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Difference between revisions of "2001 AIME I Problems/Problem 15"

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== Alternate Solution ==
 
== Alternate Solution ==
 
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 (<math>B</math> for black and <math>W</math> for white).
 
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 (<math>B</math> for black and <math>W</math> for white).
 +
 
Type I: <math>BB-WWWW-BB</math>. There are <math>4!</math> ways to arrange the black vertices and consequently two of the white vertices and <math>2!</math> ways to arrange the other two white vertices. Since the template has a period of <math>8</math>, there are <math>4!\cdot 2!\cdot 8 = 384</math> circuits of type I.
 
Type I: <math>BB-WWWW-BB</math>. There are <math>4!</math> ways to arrange the black vertices and consequently two of the white vertices and <math>2!</math> ways to arrange the other two white vertices. Since the template has a period of <math>8</math>, there are <math>4!\cdot 2!\cdot 8 = 384</math> circuits of type I.
 +
 
Type II: <math>BW-WB-BW-WB</math>. There are <math>4!</math> ways to arrange the black vertices and consequently the white vertices. Since the template has a period of <math>4</math>, there are <math>4! \cdot 4 = 96</math> circuits of type II.
 
Type II: <math>BW-WB-BW-WB</math>. There are <math>4!</math> ways to arrange the black vertices and consequently the white vertices. Since the template has a period of <math>4</math>, there are <math>4! \cdot 4 = 96</math> circuits of type II.
 +
 
Thus, there are <math>384+96=480</math> circuits satisfying the given condition, out of the <math>8!</math> possible circuits. Therefore, the desired probability is <math>\frac{480}{8!} = \frac{1}{84}</math>. The answer is <math>\boxed{085}</math>.
 
Thus, there are <math>384+96=480</math> circuits satisfying the given condition, out of the <math>8!</math> possible circuits. Therefore, the desired probability is <math>\frac{480}{8!} = \frac{1}{84}</math>. The answer is <math>\boxed{085}</math>.
  

Revision as of 14:58, 22 June 2010

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

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 adjacent to any of the numbers on the B-faces.

From here it is easiest to brute force the $10$ possibilities for the $4$ numbers left.

  • 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 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 $\boxed{085}$.

Alternate Solution

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: $BW-WB-BW-WB$. 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 $\boxed{085}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
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