AMC 10 · 2012 · #20

Grade 8 probability
Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position
probability-basicrotation-isometrycomplementary-counting complementary-countingcasework ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
A 3x3 grid has each of its 9 unit squares colored black or white independently, each color with probability 1/2. The whole grid is turned 90 degrees clockwise about its center. Then any square that is white but sits on a spot that used to hold a black square is repainted black; every other square keeps its color. Find the probability that afterward the entire grid is black.

Pick an answer.

(A)
$\frac{49}{512}$
(B)
$\frac{7}{64}$
(C)
$\frac{121}{1024}$
(D)
$\frac{81}{512}$
(E)
$\frac{9}{32}$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The turn never mixes corners with edges with the center, so the 9 squares split into three independent groups: the lone center, a loop of 4 corners, and a loop of 4 edges. Once the painting rule is restated as a simple ban on certain white pairs, each group can be counted on its own and the counts multiplied. Breaking the grid into these separate pieces is what turns a tangled picture into three tiny counting jobs.

1STEP 1

See what the turn moves

Turn the grid a quarter: the center stays put, the 4 corners ride one loop, and the 4 edge-middles ride a separate loop.

corners: TL → TR → BR → BL → TL; edges: T → R → B → L → T
2STEP 2

Restate when a square ends white

A spot ends white exactly when it and the square turning onto it both started white; all black means no such white-on-white pair.

spot ends white ⇔ (spot was white) ∧ (square turning onto it was white)
3STEP 3

Split into center, corners, edges

Rotation never mixes the three groups, so their counts multiply; the center must start black, giving 1 way there.

N_black grid = N_center × N_corners × N_edges, N_center = 1
4STEP 4

Count one 4-square loop

In a 4-loop no two neighbors may both be white: 1 all-black, 4 with one white, 2 with opposite whites, giving 7 per loop.

1 + 4 + 2 + 0 + 0 = 7 → N_corners = N_edges = 7
5STEP 5

Multiply and divide

Multiply: 1 x 7 x 7 = 49 good grids out of 512 in all, so the probability is 49/512, choice (A).

P = (1 × 7 × 7)/2⁹ = 49/512
Answer
49/512
The answer 49/512 is a fraction over 2⁹, exactly the shape you expect when every one of the 9 squares is a fair coin flip, and 49 = 7 x 7 is a whole number of favorable grids, so the count is self-consistent. The value is about 0.096, a small but not tiny chance, which fits: needing the center black plus safe patterns in both loops is demanding but far from impossible. Choice (C) 121/1024 does not sit over 512 with an integer top (it is 60.5/512), so it cannot be a whole count out of 512; (A) can.
💡Key takeaway

A quarter turn sorts the squares into a fixed center and two rings of four, so count each ring's safe patterns on its own and multiply.

  • See what the turn moves
  • Restate when a square ends white
  • Split into center, corners, edges
  • Count one 4-square loop
  • Multiply and divide

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