AMC 10 · 2007 · #16

Grade 8 countinggeometry-3d
rotation-isometrystars-and-barsspatial-visualization systematic-enumerationsymmetry-argument ↑ Prerequisites: combinations-basicspatial-visualization
📏 Medium solution 💡 3 insights
Problem
Each of a regular tetrahedron's four faces is painted one of three colours. Two paint jobs count as the same when one can be turned into the other. Count the genuinely different paint jobs.

Pick an answer.

(A)
15
(B)
18
(C)
27
(D)
54
(E)
81
How to solve
Strategy Make a Systematic List

The natural label for a coloring is its color tally: how many faces are red, white, and blue. Listing all possible tallies is a short systematic list (Tool #2). But a label only counts colorings correctly if it is honest in both directions, and only one direction is easy. Rotations obviously preserve the tally, so different tallies are never confused. The reverse claim — equal tallies force equal appearance — is the one that has to be earned, because a tetrahedron has only 12 rotations while four faces admit 24 rearrangements. So the plan is: build the list, then take an explicit inventory of the twelve rotations (Tool #10) and use a repeated color plus a half-turn (Tool #17) to close the missing half. A final pass over the answer choices (Tool #3) shows exactly which shortcut produces each trap.

1STEP 1

What a rotation can change

A rotation never changes the tally of each colour.

(r,w,b) with r+w+b=4 is unchanged by every rotation
2STEP 2

List every color tally

Listing every tally gives 15 possibilities.

3+6+3+3=15=C(6, 2)
3STEP 3

Which rearrangements are rotations

Not every rearrangement of faces is a rotation.

1+8+3=12 rotations, versus 4!=24 rearrangements of the faces
4STEP 4

A repeated color closes the gap

A repeated colour supplies the swap needed to close the gap.

4 faces, 3 colors → two faces share a color; the half-turn swapping them also swaps the other two
5STEP 5

Count the classes and answer

So the tally count is the answer, 15, choice (E).

3⁴=81, and 81/12=6.75 is not a whole number, so the answer is the tally count 15
Answer
15
The answer must sit between 81/12=6.75 and 81, since the 81 labeled colorings fall into classes of size at most 12; 15 clears that. The trap choices are the un-collapsed counts: 81=3⁴ ignores rotation entirely, and 27=81/3 divides by the wrong number. A sharper test shows the repeated-color step is load-bearing rather than decorative: run the same argument with 4 colors instead of 3. The tally count becomes C(7, 3)=35, but the honest orbit count is 36. The extra class is the coloring with all four faces different — with 4 colors no face color repeats, the free swap is unavailable, and that coloring really does come in two mirror versions that no rotation matches. With 3 colors the pigeonhole forbids exactly that situation, which is why the tally count is right here.
💡Key takeaway

Count how many faces get each color: for a tetrahedron with three colors that tally tells the whole story, because two faces sharing a color can always be swapped by a real turn.

  • What a rotation can change
  • List every color tally
  • Which rearrangements are rotations
  • A repeated color closes the gap
  • Count the classes and answer