2009 AMC 10A Problems/Problem 24

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?

$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$

First of all, number of planes determined by any three vertices of a cube is 20 (6 surface, 6 opposing parallel edges, 8 points cut by three remote vertices). Among these 20 planes only 6 surfaces will not cut into the cube. Secondly, to choose three vertices randomly, the four vertices planes each will be chosen 4 times, while the three vertices planes each will be chosen once. To conclude, the probability of a cutting in plane is (6*4+8*1)/(12*4+8*1)=32/56=4/7 (C)

Vader10

Solution $2$

We will try to use symmetry as much as possible.

Pick the first vertex $A$ , its choice clearly does not influence anything.

Pick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.

In the first case, there are $2$ faces that contain the edge $AB$ . In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$ , the entire triangle $ABC$ will be on a face. On the other hand, if $C$ 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 $C$ is $2/6 = 1/3$ .

In the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$ . Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$ .

In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$ .

Summing up all cases, the resulting probability is: $\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}$

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 3

There are $\binom{8}{3}=56$ 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 $\binom{4}{3}=4$ to choose three points from the vertices of a single face. Since there are six faces, $4 \times 6 = 24$ .

Thus, the probability of what we don't want is $\frac{24}{56} = \frac{3}{7}$ . Using complementary probability,

$1- \frac 37 = \boxed{\frac 47}$

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2009 AMC 10A Problems/Problem 24

Contents

Problem

Solution $2$

Note: (Cheap solution same approach as solution 1)

Solution 3

See Also