2026 AMC 10A Problem 21

Here is the grid from the left picture. Name each square by its row and column, for example "top-left" or "middle-right".

There are $4$ X's and $5$ O's. We need two counts: how many partitions there are in total, and how many of them win.

Part 1: there are 10 partitions.

1. Two shapes of piece. Three adjoining squares form either a straight line of three or an L (a $2 \times 2$ block with one square missing).

1. Some piece is a whole edge of the grid. The grid has $4$ corner squares. An L fits inside a $2 \times 2$ block, and each of the four $2 \times 2$ blocks of the grid contains exactly one corner square, so an L holds at most one corner. A straight piece through the middle row or middle column holds no corners. Three pieces must hold all four corners, so some piece holds two corners. That piece must be a full edge: the top row, the bottom row, the left column or the right column.

1. Each edge piece leaves 3 ways. Say the top row is a piece. What is left is a block of two rows of three. Look at the piece containing the block's top-left square. It cannot be a straight piece standing up (the block is only two squares tall), so it is either the block's whole top row (and the other row is the last piece), or one of the three L's that fit in the block's left $2 \times 2$ corner. Two of those L's leave a connected L behind; the third cuts off a single square, which cannot be a piece. So the block splits in exactly $3$ ways. Turning the picture shows the same holds for each of the four edges.

4. Count without double counting. Four edges with $3$ ways each gives $12$. But two partitions contain two edge pieces and were counted twice: the three rows (it has both the top and the bottom row) and the three columns. No partition has an edge row and an edge column together, since they would share a corner square. So the total is
$$12 - 2 = 10.$$

Part 2: exactly 4 partitions win.

1. What earns a point. A section has three squares, so "more X's than O's" means at least $2$ X's.

1. A win needs an all-O section. Two points need two sections with at least $2$ X's each, which uses at least $4$ X's. There are only $4$ X's, so the two scoring sections have exactly $2$ X's each, and the third section has no X at all.

7. Find the all-O sections. The O's are top-middle, middle-right and the whole bottom row.
- Top-middle is boxed in by X's on its left, right and below, so it cannot join other O's.
- Middle-right touches only one other O, bottom-right.

So there are just two all-O pieces: the bottom row, and the L made of middle-right, bottom-right and bottom-middle.

8. Case A: the bottom row is a piece. The top two rows split in $3$ ways (step 3):
- top row (X O X) and middle row (X X O): $2$ points, a win;
- {top-left, top-middle, middle-left} (X O X) and {top-right, middle-middle, middle-right} (X X O): $2$ points, a win;
- {top-left, middle-left, middle-middle} (X X X) and {top-middle, top-right, middle-right} (O X O): only $1$ point.

That is $2$ wins.

9. Case B: the L {middle-right, bottom-right, bottom-middle} is a piece. The squares left are the top row, middle-left, middle-middle and bottom-left. Top-right now has only one free neighbor, top-middle, so its piece is top-right, top-middle and one more square: top-left or middle-middle.
- top row (X O X) and {middle-left, middle-middle, bottom-left} (X X O): a win;
- {top-right, top-middle, middle-middle} (X O X) and the left column (X X O): a win.

That is $2$ more wins.

10. The fraction. $4$ winning partitions out of $10$:
$$\frac{4}{10} = \frac{2}{5}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{2}{5}}$.

When a win needs a lot of something scarce, count that resource first. Here two points need four X's, which forces a $2$-$2$-$0$ split and leaves only two cases to check. For the total, the four-corners argument (three pieces cannot cover four corners unless one piece takes two) guarantees that the list of partitions is complete, which is the hardest part of any "list them all" problem.

Score all $10$ partitions directly, grouped by an edge piece they contain. (The sample partition in the problem is the fifth row of the table.)

Four of the ten win, so the fraction is $\frac{2}{5}$. This takes longer than the case argument, but it doubles as a check that the list really has $10$ entries.