AMC 10 · 2018 · #15

Grade 8 counting
symmetry-argumentreflection-symmetryrotation-isometry complementary-countingsystematic-enumeration ↑ Prerequisites: symmetry-argument
📏 Medium solution 💡 2 insights
Problem
A seven by seven grid of squares is colored black and white. A coloring is called symmetric if it looks the same after any ninety degree rotation about the centre and after reflection across either diagonal or either midline. At least one square of each color must appear. Find how many symmetric colorings there are.

Pick an answer.

(A)
510
(B)
1022
(C)
8190
(D)
8192
(E)
65,534
How to solve
Strategy Visualize Spatial Relationships

The whole question is about what the eight symmetries (four rotations and four reflections) do to the squares, so Tool #17 (Visualize Spatial Relationships) leads: mentally rotate and fold the grid to see which squares are locked to the same color. Tool #1 (Draw a Diagram) pins the squares to a coordinate grid so the folding is exact. Tool #7 (Identify Subproblems) splits the squares into a few position types (center, midline arm, diagonal arm, interior), and Tool #2 (Make a Systematic List) counts how many independent color choices that leaves. Finally the words 'at least one of each color' are a Tool #16 (Change Focus / Count the Complement) cue: count every coloring, then remove the two all-one-color cases.

1STEP 1

Squares move in linked groups

The symmetries tie squares into groups.

symmetries={0^°,90^°,180^°,270^°}∪{2 diagonals, 2 midlines}=8
2STEP 2

Fold into one-eighth

Fold the grid into one eighth.

fundamental wedge={(a,b):0 ≤ a ≤ b ≤ 3}
3STEP 3

Count the independent squares

There are ten independent groups.

1+3+3+3=10 groups
4STEP 4

Two colors for each group

Two colors each gives two to the tenth.

2¹⁰=1024
5STEP 5

Remove the two banned grids

Removing the two solid grids gives 1022.

1024-2=1022 → (B)
Answer
1022
The total of 49 squares splits correctly into the 10 groups: the center is a group of 1; each midline-arm square (0,b) has a group of 4, giving 3 × 4=12; each diagonal-arm square (a,a) has a group of 4, giving 3 × 4=12; each interior square (a,b) with a < b has a group of 8, giving 3 × 8=24. Then 1+12+12+24=49, exactly the whole grid, so no square was missed or double-counted. The answer 1022 is just below 1024=2¹⁰, matching choice (B). The large choices 8190 and 8192 would come from 2¹³, which would mean 13 free groups — too many once the symmetry is applied.
💡Key takeaway

Fold the symmetric grid along its mirror lines until only 10 squares are truly free; each is black or white, giving 2¹⁰=1024 pictures, then drop the all-black and all-white ones for 1022, choice (B).

  • Squares move in linked groups
  • Fold into one-eighth
  • Count the independent squares
  • Two colors for each group
  • Remove the two banned grids