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?

Question

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.

Answer 1

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 $k = 1, 2, \ldots, \lfloor \frac{m}{2} \rfloor$.

Answer 2

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).