2024 AIME I Problems/Problem 11
Problem
Each vertex of a regular octagon is independently colored either red or blue with equal probability. The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there were originally red vertices is , where and are relatively prime positive integers. What is ?
Video Solution
https://www.youtube.com/watch?v=Nf80-0Eo72E&t=4s&ab_channel=MegaMathChannel
Solution 1
Notice that the question's condition mandates all blues to go to reds, but reds do not necessarily have to go to blue. Let us do casework on how many blues there are.
If there are no blues whatsoever, there is only one case. This case is valid, as all of the (zero) blues have gone to reds. (One could also view it as: the location of all the blues now were not previously red.) Thus, we have .
If there is a single blue somewhere, there are cases - where can the blue be? Each of these is valid.
If there are two blues, again, every case is valid, and there are cases.
If there are three blues, every case is again valid; there are such cases.
The case with four blues is trickier. Let us look at all possible subcases.
If all four are adjacent (as in the diagram below), it is obvious: we can simply reverse the diagram (rotate it by units) to achieve the problem's condition. There are possible ways to have adjacent blues, so this subcase contributes .
If three are adjacent and one is one away (as shown in the diagram below), we can not rotate the diagram to satisfy the question. This subcase does not work.
If three are adjacent and one is two away, obviously it is not possible as there is nowhere for the three adjacent blues to go.
If there are two adjacent pairs that are apart, it is not possible since we do not have anywhere to put the two pairs.
If there are two adjacent pairs that are apart, all of these cases are possible as we can rotate the diagram by vertices to work. There are of these cases.
If there is one adjacent pair and there are two separate ones each a distance of from the other, this case does not work.
If we have one adjacent pair and two separate ones that are away from each other, we can flip the diagram by vertices. There are of these cases.
Finally, if the red and blues alternate, we can simply shift the diagram by a single vertex to satisfy the question. Thus, all of these cases work, and we have subcases.
There can not be more than blues, so we are done.
Our total is . There are possible colorings, so we have and our answer is .
~Technodoggo
Solution 2
Let be the number of red vertices and be the number of blue vertices, where . By the Pigeonhole Principle, if a configuration is valid.
We claim that if , then any configuration is valid. We attempt to prove by the following:
If there are vertices, then intuitively any configuration is valid. For , we do cases:
If all the vertices in are non-adjacent, then simply rotating once in any direction suffices. If there are adjacent vertices, then WLOG let us create a set where the third is somewhere later in the set. If we assign the set as and , then intuitively, rotating it will suffice. If , then rotating it by 2 will suffice. Consider any other as simply a mirror to a configuration of the cases.
Therefore, if , then there are ways. We do count the [i]degenerate[/i] case.
Now if , we do casework on the number of adjacent vertices. 0 adjacent: . There are 4 axes of symmetry so there are only rotations of this configuration.
1 adjacent: WLOG where . Listing out the cases and trying, we get that and is the only configuration. There are ways to choose and and the rest is set, so there are ways.
2 adjacent: We can have WLOG or where . The former yields the case and by simply rotating it 2 times. The latter yields none. There are 2 axes of symmetry so there are configurations.
3 adjacent: WLOG which intuitively works. There are configurations here as can is unique.
In total, yields configurations.
There are configurations in total. There are total cases, so the probability is . Adding them up, we get .
Video Solution
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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