2004 AMC 10A Problem 16

Number the rows and columns $1$ through $5$; the black cell is at row $3$, column $3$. A $k \times k$ square is determined by its top row $i$ and left column $j$, each ranging over $1, \dots, 6 - k$. It contains the center exactly when $i \le 3 \le i + k - 1$ and the same for $j$; by symmetry the row and column counts are equal, so the answer for size $k$ is (number of valid $i$)$^2$.

• $k = 1$: $i = 3$ only. $1^2 = 1$.
• $k = 2$: $i \in \{2, 3\}$. $2^2 = 4$.
• $k = 3$: $i \in \{1, 2, 3\}$. $3^2 = 9$.
• $k = 4$: $i \in \{1, 2\}$ (top row at most $2$). $2^2 = 4$.
• $k = 5$: $i = 1$ only. $1^2 = 1$.

Total: $1 + 4 + 9 + 4 + 1 = 19$.

The answer is $\boxed{\textbf{(D)}\ 19}$.

A sub-square is located by one corner, so counting squares is counting corner positions. The containment condition is independent in the two coordinates, which turns each case into a perfect square. The symmetric list $1, 4, 9, 4, 1$ is a good self-check: small squares are constrained by having to reach the center, large ones by having to fit in the grid.

Complementary count: the grid has $25 + 16 + 9 + 4 + 1 = 55$ squares in all. Only $1 \times 1$ and $2 \times 2$ squares can avoid the center: $24$ of the unit squares and $16 - 4 = 12$ of the $2 \times 2$ squares. So $55 - 24 - 12 = 19$.

• Counting all 55 squares in the grid, or forgetting the 4 by 4 squares and getting 15.
• Miscounting the $3 \times 3$ case as $4$ or $5$ instead of $9$; every $3 \times 3$ square in a $5 \times 5$ grid contains the center.