# Inhomogeneous circle packing

In the figure, what is the diameter of the smallest circle assuming the two parallel lines are one unit apart? Note: There is at least one elegant, geometric proof.

Attribution: Mine, but wouldn't be too shocked if I only reinvented it.

The diameter of the smallest circle is

1/4 of the distance between the lines

The configuration is an instance of a

modified by the geometric transformation

Then we apply inversion in the circle centred at R

It is easy to verify that, by the properties of inversion in a circle, circle E is mapped to line OJ (circle E passes through R so it is transformed to a line; circle E is tangent to circle R at D, so the image of circle E must also be tangent to R at D). Similarly, large circle A is transformed to the line ST. Since tangency is preserved, the rest of the circles are transformed into the circles in the configuration given in the problem statement. Circle C is transformed to the smallest circle in the problem statement, and small circle A is transformed to the circle immediately below the smallest circle in the problem statement. Point B is transformed to E halfway along the radius RD, and point Q is transformed to U, one third of the way along the radius RD. That implies that UE is 1/6 of radius RD, so the ratio UE:ED=1/6:1/2=1:3, so UE:UD=1:4.

The diameter of the smallest circle is

$$\frac{1}{4}$$

Let's put the figure in a Cartesian plane: (half of the figure is omitted due to symmetry) Denote the radii of the four circles by $$R_P = \frac12$$, $$R_A$$, $$R_B$$, and $$R_C$$, respectively. We see that $$R_A$$ is the ordinate of $$A$$ and that $$1 - R_B$$ is the ordinate of $$B$$.

Since $$PA = R_A + \frac12$$ is also the distance between $$A$$ and the line $$y = -\frac12$$, $$A$$ lies on the parabola with focus $$P$$ and directrix $$y = -\frac12$$. The equation for this parabola is $$y = \frac12 x^2$$, so we let $$A = (a, \tfrac12 a^2), \quad R_A = \tfrac12 a^2.$$ Similarly, we let $$B = (b, 1 - \tfrac12 b^2), \quad R_B = \tfrac12 b^2.$$ Since $$AC$$ is a vertical line segment of length $$\frac12$$, we get $$C = (a, \tfrac12 a^2 + \tfrac12), \quad R_C = \tfrac12 - \tfrac12 a^2.$$

Now, from $$AB = R_A + R_B$$ and $$BC = R_B + R_C$$, we have the system of equations $$\begin{cases} (a - b)^2 + (\tfrac12 a^2 + \tfrac12 b^2 - 1)^2 = (\tfrac12 a^2 + \tfrac12 b^2)^2, \\ (a - b)^2 + (\tfrac12 a^2 + \tfrac12 b^2 - \tfrac12)^2 = (-\tfrac12 a^2 + \tfrac12 b^2 + \tfrac12)^2, \end{cases}$$ whose positive solution is $$\begin{cases} a = \dfrac{\sqrt{3}}{2}, \\ b = \dfrac{\sqrt{3}}{3}, \end{cases}$$ from where we arrive at the value of $$2 R_C$$.

• Interesting solution! Funnily enough I used the locus of a conic myself a few weeks ago puzzling.stackexchange.com/a/107804/73836. But I had nothing of the kind planned for this puzzle, in fact the intended solution is completely different. I'll disclose it in a few days. In the meantime, let's see what other proofs people can find. Apr 11 at 6:29

For completeness, here is my model solution which is mostly identical to @Edward Doolittle's but uses an ever so slightly more convenient map.

One thing to notice is that there are two objects, the top line and the lower middle circle which don't touch each other but every other object (parallel lines "touch at infinity"). All the others form a ring each touching two others. We just learned from Edward that this is called a Steiner configuration and we can find a map, inversion in a circle, that takes it to a more manageable sister configuration: Inversion in the dashed circle preserves E-W symmetry and maps the straight lines to congruent (because of N-S symmetry) circles. The image therefore must be We are interested in the red circle. The end points of its vertical diameter are 1 and 2 radii from the inversion centre after inversion meaning they must have been 1 and 1/2 before. The diameter is therefore 1/4 the diameter of the inversion circle which happens to equal the distance between the straight lines.