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AMC 10 · 2023B · #10

Grade 4 geometry-2d
Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position
parity-coloringspatial-visualizationsystematic-enumerationcomplementary-counting parity-coloringcomplementary-countingcasework ↑ Prerequisites: parity-coloringspatial-visualization
📏 Medium solution 💡 3 insights
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Problem
A hidden 2 × 1 domino covers two side-by-side cells of a 3 × 3 grid. Each turn, you name a cell and are told whether it is under the domino. What is the smallest number of turns that guarantees you have named at least one covered cell, no matter where the domino lies?

Pick an answer.

(A)
3
(B)
5
(C)
4
(D)
8
(E)
6

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) leads — color the 3 × 3 grid like a checkerboard with corners white. There are 5 white and 4 black cells, and because the domino covers two adjacent cells it always covers one white and one black. That single observation unlocks the upper bound: name all 4 black cells and you must hit one of the domino's two cells. Tool #16 (Change Focus / Complement) gives the matching lower bound — instead of asking "how few cells let me hit?" ask "how many cells can I leave un-named without leaving a domino-shaped gap?". The un-named cells must contain no adjacent pair, and the largest such set on a 3 × 3 grid is the 5-cell white set. So at most 5 cells can be un-named, meaning at least 9 - 5 = 4 must be named. Tool #6 (Guess and Check) acts as a sanity sweep across small strategies. Algebra is unnecessary — the picture and the complement count are decisive.

1STEP 1

Color the grid like a checkerboard

Color the 3 × 3 grid like a checkerboard (top-left white): 5 white cells (corners + center), 4 black (edge midpoints).

W & B & W ; B & W & B ; W & B & W ; counts: 5 white, 4 black
2STEP 2

See what one domino covers

Adjacent cells always differ in color, so wherever it lies the domino covers exactly one white and one black cell.

domino = (one white cell) + (one black cell)
3STEP 3

Name every black cell

Name all 4 black cells: the domino always includes a black one, so a hit is guaranteed — 4 turns suffice.

Name all 4 black cells → guaranteed hit
4STEP 4

Show 3 turns fall short

Un-named cells can have no adjacent pair; the biggest non-adjacent set is the 5 white cells, so at least 9 - 5 = 4 must be named.

un-named ≤ 5 → named ≥ 9 - 5 = 4
5STEP 5

Combine both bounds

Enough (step 3) and not enough (step 4) together pin the minimum at 4 → (C).

minimum turns = 4 → (C)
Answer
4
Three checks. (1) Strategy sanity: writing out all twelve possible domino positions (6 horizontal + 6 vertical) and checking that each one covers at least one black cell confirms the upper-bound strategy actually works. (2) Lower-bound sanity: try 3 guesses at corners, say (1,1), (1,3), (3,1). The un-named set is {(1,2), (2,1), (2,2), (2,3), (3,2), (3,3)} — and the pair (2,2), (2,3) is adjacent, so a horizontal domino fits there with no hit. So 3 really is not enough. (3) Eliminate distractors: (A) 3 is the most natural too-low guess (refuted above); (B) 5 is the white-set strategy, which works but uses more turns than needed; (D) 8 and (E) 6 are over-counts that ignore the parity insight.
💡Key takeaway

This AMC 10 problem only needs Grade 4 grid patterns you already know — color the 3 × 3 board like a checkerboard (5 white, 4 black), notice every 2 × 1 domino must cover one of each, so naming the 4 black cells always guarantees a hit and 3 guesses can never be enough.

  • Color the grid like a checkerboard
  • See what one domino covers
  • Name every black cell
  • Show 3 turns fall short
  • Combine both bounds

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