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?

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

Solution

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 iff 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: $\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{\frac 47}$

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

Page

Toolbox

Search

2009 AMC 10A Problems/Problem 24

Problem

Solution

See Also