Question: Let $n$ be a positive integer. Consider a square $S$ of side $2n$ units with sides parallel to the coordinate axes. Divide $S$ into $4n^2$ unit squares by drawing $2n−1$ horizontal and $2n−1$ vertical lines one unit apart. A circle of diameter $2n−1$ is drawn with its centre at the intersection of the two diagonals of the square $S$. How many of these unit squares contain a portion of the circumference of the circle? Source
My answer:
It's obvious that we are basically talking about the incircle of the square. So as it's four quarters are symmetrical, if we find the number of squares lying on the circumference of just one quarter, multiplying it with four must give us our desired result.
Considering any one quarter, we observe that the circle touches the side of the square at the mid-point, so number of squares touching the circumference will be equal to the number of columns falling in half of the square, which is $4 (\frac{2n-1}2)= 4n-2$.
I find my solution doubtful; Am I correct?
Any alternate solutions will be much appreciated.
Best Answer
For example, here is the case $n=10$, with squares containing a piece of circumference highlighted.
Notice that some columns contain one highlighted square in the first quadrant, others two, three or four.
But note that each highlighted square in the first quadrant is obtained from the previous one by moving either one unit right or one unit down. And the total number of steps is ...