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2001 AIME I Problems/Problem 15

Revision as of 18:12, 27 May 2016 by Benq (talk | contribs)

Problem

The numbers $1, 2, 3, 4, 5, 6, 7,$ and $8$ 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 $8$ and $1$ are considered to be consecutive, are written on faces that share an edge is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

Solution 1

Choose one face of the octahedron randomly and label it with $1$. 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 $\{3,4,5,6,7\}$, since these faces are all adjacent to $1$. There are thus $5 \cdot 4 \cdot 3 = 60$ 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 $2$ and $8$. 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 $10$ possibilities for the $4$ 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 $10$ possibilities. There are $3!=6$ permutations (more like "rotations") of each, so $60$ acceptable ways to fill in the rest of the octahedron given the $1$. There are $7!=5040$ ways to randomly fill in the rest of the octahedron. So the probability is $\frac {60}{5040} = \frac {1}{84}$. The answer is $\boxed{085}$.

Solution 2

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 ($B$ for black and $W$ for white).

Type I: $BB-WWWW-BB$. There are $4!$ ways to arrange the black vertices and consequently two of the white vertices and $2!$ ways to arrange the other two white vertices. Since the template has a period of $8$, there are $4!\cdot 2!\cdot 8 = 384$ circuits of type I.

Type II: $B-WW-BB-WW-B$. There are $4!$ ways to arrange the black vertices and consequently the white vertices. Since the template has a period of $4$, there are $4! \cdot 4 = 96$ circuits of type II.

Thus, there are $384+96=480$ circuits satisfying the given condition, out of the $8!$ possible circuits. Therefore, the desired probability is $\frac{480}{8!} = \frac{1}{84}$. The answer is $\boxed{085}$.

Solution 3

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 $5\cdot4\cdot3=60$ 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 $\frac{8\cdot60}{8!}=\frac{8\cdot5\cdot4\cdot3}{8!}=\frac{1}{84}$, from which the answer is $\boxed{085}$.

Solution 4

[asy] import three; draw((0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)); draw((1,1,0)--(1,1,1)); draw((0,1,0)--(0,1,1)); draw((0,0,0)--(1,0,0)); draw((0,0,1)--(1,0,1)); for(int i = 0; i < 2; ++i) { for(int j = 0; j < 2; ++j) { for(int k = 0; k < 2; ++k) { dot((i,j,k)); } } } // dot((0,0,1),blue); // dot((0,1,0),green); // dot((1,0,0),red); draw((0,0,0)--(1,1,0)); draw((0,1,0)--(1,0,0)); draw((0,0,1)--(1,1,1)); draw((0,1,1)--(1,0,1)); [/asy]

The probability is equivalent to counting the number of Hamiltonian cycles in this 3D graph over $7!.$ This is because each Hamiltonian cycle corresponds to eight unique ways to label the faces. Label the vertices $10,20,30,40,11,21,31,41,$ where vertices $ab$ and $cd$ are connected if $a=c$ or $b=d.$

Case 1: Four of the vertical edges are used. $6\cdot 2=12.$

Case 2: Two of the vertical edges are used. $4\cdot 3 \cdot 2\cdot 2=48.$

So, the probability is $\frac{60}{5040}=\frac{1}{84}.$ Therefore, our answer is $\boxed{85.}$

See also

2001 AIME I (Problems • Answer Key • Resources)
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