HMMT-Nov 2010 guts, Problem 2

A circle of radius 6 is drawn centered at the origin. How many squares of side length 1 and integer coordinate vertices intersect the interior of this circle?

Solution: By symmetry, the answer is four times the number of squares in the first quadrant. Let's identify each square by its coordinates at the bottom-left corner, $(x, y)$. When $x=0$, we can have $y=0 \ldots 5$, so there are 6 squares. (Letting $y=6$ is not allowed because that square intersects only the boundary of the circle.) When $x=1$, how many squares are there? The equation of the circle is $y=\sqrt{36-x^{2}}=\sqrt{36-1^{2}}$ is between 5 and 6 , so we can again have $y=0 \ldots 5$. Likewise for $x=2$ and $x=3$. When $x=4$ we have $y=\sqrt{20}$ which is between 4 and 5 , so there are 5 squares, and when $x=5$ we have $y=\sqrt{11}$ which is between 3 and 4 , so there are 4 squares. Finally, when $x=6$, we have $y=0$, and no squares intersect the interior of the circle. This gives $6+6+6+6+5+4=33$. Since this is the number in the first quadrant, we multiply by four to get 132 . $\fbox{132}$.

Answer: 132