Difference between revisions of "2003 AIME II Problems/Problem 13"

Revision as of 21:18, 6 March 2015

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

After n moves, there are $2^n$ possible paths for the ant. But the key is to realize that the number of paths that get back the the start after n moves is the same as the number of paths the do NOT get the ant to the start after $n-1$ moves. So after 1 move, there are 0 ways the get to the start. After 2 moves, there are $2^1-0$ ways to get to the start. After 3 moves, there are $2^2-(2^1-0)$ ways to get to the start. Thus after 10 moves, there are $2^9-(2^8-(2^7-(2^6-(2^5-(2^4-(2^3-(2^2-2)))))))= 342$ ways to get to the start, so the probability is $342/1024=171/512$ and the answer is $683$ .

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 $683$ .

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 $n<=10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ ways equals the number of ways to get to $B$ in $n-1$ ways plus the number of ways to get to $C$ in $n-1$ ways. $\begin{tabular}[t]{|l|ccc|} \multicolumn{4}{c}{Table}\\\hline Step&A&B&C \\\hline 1 &0 &1 &1 \\ 2 &2 &1 &1 \\ 3 &2 &3 &3\\ \vdots &\vdots&\vdots&\vdots \\ 10 &342 &341 &341 \\\hline \end{tabular}$ (Error compiling LaTeX. Unknown error_msg)

Solution 4

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. 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 $171+512=683$

@@ Line 23: / Line 23: @@
 &342 &341 &341 \\\hline
 \end{tabular}</math>
+===Solution 4===
+Let <math>P_n</math> represent the probability that the bug is at its starting vertex after <math>n</math> moves. If the bug is on its starting vertex after <math>n</math> moves, then it must be not on its starting vertex after <math>n-1</math> moves. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>.
+<math>P_0=1</math>, so now we can build it up:
+<math>P_1=0</math>,
+<math>P_2=\frac{1}{2}</math>,
+<math>P_3=\frac{1}{4}</math>,
+<math>P_4=\frac{3}{8}</math>,
+<math>P_5=\frac{5}{16}</math>,
+<math>P_6=\frac{11}{32}</math>,
+<math>P_7=\frac{21}{64}</math>,
+<math>P_8=\frac{43}{128}</math>,
+<math>P_9=\frac{85}{256}</math>,
+<math>P_{10}=\frac{171}{512}</math>,
+Thus the answer is <math>171+512=683</math>
 == See also ==
 {{AIME box|year=2003|n=II|num-b=12|num-a=14}}
 {{MAA Notice}}

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