Difference between revisions of "2009 AMC 10A Problems/Problem 24"
(New page: == Problem == Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube? <math> \m...) |
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| − | == Solution == | + | == Solution 1 == |
We will try to use symmetry as much as possible. | We will try to use symmetry as much as possible. | ||
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<cmath> | <cmath> | ||
\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47} | \frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47} | ||
| + | </cmath> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | There are 8 C 3 ways to pick three vertices from 8; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability. | ||
| + | |||
| + | There are four ways to choose three points from the vertices of a single face. Since there are six faces, 4 x 6 = 24. | ||
| + | |||
| + | Thus, the probability of what we don't want is <math>24/56 = 3/7</math>. Using complementary, | ||
| + | |||
| + | <cmath> | ||
| + | 1- \frac 37 = \boxed{\frac 47} | ||
</cmath> | </cmath> | ||
Revision as of 22:25, 7 February 2011
Contents
Problem
Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
Solution 1
We will try to use symmetry as much as possible.
Pick the first vertex 
, its choice clearly does not influence anything.
Pick the second vertex 
. With probability 
vertices and have a common edge, with probability they are in opposite corners of the same face, and with probability 
they are in opposite corners of the cube. We will handle each of the cases separately.
In the first case, there are 
faces that contain the edge 
. In each of these faces there are other vertices. If one of these 
vertices is the third vertex 
, the entire triangle 
will be on a face. On the other hand, if is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good is 
.
In the second case, the triangle will not intersect the cube iff point is one of the two points on the side that contains . Hence the probability of intersecting the inside of the cube is 
.
In the third case, already the diagonal contains points inside the cube, hence this case will be good regardless of the choice of .
Summing up all cases, the resulting probability is:
![\[\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}\]](http://latex.artofproblemsolving.com/4/b/3/4b390fbd82b19144e4b78ab84832109a96ca7bb0.png)
Solution 2
There are 8 C 3 ways to pick three vertices from 8; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
There are four ways to choose three points from the vertices of a single face. Since there are six faces, 4 x 6 = 24.
Thus, the probability of what we don't want is 
. Using complementary,
![\[1- \frac 37 = \boxed{\frac 47}\]](http://latex.artofproblemsolving.com/2/8/6/286a9362a3af3eeb6be682a1195a7352922c7174.png)
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
