Difference between revisions of "2009 AMC 10A Problems/Problem 24"
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<math>3</math> points determine a plane. therefore <math>8C3 = 56</math> ways exist to choose a plane. | <math>3</math> points determine a plane. therefore <math>8C3 = 56</math> ways exist to choose a plane. | ||
Revision as of 15:27, 20 January 2020
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 
(Easy)
(Easy)
points determine a plane. therefore 
ways exist to choose a plane.
Unfortunately, we overcounted by a factor of four, as each plane determined has 
vertices on the cube. Therefore, there are 
planes.
It is easy to see that all planes are either the cube's faces or pass through the cube. However, there are 6 faces of a cube.
Therefore, there are 
out of planes that pass through the cube, making the probability 4/7, and the correct answer is C.
Solution 
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 if 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)
Note: (Cheap solution same approach as solution 1)
This problem can be approached the same way, by picking vertices, but with a much faster and kind of cheap solution: Pick any vertex A and a face it touches. For vertex B, out of the 7 remaining vertices, 4 of them aren't on the same face as the one chosen for vertex A, so vertex C can be placed anywhere and the plane will no matter what be in the cube. Therefore, the probability of choosing a valid vertex B is 4/7.
Solution 2
There are 
ways to pick three vertices from eight total vertices; 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 
to choose three points from the vertices of a single face. Since there are six faces, 
.
Thus, the probability of what we don't want is 
. Using complementary probability,
![\[1- \frac 37 = \boxed{\frac 47}\]](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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
