Given a band of $m$ opaque squares arranged in a circle, can we find a viewpoint from which we see exactly $m/2-1$ squares? 
Given a band of $m\ge 3$ opaque squares arranged in a circle, can we find a viewpoint (i.e. a point on a sphere centered at the midpoint of the circle with a radius large enough to see the whole band from any viewpoint; the viewing direction is the vector from the viewpoint to the center of the sphere) from which we see exactly $\lfloor m/2\rfloor$ squares? Moreover, can we find a viewpoint from which we see exactly $\lfloor m/2\rfloor-1$ squares?

 
I assume that this is an easy to answer question, but I've got problems in arguing. Clearly, each square contributes $2\pi/m\text{ rad}$ to the angle sum of the circle. I don't know how I need to phrase this, but it's also clear that we can see any square whose normal $n$ takes an angle in $(-\pi/2,\pi/2)$ with $-v$, where $v$ is the viewing direction. So, I guess the answer has something to do with $$\frac{2\pi\text{ rad}}m\stackrel!=\frac{\pi\text{ rad}}k\Leftrightarrow k=\frac m2\tag 1\;.$$
But as I said before, I need help to find a rigorous argumentation.
 A: The outward normal vectors point in directions $\frac{2\pi k}{m}$ ($k = 0, \ldots, m-1$), as you observe. Let's assume that the face-centers are at locations 
$$
P_k = (\cos(2\pi k/m), \sin (2 \pi k / m))
$$
as well, i.e., on a unit circle. 
Letting $\alpha = \frac{2\pi}{m}$, we have
$$
P_k = (\cos k\alpha, \sin k \alpha) \\
n_k = [\cos k\alpha, \sin k \alpha]
$$
where $n_k$ is the normal to the $k$th face. 
Consider a view from a point far along the ray in direction $\beta = \frac{\pi}{2} + \frac{\alpha}{2}$, i.e.
$$
Q = (s \cos \beta, s \sin \beta)
$$
Now face $k$ is visible from $Q$ if $ (Q - P_k) \cdot n_k > 0$, which simplifies to 
$$
Q \cdot n_k > P_k \cdot n_k.
$$
Since $P_k$ and $n_k$ are the same (in coordinates) and are points of the unit circle, this condition becomes 
$$
Q \cdot n_k > 1.
$$
And then
$$
Q \cdot n_k  = [s \cos \beta , s \sin \beta] \cdot  [\cos k\alpha, \sin k \alpha]\\
=  s \cos \beta \cos k \alpha + s \sin \beta \sin k \alpha.
$$
Letting $u = k \alpha$ for a moment, this is 
$$
Q\cdot n_k  =   s(\cos \beta \cos u +  \sin \beta \sin u)\\
= s \cos (\beta - u)
$$
As long as $\beta - u$ has positive cosine, we can choose $s$ large enough to make this number $> 1$. Let's look at $\beta - u = \beta - k \alpha$. 
For the cosine to be positive, we need
\begin{align}
-\pi/2 &< \beta - k \alpha < \pi/2 \\
-\pi &< \pi + \alpha - 2k \alpha < \pi \\
-\pi &< \pi + (1-2k)\alpha  < \pi \\
-\pi &< \pi + (1-2k)\frac{2\pi}{m}  < \pi \\
-1 &< 1 + (1-2k)\frac{2}{m}  < 1 \\
-m &< m + 2(1-2k)  < m \\
-2m &< 2(1-2k)  < 0 \\
0 &< 2(2k-1)  < 2m \\
0 &< 2k-1  < m 
\end{align}
which has the $\frac{m}{2} - 1$ solutions $k = 1, 2, \ldots, \lfloor \frac{m}{2} \rfloor$. 
A: Adapting the idea of John Hughes, I came up with the following solution: Identifying $\mathbb C\cong\mathbb R^2$, we can assume that the centers of the squares are at the locations $$c_k:=e^{k\alpha\rm i}\;\;\;\text{for }k\in\left\{0,\ldots,m-1\right\}$$ with $\alpha:=2\pi/m$, i.e. the centers of the squares are the $m$th roots of unity lying on the unit circle. The corresponding outward unit normals are $n_k:=c_k\color{blue}{-0}$. Let $v$ be a view point at angle $\beta:=\pi/2$, i.e. $$v:=ce^{\frac\pi2{\rm i}}$$ for some $c>0$ large enough to see the entire band (but since that's not important, we will assume $c=1$). Notice that the viewing direction corresponding to $v$ is $0-v$ (again, that's not important, but nice to know). We can see the $k$th square from $v$ iff $$\sphericalangle(c_k,v)\in\left(-\frac\pi2,\frac\pi2\right)\;,$$ i.e. iff $$0<\cos\sphericalangle(c_k,v)=\langle c_k,v\rangle=\cos(k\alpha)\cos\frac\pi2+\sin(k\alpha)\sin\frac\pi2=\sin(k\alpha)\;,$$ i.e. iff $$k\alpha\in(0,\pi)\Leftrightarrow k\in\left(0,\frac m2\right)\;.$$

So, there's a viewpoint from which we see exactly $\lfloor\frac m2\rfloor-1$ squares. Replacing $\beta$ by $(\alpha+\pi)/2$, the same argumentation yields that there's a viewpoint from which we see exactly $\lfloor\frac m2\rfloor$ squares (see John Hughes' answer).
