Revival of Sangaku

You have to be careful with the geometry and algebra of this problem! The largest circle (red) has a radius of $\frac{1}{4}$ .

The radii of the upward circles follow the sequence $\frac{1}{2n(n-1)}$ starting from $n = 3$ . Let us call this Class 1.

The radii of the side circles are symmetric at both sides, which follow the sequence $\frac{1}{{n}^{2}}$ , once again starting at $n = 3$ . Let us call this Class 2.

Therefore, the total area is (Class 1) + 2(Class 2), which under close inspection exploits the zeta function and telescoping series. The subsequent answer is $\frac{8{\pi}^{5} + 30{\pi}^{3} - 1035\pi}{360}.$

I didn't know how the radii of a circle in this situation compared to the ones before it in a sequence, so I had to solve this problem a bit different than Steven Zheng described.

I first used the geometry of the problem to generate a set of recursive functions:

For the top circles, define $c_{i} = 1 - 2\Sigma_{j=1}^{i-1}r_{j}$ $r_{i} = \frac{c_{i}^2}{2(1+c_{i})}$ $c_{0} = 1$ $r_{0} = \frac{1}{4}$

For the symmetric circles, define $x_{i} = x_{i-1}\left(\frac{1-\sqrt{r_{i-1}}}{1-r_{i-1}} \right)$ $r_{i} = \frac{x_{i}^2}{4}$ $x_{0} = 1$ $r_{0} = \frac{1}{4}$

Note that $x_{i}$ is the horizontal distance between the closet edge and circle $i$ . One can then use these recursive equations to obtain each $r_{i}$ for the top and symmetric circles. The final equation for the area of the circles is then:

$A_{total} = A_{0} + \Sigma_{i=1}^{\infty}(A_{i}^{t} + 2A_{i}^{s})$ $A_{total} = \pi\left(r_{0}^2 + \Sigma_{i=1,t}^{\infty} r_{i}^{2} + 2\Sigma_{i=1,s}^{\infty} r_{i}^2\right)$

Where $t$ denotes the top circle series and $s$ denotes the symmetric circles series.

One is then able to just compute the series using some programming language until you see no change in the value of $A_{total}$ , as you will have hit the limits of finite precision.

Cheers.