AMC 10 · 2009 · #24
Grade 7 geometry-3dThree 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?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Pick $3$ different corners of a cube at random. Three corners always fix exactly one flat plane. Find the chance that this plane passes through the inside of the cube instead of just skimming along its surface.
Givens: A cube has $8$ vertices (corners).; Three different corners are chosen at random, and every group of three corners is equally likely.; Any three points that are not in a straight line fix exactly one plane.
Unknowns: The probability that the plane through the three chosen corners cuts through the inside of the cube.
Understand
Restated: Pick $3$ different corners of a cube at random. Three corners always fix exactly one flat plane. Find the chance that this plane passes through the inside of the cube instead of just skimming along its surface.
Givens: A cube has $8$ vertices (corners).; Three different corners are chosen at random, and every group of three corners is equally likely.; Any three points that are not in a straight line fix exactly one plane.
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #17 Visualize Spatial Relationships, #2 Make a Systematic List
Counting every triple of corners whose plane cuts the interior is messy, but the opposite is tiny and clean, so tool #16 (Change Focus / Count the Complement) is the main move: count the triples that DON'T cut, then subtract. Tool #17 (Visualize Spatial Relationships) supplies the key picture, that a plane misses the inside only when its three corners share one flat face. Tool #2 (Make a Systematic List) does the actual counting, $\binom{8}{3}$ for the total and $6\times\binom{4}{3}$ for the skimming faces.
Execute — Answer: C
7.SP.C.8 Step 1 Count all corner triples
- Start with the whole picture.
- The cube has $8$ corners and we choose $3$ of them.
- The number of ways to pick $3$ things out of $8$ is $\binom{8}{3}=\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=56$.
- This $56$ is the total number of equally likely choices, the bottom of our probability.
💡 Counting every equally likely choice once gives a fair denominator for the probability.
6.G.A.4 Step 2 See which planes skim
- Picture the cube.
- If the three chosen corners all sit on one flat face, their plane is just that face and only skims the surface, missing the inside.
- If the three corners do NOT share a face, the triangle they make has to stretch across the box, so the plane dives through the interior.
- So the only planes that avoid the inside are the six flat faces.
💡 A cube's flat faces are the only ways three corners can share a plane without reaching inside.
7.SP.C.8 Step 3 Count the skimming triples
- Now count the opposite of what we want: triples whose plane skims.
- Each face is a square with $4$ corners, and any $3$ of those $4$ still lie on that face, giving $\binom{4}{3}=4$ triples per face.
- The cube has $6$ faces, so the number of skimming triples is $6\times 4=24$.
💡 Every skimming plane is one of the six faces, so counting faces counts all the non-cutting triples.
7.SP.C.7 Step 4 Take the complement
- The chance of a skimming (bad) plane is $\frac{24}{56}=\frac{3}{7}$.
- Since a plane either skims or cuts, the chance we want is everything left over: $1-\frac{3}{7}$.
💡 When events split into just two kinds, one probability is $1$ minus the other.
7.SP.C.5 Step 5 Finish the probability
- Subtract: $1-\frac{3}{7}=\frac{4}{7}$.
- You can check it by counting favorable triples directly: $56-24=32$, and $\frac{32}{56}=\frac{4}{7}$.
- This matches choice $\textbf{(C)}$.
💡 Subtracting the easy skimming chance from $1$ hands you the cutting chance directly.
7.SP.C.8 Start with the whole picture. The cube has $8$ corners and we choose $3$ of them 6.G.A.4 Picture the cube. If the three chosen corners all sit on one flat face, their pl 7.SP.C.8 Now count the opposite of what we want: triples whose plane skims. Each face is 7.SP.C.7 The chance of a skimming (bad) plane is $\frac{24}{56}=\frac{3}{7}$. Since a pla 7.SP.C.5 Subtract: $1-\frac{3}{7}=\frac{4}{7}$. You can check it by counting favorable tr Review
Reasonableness: The favorable count is $56-24=32$, and $\frac{32}{56}=\frac{4}{7}\approx 0.57$, a legal probability between $0$ and $1$. It makes sense that the answer is more than half: most random corner triples do not happen to land on a single face, so most planes cut inside, and $\frac{4}{7}$ is indeed just above $\frac{1}{2}$. Choice (C) is $\frac{4}{7}$, which matches.
Alternative: Count the cutting cases directly with casework. Fix the first corner, then sort the second corner by distance: sharing an edge, across a face diagonal, or across the cube's long diagonal. Finding, in each case, the chance the third corner makes the plane cut the interior and adding them gives the same $\frac{4}{7}$. The complement method is shorter because the only non-cutting planes, the six faces, are so easy to count.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $56$ total corner triples and the $24$ triples that lie on a single face.)6.G.A.4Represent three-dimensional figures using nets and find surface area (Seeing that a plane skips the cube's interior only when its three corners share one flat face.)7.SP.C.7Develop probability models and use them to find probabilities of events (Using the complement rule, that skimming and cutting are the only two outcomes.)7.SP.C.5Understand that the probability of a chance event is between 0 and 1 (Computing $1-\frac{3}{7}=\frac{4}{7}$ and checking it is a valid probability.)
⭐ Three corners of a cube skip the inside only when they all sit on one flat face, and that lucky lineup is rare, so most planes cut through.
⭐ Three corners of a cube skip the inside only when they all sit on one flat face, and that lucky lineup is rare, so most planes cut through.
More like this
Same archetype — closest grade level first.