2009 AMC 10A Problem 24

Question

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?

\dfrac{1}{4}

\dfrac{3}{8}

\dfrac{4}{7}

\dfrac{5}{7}

\dfrac{3}{4}

Answer: C

Concepts:complementary counting cube geometry combinations

Difficulty rating: 1860

Small hint:

The plane avoids the interior exactly when all three vertices lie on one face

Big hint:

Count the same-face triples and subtract from the total

\binom{8}{3}

Solution:

Three vertices determine a plane that cuts through the interior unless all three lie on a single face.

Each of the

6

faces gives

\binom{4}{3} = 4

triples, so

6 \cdot 4 = 24

triples lie on a face out of

\binom{8}{3} = 56

total.

The probability of hitting the interior is

1 - \dfrac{24}{56} = \dfrac{4}{7}.

Thus, the correct answer is C.