Find Interior Angles of Irregular Symmetrical Polygon

Question

Apologies if this is has an obvious answer, but I've been stuck on this for a bit now.

I've been trying to figure out how to make a symmetrical polygon with a base of m length, with n additional sides of s length using just those values, and I've gotten stuck. I'm not looking for an answer per se (although it would be appreciated). I'm looking for a next step that I might have overlooked.

This shape has some given values and rules.

1. n is the number of sides excluding the base, and must be at least 2
2. m can be any length between $0$ and $n \cdot s$
3. The shape is symmetrical, the line of symmetry is perpendicular to the base and $\frac{m}{2}$ from either endpoint of the base.
4. In practice, s, n, and m are known values

Here is a picture of what I am describing:

In this example I've built the shape in reverse. I set $n=5$, $s=3$, and $\theta = 110 ^\circ$. I've also found a relationship between the angles using the angle sum of a polygon: $$ \theta = 180 - \frac{2\zeta}{n-1}, \zeta= \frac{(180 - \theta)(n-1)}{2}$$ Using that (and Geogebra) I found that $\zeta = 140 ^\circ$. But ideally, I would like to calculate $\theta$ from n, s, and m without setting $\theta$ beforehand.

I'm looking for an equation for $\theta$ and $\zeta$. I know their values relative to each other, but I haven't been able to figure out how to properly find these angles, aside from making triangles out of the entire thing, which I'm still unsure on where to start using just those values. Any help would be appreciated.

Answer 1

Map everything to complex numbers.

Let $\psi = \pi - \theta$ be the common external angle among the $n$ segments of length $s$.

Let $z = e^{i\psi}$. This is the "rotation" one need to rotate the $k^{th}$ segment of length $s$ to the orientation of $(k+1)^{th}$ segment.

What we need to do is find a $\psi$ such that $s|1 + z + \cdots + z^{n-1}| = m$. i.e. a $\psi$ which make the endpoints of the polyline of $n$ segments at a distance $m$ apart. This leads to

$$s \left|\frac{z^n-1}{z-1}\right| = m \quad\iff\quad s \sin\frac{n\psi}{2} - m \sin\frac{\psi}{2} = 0 \tag{*1}$$

Since $0 < m < ns$, the equation on RHS has real solutions. The smallest positive solution of $\psi$ is the one we need (other solutions give us self-intersecting polygons).

One can rewrite RHS of ($*1$) in terms of Chebyshev polynomials of 2nd kind:

$${\rm RHS}(*1) \quad\iff\quad U_{n-1}\left(\cos\frac{\psi}{2}\right) = \frac{m}{s}$$

For small $n$ (even $n \le 4$ and odd $n \le 9$), this allow us to express $\sin\frac{\theta}{2} = \cos\frac{\psi}{2}$ as radicals and hence $\theta$ in terms of elementary functions. For other $n$, we have no choice but to solve $\psi$ numerically.

Answer 2

The case $n=5$.

In your picture let $ABCDEF$ be our polygon, $AF=m$, $AB=BC=CD=DE=EF=s$, $\measuredangle ABC=\theta$ and you got already that $\zeta=\measuredangle BAF=360^{\circ}-2\theta$.

Thus, $\theta>90^{\circ},$ otherwise, we have no a polygon, $$\measuredangle AFD=\measuredangle CAF=270^{\circ}-\frac{3}{2}\theta$$ and $$\measuredangle ACD=\measuredangle CDF=\frac{3}{2}\theta-90^{\circ}.$$ Since $$270^{\circ}-\frac{3}{2}\theta+\frac{3}{2}\theta-90^{\circ}=180^{\circ},$$ we see that $ACDF$ is a trapezoid and consider two cases.

1. $s>m$.

Thus,$$270^{\circ}-\frac{3}{2}\theta>\frac{3}{2}\theta-90^{\circ},$$ which gives $$90^{\circ}<\theta<120^{\circ}.$$ Now, since $$AC=FD=2s\sin\frac{\theta}{2},$$ we obtain: $$\cos\left(\frac{3}{2}\theta-90^{\circ}\right)=\frac{\frac{s-m}{2}}{2s\sin\frac{\theta}{2}},$$ or $$\sin\frac{\theta}{2}\sin\frac{3\theta}{2}=\frac{s-m}{4s}$$ or $$\cos\theta-\cos2\theta=\frac{s-m}{2s}$$ or $$2\cos^2\theta-\cos\theta-\frac{s+m}{2s}=0,$$ which gives $$\theta=\arccos\frac{1-\sqrt{1+\frac{4(s+m)}{s}}}{4}.$$ The case $m\geq s$ is a similar.

I hope it will help in the general.

Answer 3

Small note1: the problem is clearly presented but the example is a bit misleading. The problem gives lengths $m$ and $s$. The example gives length $s$ and angle $\theta$ and does not mention length $m$.

Small note2: the title suggests the aim of the problem is to find interior angles; the second paragraph suggests it is to construct the polygon; and the last paragraph suggests it is to find a relationship between two angles. The first two are equivalent but the last involves additional assumptions. Here I ignore the last statement about finding the relationship between two angles.

Short answer: the question has an infinite number of solutions.

Explanation: without loss of generality, here I demonstrate the above argument for $n=5$ , $m=6$ and $s=18$. Here is one possible way to construct the polygon $ABCDEF$, where the length of $AF$ is $m$ :

1. Construct a regular polygon with $n+1$ sides of length $s$ . For our example case, vertices $B$, $C$, $D$ and $E$ of the regular polygon are shown in the figure above. Let the base of this polygon be horizontal.
2. Draw $Circle1$ circumscribing the regular polygon constructed in step 1, and draw the vertical diameter of $Circle1$.
3. Draw $Circle2$ with center $B$ and radius $s$ , and $Circle3$ with center $E$ and radius $s$.
4. Draw two vertical lines on the sides of the vertical diameter of $Circle1$, each at a distance of $\frac{m}{2}$ from the diameter. These two lines meet $Circle2$ and $Circle3$ and $A$ and $F$, respectively.

The polygon $ABCDEF$ is an answer to the problem (why?).

Now, consider the sides of the polygon as rods with pivot joints at vertices. This way, if we can deform the shape, the polygon's angles change but its side lengths remain unchanged. Specifically, suppose we pull the side $CD$ upwards by an arbitrary distance, such that $CD$ remains horizontal and its midpoint remains on the perpendicular bisector of $AF$, as shown below. Then the resulting polygon still satisfies the requirements of the problem (why?) and therefore it is an answer. Note that this displacement was arbitrary. Therefore, the problem has an infinite number of solutions.