AMC 10 · 2009 · #24

Grade 7 geometry-3d
Archetype Overlap / Complement Counting └ Sub-archetype Complement by Subtracting Forbidden Arrangements
combinations-basiccomplementary-countingspatial-visualizationprobability-basic complementary-countingsystematic-enumeration ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 3 insights
Problem
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.

Pick an answer.

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

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

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, C(8, 3) for the total and 6×C(4, 3) for the skimming faces.

1STEP 1

Count all corner triples

Pick 3 of the cube's 8 corners: C(8,3)C(8,3) gives 56 equally likely triples, the denominator of the probability.

C(8, 3)=(8 · 7 · 6)/(3 · 2 · 1)=56
2STEP 2

See which planes skim

Three corners on one flat face give a plane that only skims; otherwise the triangle stretches across the box and cuts the interior.

3STEP 3

Count the skimming triples

Each square face holds C(4,3)=4C(4,3)=4 such triples, and the cube has 6 faces, so 24 triples only skim.

6×C(4, 3)=6 × 4=24
4STEP 4

Take the complement

A skimming plane has chance 2456\frac{24}{56}, which is 37\frac{3}{7}; since a plane either skims or cuts, subtract that from 1.

P(skim)=24/56=3/7, P(cut)=1-3/7
5STEP 5

Finish the probability

Subtracting gives 47\frac{4}{7}, and the direct count agrees: 56−24=3256-24=32 favorable out of 56, matching choice (C).

P(cut)=1-3/7=4/7=32/56 → (C)
Answer
4/7
The favorable count is 56-24=32, and 32/56=4/7≈ 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 4/7 is indeed just above 1/2. Choice (C) is 4/7, which matches.
💡Key takeaway

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.

  • Count all corner triples
  • See which planes skim
  • Count the skimming triples
  • Take the complement
  • Finish the probability

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