Four Circles In A Parallelogram

In the following figure, circles with centers $A, Z$ have the same radius $a$ , circle with center $B$ has radius $b$ , and circle with center $C$ has radius $c$ . We'll assume that $a=1$ for simplification.

First:

$DA = \sqrt{{((c+a)}^2 -{(c-a)}^2} = 2\sqrt{a c}$
$EB = \sqrt{{((c+b)}^2 -{(c-b)}^2} = 2\sqrt{a b}$
$FB = EB-DA$
$FA=2c -a-b$
$\sqrt{{FB}^2+{FA}^2}=a+b$

which gets us

$c = 2\sqrt{ab}$

Solving for $b$ , we have

$b=\dfrac{c^2}{4a}$

Now, all of the following triangles are similar:

$\Delta CDZ$
$\Delta ZHG$
$\Delta KJC$
$\Delta BNQ$
$\Delta LMA$

where

$\dfrac{CD}{ZD} = \dfrac{c-a}{2\sqrt{ac}} = Tan(\theta) = \dfrac{1}{Cot(\theta)}$

Because $GK = LQ$ , we have

$a Cot( \theta) + 2\sqrt{ac} + c Tan(\theta) = a Tan( \theta) + 2\sqrt{ab} + b Cot(\theta)$

Letting $a=1$ and $b = \dfrac{c^2}{4}$ , and eliminating the denominator, we're left with

$2 c^2 \sqrt{c}- c^2 - 2c\sqrt{c} -3c +1 = 0$

Letting $c = d^2$ , we have

$2d^5 -d^4 -2d^3 -3d^2 +1 = 0$

which factors into

$(2d-1)(d^2+d+1)(d^2 -d -1) =0$

which has only one real solution where $d>1$

$d = \dfrac{1}{2}(1+\sqrt{5})$

$c = d^2 = \dfrac{1}{2}(3+\sqrt{5})$

Since

$Tan(\theta) = \dfrac{c-a}{2\sqrt{ac}}$

we can let $a = 1$ , and find the tangent of twice the angle

$Tan(2ArcTan( \dfrac{c-a}{2\sqrt{ac}})) = \dfrac{4}{3}$

Another nice problem by Mr. Mendrin! For reference, if the radius of the large circle is 1, then the radius of the two smaller blue circles is $\dfrac{3 - \sqrt{5}}{2}$ , and the radius of the purple circle is $\dfrac{3 + \sqrt{5}}{8}$ .

I tried solving the problem properly, but quickly got bogged down in the calculations. I then resorted to adjusting the variables, until the diagram looked right, and guessed. Perhaps Mr. Mendrin can furnish a good solution.

I thought of slope or Tan(half of acute angle) =(1/2) which fits into the scheme. Therefore, Answer=4/3, Formal answer gets you in few triangular and algebraic relations. Did a geometric construction to verify.