How to approximate a parameter that gives a tangent line to three circles?

Question

Three disks are placed on the ground like this:

From left to right, their radii are $\frac{1}{x-1}, \frac{1}{x}, \frac{1}{x+1}$ metres. They lie in a plane perpendicular to the ground. The middle disk touches the other two disks.

Using only paper and pen, approximate the value of $x$ such that the middle disk is tangent to the line that is tangent to and above the other two disks. You may assume that the earth is a sphere of radius $R$ metres.

(Before you read the last sentence, it seems like there's something wrong with the question, because it seems like the middle disk should never touch the line. But the ground is actually a circular arc of the earth, so the middle disk is "pushed up" and touches the line for some value of $x$.)

The answer turns out to be, elegantly, $x\approx R/2$. But the algebra seems to be horrendous and I needed to use my computer to find the answer.

My attempt

Call the angles at the centre of the middle disk $A, B, C, D, E$ with $A$ at the lower-left and going clockwise.

$A=\arccos{\left(\dfrac{\left(\frac{1}{x}+\frac{1}{x-1}\right)^2+\left(R+\frac{1}{x}\right)^2-\left(R+\frac{1}{x-1}\right)^2}{2\left(\frac{1}{x}+\frac{1}{x-1}\right)\left(R+\frac{1}{x}\right)}\right)}$

$B=\arcsin{\left(\dfrac{\frac{1}{x-1}-\frac{1}{x}}{\frac{1}{x-1}+\frac{1}{x}}\right)}$

$C=\dfrac{\pi}{2}$

$D=\arccos{\left(\dfrac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}\right)}$

$E=\arccos{\left(\dfrac{\left(\frac{1}{x}+\frac{1}{x+1}\right)^2+\left(R+\frac{1}{x}\right)^2-\left(R+\frac{1}{x+1}\right)^2}{2\left(\frac{1}{x}+\frac{1}{x+1}\right)\left(R+\frac{1}{x}\right)}\right)}$

We assume that the middle disk is tangent to the line that is tangent to and above the other two disks. This implies:

$$A+B+C+D+E=2\pi$$

I am utterly unable to approximate $x$ without a computer, even after attempting to simplify it. And yet the computer-assisted answer is just $x\approx R/2$. Can $x$ be approximated without a computer?

(This question was inspired by a frame challenge.)

Answer 1

They simplify to $$A=\frac\pi2+\arcsin\frac{\left(R+\frac1x\right)-2}{\left(R+\frac1x\right)(2x-1)}\\ B=\arcsin\frac1{2x-1}\\ C=\frac\pi2\\ D=\frac\pi2-\arcsin\frac1{2x+1}\\ E=\frac\pi2-\arcsin\frac{\left(R+\frac1x\right)+2}{\left(R+\frac1x\right)(2x+1)} $$ Suppose $x=O(R)$.
All four key angles turn out to be $O(1/R)$, which is small enough that I can ignore the arcsine at the precision I need, which is $O(1/R^3)$. Since the $O(1/R)$ terms cancel exactly, the nonlinear $O(1/R^3)$ terms in arcsines will cancel as well.
$$A\approx\frac\pi2+B-\frac2{\left(R+\frac1x\right)(2x-1)}\\ B+D\approx\frac\pi2+\frac2{4x^2-1}\\ E\approx D-\frac2{\left(R+\frac1x\right)(2x+1)}\\ A+B+C+D+E\approx2\pi+\frac4{4x^2-1}\left(1-\frac{2x}{R+\frac1x}\right)$$ So my leading order estimate is $$x\approx\frac R2+\frac1R$$

Answer 2

Using the solution given by @Blue in comments, it is not difficult to solve exactly the cubic equation in $y$ provided that we use the hyperbolic solution.

Using only pen and paper, the exact result is $$\color{blue}{y=\frac 16 +\frac{\sqrt{r^2+6}}{3 r}\cosh \left(\frac{1}{3} \text{sech}^{-1}\left(\frac{2 r \left(r^2+6\right)^{3/2}}{2 \left(r^2+9\right) r^2+27}\right)\right)}$$ which must be very close to an hyperbola.

Using a computer, expanding for large values of $r$

$$y=\frac 12 + \frac 1 {r^2}\left( 1-\frac{3}{2 r^2}+\frac{5}{r^4}+O\left(\frac{1}{r^6}\right)\right)$$ that is to say $$x=\frac r2 + \frac 1 {r}\left( 1-\frac{3}{2 r^2}+\frac{5}{r^4}+O\left(\frac{1}{r^6}\right)\right)$$

as already given by @Empy2 using simple and legitimate approximations.