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Difference between revisions of "1985 AIME Problems/Problem 12"

m (→‎Solution 7 (Algebraic))
(→‎Solution 7 (Algebraic): Reformatted Sol 7)
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where <math>S_n=A_n+B_n+C_n+D_n.</math>
 
where <math>S_n=A_n+B_n+C_n+D_n.</math>
  
−
Note that
+
More generally, we get <cmath>A_{n+2}=A_n+2S_n. \qquad\qquad (\spadesuit)</cmath>
 +
For <math>S_n,</math> note that
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li>Base Case:</li><p>
 
   <li>Base Case:</li><p>
Line 76: Line 77:
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
</ol>
 
</ol>
−
Clearly, <math>S_1,S_2,S_3,\cdots</math> is a geometric sequence with the first term <math>S_1=3</math> and the common ratio <math>3.</math> Therefore, its explicit formula is <cmath>S_n=3^n. \qquad\qquad (\star)</cmath>
+
Clearly, <math>S_1,S_2,S_3,\cdots</math> is a geometric sequence with the first term <math>S_1=3</math> and the common ratio <math>\frac{S_{n+1}}{S_n}=3.</math> Therefore, its explicit formula is <cmath>S_n=3^n. \qquad\qquad (\clubsuit)</cmath>
−
By <math>(\star),</math> we can rewrite <math>A_7,</math> then proceed recursively:
+
By <math>(\spadesuit)</math> and <math>(\clubsuit),</math> we can rewrite <math>A_7,</math> then proceed recursively:
−
<cmath>\begin{alignat*}{6}  
+
<cmath>\begin{align*}
−
A_7 &= A_5+2S_5 &&= A_5+2\left(3^5\right), &\qquad\qquad (1) \\
+
A_7 &= A_5+2S_5 \\
−
A_5 &= A_3+2S_3 &&= A_3+2\left(3^3\right), &\qquad\qquad (2) \\
+
&=\left(A_3+2S_3\right)+2S_5 \\
−
A_3 &= \underbrace{A_1}_{0}+2S_1 &&= 2(3). &\qquad\qquad (3)
+
&=\bigl(\bigl(\ \underbrace{A_1}_{0}+2S_1\bigr)+2S_3\bigr)+2S_5 \\
−
\end{alignat*}</cmath>
+
&=2S_1+2S_3+2S_5 \\
−
Substituting <math>(2)</math> and <math>(3)</math> into <math>(1)</math> gives <cmath>A_7=2\left(3^5\right)+2\left(3^3\right)+2(3).</cmath>
+
&=2(3)+2\left(3^3\right)+2\left(3^5\right).
−
 
+
\end{align*}</cmath>
 
<u><b>Answer</b></u>
 
<u><b>Answer</b></u>
  
−
The requested probability is <cmath>p=\frac{A_7}{3^7}=\frac{2\left(3^5\right)+2\left(3^3\right)+2(3)}{3^7}=\frac{2\left(3^4\right)+2\left(3^2\right)+2(1)}{3^6}=\frac{182}{729},</cmath> from which <math>n=\boxed{182}.</math>
+
The requested probability is <cmath>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},</cmath> from which <math>n=\boxed{182}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 05:30, 27 April 2021

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

Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$, we have $P(n + 1) = \frac13 (1 - P(n))$. We also know $P(0) = 1$, so we can quickly compute $P(1)=0$, $P(2) = \frac 13$, $P(3) = \frac29$, $P(4) = \frac7{27}$, $P(5) = \frac{20}{81}$, $P(6) = \frac{61}{243}$ and $P(7) = \frac{182}{729}$, so the answer is $\boxed{182}$. One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$.

Solution 2

We can find the number of different times the bug reaches vertex $A$ before the 7th 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 2nd 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 $f(1) = 0$, $f(6)$ cannot be used since on the 7th 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 $3^7$: $\frac{546}{3^7} = \frac{182}{3^6}$.

Solution 3

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 solution 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.

Solution 4 (Cheap Solution)

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 a 1/4 probability of coming back to A after 7 moves. Dividing 729 by 4 gives a number between 182 and 183. If the ant continuously alternates his location from A to some other vertice, in the end, it will not be at A. Therefore we chose the smaller number, 182. - Rocket123

Solution 5

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\\ &=3^6-3^5+3^4-...+a_1\\ &=546 \end{align*} Thus $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$.

~ Nafer

Solution 6

Draw an organized table and keep track of how many ways are there to get to each vertex after 1, 2, 3, ..., 7 steps. Thus you get the answer of $p=\frac{546}{3^7}=\frac{182}{729}\Longrightarrow n=\boxed{182}$. -pi_is_3.141

Solution 7 (Algebraic)

Denominator

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

Numerator (Recursion)

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

Since the bug must crawl to $B,C,$ 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 $S_n=A_n+B_n+C_n+D_n.$

More generally, we get \[A_{n+2}=A_n+2S_n. \qquad\qquad (\spadesuit)\] For $S_n,$ 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_{n+1}&=A_{n+1}+B_{n+1}+C_{n+1}+D_{n+1} \\ &=(B_n+C_n+D_n)+(A_n+C_n+D_n)+(A_n+B_n+D_n)+(A_n+B_n+C_n) \\ &=3S_n. \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 $\frac{S_{n+1}}{S_n}=3.$ Therefore, its explicit formula is \[S_n=3^n. \qquad\qquad (\clubsuit)\] By $(\spadesuit)$ and $(\clubsuit),$ we can rewrite $A_7,$ then proceed recursively: \begin{align*} A_7 &= A_5+2S_5 \\ &=\left(A_3+2S_3\right)+2S_5 \\ &=\bigl(\bigl(\ \underbrace{A_1}_{0}+2S_1\bigr)+2S_3\bigr)+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

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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