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2012 AMC 10A Problem 20

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2012 AMC 10A Problem 20AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 7 min · position P16-20

A 3×33 \times 3 square is partitioned into 99 unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated 90 ∘90\,^{\circ} clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?

  1. A)

    49512\frac{49}{512}

  2. B)

    764\frac{7}{64}

  3. C)

    1211024\frac{121}{1024}

  4. D)

    81512\frac{81}{512}

  5. E)

    932\frac{9}{32}

Answer

A

Key insight

The rotation splits the grid into the center and two 4-cycles; a square stays white only if its predecessor was white, so each 4-cycle needs no adjacent whites.

Solution

After the rotation, a position ends up black if the square that moved there is black, or if the square originally there was black (a white arrival is repainted). So a position stays white only when both it and the square rotating into it were white.

The rotation moves the four corners around in a cycle, moves the four edge squares around in a cycle, and fixes the center. These three groups are independent.

  • Center: must be black, probability 12\frac12.
  • Corners: label them in rotation order 1→2→3→4→11 \to 2 \to 3 \to 4 \to 1. Position kk ends white exactly when squares kk and k−1k - 1 are both white. So all four end black if and only if no two cyclically adjacent corners are both white. Count such colorings: zero whites (11 way), one white (44 ways), two whites that are opposite (22 ways); three or more whites always include an adjacent pair. That is 77 of the 1616 colorings, probability 716\frac{7}{16}.
  • Edges: the same cycle structure, probability 716\frac{7}{16}.

Multiply:
12⋅716⋅716=49512.\frac12 \cdot \frac{7}{16} \cdot \frac{7}{16} = \frac{49}{512}.

The answer is (A) 49512\boxed{\textbf{(A)}\ \frac{49}{512}}.

Why this works

A symmetry operation partitions the squares into orbits, and the repainting rule only compares a square with its predecessor in the orbit. That converts a 9-square problem into two identical 4-cycle problems and a fixed point. Counting "no two adjacent whites around a cycle" is a small casework on the number of whites; recognizing that the condition is on adjacent pairs, not "at most one white," is the whole problem.

The trap

Requiring each 4-cycle to contain at most one white square (5 of 16) instead of allowing two non-adjacent whites, which gives 25/512 instead of 49/512.

Common mistakes

  • Requiring each 4-cycle to contain at most one white square (5 of 16) instead of allowing two non-adjacent whites, which gives 25/512 instead of 49/512.
  • Forgetting the center square, which contributes the factor 12\frac12 and turns 49256\frac{49}{256} into 49512\frac{49}{512}.

Techniques

Split into exhaustive cases and handle each · Exploit symmetry to reduce work or pair up objects