Geometry and the Discrete Fourier Transform

Elementary plane geometry was long a favorite of mathematicians and amateurs alike, but has fallen a bit into disfavor nowadays. In the nineteenth century it was so popular a subject that the comprehensive index of scientific papers put out by the Royal Society in London, England, explicitly excluded over 600 papers on triangles and conic sections from its index volume for Pure Mathematics which already comprised over 32,000 items. It is of interest that one can view many of the theorems of plane geometry from a point of view provided by the Discrete Fourier Transform (DFT) and that this leads on to some other results on circulant matrices, smooth interpolation of point distributions in the plane, zeroes of polynomials, and even a connection with quantum bits and the Heisenberg-Weyl algebra as it turns up in signal processing. In addition the geometric use of the DFT provides some beautiful diagrams for mechanical problems.

What is plane geometry? Start with the two-dimensional plane, which we'll think of as having as coordinates complex numbers $x + \ii y \in \cplx$. Since theorems about intervals defined by their two end-points are perhaps not too interesting, we'll start with the geometry of triangles. There are several common classes of theorems about triangles. I'll distinguish three types to begin with: Coincidence, Collinearity and Equilaterality.

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Theorems of Coincidence

Triangle Medians

The first example is a triangle and the lines from the vertices to the bisection points of opposite sides, which lines are called the medians. Pairwise, they intersect in 3 new points. The first example theorem says these three points coincide. The point of common intersection is called the centroid of the triangle. The figure below shows this.

The figures will often be accompanied by three options, Show, Hide and Pop Up. The figure below is shown and can be hidden; if a figure is by default hidden, it can be shown by clicking appropriately. Clicking Pop Up opens the interactive form of a diagram in a separate small window. This can be left open along with the article or closed in the way usual for a browser window. Note that the diagrams were created with Geogebra and are Java applets, so that you must have a suitable version of Java running to view them. In this first case we shall actually show the interactive diagram to give a taste of what can be done. But later the illustrations will be fixed images to make this article load faster.
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Triangle Altitudes

Again we have a coincidence for the altitudes of a triangle, the lines drawn from the vertices perpendicular to the opposite sides. The point of coincidence is called the orthocenter.
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Triangle Cevians

Just to show that not every natural construction leads to a coincidence of pairwise intersetions, note this is not so for arbitrary Cevians of a triangle, which are lines from the vertices to points on the opposite sides.
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Coincidence theorems

The prime example of a coincidence theorem is probably that concerning the Euler line. The remarkable fact, first noted by Euler, is that the circumcenter $C$, the centroid $G$ and the orthocenter $O$ are collinear for any triangle with vertices $A_1$, $A_2$ and $A_3$. The medians are in green and the perpendicular bisectors in blue. The line through them is shown in orange so as to stand out a bit in the figure below.
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Equilaterality theorems

Napoleon's Theorem

Napoleon's theorem, though probably not directly due to the emperor himself, has long been eponymous with him. In its simplest form it says that if one constructs on each side of a nondegenerate plane triangle, and external to it, an equilateral triangle of base equal to the side, and then takes the three centroids of the resulting triangles, then these points are the vertices of an equilateral triangle. That is you start with an arbitrary triangle, construct three new points in a specified way and the resulting points are the vertices of an equilateral triangle. This is illustrated in the accoompanying diagram, which actually shows that a similar construction using triangles on the other sides of the triangle sides also leads to a different equilateral triangle.
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There is a rather more general form of the theorem which will be seen as a consequence of what will be proved below. It states that the results of a whole class of constructions which are specified to be made on the edges of an arbitrary nondegenerate plane $N$-gon will be a regular $N$-gon of a specific type.

Morley's Theorem

What was hailed when it was discovered as the first really new advance in plane geometry since the Greeks became known as Morley's Miracle. It was found in about 1900 but not really published by its discoverer Frank Morley until many years later. Many proofs of it have appeared, a notable recent one of great generality being due to Alain Connes. It can be thought to be related to the approach explored below.

Morley's basic concistruction is: F or any non-degenerate triangle consider the internal angle trisectors. Then consider the pairwise intersections of adjacent ones. The resulting points are the vertices of an equilateral triangle. Again a diagram shows the remarkable assertion nicely.
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There is even more to be found from considering angle trisectors of which there are two for each angle; indeed at each vertex there are two angles the internal and the external. By considering all suitable combinations of angles and trisectors, in fact one finds $18$ equilateral Morley triangles assocciated to a given triangle. Liouville and others later went on to consider other fractions of polygon angles and more general constructions. Again a diagram can help.
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A pentagon construction

A new twentieth-century type of construction, that was devised probably around 1940 but only published in 1960 in a piece in honor of a colleague Jekuthiel Ginsburg, is due to the American Fields medalist Jesse Douglas.

Consider a pentagon in the plane. For that matter, we could let the points be in a space of dimension up to 4, since 5 points can only subtend a space of at most dimension 4. However, the formalism is simpler for the plane so that's where we'll start. The full situation will be considered later. We are viewing the 2-dimensional plane as given by a single complex coordinate $z = x + \ii y$. Suppose the five vertex points are $ p_1$, $p_2$, $p_3$, $p_4$, and $p_5$; so we are given the complex vector in ${\mathbb C}^5$, a list of points, $$ p=(p_1,p_2,p_3,p_4,p_5).$$ Since no one point of the five is distinguished, we shall consider the indices $i$ of $p_i$ modulo $5$: sums in the index are cyclic, so that, for instance, $p_{3+4} = p_2$. Douglas suggests: Join each vertex $p_i$ to the midpoint of the edge opposite between the points $p_{i+2}$ and $p_{i+3}$ ; then extend that line by an extra piece of length $(1/\sqrt{5})$ of the original piece to create a new vertex $q_i$. Douglas observed that the new pentagon $(q_1, q_2, q_3, q_4, q_5)$ is always the affine image of a convex regular pentagon. He also remarked that if, instead, the lines to the opposite midpoints are shortened by a similar decrement ratio of $(1/\sqrt{5})$ to give new vertices $r_i$, then $(r_1, r_2, r_3, r_4, r_5)$ is the affine image of a regular star pentagon, commonly called a pentagram. We interpret this in coordinates and plot sample results. We start with, say, $ p=(p_1,p_2,p_3,p_4,p_5)$. The plots are done using an arbitrarily chosen example, but as usual the vertices can be dragged about in the figure. $$ p=(1.3 + 2.2 \ii, 4.4 + 1.5 \ii, 6 + 5.5 \ii, 1.5 + 4.1 \ii, 5.6 + 6.4 \ii). $$
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We see $$q_i = p_i + \left(1+\frac1{\sqrt{5}}\right)\left({\frac{1}{2}}(p_{i+2}+p_{i+3}) - p_i\right) ,$$
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and similarly that $$r_i = p_i + \left(1-\frac1{\sqrt{5}}\right)\left({\frac{1}{2}}(p_{i+2}+p_{i+3}) - p_i\right) .$$
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Putting all this into a diagram with all the lines drawn, which it is amusing to pull about to see the effectiveness of the construction, we get
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Geometry with the Discrete Fourier Transform

The pentagon construction explained

Why does the above work? It can be explained by examining this construction with a really simple example of what can be thought of as a Geometrical Fourier Transform. In this case it's the Discrete Fourier Transform of order 5. Let $\omega_k$ be the simplest primitive $k$th root of unity: $$ \omega_k = \cos\frac {2\pi}k+ \sin\frac {2\pi}k \ii. $$ The standard convex regular pentagon can be taken to be the ordered set of points $$ (\omega_5^0,\omega_5^1,\omega_5^2,\omega_5^3,\omega_5^4) =(1,\omega_5^{\strut},\omega_5^2,\omega_5^3,\omega_5^4). $$ The standard regular star pentagon, or pentagram, can be taken to be the ordered set of points $$ (\omega_5^{2\times0},\omega_5^{2\times1},\omega_5^{2\times2}, \omega_5^{2\times3},\omega_5^{2\times4}) =(1,\omega_5^2,\omega_5^4,\omega_5^1,\omega_5^3), $$ which is the result of taking every other point starting from $1$. Note that complex conjugation acts as $\overline{\omega_5^j} = \omega_5^{5-j}$. We'll do the case $n=5$ of the starting construction first, so take $\omega := \omega_5^{\strut}$ until further notice. Remark also that $$\cos\frac{2}{\pi}5 = \cos 72^\circ = \frac{1}{4} (-1+\sqrt5)\simeq 0.309, \quad \sin\frac{2}{\pi} 5 =\cos 18^\circ = \frac{1}{4}\sqrt{10+2\sqrt5} \simeq 0.951 . $$ In fact, in terms of square roots, $$ \left( \begin{array}{c} 1\\ \omega_5^{\phantom{1}}\\ \omega_5^2\\ \omega_5^3\\ \omega_5^4 \end{array}\right) = \frac{1}{4} \left( \begin{array}{c} 4\\ -1+\sqrt5 + \ii \sqrt{10+2\sqrt5}\\ -1-\sqrt5 + \ii \sqrt{10-2\sqrt5}\\ -1-\sqrt5 - \ii \sqrt{10-2\sqrt5}\\ -1+\sqrt5 - \ii \sqrt{10+2\sqrt5} \end{array} \right) $$ $$= \frac{1}{4} \left\{\, \left[ -\left( \begin{array}{c} 1\\ 1\\ 1\\ 1\\ 1 \end{array} \right) +\sqrt5\, \left( \begin{array}{c} \sqrt5\\ +1\\ -1\\ -1\\ +1 \end{array} \right) \,\right] + \frac{\ii}{2} \sqrt{10+2\sqrt5}\, \left[\,\, \left( \begin{array}{c} 0\\ 2\\ 1\\ -1\\ -2 \end{array} \right) +\sqrt5\, \left( \begin{array}{c} 0\\ 0\\ -1\\ 1\\ 0 \end{array} \right) \, \right]\, \right\} $$ which seems complicated enough.

Going back to examine the coordinates for $q_i$ we see $$q_i = -\frac{\sqrt5}5 p_i + \frac{5+\sqrt5}{10} p_{i+2}+ \frac{5+\sqrt5}{10} p_{i+3}.$$ To establish properties of the shape of a pentagon we try to see what components of the various standard types it has. The general fact is that any plane pentagon can be written as a linear superposition, with complex coefficients, of the standard polygonal forms with $5$ points. The 5 basic forms are:

• a degenerate collection of five superposed points,
• the standard regular convex pentagon,
• the standard regular star pentagon,
• the reversed regular star pentagon,
• the reversed regular convex pentagon.

‘Standard’ means consisting of points on the unit circle, and ‘reversed’ means the standard form described in the opposite orientation. We will also refer to the reversed forms of the figures as their opposites, since they have the opposite orientation. The four non-degenerate types are illustrated below.

The coordinates of the five basic forms of a pentagon are: $$ ( 1, 1, 1, 1, 1 ) \,,\quad( 1, \omega_5^\strut, \omega_5^2, \omega_5^3, \omega_5^4 ) \,,\quad( 1, \omega_5^2, \omega_5^4, \omega_5^\strut, \omega_5^3 ) \,,\quad( 1, \omega_5^3, \omega_5^\strut, \omega_5^4, \omega_5^2 ) \,,\quad( 1, \omega_5^4, \omega_5^3, \omega_5^2, \omega_5^\strut ) $$ We shall consider a pentagon $p$ as a function on the cyclic group ${\mathbb Z}/5 {\mathbb Z}$. In fact that's what we began to do as soon as we adopted the convention that the indices were to be viewed modulo $5$. The decomposition of a given pentagon into its standard components is just an example of harmonic analysis for the cyclic group ${\mathbb Z}_{/5} := {\mathbb Z}/5 {\mathbb Z}$. This is known as the Discrete Fourier Transform of order $5$. The 5-dimensional space, ${\mathbb C}{\mathbb Z}_{/5} $, known as ${\mathbb C}{\mathbb Z}_{/5}$'s complex group algebra, of ${\mathbb C}$-valued functions on ${\mathbb Z}_{/5}$ has as a basis the functions $$\begin{eqnarray} \chi_0&=&( 1, 1, 1, 1, 1 ) \\ \chi_1&=&( 1, \omega_5^\strut, \omega_5^2, \omega_5^3, \omega_5^4 ) \\ \chi_2&=&( 1, \omega_5^2, \omega_5^4, \omega_5^\strut, \omega_5^3 ) \\ \chi_3&=&( 1, \omega_5^3, \omega_5^\strut, \omega_5^4, \omega_5^2 ) \\ \chi_4&=&( 1, \omega_5^4, \omega_5^3, \omega_5^2, \omega_5^\strut ). \end{eqnarray} $$ It is also a space with a natural Hermitian inner product. Namely for $u,v\in{\mathbb C}{\mathbb Z}_{/5}$ $$ \langle u | v\rangle := \frac{1}{|{\mathbb Z}_{/5}|}\sum_{g\in{\mathbb Z}_{/5}} \overline{u(g)}v(g) = \frac{1}{5}\sum_{1\leq j\leq 5}\overline{u_j}v_j. $$ So, if we note that the numbering of the basis functions above is like the characters of ${\mathbb Z}_{/5}$ that they are, according to $$ \chi_k = (\omega_5^{k\cdot0},\omega_5^{k\cdot1},\omega_5^{k\cdot2},\omega_5^{k\cdot3}, \omega_5^{k\cdot4})\quad\quad 0\leq k\leq4, $$ then the expansion of an arbitrary function $f\in{\mathbb C}{\mathbb Z}_{/5}$ is $$ f = \sum_{k=0}^{4}\langle\chi_k | f\rangle \chi_k=\sum_{k=0}^{4}\hat f_k \chi_k, $$ where $\hat f_{0}$ is the center of gravity of the value set of $f$, and $\hat f_{1},\hat f_{2},\hat f_{3},\hat f_{4}$ are the coefficients in the expression of the arbitrary pentagon $f$ as a linear sum of standard ones. The $(\chi_k)_{0\le k\le 4}$ form an orthonormal basis, and $\chi_0$ is the trivial (or degenerate) character of the trivial representation. Now we examine the pentagon $q$ that Douglas derived in this connection: The center of gravity is just an expression of where the geometric object is located, so we don't expect any special properties for it. But for $q_i = -\frac{\sqrt5}5 p_i + \frac{5+\sqrt5}{10} p_{i+2}+ \frac{5+\sqrt5}{10} p_{i+3}$ look at $\hat q_2$, or at $$ \begin{eqnarray} 5\hat {q_2} &=&5 \langle \chi_2 | q\rangle \\ &=& 1\cdot q_1 + \overline{ \omega_5^{2}}\cdot q_2 + \overline{ \omega_5^{4}}\cdot q_3 + \overline{ \omega_5^{1}}\cdot q_4 + \overline{ \omega_5^{3}}\cdot q_5\\ &=& q_1 +\omega_5^{3} q_2 + \omega_5^{1} q_3 + \omega_5^{4} q_4 + \omega_5^{2} q_5\\ &=& -\frac{\sqrt5}5( p_1 +\omega_5^{3} p_2 + \omega_5^{1} p_3 + \omega_5^{4} p_4 + \omega_5^{2} p_5) +\\ & & \frac{5+\sqrt5}{10}( p_3 +\omega_5^{3} p_4 + \omega_5^{1} p_5 + \omega_5^{4} p_1 + \omega_5^{2} p_2) +\\ & & \frac{5+\sqrt5}{10}( p_4 +\omega_5^{3} p_5 + \omega_5^{1} p_1 + \omega_5^{4} p_2 + \omega_5^{2} p_3)\\ &=& (-\frac{\sqrt5}5\cdot1 + \frac{5+\sqrt5}{10} \omega_5^{4} + \frac{5+\sqrt5}{10} \omega_5^{1})p_1 +\\ & & (-\frac{\sqrt5}5 \omega_5^{3} + \frac{5+\sqrt5}{10} \omega_5^{2} + \frac{5+\sqrt5}{10} \omega_5^{4})p_2 +\\ & & (-\frac{\sqrt5}5 \omega_5^{1} + \frac{5+\sqrt5}{10} \cdot1 + \frac{5+\sqrt5}{10} \omega_5^{2})p_3 +\\ & & (-\frac{\sqrt5}5 \omega_5^{4} + \frac{5+\sqrt5}{10} \omega_5^{3} + \frac{5+\sqrt5}{10} \cdot1)p_4 +\\ & & (-\frac{\sqrt5}5 \omega_5^{2} + \frac{5+\sqrt5}{10} \omega_5^{1} + \frac{5+\sqrt5}{10} \omega_5^{3})p_5 . \end{eqnarray} $$ The thing to note is that the coefficients of the points $p_i$ are all multiples by powers of $\omega_5$ of the coefficient of $p_1$. Furthermore the coefficient of $p_1$ is a real number since $\overline{\omega_5^{4}} = \omega_5^{1}$. But $$ \begin{eqnarray} - \frac{\sqrt5}5\cdot1 + \frac{5+\sqrt5}{10} \omega_5^{4} + \frac{1+\sqrt5}{2}\omega_5^{1} &=& -\frac{\sqrt5}5 + \frac{5+\sqrt5}{10}\left(\omega_5^{4} + \omega_5^{1}\right) \\ &=& -\frac{\sqrt5}5 + \frac{5+\sqrt5}{10} 2 \Re{ \omega_5^{1}} \\ &=& -\frac{\sqrt5}5 + \frac{(5+\sqrt5)(-1+\sqrt5)}{10\times 2} \\ &=& -\frac{\sqrt5}5 + \frac{-5+5-\sqrt5+5\sqrt5}{10\times 2} \\ &=& -\frac{\sqrt5}5 + \frac{4\sqrt5}{20} = 0 . \end{eqnarray} $$ Thus $\hat q_2$ vanishes indentically. So there is no regular pentagram component of the polygon $q$. Similarly look at $\hat q_3$, or at $$ \begin{eqnarray} 5\hat { q_3} &=& 5\ \langle \chi_3 | q\rangle \\ &=& 1\cdot q_1 + \overline{ \omega_5^{3}}\cdot q_2 + \overline{ \omega_5^{1}}\cdot q_3 + \overline{ \omega_5^{4}}\cdot q_4 + \overline{ \omega_5^{2}}\cdot q_5\\ &=& q_1 +\omega_5^{2} q_2 + \omega_5^{4} q_3 + \omega_5^{1} q_4 + \omega_5^{3} q_5\\ &=& 5\ \langle \overline{\chi_2} | q\rangle \\ &=& 5\ \overline{\langle \overline q,\chi_2\rangle} \\ &=& 0 . \end{eqnarray} $$ This is because the reason that $\langle \chi_3 | q\rangle = 0$ has little to do with the properties of the $q_i$ as complex numbers, but was entirely the result of the interaction between the $5$th roots of unity, and the real coefficients in the specific make-up of the $q_i$ starting from arbitrary complex numbers $p_i$. If this is unconvincing, then a similar argument to the previous one goes through but involving the verification $$ - \frac{\sqrt5}5\cdot1 + \frac{5+\sqrt5}{10} \omega_5^{1} + \frac{5+\sqrt5}{10}\omega_5^{4} = 0. $$ We have thus shown that $q = {\hat q}_0 + {\hat q}_1 \chi_1 + {\hat q}_4\chi_4={\hat q}_0 + {\hat q}_1 \chi_1 + {\hat q}_4 \overline{\chi_1}$. This means that $q$ is an affine image of the regular convex pentagon, as will be discussed below. We could have convinced ourselves geometrically that the coefficients vanish by looking at the diagrams for the pentagon in the unit circle, but it seemed useful to do the simple algebra. Next consider the other similarly constructed pentagon $r$, the one whose points are given by $$ r_i = \frac{\sqrt5}5 p_i + \frac{5-\sqrt5}{10} p_{i+2}+ \frac{5-\sqrt5}{10} p_{i+3}. $$ For this case we examine correspondingly the components ${\hat r}_1$ and ${\hat r}_4$. Again we can group the coefficients of the points $p_i$, and see that they are all multiples of any one of them. For instance, the coefficient of $p_1$ is $$ \frac{\sqrt5}5\cdot1 + \frac{5-\sqrt5}{10} \omega_5^{2} + \frac{5-\sqrt5}{10} \omega_5^{3}= \frac{\sqrt5}5 + \frac{5-\sqrt5}{10}\frac{-1-\sqrt5}{2} = 0, $$ just as before, up to a couple of sign changes. Also as before, we can conclude that ${\hat r}_1 = {\hat r}_4 = 0$, and say that the polygon $r$ is the affine image of a regular pentagram.

The general Discrete Fourier Transform

Let us fix notation for the general case of what we used in the foregoing explanation. We consider an $N$-gon $p$, for a natural number $N>0$, as a function on the cyclic group $\intg_{/N} = \intg / N\intg $. We define $\omega_N := \ee^{\ii 2 \pi/N}$. Then we wish to decompose the $N$-dimensional space, $\cplx\intg_{/N}$, of $\cplx$-valued functions on $\intg_{/N}$ with respect to the basis of characters $$(\chi_k)_{k\in \intg} \quad{\rm where} \quad \chi_k = (\omega_N^{k\cdot j})_{0\le j \le N} = (\omega_N^{k\cdot0}, \omega_N^{k\cdot1},\dots, \omega_N^{k\cdot N-1}),$$ which are orthonormal for the natural Hermitian inner product, namely for $u,v\in\cplx\intg_{/N}$ $$ \langle u | v\rangle = \frac{1}{|\intg_{/N}|}\sum_{g\in\intg_{/N}} \overline{u(g)}v(g) = \frac{1}{N}\sum_{0\leq j< N-1}\overline{u_j}v_j, $$ and $$ \langle \chi_j | \chi_k\rangle = \loglb j==k\logrb = \begin{cases}1 \text{ if } j=k | k\cr 0 \text{ otherwise.}\end{cases} $$ The expansion of an arbitrary function $f=(f_1,\dots,f_N)\in\cplx\intg_{/N}$ is $$f = \sum_{k=0}^{N-1}\langle\chi_k | f\rangle \chi_k=\sum_{k=0}^{N-1}\hat f_k \chi_k,$$ where $\hat f_{0}$ is the center of gravity (a.k.a. centroid, barycenter or mean) of the value set of $f$, $$ \hat f_{0} = \frac{1}{N} (f_1 + \cdots + f_n) $$ and $\hat f_{1},\dots,\hat f_{N-1}$ are the coefficients in the expression of the arbitrary function $f$, an $N$-gon, as a linear sum of standard ones.

There can be a need, in the course of an argument, to emphasize the number of points involved, $N$, as well as the order, $k$ of a character. In that case we shall write, as needed, $$ \chi_{N,k} = (\omega_N^{k\cdot j})_{0\le j < N} = (\omega_N^{k\cdot j}, j\in{\bar N}) = (\omega_N^{k\cdot0}, \omega_N^{k\cdot1},\dots \omega_N^{k\cdot N-1}),$$ where we have written $\bar N = 0..N-1$ for an integer interval; similarly we may write $\langle u | v\rangle_N$.

It is clear then that $\chi_{N,1}$ lists the vertices of the standard convex regular positively oriented $N$-gon inscribed on the unit circle. Similarly $\chi_{N,N-1}$ gives the standard negatively oriented unit $N$-gon. The combination $\lambda \chi_{N,1} +\mu \chi_{N,N-1}$ with $\lambda, \mu\in\cplx$ then gives a general affine convex $N$-gon.

When $ 1 < k < N-1$ the $N$-gon given by $\chi_{N,k}$ will have a multiplicity if $k|N$ ($k$ divides $N$) and that will be $N/k$. Otherwise it will be sort of ‘star’ polygon if $k$ is prime to $N$, i.e., $(k,N)=1$, their greatest common divisor is 1. For instance, $\chi_{6,3}$ is the standard hexagon's horizontal equator (a ‘di-gon’) described back and forth 3 times. $\chi_{6,4}$ is the standard unit equilateral triangle $\chi_{3,1}$ listed twice in reverse, or you can think of it as $\chi_{3,2}$ twice.

Triangle geometry

Since the claim is that almost all classical plane geometry has a natural expression from the point of view outlined above, we start by looking at the standard types of triangles and their natural expressions in this Discrete Fourier Transform notation. Consider first of all triangles whose centroids are at the origin; after all, it is easy enough to move any triangle into this simplifying position by a translation.

The first example is the simple average of a standard equilateral triangle and its reverse, namely $\sfrac12(\chi_1+\chi_2)$. The points are thus $(1,-\sfrac12,-\sfrac12)$; this is a degenerate triangle of points with the second and third vertices coincident at $-\sfrac12$. It is a digon with a double point at the left-hand end.

If we separate the two triangles by making the standard one larger, say lying on a circle of radius $\lambda>1$ then we get $$\begin{eqnarray*} \sfrac12(\lambda\chi_1+\chi_2)&=& \sfrac12\left\{\lambda(1,-\sfrac12+\sfrac{\sqrt3}2\ii,-\sfrac12-\sfrac{\sqrt3}2\ii) +(1,-\sfrac12-\sfrac{\sqrt3}2\ii,-\sfrac12+\sfrac{\sqrt3}2\ii)\right\}\\ &=&\sfrac12\left(\lambda+1,-\sfrac12(\lambda+1)+ \sfrac{\sqrt3}2(\lambda-1)\ii,-\sfrac12(\lambda+1)-\sfrac{\sqrt3}2(\lambda-1)\ii\right) \end{eqnarray*} $$ which shows the triangle is an isosceles one with height $\sfrac34(\lambda+1)$ lying along the real axis.

If we separate the two basic triangles $\chi_1$ and $\chi_2$ by turning them in opposite directions by a turn $\ee^{\ii \alpha}$ (in the attractive terminology of Frank Morley for complex numbers of modulus 1) to make $$ \begin{eqnarray} \sfrac12(\ee^{\ii \alpha}\chi_1+\ee^{-\ii \alpha}\chi_2)&=& \sfrac12\left\{\ee^{\ii \alpha}(1,-\sfrac12+\sfrac{\sqrt3}2\ii,-\sfrac12-\sfrac{\sqrt3}2\ii) +\ee^{-\ii \alpha}(1,-\sfrac12-\sfrac{\sqrt3}2\ii,-\sfrac12+\sfrac{\sqrt3}2\ii)\right\}\cr &=& \left(\sfrac12(\ee^{\ii \alpha}+\ee^{-\ii \alpha}), -\sfrac14(\ee^{\ii \alpha}+\ee^{-\ii \alpha})+ \sfrac{\sqrt3}4\ii(\ee^{\ii \alpha}-\ee^{-\ii \alpha}) ,-\sfrac14(\ee^{\ii \alpha}+\ee^{-\ii \alpha})- \sfrac{\sqrt3}4\ii(\ee^{\ii \alpha}-\ee^{-\ii \alpha})\right)\cr &=&(\cos \alpha, -\sfrac12\cos \alpha-\sfrac{\sqrt3}2\sin \alpha, -\sfrac12\cos \alpha+\sfrac{\sqrt3}2\sin \alpha) ) \end{eqnarray} $$ then we have achieved a splitting of the double point at $-\sfrac12$ in $\sfrac12(\chi_1+\chi_2)$ but still have a degenerate triangle on the real axis, with the second vertex at $z_1=-\sfrac12\cos \alpha-\sfrac{\sqrt3}2\sin \alpha$ and the third symmetrically placed about $-\frac12\cos\alpha$ at $z_2=-\sfrac12\cos \alpha+\sfrac{\sqrt3}2\sin \alpha$.

Continuing from:

Interpolation with the Discrete Fourier Transform

An algebraic point of view on the Discrete Fourier Transform

Continuing by:

Polynomials and other relationships