<s>[INST] Geometry problem involving infinite number of circles What is the sum of the areas of the grey circles? I have not made any progress so far.

 [/INST] Consider the picture below.

On the left hand side we have the original picture which we have put in the complex plane, and on the right hand side is its image under the Möbius transformation $f(z) = \frac{z}{2-z}$. The inverse transformation is $g(z) = \frac{2z}{z+1}$. In the new coordinates the green circles have radius $\frac{1}{2}$ and center at $C_k = \frac{1}{2} + i(k + \frac{1}{2})$, $k=0,1,2,\dots$.
We have $\text{Jac}\, g(z) = |g'(z)|^2 = \frac{4}{|z+1|^4}$. Therefore the area of the disks in the original picture is
$$\sum_{k=0}^\infty \int_{B_k} \frac{4}{|(x+iy)+1|^4} \, dx \, dy,$$
where $B_k = \{z : |z - C_k| \le \frac{1}{2}\}$ are the discs in the second picture. Calculating the integral inside the sum turned out to be tedious, however, so I opted for another route. EDIT: achille hui showed how to calculate this integral in an answer to my question at Integral related to a geometry problem. This yields a shorter way to get the answer.
The points where a green circle touches the red line or the blue line are at $i(k + \frac{1}{2})$ and $1 + i(k + \frac{1}{2})$ respectively. Therefore in the original picture they are at
$$A = g(i(k+\frac{1}{2})) = \frac{2(k+\frac{1}{2})^2}{1 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{1 + (k + \frac{1}{2})^2}$$
and
$$B = g(1 + i(k + \frac{1}{2})) = \frac{4 + 2(k+\frac{1}{2})^2}{4 + (k + \frac{1}{2})^2} + i \frac{2(k+\frac{1}{2})}{4 + (k+\frac{1}{2})^2}.$$
Now to find the center of the green circle we calculate the intersection of the lines $1 + t(A-1)$ and $\frac{3}{2} + s(B - \frac{3}{2})$. Real and imaginary part give us two linear equations for $s$ and $t$, and we end up with the solution
$$s = \frac{4k^2 + 4k + 17}{4k^2 + 4k + 9}, \quad t = \frac{4k^2 + 4k + 5}{4k^2 + 4k + 9}.$$
Thus the center is at
$$1 + t(A-1) = \frac{8k^2 + 8k + 6}{4k^2 + 4k + 9} + i \frac{8k + 4}{4k^2 + 4k + 9}.$$
The radius is then
$$|A - 1 - t(A-1)| = |A-1| |t-1| = \frac{4}{4k^2 + 4k + 9}.$$
Hence the answer to the problem is
$$\sum_{k=0}^\infty \pi \frac{16}{(4k^2 + 4k + 9)^2},$$
for which Wolfram Alpha gives the closed form
$$\frac{1}{16} \pi^2 (\sqrt{2} \tanh(\sqrt{2}\pi) - 2 \pi \text{sech}^2(\sqrt{2}\pi)) \approx 0.8699725\dots$$
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