Difference between revisions of "2003 AIME II Problems/Problem 13"
(→Solution 1) |
Mathgeek2006 (talk | contribs) m (→Solution 3) |
||
Line 28: | Line 28: | ||
===Solution 3=== | ===Solution 3=== | ||
Label the vertices of the triangle <math>A,B,C</math> with the ant starting at <math>A</math>. We will make a table of the number of ways to get to <math>A,B,C</math> in <math>n</math> moves <math>n<=10</math>. The values of the table are calculated from the fact that the number of ways from a vertex say <math>A</math> in <math>n</math> ways equals the number of ways to get to <math>B</math> in <math>n-1</math> ways plus the number of ways to get to <math>C</math> in <math>n-1</math> ways. | Label the vertices of the triangle <math>A,B,C</math> with the ant starting at <math>A</math>. We will make a table of the number of ways to get to <math>A,B,C</math> in <math>n</math> moves <math>n<=10</math>. The values of the table are calculated from the fact that the number of ways from a vertex say <math>A</math> in <math>n</math> ways equals the number of ways to get to <math>B</math> in <math>n-1</math> ways plus the number of ways to get to <math>C</math> in <math>n-1</math> ways. | ||
− | < | + | <cmath>\begin{array}{|l|ccc|} |
− | \multicolumn{4}{c}{Table}\\\hline | + | \multicolumn{4}{c}{\text{Table}}\\\hline |
− | Step&A&B&C \\\hline | + | \text{Step}&A&B&C \\\hline |
1 &0 &1 &1 \\ | 1 &0 &1 &1 \\ | ||
2 &2 &1 &1 \\ | 2 &2 &1 &1 \\ | ||
Line 36: | Line 36: | ||
\vdots &\vdots&\vdots&\vdots \\ | \vdots &\vdots&\vdots&\vdots \\ | ||
10 &342 &341 &341 \\\hline | 10 &342 &341 &341 \\\hline | ||
− | \end{ | + | \end{array}</cmath> |
== See also == | == See also == | ||
{{AIME box|year=2003|n=II|num-b=12|num-a=14}} | {{AIME box|year=2003|n=II|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:21, 10 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 where and are relatively prime positive integers, find
Solution
Solution 1
Let represent the probability that the bug is at its starting vertex after moves. If the bug is on its starting vertex after moves, then it must be not on its starting vertex after moves. At this point it has chance of reaching the starting vertex in the next move. Thus . , so now we can build it up:
, , , , , , , , , ,
Thus the answer is
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., . Since , it is only possible that .
In the first case, we pick out of the ant's steps to be clockwise, for a total of paths. In the second case, we choose of the steps to be clockwise, and in the third case we choose to be clockwise. Hence the total number of ways to return to the original vertex is . Since the ant has two possible steps at each point, there are routes the ant can take, and the probability we seek is , and the answer is .
Solution 3
Label the vertices of the triangle with the ant starting at . We will make a table of the number of ways to get to in moves . The values of the table are calculated from the fact that the number of ways from a vertex say in ways equals the number of ways to get to in ways plus the number of ways to get to in ways.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.