2001 AIME I Problems/Problem 15
The numbers and are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where and are considered to be consecutive, are written on faces that share an edge is where and are relatively prime positive integers. Find
Choose one face of the octahedron randomly and label it with . There are three faces adjacent to this one, which we will call A-faces. There are three faces adjacent to two of the A-faces, which we will call B-faces, and one face adjacent to the three B-faces, which we will call the C-face.
Clearly, the labels for the A-faces must come from the set , since these faces are all adjacent to . There are thus ways to assign the labels for the A-faces.
The labels for the B-faces and C-face are the two remaining numbers from the above set, plus and . The number on the C-face must not be consecutive to any of the numbers on the B-faces.
From here it is easiest to brute force the possibilities for the numbers on the B and C faces:
- 2348 (2678): 8(2) is the only one not adjacent to any of the others, so it goes on the C-face. 4(6) has only one B-face it can go to, while 2 and 3 (7 and 8) can be assigned randomly to the last two. 2 possibilities here.
- 2358 (2578): 5 cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so just 1 possibility here.
- 2368 (2478): 6(4) cannot go on any of the B-faces, so it must be on the C-face. 3 and 8 (2 and 7) have only one allowable B-face, so 1 possibility here.
- 2458 (2568): All of the numbers have only one B-face they could go to. 2 and 4 (6 and 8) can go on the same, so one must go to the C-face. Only 2(8) is not consecutive with any of the others, so it goes on the C-face. 1 possibility.
- 2378: None of the numbers can go on the C-face because they will be consecutive with one of the B-face numbers. So this possibility is impossible.
- 2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, so this possibility is impossible.
There is a total of possibilities. There are permutations of each, so acceptable ways to fill in the rest of the octahedron given the . There are ways to randomly fill in the rest of the octahedron. So the probability is . The answer is .
Consider the cube formed from the face centers of the regular octahedron. Color the vertices in a checker board fashion. We seek the number of circuits traversing the cube entirely composed of diagonals. Notice for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits ( for black and for white).
Type I: . There are ways to arrange the black vertices and consequently two of the white vertices and ways to arrange the other two white vertices. Since the template has a period of , there are circuits of type I.
Type II: . There are ways to arrange the black vertices and consequently the white vertices. Since the template has a period of , there are circuits of type II.
Thus, there are circuits satisfying the given condition, out of the possible circuits. Therefore, the desired probability is . The answer is .
As in the previous solution, consider the cube formed by taking each face of the octahedron as a vertex. Let one fixed vertex be A. Then each configuration (letting each vertex have a number value from 1-8) of A and the three vertices adjacent to A uniquely determine a configuration that satisfies the conditions, i.e. no two vertices have consecutive numbers. Thus, the number of desired configurations is equivalent to the number of ways of choosing the values of A and its three adjacent vertices.
The value of A can be chosen in 8 ways, and the 3 vertices adjacent to A can be chosen in ways, since they aren't adjacent to each other, but they can't, after all, be consecutive values to A. For example, if A=1, then the next vertex can't be 1,2 or 8, so there are 5 choices. However, the next vertex also adjacent to A can be chosen in 4 ways; it can't be equal to 1,2,8, or the value of the previously chosen vertex. With the same reasoning, the last such vertex has 3 possible choices.
The total number of ways to choose the values of the vertices of the cube independently is 8!, so our probability is thus , from which the answer is .
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