2025 AMC 8 Problem 11

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In This Page:

To help you master 2025 AMC 8 Problem 11, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.

2025 AMC8 Problem 11

Solution

Solution 1

Direct Construction and Corner Logic

Place the required $S$ tile in the grid safely away from the corners, and identify the tetromino shapes formed by the remaining empty squares.

When filling a grid, an empty square must connect to enough other empty squares to fit a tile. If a piece creates a ‘dead end’ (an empty square with no empty neighbors), the puzzle cannot be solved.

Step 1: Understand the grid and the pieces

A $3\times 4$ rectangle has a total of 12 squares. Since each tetromino is made of 4 squares, exactly three tetrominoes will completely cover the grid without overlapping. We are told one of these is an $S$ tile.

$$3\times 4=12\text{total squares}$$

$$12÷4=3\text{tetrominoes}$$

Step 2: Analyze the dangers of the corners

The hardest parts of any grid to fill are the corners. The $S$ tile has a zigzag shape with empty spaces in the corners of its rectangular bounding box. If we push the $S$ tile tightly into a corner of the grid, its solid squares will surround the grid’s corner square. This creates a 1-square hole that no 4-square tetromino can possibly fill.

Step 3: Place the S tile safely in the middle

To avoid trapping corner squares, we should place the $S$ tile in the center of the grid. Let’s test placing a vertical $S$ tile (which is 3 squares tall and 2 squares wide) exactly in the middle two columns of the $3\times 4$ grid. Suppose the grid has 3 rows and 4 columns. The $S$ tile will occupy Row 1 Column 2, Row 2 Column 2, Row 2 Column 3, and Row 3 Column 3.

Step 4: Identify the remaining left-side shape

With the $S$ tile in the middle, the remaining 8 empty squares are divided into two separate regions on the left and right. On the left side, the entire first column (3 squares) is empty. The only other empty square on the left is at Row 3 Column 2. Connecting these 4 squares perfectly forms an $L$ tetromino.

Step 5: Identify the remaining right-side shape

On the right side, the entire fourth column (3 squares) is empty. The only other empty square on the right is at Row 1 Column 3. Connecting these 4 squares also perfectly forms an $L$ tetromino (rotated).

Step 6: Conclude the matching answer choice

We have successfully tiled the entire grid using one $S$ tile, one $L$ tile, and another $L$ tile. Looking at the answer choices, the combination of the ‘other two tiles’ is $L$ and $L$. Since this is a multiple-choice math competition, this single valid construction is enough to confirm the correct answer.

Correct answer: C. $L$ and $L$
The other two tiles are L and L. This matches answer choice C. A quick check of the area confirms that 4 (for S) + 4 (for L) + 4 (for L) = 12 squares, exactly filling the 3×4 grid.

Solution 2

Checkerboard Coloring and Elimination

Color the grid like a checkerboard. The $T$ tile covers an unbalanced number of black and white squares, which makes it impossible to use in certain combinations.

Checkerboard Coloring: A standard $3\times 4$ checkerboard has exactly 6 black and 6 white squares. Any set of tiles used to cover it must collectively cover exactly 6 black and 6 white squares.

Step 1: Color the grid and count colors

Imagine coloring the $3\times 4$ grid with an alternating black and white pattern. Because the grid has an even number of total squares (12), it will contain exactly 6 black squares and 6 white squares.

Step 2: Analyze the color parity of the tiles

Examine the pieces: the $I$, $O$, $L$, and $S$ tiles each cover exactly 2 black and 2 white squares, no matter how they are rotated or flipped. However, the $T$ tile is composed of a central square and three neighbors. On a checkerboard, neighbors always have the opposite color. Therefore, a $T$ tile covers either 3 black and 1 white, or 1 black and 3 white squares.

Step 3: Eliminate choices with the T tile

If we were to use exactly one $T$ tile alongside two other non-$T$ tiles, the total colors covered would be $3+2+2=7$ of one color and $1+2+2=5$ of the other. This gives 7 black and 5 white (or vice versa), which cannot match our grid’s 6 black and 6 white squares. Thus, we cannot use a $T$ tile. This eliminates choices B ($I$ and $T$) and E ($O$ and $T$).

$$\text{Non-}T\text{pieces}:2B+2W$$

$$T\text{piece}:3B+1W\text{(or}1B+3W\text{)}$$

$$\text{Total}:7B+5W\ne 6B+6W$$

Step 4: Analyze the I tile

Consider choice A ($I$ and $L$). The grid only has 3 rows. Because the $I$ tile is 4 squares long, it cannot be placed vertically. It must be placed horizontally, taking up an entire full row. This leaves a smaller $2\times 4$ grid to be filled by the $S$ and $L$ tiles.

Step 5: Eliminate choice A

The $S$ tile has a bounding box that is 3 squares wide and 2 squares tall. If we place it into a $2\times 4$ grid, it is forced to touch the left or right edge. Because the $S$ shape has ‘missing’ corners, pushing it against the edge of a narrow grid will always trap a 1-square empty corner. Thus, an $S$ tile cannot fit in a $2\times 4$ grid, eliminating choice A.

Step 6: Conclude between the remaining choices

We are left with C ($L$ and $L$) and D ($L$ and $S$). As discovered in Method 1, every valid placement of a single $S$ tile perfectly forces the remaining squares into two $L$ shapes. It is impossible to fit a second $S$ tile without trapping empty squares, so choice C is the only correct answer.

Correct answer: C. $L$ and $L$
The only viable pieces are two L tiles, which corresponds to choice C. Both methods independently eliminate other possibilities and verify L and L as the correct set.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2025 Question 11. These real AMC 8 problems test the same key concept—Coloring, Tiling & Covering. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 8 Problem 21

2024 AMC 8 Problem 7

2011 AMC 8 Problem 3

2010 AMC 8 Problem 7

2009 AMC 8 Problem 18

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Checkerboard Coloring for Tiling:

Valid for rectangular grids and standard polyominoes (like tetrominoes or dominoes). The $T$ tile is the only tetromino that covers an uneven parity of colors (3 of one, 1 of the other). The grid has an even parity (6 and 6), which mathematically proves one $T$ tile cannot be used.

• Dead End / Corner Isolation Logic:

Applies to any space-filling or packing puzzle. The $S$ tile has indented corners, which easily surround and trap grid corners, making corner placements impossible.

Key Takeaway: Remember This Strategy for Next Time

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