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Difference between revisions of "2024 AIME I Problems/Problem 11"

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==Problem==
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The vertices of a regular octagon are coloured either red or blue with equal probability. The probability that the octagon can be rotated in such a way that all blue vertices end up at points that were originally red is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
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==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.  
 
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.  
  

Revision as of 18:23, 2 February 2024

Problem

The vertices of a regular octagon are coloured either red or blue with equal probability. The probability that the octagon can be rotated in such a way that all blue vertices end up at points that were originally red is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

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 $1$.

If there is a single blue somewhere, there are $8$ cases - where can the blue be? Each of these is valid.

If there are two blues, again, every case is valid, and there are $\dbinom82=28$ cases.

If there are three blues, every case is again valid; there are $\dbinom83=56$ 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 $4$ units) to achieve the problem's condition. There are $8$ possible ways to have $4$ adjacent blues, so this subcase contributes $8$. [asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,0,1,1,1,1}; oct11(sus); [/asy]

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.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,1,0,1,1,1}; oct11(sus); [/asy]

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.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,0,1,1,0,1,1}; oct11(sus); [/asy]

If there are two adjacent pairs that are $1$ apart, it is not possible since we do not have anywhere to put the two pairs.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,0,1,1,1}; oct11(sus); [/asy]

If there are two adjacent pairs that are $2$ apart, all of these cases are possible as we can rotate the diagram by $2$ vertices to work. There are $4$ of these cases.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,1,0,0,1,1}; oct11(sus); [/asy]

If there is one adjacent pair and there are two separate ones each a distance of $1$ from the other, this case does not work.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,1,0,1,1}; oct11(sus); [/asy]

If we have one adjacent pair and two separate ones that are $2$ away from each other, we can flip the diagram by $4$ vertices. There are $8$ of these cases.

[asy] import graph; void oct11(int[] pts) { pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)}; draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.707)--cycle); for (int i = 0; i < 8; i+=1) { if (pts[i] == 0) { dot(vertices[i], blue); } if (pts[i] == 1) { dot(vertices[i], red); } } }; int[] sus = {0,0,1,0,1,1,0,1}; oct11(sus); [/asy]

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 $2$ subcases.

There can not be more than $4$ blues, so we are done.

Our total is $1+8+28+56+8+4+8+2=115$. There are $2^8=256$ possible colorings, so we have $\dfrac{115}{256}$ and our answer is $115+256=\boxed{371}$.

~Technodoggo

Video Solution (FASTEST)

https://www.youtube.com/watch?v=Nf80-0Eo72E&t=4s&ab_channel=MegaMathChannel

See also

2024 AIME I (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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