AMC 8 · 2025 · #11

• 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.

• 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.

• 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.

• 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.

• 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.