How high can a loose rope around a circle pulled above ground Assume we have a circle of radius $r$ and thus circumference of $2 \pi r$. Assume there is a rope $\mu$ fitted around the circumference of the circle.
Now, we increase the length of $\mu$ by one.
It is a well know fact that if the rope is lifted above the circle at every point, ie. we make a new circle whose circumference is $2\pi r + 1$, then the difference between the radii of the old and new circle is a constant, $\frac{1}{2 \pi}.$
This question is often posed so that we have a rope around the equator.
If this scenario was real, I, personally, wouldn't pull it at every point, but rather at a single point. The result would look something like this picture: 
It seems evident that the angle between the ground and the rope would be $\pi/2$, for if it is lower, you can pull more and just the angle would get bigger, but if it were bigger, it would add stress on the ground, which won't move. But my geometry is not so great, so proving this would be nice.
The question is, what is $x$.
Attempt to solve $x$:
We get the following equations: 
$
\left. 
\begin{aligned}
 (r+x)^2 &= r^2+y^2 \ \ \ \ (\text{Pythagorean Theorem})\\
 \frac{\sin(\theta/2)}{y} &= \frac{1}{x+r} \ \ \ \ (\text{Law of Sines}) \\
 2y+r(2\pi-\theta) &= 2\pi r +1 \ \ \ \ (\text{Length of Arc})
\end{aligned} 
\right\} \qquad$
Now, from the third we get $$y = \frac{2\pi r+1-r(2\pi -\theta)}{2} = \frac{1}{2} (1+r \theta)$$
and from the first $$y = \sqrt{(r+x)^2-r^2}.$$
These together give us $$\theta = \frac{2 \sqrt{x (2 r+x)}-1}{r}.$$
Place the last two results to the second and we get $$\frac{\sin \left(\frac{2 \sqrt{x (2 r+x)}-1}{2 r}\right)}{\sqrt{(r+x)^2-r^2}}=\frac{1}{r+x}.$$
This doesn't look like it can be solved analytically.
So, I ask, is there a way to solve $x$ analytically and is the image I drew of the situation correct?
 A: A hint:
It is true that an exact solution of the problem leads to equations that cannot be solved explicitly. But you should use that "in practice" the increment $s$ of the rope (put to $1$ in your setting) is very small compared to $r$, which results in a $\theta\ll1$.
So I suggest to replace your third equation by
$$2y+r(2\pi -\theta)=2\pi r+ s\ ,$$
and then get
$${s\over r}=2\tan{\theta\over2} -\theta\doteq{\theta^3\over12}\qquad(\theta\ll1)\ .$$
Now proceed from here, and avoid taking $\sin$ from complicated arguments.
A: Rather than trying to solve the full system in one hit, let's focus on $y$ and $\theta$.
$$
y=r\tan(\theta/2)
$$
From this, we have
$$
2r\tan(\theta/2) + r(2\pi-\theta)=2\pi r+1
$$
or
$$
\tan\left(\frac{\theta}{2}\right) - \frac{\theta}2 =\frac1{2r}
$$
While there is no nice closed-form solution to this equation, it is much simpler than the direct solution for $x$. And once you have $\theta$, the rest falls into place easily. Specifically,
$$
x = r[\cos(\theta/2)-1]
$$
It's much nicer if we don't have to do another trig function, though, so we write it as
$$
x = \frac{2r}{\sqrt{\left(\frac1r+\theta\right)^2+4}}-r
$$
This does make it less messy, in my opinion.
