Computing $r$ for a $1\times r$ strip inside (cut from) a $1\times (1-r)$ rectangle 
Cut a $1\times r$ strip from a $1\times 1$ square and place it inside the remaining $1\times (1-r)$ rectangle such that every corner of the strip touches the interior of a different side of the rectangle. Compute $r$.

I tried to solve this problem naively by assigning variables to the relevant lengths and investigating the relationships among related sides of similar triangles and whatnot, but swiftly found myself mucking about in a bizarre morass of awkward equations in too many variables- it doesn't even seem worth describing the ugly results I got (though I can add it in an edit if anyone thinks it would help). I asked a friend of mine who's very good at math, and she also wound up with a shocking number of equations and (more worryingly, in my opinion) variables. I then tried to draw it manually in GeoGebra and a web version of the Geometer's Sketchpad$^1$, and got nowhere with either of those either. Can anyone solve this at least semi-elegantly, without extendedly hacking through a forest of bizarrely many equations and variables?
I came up with this problem spontaneously a few days ago, while considering paper-folding problems from my old high school's math contests. I'm pretty confident that there should be only one solution (as we let the remaining rectangle get wider, the cut strip should also get wider, with a crossover point when the sum of their widths is exactly $1$), but beyond that I really don't know.
$^1$ Is there a way to get a full version of the application anymore? I feel like I unfortunately got into the idea of the application just after the makers stopped developing it/allowing licenses, but I think it could still be a really cool tool if I could access the first version.
 A: 
Let $\theta$ be the angle that the strip is rotated, which is the angle between the $1$-sides of the strip and the remaining rectangle.
Consider the horizontal and vertical sides of the outer rectangle,
$$\left\{\begin{align*}
r\sin\theta + 1\cos\theta &= 1 &&\text{(Horizontally)} (1)\\
r\cos\theta + 1\sin\theta &= 1-r && \text{(Vertically)} (2)
\end{align*}\right.$$
WolframAlpha solves the system of equations and gives one solution in valid range, and another invalid solution where a $1\times \frac12$ strip touching all four corners of the remaining rectangle:
$$(r,\theta) = \left(2-\sqrt 3, \frac\pi6\right)\text{ or } \left(\frac12, 0\right)\text{ (rejected)}$$

Solving the equations by hand: scaling eq. $(1)\times(1+\cos\theta)$ and eq. $(2)\times \sin\theta$, then eliminating $r$,
$$\begin{align*}
r\sin\theta &= 1-\cos\theta & r+r\cos\theta &= 1-\sin\theta\\
r\sin\theta(1+\cos\theta) &= 1-\cos^2\theta & r(1+\cos\theta)\sin\theta &= (1-\sin\theta)\sin\theta\\
\end{align*}$$
$$\begin{align*}
1-\cos^2\theta &= (1-\sin\theta)\sin\theta\\
\sin^2\theta &= (1-\sin\theta)\sin\theta\\
(2\sin\theta - 1)\sin\theta &= 0\\
\end{align*}$$
$$\begin{align*}
\sin\theta &= \frac12 &&\text{or} & \sin\theta &= 0\\
\cos\theta &= \frac{\sqrt3}2 &&\text{or} & \cos\theta &= 1\\
r &= 2-\sqrt3 &&\text{or} & r &= \frac12
\end{align*}$$
