2016 AMC 10B Problem 14

The region is the triangle-like area with $y \ge -0.1$, $x \le 5.1$ and $y \le \pi x$. For lattice squares the first two conditions simply say $y \ge 0$ and $x \le 5$.

Describe a square by its lower-left corner $(a, b)$ and side $s$. Since the line $y = \pi x$ rises to the right, the corner of the square most likely to cross it is the top-left corner $(a, b+s)$. So the square fits exactly when
$$b \ge 0, \qquad a + s \le 5, \qquad b + s \le \pi a.$$
Because $b + s$ is an integer, the last condition is $b + s \le \lfloor \pi a \rfloor$, and $\lfloor \pi a\rfloor = 3, 6, 9, 12$ for $a = 1, 2, 3, 4$ (and $a = 0$ gives nothing).

For each $a$ and $s$ the number of valid $b$ is $\lfloor \pi a \rfloor - s + 1$ when positive:

• $s = 1$: $a = 1,2,3,4$ give $3 + 6 + 9 + 12 = 30$.
• $s = 2$: $a = 1,2,3$ give $2 + 5 + 8 = 15$.
• $s = 3$: $a = 1,2$ give $1 + 4 = 5$.
• $s = 4$: $a = 1$ would need $b + 4 \le 3$, impossible.

Total: $30 + 15 + 5 = 50$.

The answer is $\boxed{\textbf{(D)}\ 50}$.

"Lies entirely within" a region cut by a line only needs to be checked at the one vertex that is extreme in the direction of the constraint; for a line with positive slope that is the top-left corner. After that the problem is pure counting, organized by side length so each case is an arithmetic sum. The odd numbers $-0.1$ and $5.1$ are there just to make the boundary lattice points allowed.

• Counting only unit squares, which gives $30$ (choice A); squares of side $2$ and $3$ also fit.
• Testing the top-right corner $(a+s, b+s)$ against the line instead of the top-left one, which admits squares whose left edge sticks out above the line.