AMC 8 · 2025 · #11

Grade 3 geometry-2dcounting

Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Fixed Piece Set — Feasibility by Elimination

area-rectanglesparity-coloringspatial-visualizationparity parity-coloringcaseworksystematic-enumeration ↑ Prerequisites: area-rectanglesparity

📏 Long solution 💡 3 insights 📊 Diagram

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Problem

A 3 × 4 rectangle is completely covered by three tetrominoes chosen from the five shapes I, O, L, T, S (rotations and reflections allowed). One of the three tiles must be an S tile. Which pair of tiles makes up the other two?

Pick an answer.

(A)

I and L

(B)

I and T

(C)

L and L

(D)

L and S

(E)

O and T

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tiling problems live on a picture, so Tool #1 (Draw a Diagram) is the starting move: color the 3 × 4 board like a chessboard and count how many black/white squares each tetromino must cover. That single picture turns a hard geometry question into easy parity arithmetic. Tool #3 (Eliminate Possibilities) then knocks out the choices that violate the black/white balance — this is the classic AMC multiple-choice move. Finally, Tool #10 (Physical Representation) finishes the job: with only three options left, cut out paper tetrominoes (or shade cells on graph paper) and actually try to place an S together with the candidates. The combination that fits is the answer.

1STEP 1

Color the 3 × 4 board like a chessboard: the 12 squares split into 6 black and 6 white.

3 × 4 = 12, 12 ÷ 2 = 6 black + 6 white

2STEP 2

On a chessboard, an I, O, L, or S tile covers 2 black + 2 white, but a T tile covers 3 of one color and 1 of the other.

I, O, L, S → 2B+2W ; T → 3B+1W or 1B+3W

3STEP 3

The S tile takes 2 black + 2 white, leaving 4 black and 4 white; a T pairs only with another T — eliminate (B) and (E).

Remaining after S: 6-2=4 B, 6-2=4 W

4STEP 4

(A)'s I-tile forces a 2 × 4 strip that can't hold an S, and (D)'s S + S + L always leaves a gap, so both (A) and (D) fail.

(A) fails: I forces a full row → 2 × 4 strip cannot be L + S

5STEP 5

Placing the S tile bottom-left splits the eight empty cells into two L-shaped regions — a real S + L + L tiling exists, giving (C).

S at {(1,2),(2,2),(2,1),(3,1)} + L at {(1,1),(1,3),(2,3),(3,3)} + L at {(3,2),(4,1),(4,2),(4,3)} → (C)

Answer

L and L

Three tiles of 4 squares cover 12 squares, which matches the 3 × 4 = 12 area of the board — the area arithmetic checks out. The chessboard parity (6 black, 6 white) is preserved by the S tile (2+2) and by the two L tiles (2+2 each), summing to 6+6. The explicit placement of S, L, L given above tiles the rectangle with no overlaps and no holes (one can verify by shading the cells on graph paper), so the answer (C) L and L is both internally consistent and constructively demonstrated.

💡Key takeaway

This AMC 8 problem only needs Grade 3 unit-square counting (and a clever chessboard picture) that you already know!

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