Difference between revisions of "2009 AMC 10B Problems/Problem 25"
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{{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}} | {{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #25]] and [[2009 AMC 12B Problems|2009 AMC 12B #17]]}} | ||
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== Solution == | == Solution == | ||
Revision as of 09:28, 27 May 2021
- The following problem is from both the 2009 AMC 10B #25 and 2009 AMC 12B #17, so both problems redirect to this page.
Contents
NO MERCY
Solution
Solution 1
There are two possible stripe orientations for each of the six faces of the cube, so there are 
possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripe orientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of 
stripe combinations on the cube result in a continuous stripe around the cube. The required probability is 
.
Solution 2
Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form an encircling stripe. The probability of the first case is 
, and the probability of the second case is 
. The cases are disjoint, so the probabilities sum 
.
Solution 3
There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probability of each such stripe alignment is . Since there are three such possibilities and they are disjoint, the total probability is 
. The answer is 
.
Solution 4
Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encircling the cube. The probability that the stripes aren't aligned is 
, since for each alignment of one stripe, there is one and only one way to align the other two stripes out of four total possibilities. therefore there is a probability of 
that two stripes are aligned.
Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connected to make a continuous stripe. There is a probability of 
that they are aligned, so there is a probability of 
that there is a continuous stripe.
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2009 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
