Semicircle Hanging Out of a Rectangle Let $ABCD$ be a rectangle so that $AB=CD=2$ and $BC=AD=x$, where $x<1$. A semicircle of radius $1$ and diameter $AB$ is drawn so that the semicircle intersects $CD$ at points $M$ and $N$. If the minor arc between $M$ and $N$ can be fit snugly in $ABCD$ with $MN$ on $AD$ (i.e. the minor arc can be fit tangent to the semicircle and has $MN$ on $AD$) then what is the value of $x$?
Apologies if this is difficult to decipher what I am exactly saying. I do not know how to draw a diagram, but if you have any clarifying questions I will be happy to answer it. 
I have tried rotating the rectangle to form some sort of tangency argument, but to no avail...so like, draw the tangent minor arc fitting snugly in the rectangle as described. Then we "complete the semicircle" by drawing the rest of it underneath the original semicircle. Then we see we can make another copy of $ABCD$ which is a rotation of the first. Darn I wish I knew how to use geogebra lol...
Any ideas?

 A: Not an answer, but I don't think we can put an image in a comment.  Here is an image.  The arc starting at $D$ is supposed to be the same size as the arc $MN$ and the two arcs are tangent.  Now I believe there is enough information to find $x$.

A: I imagine a configuration like the one pictured below.  Algebraically, you have to intersect a pair of conic sections (quadratic equations) to find the various coordinates.  This reduces to solving a quartic equation.  Here's how.
Choose coordinates so that the original semicircle is centered at the origin.  Let $x$ denote the height of the rectangle and $2y$ the length of the chord.  Call the point of tangency $(-u, v)$, so the center of the other circle is at $(-2u, 2v)$.  The given assumptions imply the following system of equations.
$$
\left\{ \begin{align}
x^2 + y^2 &= 1 \\
u^2 + v^2 &= 1 \\
2u &= x + 1 \\
2v &= x - y
\end{align} \right. 
$$

The height of the rectangle (obtained using a computer algebra system) is
$$
x = -\frac{1}{5}+\frac{1}{10 \sqrt{\frac{3}{52+5 \sqrt[3]{7424-384 \sqrt{78}}+20 \sqrt[3]{2 \left(58+3 \sqrt{78}\right)}}}}+\frac{1}{2} \surd \left(\frac{104}{75}-\frac{1}{15} \sqrt[3]{7424-384 \sqrt{78}}-\frac{4}{15} \sqrt[3]{2 \left(58+3 \sqrt{78}\right)}+\frac{304}{25} \sqrt{\frac{3}{52+5 \sqrt[3]{7424-384 \sqrt{78}}+20 \sqrt[3]{2 \left(58+3 \sqrt{78}\right)}}}\right),
$$
which is about $0.924646$.
A: If I understand your question correctly, I think you're saying place the semicircle in a box and cut off the minor arc that doesn't fit, but try to fit it back in the box so that it's tangent to the original semicircle.  Now, in order to do this, you need to translate $M$ and $N$ to $M'$ and $N'$.  Let's say $M'=D$, and $N'$ lies on $\overline{AD}$ so that $\overline{M'N'}=\overline{MN}$.  Now, we essentially have two circles of equal radius that we are trying to move around so they are tangent at one point.  This means that we want their centers to be a distance of $r_1+r_2=2$ apart.  
If we call $O_1$ the midpoint of $\overline{AB}$, let $O_1$ lie on the origin in the Euclidean plane.  Then $M=(-\sqrt{1-c^2},-c)$, $N=(\sqrt{1-c^2},-c)$, and the length of $\overline{MN}$ is $2\sqrt{1-c^2}$.  We then find that, since $M'=(-1,-c)$, $N'=(-1,2\sqrt{1-c^2}-c)$.  Since $M'$ and $N'$ lie on a circle with radius one, we can find the circle center on the perpendicular bisector of segment $\overline{M'N'}$.
Calling the second center $O_2$ and the midpoint of $\overline{M'N'}$ $X$, $\Delta M'XO_2$ is a right triangle with $\overline{M'O_2}=1$, $\overline{M'X}=\sqrt{1-c^2}$, so $\overline{O_2X}=c$.  Thus, $O_2=(-1-c,\sqrt{1-c^2}-c)$.
Now, we solve for $d(O_1,O_2)=2$. $(1+c)^2+(\sqrt{1-c^2}-c)^2=4$.  With some bash, this simplifies to $5c^4+4c^3-4c^2-8c+4=0$, and $c\approx 0.92464593203170915717\ldots$
