AMC 8 · 2025 · #11
Grade 3 geometry-2dcountingProblem
A tetromino consists of four squares connected along their edges. There are five possible tetromino shapes, I, O, L, T, and S, shown below, which can be rotated or flipped over. Three tetrominoes are used to completely cover a 3×4 rectangle. At least one of the tiles is an S tile. What are the other two tiles?
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A $3 \times 4$ 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?
Givens: Board is a $3 \times 4$ rectangle (area $= 12$ unit squares); Three tetrominoes are used, each made of $4$ unit squares; Tile types allowed: $I, O, L, T, S$ (rotations and reflections allowed); Exactly one of the three tiles is an $S$ tile; Answer choices: (A) I and L, (B) I and T, (C) L and L, (D) L and S, (E) O and T
Unknowns: The pair of tetromino shapes that join the mandatory $S$ tile to fully cover the $3 \times 4$ board
Understand
Restated: A $3 \times 4$ 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?
Givens: Board is a $3 \times 4$ rectangle (area $= 12$ unit squares); Three tetrominoes are used, each made of $4$ unit squares; Tile types allowed: $I, O, L, T, S$ (rotations and reflections allowed); Exactly one of the three tiles is an $S$ tile; Answer choices: (A) I and L, (B) I and T, (C) L and L, (D) L and S, (E) O and T
Plan
Primary tool: #1 Draw a Diagram
Secondary: #3 Eliminate Possibilities, #10 Create a Physical Representation
Tiling problems live on a picture, so Tool #1 (Draw a Diagram) is the starting move: color the $3 \times 4$ 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.
Execute — Answer: C
3.MD.C.6 Step 1 - Color the $3 \times 4$ board like a chessboard.
- Because $3 \times 4 = 12$ and the colors alternate, there are exactly $6$ black squares and $6$ white squares on the board.
💡 Counting unit squares of each color on a grid is exactly the Grade 3 "measure area by counting unit squares" skill.
1.G.A.2 Step 2 - Check the black/white footprint of each tetromino on a chessboard.
- Place each shape mentally and count: an $I$, $O$, $L$, or $S$ tile always covers $2$ black and $2$ white squares no matter how it is rotated or flipped.
- A $T$ tile is the odd one out — its three-in-a-row body and one foot in the middle always cover $3$ of one color and $1$ of the other.
💡 Composing tetrominoes from $4$ unit squares and reading off their color counts is the Grade 1 "compose 2D shapes" idea.
2.OA.C.3 Step 3 - Use the color totals to rule out choices.
- The mandatory $S$ tile uses $2$ black and $2$ white, leaving $4$ black and $4$ white for the other two tiles.
- If one of them is a $T$, that $T$ uses $3$ of one color and $1$ of the other, so the third tile would need to use $1$ of the first color and $3$ of the second — which forces the third tile to also be a $T$.
- So a $T$ can only appear paired with another $T$.
- Choices (B) I and T and (E) O and T mix a $T$ with a non-$T$, so they break the color balance and are eliminated.
💡 Checking that 'black count = white count' is even-vs-odd / balance reasoning — a Grade 2 odd-or-even style argument.
1.G.A.2 Step 4 - Three choices remain: (A) I and L, (C) L and L, (D) L and S.
- Try each by cutting paper tetrominoes (or shading squares on graph paper).
- For (A), the $I$ tile is $1 \times 4$, so in a $3 \times 4$ board it must occupy a full row; the leftover is a $2 \times 4$ strip that has to be tiled by one $L$ and one $S$, and no placement of an $L$ in a $2 \times 4$ strip leaves an $S$-shaped hole — every leftover is itself another $L$, never an $S$.
- For (D) with $S, S, L$, every placement of the first $S$ either creates an isolated square or leaves a region that cannot be split into a second $S$ plus an $L$.
- Both (A) and (D) fail.
💡 Physically arranging the small shapes to compose the rectangle is hands-on Grade 1 shape composition.
1.G.A.2 Step 5 - Now place choice (C): one $S$ tile plus two $L$ tiles.
- Put the $S$ (in its reflected form) so it covers the cells $(1,2), (2,2), (2,1), (3,1)$ in (column, row) coordinates from the bottom-left.
- The eight uncovered cells split cleanly into two $L$-shaped pieces: the left/top region $(1,1), (1,3), (2,3), (3,3)$ is one rotated $L$, and the right region $(3,2), (4,1), (4,2), (4,3)$ is another rotated $L$.
- A valid tiling exists, so the partners of the $S$ tile are two $L$ tiles, giving choice $\textbf{(C)}$.
💡 Building the $3 \times 4$ rectangle out of three pre-cut tetrominoes is exactly the kindergarten/Grade 1 "compose a larger shape from smaller shapes" idea.
3.MD.C.6 Color the $3 \times 4$ board like a chessboard. Because $3 \times 4 = 12$ and th 1.G.A.2 Check the black/white footprint of each tetromino on a chessboard. Place each sh 2.OA.C.3 Use the color totals to rule out choices. The mandatory $S$ tile uses $2$ black 1.G.A.2 Three choices remain: (A) I and L, (C) L and L, (D) L and S. Try each by cutting 1.G.A.2 Now place choice (C): one $S$ tile plus two $L$ tiles. Put the $S$ (in its refle Review
Reasonableness: Three tiles of $4$ squares cover $12$ squares, which matches the $3 \times 4 = 12$ 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.
Alternative: Tool #2 (Make a Systematic List): list every essentially different placement of the $S$ tile inside the $3 \times 4$ board (there are only a handful up to symmetry). For each placement, look at the $8$ uncovered cells and ask 'can this leftover be split into two tetrominoes?' This brute-force enumeration is slower but doesn't require any clever coloring — it directly proves that the only successful pairing is two $L$ tiles.
CCSS standards used (min grade 3)
3.MD.C.6Measure areas by counting unit squares (Counting that the $3 \times 4$ board has $12$ unit squares which split into $6$ black and $6$ white squares under a chessboard coloring.)1.G.A.2Compose two-dimensional shapes or three-dimensional shapes (Composing each tetromino from $4$ unit squares to read its black/white footprint, and composing the $3 \times 4$ rectangle from one $S$ plus two $L$ tetrominoes to demonstrate the tiling.)2.OA.C.3Determine whether a group of objects has an odd or even number (Using even-vs-odd balance ($4$ black $=$ $4$ white remaining after the $S$ tile) to eliminate choices that pair a $T$ tile with a non-$T$ tile.)
⭐ This AMC 8 problem only needs Grade 3 unit-square counting (and a clever chessboard picture) that you already know!
⭐ This AMC 8 problem only needs Grade 3 unit-square counting (and a clever chessboard picture) that you already know!
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