Kissing points of circle packing *on a circle*?

Question

Recently, when playing about with circle packings while looking at Kleinian groups and this proof of the Fold and Cut Theorem with a friend, we noticed something intriguing.

It seemed that when we drew four circles (of potentially different sizes), such that they touched each other at 4 points in a "ring-like" formation (like in the image below), that the kissing-points made a cyclic quadrilateral. We then put this into GeoGebra and observed that this was true, seemingly, no matter the sizes of the various circles.

Thus, I have multiple questions:

• Is this true? What's the proof?
• Is this true for "ring-like" formations of 5, 6 or $n$ circles?
• If the above are true, as I suspect, then one might observe that for every circle packing we can naturally construct the inverse circle packing, by replacing each "ring-like formation" in the original circle packing with a single new circle going through their kissing-points. Does this inverse circle packing have any interesting properties?

Answer 1

The observation is true in general because each interior angle of the quadrilateral is the sum of the angles between the adjacent sides and the common tangent of the kissing circles, and the angle between a side and the tangent is half the angle of the arc cut off by that side. That is, if the angles of the shorter arcs between kissing points on each circle (that is, the central angles subtended by the kissing points) are $\alpha, \beta, \gamma, \delta$ in counterclockwise sequence, as illustrated in the figure below, then the interior angles of the quadrilateral are $\frac12(\alpha + \beta),$ $\frac12(\beta+ \gamma),$ $\frac12(\gamma+ \delta),$ and $\frac12(\delta+ \alpha)$ in counterclockwise sequence. We observe that the sum of each pair of opposite angles is $\frac12(\alpha + \beta + \gamma+ \delta),$ and since the two sums are equal, each is $\pi$ and the quadrilateral is cyclic.

It is relatively easy to construct a counterexample for any $n > 4.$ Just place four consecutive kissing circles and construct the circle through the three kissing points. The circle through the kissing points intersects the first and fourth circle at specific points. If $n = 5,$ only one circle is tangent to the first and fourth circles at those two points; choose a different size circle, and its kissing points will not be on the same circle as the first three kissing points. For $n > 5,$ you can make the fifth circle tangent to the fourth anywhere except the point on the circle through the first three kissing points, then follow with as many additional circles as you need in order to build a cycle of $n.$

For the third part, consider four congruent circles in a cycle, whose kissing points form a non-square rhombus. Reflect the figure across the line through the center of two adjacent circles, forming a new cycle of four whose kissing points also form a rhombus. The circumcenters of the two rhombuses will then not be collinear with the single shared vertex of the two rhombuses, and therefore their circumcircles (which intersect at that point) are not tangent. So the result of substituting circumcircles of the sets of kissing points in one "packing" is not another "packing."

Answer 2

The statement is true for $n = 4$.

Let's say we have 4 circles $C_1, C_2, C_3, C_4$ in a ring-link formation with $C_1$ touching $C_2$ at $P$, $C_2$ touching $C_3$ at $Q$, $C_3$ touching $C_4$ at $R$ and finally $C_4$ touching $C_1$ at $O$.

Perform a circle inversion with respect to $O$. Since $C_1$ and $C_4$ are tangent to each other at $O$, they get mapped to a pair of parallel lines $\ell_1$ and $\ell_4$. Let $C_2', C_3', P', Q', R'$ be the image of $C_2, C_3, P, Q, R$ under inversion. Since $O$ doesn't lie on $C_2, C_3$, their image $C_2'$ and $C_3'$ is a pair of circles. Furthermore, line $\ell_1$ is touching $C_2'$ at $P'$, circle $C_2'$ is touching $C_3'$ at $Q'$, circle $C_3'$ is touching $\ell_4$ at $R'$.

Let $S$ and $T$ be the centers of $C_2'$ and $C_3'$. Let $\ell$ be the line passing through $S, Q'$ and $T$. Let $\ell$ intersect $\ell_1$ at $U$. If $\angle P'UQ'$ is $\theta$, it is easy to verify $\angle P'Q'S = \angle R'Q'T = \frac{\pi}{4} - \frac{\theta}{2}$. This implies the segment $P'Q'$ and $Q'R'$ belongs to same line. Since $P', Q', R'$ lie on a line, under inverse circle inversion, they get mapped back to a circle through $O$.

As a result, $P, Q, R, O$ are cyclic.

For $n > 4$, $n-2$ circles get mapped to circles under circle inversion. It is easy to arrange the $n-2$ image circles to make their contact points not lying on any line. This means the statement is false for $n > 4$ circles in a generic "ring-like" configuration.