Applonius Circle/ Ford Circle / Infinite GP / Circle Packing

Question

All the smaller circles are mutually tangent and continue to infinity. What is sum of radii of all the smaller circles?

Answer 1

Finding the radius of the first circle

Initially you have the outer circle (of diameter $k+1$), the right inner circle (of diameter $k$), and the horizontal line. Placing my origin into the point to the very right, you can formulate coordinates in Lie sphere geometry for these three circles.

\begin{align*} C_o &= \begin{pmatrix} k + 1 \\ 0 \\ -1 \\ 1 \\ k + 1 \end{pmatrix} & C_k &= \begin{pmatrix} k \\ 0 \\ -1 \\ 1 \\ -k \end{pmatrix} & C_h &= \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{align*}

Now you are looking for the first circle to the left, which is a circle tangent to the three given objects. You look for an object on the Lie quadric which also has vanishing inner product with these three vectors. The solution looks like this, according to my computer algebra system:

$$C=\begin{pmatrix} 8 \, k^{3} + 12 \, k^{2} + 4 \, k \\ 4 \, k^{2} + 4 \, k \\ 4 \, k^{4} + 8 \, k^{3} - 4 \, k - 1 \\ 4 \, k^{4} + 8 \, k^{3} + 8 \, k^{2} + 4 \, k + 1 \\ 4 \, k^{2} + 4 \, k \end{pmatrix}$$

The point here is that you can scale this vector such that $C_4-C_3=1$ and then you can read off the radius of your first circle on the left from the last coordinate $C_5$ as

$$ r_1 = \frac{2\,(k^2 + k)}{4\,k^2 + 4\,k + 1} $$

Finding the other radii

Now that you have that first radius, you can obtain others using Descartes' theorem:

$$ \frac1{r_{i+1}} = \frac1{r_i} + \frac2k - \frac2{k+1} \pm 2\sqrt{\frac2{r_i\,k} - \frac4{k\,(k+1)} - \frac2{r_i\,(k+1)}} $$

Of the two solutions, you'd want the one where you add the square root, since that will lead to a smaller result and hence result in the right iteration direction. The other solution would iterate in the opposite direction.

Iterating

Of course, the above doesn't tell you the solution for that infinite sum of radii you were asking about. But it should give you a good start. If someone wants to update this answer with a suitable computation for this, feel free to do so. Or write your own answer building on this.

Until then, you can obtain numeric results for a given $k$ using the above, even without a closed formula for the final sum. For example, I got the following sums using simple iteration with 20,000 steps for each $k$:

Answer 2

This is similar to the Pappus chain in the arbelos and can be answered with circle inversion the same way as that problem is analyzed at Cut The Knot.

Inversion of the figure through a circle centered at $A$ through $C$ (dashed red) takes the semicircles on $AB$ and $AC$ to parallel lines perpendicular to $AC$. The images of all of the inscribed circles are tangent to both lines and so are all congruent.

Given $AB=k, BC=1$, the radius of inversion is $AC=k+1$. Hence the image of $B$ satisfies $$ AB\cdot AB' = AC^2 = (k+1)^2\\ AB' = \frac{(k+1)^2}{k} $$ and the common radius of the image circles is $$ r' = CM=\frac{AB'-AC}{2} = \frac{(k+1)^2-k(k+1)}{2k} = \frac{k+1}{2k} $$

Let the inscribed circles have centers at $O_1,O_2,\ldots$ with radii $r_1,r_2,\ldots$, and their images have centers at $O_1', O_2',\ldots$ with common radius $r'$. Then since all image circles are tangent along the line $MO_1'$ it is immediate that $MO_n' = (2n-1)r'$.

Take a common tangent to the $n$th circle and its image meeting them at $T_n$ and $T_n'$ respectively. Then $AT_nO_n$ is similar to $AT_n'O_n'$ and hence $$ r_n = r' \frac{AT_n}{AT_n'} $$ but $T_n$ and $T_n'$ are also inverses and so satisfy $$ AT_n \cdot AT_n' = AC^2 = (k+1)^2 $$ and by the Pythagorean theorem $$ (AT_n')^2 + (r')^2 = (AO_n')^2 = AM^2+(MO_n'^2) \\ (AT_n')^2 = -\left(\frac{k+1}{2k}\right)^2+\left(\frac{(k+1)(2k+1)}{2k}\right)^2+\left(\frac{(2n-1)(k+1)}{2k}\right)^2 \\ (AT_n')^2 = \frac{(k+1)^2}{4k^2}\left(-1+(2k+1)^2+(2n-1)^2\right) $$ and putting it all together $$ \begin{align} r_n &= r' \frac{AT_n}{AT_n'} \\ &= \frac{k+1}{2k} \frac{(k+1)^2}{(AT_n')^2} \\ &= \frac{2k(k+1)}{4k(k+1)+(2n-1)^2} \end{align} $$

Then with help from WolframAlpha the sum of the radii is $$ \begin{align} \sum_{n=1}^\infty r_n &= 2k(k+1) \sum_{n=1}^\infty \frac{1}{(2n-1)^2+4k(k+1)} \\ &= 2k(k+1) \frac{\pi \tanh(\pi \sqrt{4k(k+1)}/2)}{4\sqrt{4k(k+1)}}\\ &= \frac{\pi}{4}\sqrt{k(k+1)} \tanh(\pi\sqrt{k(k+1)}) \end{align} $$

QED.

Let $P_1$ be the foot of the perpendicular from $O_1$ to $AC$. As a check we can observe that the semicircle on $AP_1$ that is the inverse of the ray $MO_1'$ (show green dashed in the diagram) passes through the points of tangency between the inscribed circles and near their centers, so is an approximation (and lower bound) for twice the sum of the radii.

Since $AP_1O_1$ is similar to $AMO_1'$ $$ AP_1 = AM\frac{r_1}{r'} = \frac{2k(k+1)}{2k+1} $$ and half the arc length gives the bound $$ \sum_{n=1}^\infty r_n > \frac{\pi AP_1}{4} = \frac{\pi k(k+1)}{4k+2} $$ for example, when $k=50$ this bound gives $\sum r_n>39.6587\cdots$, whereas the exact value from the expression above is $\sum r_n=39.6606\cdots$.