The diameter of a circle is a chord subtending an obtuse central angle in another. How to inscribe a square in the composite boundary? Below is a curve (outlining the red and blue shaded areas) formed by impacting two circles (red and blue) of different diameters into one another. 

$AB$ is the diameter of the red shaded circle and $O$ is the centre of the blue circle and $\angle AOB$ is greater than $90^\circ$.  

How does one find an inscribed square in this composite curve so that it is not a counterexample of the Toeplitz conjecture that every simple closed curve admits such a square? (See the Inscribed Square Problem.)

This second image is an approximate inscribed square (yellow) that I eyeballed, but how would I find it or/ prove that it is correct mathematically?

 A: I think in this case a straightforward geometric construction and calculation can allow you to find an inscribed square symmetric with respect to the line passing through the centers of the two circles (like the yellow one on your second picture). 
Since the curve formed by the two circles is symmetric with respect to the line passing through the centers of the circles, it makes it easy to look for a square that is also symmetric with respect to this line. If $M$ is the midpoint of $AB$, then $M$ is the center of the red circle. Let the line $MO$ intersects the exterior red circular arc at point $C$ and the exterior blue arc at point $D$. If you have the length of $AB$ and the angle $\alpha$, the radius of the red circle is 
$r = \frac{|AB|}{2}$ and the radius of the blue one is $R = \frac{|AB|}{2\sin(\frac{\alpha}{2})}$. The distance between the centers of the two circles is then 
$h = \frac{|AB|}{2}\cot(\frac{\alpha}{2})$. Therefore $CD = r + R + h$. The vertices of the inscribed square in question are denoted by $E, F$ on the red circle and $K, L$ on the blue one, where $E$ and $F$ are symmetric with respect to $CD$ and $K$ and $L$ are also symmetric with respect to $CD$. Thus, $CD$ is the orthogonal bisector of both chords $EF$ and $KL$. Let $G = EF \cap CD$ and $H = KL \cap CD$. Then denote the segments $CG = x$ and $DH = y$. If you want $E,F,K,L$ to be positioned so that they form a square, then you have to calculate what $x$ and $y$ should be and what the edge-length $a$ of the square should be. The intersecting chord theorem applied to the line $CD$ and the symmetric chords $EF$ and $KL$ respectively, yields the following system of three quadratic equations
\begin{align}
&x(2r - x) = \frac{a^2}{4}\\
&y(2R-y) = \frac{a^2}{4}\\ 
&r + h + R - x - y = a 
\end{align}
When you solve it for $x, y, a$ you get the square you want.
