Difference between revisions of "2001 AIME I Problems/Problem 15"
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*2468: Both 4 and 6 cannot go on any B-face. They cannot both go on the C-face, 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 <math>10</math> possibilities. There are <math>3!=6</math> permutations of each, so <math>60</math> acceptable ways to fill in the rest of the octahedron given the <math>1</math>. There are <math>7!=5040</math> ways to randomly fill in the rest of the octahedron. So the probability is <math>\frac {60}{5040} = \frac {1}{84}</math>. The answer is <math>\boxed{085}</math>. | + | There is a total of <math>10</math> possibilities. There are <math>3!=6</math> permutations of each, so <math>60</math> acceptable ways to fill in the rest of the octahedron given the <math>1</math>. There are <math>7!=5040</math> ways to randomly fill in the rest of the octahedron. So the probability is <math>\frac {60}{5040} = \frac {1}{84}</math>. The answer is <math>\boxed{085}</math>. |
| + | |||
| + | == 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). | ||
| + | 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. | ||
| + | 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>. | ||
== See also == | == See also == | ||
Revision as of 14:57, 22 June 2010
Problem
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 
Solution
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 adjacent to any of the numbers on the B-faces.
From here it is easiest to brute force the 
possibilities for the 
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 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 
.
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 (
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 .
See also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
