Braided Geometrical Sequences of Circles

Suppose that distance of the points of tangency of the two circles of radius $r$ and $R$ from the vertex are $x$ and $X$ respectively. Considering the two right-angled trapezia (marked in orange) we deduce that $X\sqrt{2} - x \; = \; x\sqrt{2} - X \; = \; 2\sqrt{\sqrt{2}rR} \; = \; 2^{\frac54}\sqrt{rR}$ so that $x \; = \; X \; = \; \frac{2^{\frac54}}{\sqrt{2}-1}\sqrt{rR} \hspace{6.5cm} (1)$ Considering the two pairs of congruent right-angled triangles (marked with dotted lines) we deduce that $\theta \; = \; 2\tan^{-1}\big(\tfrac{r}{X}\big) + 2\tan^{-1}\big(\tfrac{R}{X}\big) \hspace{5.5cm} (2)$ The fact that the line $PQ$ has length $R(1+\sqrt{2})$ means, by choosing a suitable coordinate system, that $X^2(3 - 2\sqrt{2}\cos\theta) - 4\sqrt{2}XR\sin\theta \; = \; 2\sqrt{2}R^2(1 - \cos\theta) \hspace{3cm} (3)$ Equation (1) tells us that the circles are all tangent to the two straight lines. Equation (2) tells us that a pair of red and green circles bridges the gap between the two lines, while equation (3) ensures that successive green circles are tangent with each other.

Eliminating variables from equations (1), (2) and (3) gives us a quadratic equation for $p = \big(\tfrac{R}{X}\big)^2$ , namely $128\sqrt{2}p^2 - 32(3 -2\sqrt{2})p - (3-2\sqrt{2}) \;=\: o$ which has one positive and one negative real root, and hence there is a unique positive value for $\tfrac{R}{X}$ . Moreover $\tfrac{r}{X}$ and $\theta$ are determined uniquely by $\tfrac{R}{X}$ . Solving we obtain $\frac{R}{X} \; = \; \frac14\sqrt{3\sqrt{2} - 4 + \sqrt{30 - 21\sqrt{2}}}$ and so $\tfrac{r}{X} \; = \; 0.136346 \hspace{2cm} \tfrac{R}{X} \; = \; 0.22245 \hspace{2cm} \theta \; = \; \boxed{40.6108^\circ}$

Let $a, 1, ar, r, ar^2, r^2...$ be the radii of the circles, where $r$ is the geometric ratio, here equal to $\frac{1}{\sqrt{2}}$ .
Let $s$ be the slope of the line between centers of circles $1, r$ .
The distances between the points where circles $a, r$ and $1, ar$ are tangent to the wedge are $2\sqrt{ar}$ .
The distances between the points where circles $r, ar^2$ and $ar, ar^3$ are tangent to the wedge are $2r\sqrt{ar}$ .
Etc. Hence, the triangles formed by the apex of the wedge and where pairs of circles are tangent to it are isosceles.
By symmetry, the tangents of circles $1, r, r^2,...$ are colinear.
Given an arbitrary slope $s$ , the coordinates of centers of circles $1, r, r^2, ...$ can be found.
From this, we can find the distance between the tangents of circles $1, r, r^2..$ to the wedge can be found.
From this, the radius $a$ can be determined, and the coordinates of the circle $ar$ can be found.
From this, the distance between centers of circles $r, ar$ can be found, and compared with length $r + ar$ .
By such means, an exact answer for $s$ is possible.
Once we have $s$ , we can find the apex angle of the wedge.