Inspired by Jeff Chen

Consider a circle $\mathfrak{C}$ with center $P$ and radius equal to the diameter of these congruent circles. Invert with respect to this circle $\mathfrak{C}$ .

Take an original circle $C_i$ . It intersects $\mathfrak{C}$ at a point $P_i$ , and the inverted circle will pass through $P_i$ . Since the point $P$ is inverted to the point at infinity, this means that the circle is inverted into a line that is tangential to the circle $\mathfrak{C}$ at the point $P_ i$ .

We now have a set of 1000 lines that are tangential to a circle $\mathfrak{C}$ . It is clear that no 3 of these lines can intersect at a same point.

Claim: If we have $n$ lines, no 3 of which are concurrent, then the number of regions is equal to $1 + n +$ the number of points of intersection.

Proof: Take a line $\ell$ that is no parallel to any of our $n$ lines. Rotate our system so that line $\ell$ is horizontal. Now, for most of the regions, they will have a "lowest point", which is uniquely determined since none of the $n$ lines is parallel to $\ell$ . This point must be a point of intersection of 2 of the $n$ lines. Conversely, given any point of intersection, it corresponds to a region of which this is the lowest point. This thus establishes the bijection with "number of points of intersection".

The regions that we miss out on, are those with no lowest points. These are regions that are unbounded below. Take a horizontal line that is far below any point of intersection. This horizontal line intersects each of the other lines exactly once. We see that there is a bijection between (almost all) unbounded regions and the right most point on this horizontal line. This estabilishes the bijection with "n". The only region that is missed out, is the region that is unbounded below and unbounded to the right. There is only 1 such region.

Hence, the total number of regions is $1 + n +$ the number of points of intersection. $_ \square$

So, the goal is now to minimize the number of points of intersection. Of course, the best that we want, is for the lines to be parallel to each other. However, given that these lines are tangential to $\mathfrak{C}$ , each line is parallel to at most 1 other. Hence, each of the 1000 lines must intersect at least 998 lines, and so by double counting there are at least $\frac{ 1000 \times 998} { 2} = 499000$ points of intersection.

Thus, from our lemma, there are at least $1 + 1000 + 499000 = 500001$ regions. This can be achieved with 500 pairs of circles where each pair intersect only at $P$ .