AMC 10 · 2023 · #10

Grade 4 geometry-2d

Archetype Casework by Position / Vertex Type

parity-coloringspatial-visualizationsystematic-enumerationcomplementary-counting parity-coloringcomplementary-countingcasework ↑ Prerequisites: parity-coloringspatial-visualization

📏 Medium solution 💡 3 insights

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Problem

You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: A hidden $2 \times 1$ domino covers two side-by-side cells of a $3 \times 3$ 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?

Givens: Grid is $3 \times 3$ (nine cells total); The hidden domino is $2 \times 1$ and covers two edge-adjacent cells (horizontal or vertical); You only learn the answer for the cell you name on each turn; Goal: guarantee a hit no matter where the domino was placed; Answer choices: (A) $3$, (B) $5$, (C) $4$, (D) $8$, (E) $6$

Unknowns: The minimum number of turns that always forces at least one hit

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #16 Change Focus / Count the Complement, #6 Guess and Check

Tool #1 (Draw a Diagram) leads — color the $3 \times 3$ 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 $3 \times 3$ 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.

Execute — Answer: C

#1 Draw a Diagram 2.G.A.2 Step 1

Draw the $3 \times 3$ grid and color it like a checkerboard, with the top-left corner white.
Count the cells of each color.
There are $5$ white cells (the four corners plus the center) and $4$ black cells (the four edge midpoints).

$$\begin{array}{|c|c|c|} \hline W & B & W \\ \hline B & W & B \\ \hline W & B & W \\ \hline \end{array} \quad \text{counts: } 5 \text{ white}, \; 4 \text{ black}$$

💡 Partitioning the rectangle into rows and columns of unit squares to count each color — Grade 2 grid partitioning.

#1 Draw a Diagram K.G.B.4 Step 2

Any two cells that share an edge have opposite colors on a checkerboard.
The hidden $2 \times 1$ domino covers two edge-adjacent cells, so it must cover exactly one white and one black cell, no matter where it lies.

$$\text{domino} = (\text{one white cell}) + (\text{one black cell})$$

💡 Comparing two adjacent cells of the checkerboard — Kindergarten-level shape comparison.

#6 Guess and Check K.G.B.6 Step 3

Strategy: name every black cell.
Since the domino must include one black cell, one of the four turns is guaranteed to land on a covered cell.
So $4$ turns are enough.

$$\text{Name all } 4 \text{ black cells} \;\Rightarrow\; \text{guaranteed hit}$$

💡 Picking the smaller of the two colors as the guess set — Kindergarten compose/select.

#16 Change Focus / Count the Complement 4.OA.C.5 Step 4

Now check that $3$ turns are not enough.
If you name only $3$ cells, $6$ cells stay un-named.
For your strategy to guarantee a hit, those $6$ un-named cells must contain no two adjacent cells (otherwise the domino could hide on that adjacent pair).
The biggest collection of cells on the $3 \times 3$ grid with no two adjacent is the set of $5$ white cells.
So at most $5$ cells can be un-named — meaning you must name at least $9 - 5 = 4$.

$$\text{un-named} \le 5 \;\Rightarrow\; \text{named} \ge 9 - 5 = 4$$

💡 Switching to counting the un-named cells (the complement) gives the lower bound — Grade 4 generating a pattern (a maximum no-adjacent set).

#6 Guess and Check 1.OA.A.1 Step 5

Combine.
From step 3, $4$ turns suffice.
From step 4, $3$ turns do not suffice.
Therefore the minimum number of turns is $4$.

$$\text{minimum turns} = 4 \;\Rightarrow\; \textbf{(C)}$$

💡 Combining "enough" and "not enough" to pin the smallest count — Grade 1 word-problem reasoning.

[1] #1 2.G.A.2 Draw the $3 \times 3$ grid and color it like a checkerboard, with the top-left c

[2] #1 K.G.B.4 Any two cells that share an edge have opposite colors on a checkerboard. The hid

[3] #6 K.G.B.6 Strategy: name every black cell. Since the domino must include one black cell, o

[4] #16 4.OA.C.5 Now check that $3$ turns are not enough. If you name only $3$ cells, $6$ cells s

[5] #6 1.OA.A.1 Combine. From step 3, $4$ turns suffice. From step 4, $3$ turns do not suffice.

Review

Reasonableness: 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.

Alternative: Tool #2 (Systematic List): list all $12$ domino positions in the $3 \times 3$ grid, then look for the smallest set of cells that intersects every position. Brute-force checking shows $4$ cells work (the four edge midpoints) and no $3$-cell set works. Same answer, but more bookkeeping than the checkerboard-coloring shortcut.

CCSS standards used (min grade 4)

2.G.A.2 Partition a rectangle into rows and columns of same-size squares (Drawing the $3 \times 3$ grid and laying out the checkerboard coloring to count the white and black cells.)
K.G.B.4 Analyze and compare two- and three-dimensional shapes (Observing that two cells sharing an edge always have opposite checkerboard colors — a basic same/different shape-property comparison.)
K.G.B.6 Compose simple shapes to form larger shapes (Picking the $4$-cell black set as the guess set that any $2 \times 1$ domino must hit.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Identifying the maximum no-adjacent pattern of cells on the grid (the $5$ white cells) to bound the size of the un-named set.)
1.OA.A.1 Solve addition and subtraction word problems within 20 (Putting the two bounds together — $4$ turns suffice and at least $4$ turns are required, so the minimum is $4$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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