all $k\geq1,$ note that after $k-1$ crawls:

  1. The probability that the bug is at vertex $A$ is $P(k-1),$ and the probability that it crawls to vertex $A$ on the next move is $0.$

  2. The probability that the bug is not at vertex $A$ is $1-P(k-1),$ and the probability that it crawls to vertex $A$ on the next move is $\frac13.$

Together, the recursive formula for $P(k)$ is \[P(k) = \begin{cases} 1 & \mathrm{if} \ k=0 \\ \frac13(1-P(k-1)) & \mathrm{if} \ k\geq1 \end{cases}.\] Two solutions follow from here:

Solution 1.1 (Recursive Formula)

We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac{182}{729}. \end{alignat*} Therefore, the answer is $n=\boxed{182}.$

~Azjps (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 1.2 (Explicit Formula)

Let $P(k)=Q(k)+c$ for some function $Q(k)$ and constant $c.$ For all $k\geq1,$ the recursive formula for $P(k)$ becomes \[Q(k)+c=\frac13(1-(Q(k-1)+c))=\frac13-\frac13Q(k-1)-\frac13c.\] Solving for $Q(k),$ we get \[Q(k)=\frac13-\frac13Q(k-1)-\frac43c.\] For simplicity purposes, we set $c=\frac14,$ which gives \[Q(k)=-\frac13Q(k-1).\] Clearly, $Q(0),Q(1),Q(2),\cdots$ is a geometric sequence with the common ratio $\frac{Q(k)}{Q(k-1)}=-\frac13.$ Substituting $P(0)=1$ and $c=\frac14$ into $P(0)=Q(0)+c$ produces $Q(0)=\frac34,$ the first term of the geometric sequence.

So, the explicit formula for $Q(k)$ is \[Q(k)=\frac34\left(-\frac13\right)^k,\] and the explicit formula for $P(k)$ is \[P(k)=\frac34\left(-\frac13\right)^k+\frac14.\] Finally, the requested probability is $p=P(7)=\frac{182}{729},$ from which $n=\boxed{182}.$

~MRENTHUSIASM

Solution 2 (Multivariable Recursion by Algebra)

Denominator

There are $3^7$ ways for the bug to make $7$ independent crawls without restrictions.

Numerator (Recursion)

Let $V_k$ denote the number of ways for the bug to crawl exactly $k$ meters starting from vertex $V$ and ending at vertex $A,$ where $V\in\{A,B,C,D\}$ and $k$ is a positive integer. We wish to find $A_7.$

Since the bug must crawl to vertex $B,C,$ or $D$ on the first move, we have \begin{align*} A_7&=B_6+C_6+D_6 \\ &=(A_5+C_5+D_5)+(A_5+B_5+D_5)+(A_5+B_5+C_5) \\ &=A_5+2(A_5+B_5+C_5+D_5) \\ &=A_5+2S_5, \end{align*} where $S_k=A_k+B_k+C_k+D_k.$

More generally, we get \[A_{k+2}=A_k+2S_k. \qquad\qquad (\spadesuit)\] For $S_k,$ note that

  1. Base Case:

    2. \begin{align*} S_1&=A_1+B_1+C_1+D_1 \\ &=0+1+1+1 \\ &=3. \end{align*}

  2. Recursive Case:

    3. \begin{align*} S_{k+1}&=A_{k+1}+B_{k+1}+C_{k+1}+D_{k+1} \\ &=(B_k+C_k+D_k)+(A_k+C_k+D_k)+(A_k+B_k+D_k)+(A_k+B_k+C_k) \\ &=3(A_k+B_k+C_k+D_k) \\ &=3S_k. \end{align*}

Clearly, $S_1,S_2,S_3,\cdots$ is a geometric sequence with the first term $S_1=3$ and the common ratio $\frac{S_{k+1}}{S_k}=3.$ Therefore, its explicit formula is \[S_k=3^k. \qquad\qquad (\clubsuit)\] Recall that $A_1=0.$ By $(\spadesuit)$ and $(\clubsuit),$ we rewrite $A_7$ recursively: \begin{align*} A_7 &= A_5+2S_5 \\ &=\left(A_3+2S_3\right)+2S_5 \\ &=\left(\left(A_1+2S_1\right)+2S_3\right)+2S_5 \\ &=2S_1+2S_3+2S_5 \\ &=2(3)+2\left(3^3\right)+2\left(3^5\right). \end{align*} Answer

The requested probability is \[p=\frac{A_7}{3^7}=\frac{2(3)+2\left(3^3\right)+2\left(3^5\right)}{3^7}=\frac{2(1)+2\left(3^2\right)+2\left(3^4\right)}{3^6}=\frac{182}{729},\] from which $n=\boxed{182}.$

~MRENTHUSIASM

Solution 3 (Multivariable Recursion by Table)

Define notation $V_k$ as Solution 2 does.

In fact, we can generalize the following relationships for all nonnegative integers $k:$ \begin{align*} A_0&=1, \\ B_0&=0, \\ C_0&=0, \\ D_0&=0, \\ A_{k+1}&=B_k+C_k+D_k, \\ B_{k+1}&=A_k+C_k+D_k, \\ C_{k+1}&=A_k+B_k+D_k, \\ D_{k+1}&=A_k+B_k+C_k. \\ \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c} \hspace{7mm}&\hspace{6.5mm}&\hspace{6.5mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} & \boldsymbol{8} \\ \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{A_k} &&&&&&&& \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{B_k} &&&&&&&& \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{C_k} &&&&&&&& \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{D_k} &&&&&&&& \\ \end{array}\] ~MRENTHUSIASM

Solution 4 (Single Variable Version of Solution 2)

Let $a_n$ denotes the number of ways that the bug arrives at $A$ after crawling $n$ meters, then we have $a_1=0$.

Notice that there is respectively $1$ way to arrive at $A$ for each of the different routes after the previous $n-1$ crawls, excluding the possibility that the bug ends up at $A$ after the $(n-1)$th crawl (as it will be forced to move somewhere else.). Thus, we get the recurrence relation \[a_n=3^{n-1}-a_{n-1}.\] Quick calculations yield \begin{align*} a_7&=3^6-a_6\\ &=3^6-\left(3^5-3^4+3^3-3^2+3-a_1\right)\\ &=546. \end{align*} Thus, $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$.

~Nafer

Solution 5 (Partitions)

We can find the number of different times the bug reaches vertex $A$ before the $7$th move, and use these smaller cycles to calculate the number of different ways the bug can end up back at $A.$

Define $f(x)$ to be the number of paths of length $x$ which start and end at $A$ but do not pass through $A$ otherwise. Obviously $f(1) = 0.$ In general for $f(x),$ the bug has three initial edges to pick from. From there, since the bug cannot return to $A$ by definition, the bug has exactly two choices. This continues from the $2$nd move up to the $(x-1)$th move. The last move must be a return to $A,$ so this move is determined. So $f(x) = 2^{x-2}3.$

Now we need to find the number of cycles by which the bug can reach $A$ at the end. Since $f(1) = 0,$ we know that $f(6)$ cannot be used, as on the $7$th move the bug cannot move from $A$ to $A.$ So we need to find the number of partitions of $7$ using only $2,3,4,5,$ and $7.$ These are $f(2)f(2)f(3),f(2)f(5),f(3)f(4),$ and $f(7).$ 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: \[{3\choose1}f(2)f(2)f(3) + {2\choose1}f(2)f(5) + {2\choose1}f(3)f(4) + f(7) = 3(3)(3)(2\cdot3) + 2(3)(2^33) + 2(2\cdot3)(2^23) + (2^53) = 546.\] Finally, this is a probability question, so we divide by $3^7,$ or $\frac{546}{3^7} = \frac{182}{3^6}.$ The answer is $n=\boxed{182}.$

Solution 6 (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 $P(n)$ to be very close to $\frac 14$ for decently-sized $n$, for example $n = 7$. (In fact, from looking at the previous solutions we can see that it is already close when $n = 3$, 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 $\frac n{729}$, we realize that $n$ must be very close to $\frac{729}{4} = 182.25$, as indeed it is. The answer is $n=\boxed{182}$.

Solution 7 (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 $4$ vertices of a tetrahedron, there is approximately an $1/4$ probability of coming back to $A$ after $7$ moves. Dividing $729$ by $4$ gives a number between $182$ and $183.$ If the bug continuously alternates its location from $A$ to some other vertex, in the end, it will not be at $A.$ Therefore we choose the smaller number, $\boxed{182}.$

~Rocket123

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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