Question

Each of the $9$ squares in a $3 \times 3$ grid is to be colored red, blue, or yellow in such a way that each red square shares an edge with at least one blue square, each blue square shares an edge with at least one yellow square, and each yellow square shares an edge with at least one red square. Colorings that can be obtained from one another by rotations and/or reflections are to be considered the same. How many different colorings are possible?

$3$

$9$

$12$

$18$

$27$

Answer: C

Concepts:Burnside’s Lemma casework

Difficulty rating: 2100

Solution:

First count colorings of a grid whose positions are distinguished. Fix the center square as red and list the four edge-middle colors cyclically. Up to a rotation or reflection, the only possible edge patterns are $YRBR$ and $YRBY.$ The first has $4$ placements and $3$ possible cyclic corner strings, $BRYB, BYRB,$ and $BYYB;$ the second has $8$ placements and $2$ possible corner strings, $BYBB$ and $BYRB.$ Thus there are $4 \cdot 3 + 8 \cdot 2 = 28$ colorings with a red center. The center has $3$ possible colors, so there are $84$ labeled colorings.

Now apply Burnside's lemma. The identity fixes all $84$ colorings. No nonidentity rotation fixes a valid coloring. Each of the two reflections across a horizontal or vertical axis fixes $6$ colorings, while each diagonal reflection fixes none. Therefore, the number of colorings up to rotations and reflections is $\dfrac{84 + 6 + 6}{8} = 12.$ Thus, C is the correct answer.

2025 AMC 10B Problem 21

Answer: C

Difficulty rating: 2100

Solution:

Problem 21 in Other Years