2007 AMC 12B Problem 16

Attempt Problem 16 of the 2007 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?
1515
1818
2727
5454
8181
Answer: A
Concepts:Burnside’s Lemmasymmetry
Small hint:
The tetrahedron has 1212 rotational symmetries; apply Burnside’s lemma
Big hint:
Count colorings fixed by the identity, the 88 vertex rotations, and the 33 edge rotations
Solution:
The rotation group of the tetrahedron has 1212 elements: the identity, 88 rotations of order 33 about a vertex-face axis, and 33 rotations of order 22 about an edge-midpoint axis.
The identity fixes all 34=813^4=81 colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes 32=93^2=9 colorings; likewise each edge rotation swaps two pairs of faces and fixes 32=9.3^2=9.
By Burnside’s lemma the number of distinguishable colorings is 81+8⋅9+3⋅912=18012=15. \dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.
Thus, the correct answer is A.
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