2009 AMC 10A Problem 24

P21-25~4 minPrint

Basic Probability Solid Geometry Count the complement and subtract from the total

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?

Show solution without answering

Key insight

Count the bad choices: the plane misses the inside only when all three corners lie on one face. That is 6 faces times 4 choices, 24 out of 56.

Solution

The good cases come in many shapes, but the bad cases are simple, so count the bad ones and subtract.

1. Total ways to choose $3$ of the $8$ corners: $\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ .
2. Bad choices. If all three corners are on the same face, the plane through them is the plane of that face. It only touches the surface of the cube, not the inside.
3. Each face has $4$ corners, and there are $4$ ways to choose $3$ of them (leave one out). With $6$ faces that is $6 \times 4 = 24$ bad choices. None is counted twice, since two faces share only $2$ corners.
4. Every other choice is good. Put the cube with corners at coordinates made of $0$ s and $1$ s, like $(0,0,0)$ and $(1,0,1)$ . Take the average of the three chosen corners; this point lies in their plane. An average coordinate is $0$ or $1$ only when all three corners have the same value there, which means they are on one face. Otherwise each average coordinate is $\frac13$ or $\frac23$ , so the point is strictly inside the cube.
For example, $(1,0,0)$ , $(0,1,0)$ , $(1,1,1)$ are not on one face, and their average $(\frac23, \frac23, \frac13)$ is inside.
5. Good choices: $56 - 24 = 32$ . Probability: $\dfrac{32}{56} = \dfrac{4}{7}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$ .

Why this works

When the cases you want are varied and messy but the cases you do not want are few and neat, count the unwanted ones and subtract from the total. Here the only planes that stay outside the cube are the six face planes, so the counting takes one line.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

There are $\binom{8}{3} = 56$ equally likely triples of vertices, and any three vertices of a cube are non-collinear, so each triple determines a plane.

Count the complement: triples whose plane stays outside the interior. If the three vertices all lie on one face, their plane is that face's plane and touches only the boundary. Each of the $6$ faces has $4$ vertices, giving $\binom{4}{3} = 4$ triples per face, and no triple lies on two faces (two faces share at most $2$ vertices). That is $6 \cdot 4 = 24$ "bad" triples.

Conversely, if the three vertices are not all on one face, their plane enters the interior. Put the cube at $[0,1]^3$ . The centroid of the three vertices lies in their plane, and each of its coordinates is $0$ or $1$ only if all three vertices share that coordinate, i.e. lie on a common face. Otherwise every coordinate is $\frac13$ or $\frac23$ , so the centroid is strictly inside the cube.

So $56 - 24 = 32$ triples work, and the probability is $\frac{32}{56} = \frac{4}{7}$ .

The answer is $\boxed{\textbf{(C)}\ \frac{4}{7}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques