Difference between revisions of "2005 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | We can consider the orientation of each of the individual cubes independently. |
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| + | {{image}} | ||
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| + | The unit cube at the center of our large cube has no exterior faces, so all of its orientations work. | ||
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| + | For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This surely happens in <math>\frac{4}{6} = \frac{2}{3}</math> of all orientations, so from these cubes we gain a factor of <math>\left(\frac{2}{3}\right)^6</math>. | ||
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| + | The twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge. Thus, we need to know the number of such pairs that are both painted orange. We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces. Thus, the probability that two orange faces show for one of these cubes is <math>\frac{5}{12}</math>, so from all of these cubes we gain a factor of <math>\left(\frac{5}{12}\right)^{12} = \frac{5^{12}}{2^{24}3^{12}}</math>. | ||
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| + | Finally, we need to orient the eight corner cubes. Each such cube has 3 faces showing, and these three faces share a common vertex. Thus, we need to know the number of vertices for which all three adjacent faces are painted orange. There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of <math>\frac{2}{8} = \frac{1}{4}</math> and all the corner cubes together contribute a probability of <math>\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}</math> | ||
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| + | Since these probabilities are independent, the overall probability is just their product, <math>\frac{2^6}{3^6} \cdot \frac{5^{12}}{2^{24}3^{12}} \cdot \frac{1}{2^{16}} = \frac{5^{12}}{2^{34}\cdot 3^{18}}</math> and so the answer is <math>2 + 3 + 5 + 12 + 34 + 18 = 074</math>. | ||
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| + | As an aside, note that the placement of the two unpainted faces is in fact of vital importance: if they were on opposite faces, the answer would be 0 because any placement of such a cube in the corner of the large cube would show one unpainted face. | ||
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== See also == | == See also == | ||
* [[2005 AIME I Problems/Problem 8 | Previous problem]] | * [[2005 AIME I Problems/Problem 8 | Previous problem]] | ||
* [[2005 AIME I Problems/Problem 10 | Next problem]] | * [[2005 AIME I Problems/Problem 10 | Next problem]] | ||
* [[2005 AIME I Problems]] | * [[2005 AIME I Problems]] | ||
Revision as of 12:00, 17 January 2007
Problem
Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a 
cube. Given the probability of the entire surface area of the larger cube is orange is 
where 
and 
are distinct primes and 
and 
are positive integers, find 
Solution
We can consider the orientation of each of the individual cubes independently.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
The unit cube at the center of our large cube has no exterior faces, so all of its orientations work.
For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This surely happens in 
of all orientations, so from these cubes we gain a factor of 
.
The twelve unit cubes along the edges of the large cube have two faces showing, and these two faces are joined along an edge. Thus, we need to know the number of such pairs that are both painted orange. We have a pair for each edge, and 7 edges border one of the unpainted faces while only 5 border two painted faces. Thus, the probability that two orange faces show for one of these cubes is 
, so from all of these cubes we gain a factor of 
.
Finally, we need to orient the eight corner cubes. Each such cube has 3 faces showing, and these three faces share a common vertex. Thus, we need to know the number of vertices for which all three adjacent faces are painted orange. There are six vertices which are a vertex of one of the unpainted faces and two vertices which have our desired property, so each corner cube contributes a probability of 
and all the corner cubes together contribute a probability of 
Since these probabilities are independent, the overall probability is just their product, 
and so the answer is 
.
As an aside, note that the placement of the two unpainted faces is in fact of vital importance: if they were on opposite faces, the answer would be 0 because any placement of such a cube in the corner of the large cube would show one unpainted face.
