Circles overlapping a central point

Question

If I have a circle x with radius r. How many circles can I add around it with same radius such that these circles overlap the center point of circle x without overlapping any other circles' center point? Here is a valid example with five circles. Is this a known problem?

Answer 1

If you "just barely" allow the circles to contain their neighbors' centers (i.e., the neighbor center lies on the boundary circle, rather than inside the disk that the circle bounds), then the answer is "1 central circle, 6 more around it". (Your example shows one central and four around it).

If you say "nope, you really have to MISS the other circles' centers", then the answer is "one central circle, 5 around it."

Reason: for two of the "surrounding" circles to have just-overlapping centers requires that the angle A-O-B, where $A$ is the center of one, $B$ the center of the other, and $O$ the center of the drawing, be exactly 60 degrees, so you can fit at most 6. If you insist on non-overlapping, then the angle between adjacent surrounding circles must be more than 60, and hence there can be no more than 5 of them.

Added details: Let's assume that the radii of all circles is 1.

I'll use "circle" to mean "points at distance 1 from the center" and disk to mean a "filled-in" circle, i.e., points at distance less than or equal to 1 from the center.

Call the original circle $C$, and the surrounding circles $D_1, D_2, \ldots$, with centers $d_1, d_2, \ldots$. Call the center of the original circle $c$.

Since each surrounding circle $D_i$ contains $c$, the distance from $d_i$ to $c$ is $1$; that means that $d_i$ is a point of $C$. So all "surrounding circle" centers are on $C$.

Each point of $C$ is at some angle, measured counterclockwise from the $x$-axis. That means that to each circle-center $d_i$, there's an associated angle $a_i$ between $0$ and $2\pi$. By renumbering, we can assume that $$ 0 \le a_1 \le a_2 \le \ldots < 2\pi $$

Observation 1: No two surrounding circles have the same center. For if they did, then the center of each would be in the disk corresponding to the other, which is not allowed. Thus we know that $$ 0 \le a_1 < a_2 < \ldots < 2\pi. $$

Observation 2: If $a_{i+1} - a_i \le 2\pi/6$, then the center $d_{i+1}$ of circles $i_1$ lies within the disk for circle $i$.

Reason: consider the triangle with vertices $c, d_i, d{i+1}$. The edges $cd_i$ and $cd_{i+1}$ are both length 1, hence the circle is isoceles. The angle at $c$ is exactly $a_{i+1} - a_i$, which we are assuming is less than $2\pi/6 = \pi/3$, hence the third side has length less than one. But that means that the distance from $d_{i+1}$ to $d_i$ is less than $1$, i.e., that $d_{i+1}$ is within the radius-1 disk around $d_i$, as claimed.

Since the hypotheses tell us that no surrounding-circle-center lies within another surrounding-circle's disk, we know that $a_{i+1} - a_i$ is greater than $\pi/3$ for every $i$. Thus the angles must all satisfy relations like these: $$ a_2 > a_1 + \pi/3 \\ a_3 > a_2 + \pi/3 > a_1 + 2\pi/3 \\ a_4 > a_3 + \pi/3 > a_1 + 3\pi/3 \\ a_5 > a_4 + \pi/3 > a_1 + 4\pi/3 \\ a_6 > a_5 + \pi/3 > a_1 + 6\pi/3 $$ Now if there were six angles, we'd have $$ 0 < a_1 < \ldots < a_6 < 2\pi $$ But since $a_6 > a_1 + 2\pi$, from the chain of inequalities above, and $a_1 > 0$, but the line just above, we know that $$ a_6 > 0 + 2\pi = 2\pi $$ which contradicts the conclusion that $a_6 < 2\pi$. Hence there cannot be as many as six surrounding-circle-centers.