Geometric constructibility of polygons

# Geometric constructibility of polygons lying on a circular arc

Delbrin Ahmed University of Szeged, Bolyai Institute  and the University of Duhok Gábor Czédli University of Szeged, Bolyai Institute. Szeged, Aradi vértanúk tere 1, HUNGARY 6720  and  Eszter K. Horváth University of Szeged, Bolyai Institute. Szeged, Aradi vértanúk tere 1, HUNGARY 6720
October 24, 2017, 17:43
nk
2000 Mathematics Subject Classification. Primary 51M04, secondary 12D05
###### Abstract.

For a positive integer , an -sided polygon lying on a circular arc or, shortly, an -fan is a sequence of points on a circle going counterclockwise such that the “total rotation” from the first point to the last one is at most . We prove that for , the -fan cannot be constructed with straightedge and compass in general from its central angle and its central distances, which are the distances of the edges from the center of the circle. Also, we prove that for each fixed in the interval and for every , there exists a concrete -fan with central angle that is not constructible from its central distances and . The present paper generalizes some earlier results published by the second author and Á. Kunos on the particular cases and .

###### Key words and phrases:
Geometric constructibility, circular arc, inscribed polygon, cyclic polygon, compass and ruler, straightedge and compass
This research of the second and third authors is supported by the Hungarian Research Grant KH 126581

## 1. Introduction and our results

### A short historical survey

With the exception of squaring the circle, not much research interest was paid to geometric constructibility problems for one and a half centuries after the Gauss–Wantzel Theorem in [7], which completely described the constructible regular -gons. This can be well explained by the fact that most of the ancient constructibility problems as well as constructing triangles from various given data are too elementary and, furthermore, nowadays it does not require too much skill to solve them with the help of computer algebra in few minutes. This is exemplified by the textbooks Czédli [2] and Czédli and Szendrei [4], where more than a hundred constructibility problems are solved.

It was Schreiber [6] who revitalized the research of geometric constructibility by an interesting non-trivial problem, the constructibility of cyclic (also known as inscribed) polygons from their side lengths. Furthermore, he pointed out that this problem requires a variety of interesting tools from algebra and geometry. The first complete proof of his theorem on the non-constructibility of cyclic -gons from their side lengths for every used some involved tools even from analysis; see Czédli and Kunos [5].

### Polygons on a circular arc

Let . By an -sided polygon lying on a circular arc or, shortly, by an -fan we mean a planar polygon such that the vertices , in this order, lie on the same circular arc, see on the right of Figure 1. The short name “-fan” is explained by the similarity with a not fully open hand fan. Some important real numbers that determine an -fan are also given in Figure 1; the central distances of the sides from the center of the circular arc, the central angle , and the radius are worth separate mentioning here. Like in the earlier papers Schreiber [6], Czédli and Kunos [5], and Czédli [3], an easy argument based on properties of continuous real functions shows that the ordered tuple determines the -fan up to isometry, provided that or . We denote by the -fan determined by this tuple; the subscript comes from “central distances”. For the central angle , we always assume that . Furthermore, we always assume that our -fans are convex in the sense that the angle at contains for . If , then we assume also that the angle at contains . Convexity means that “zigzag polygons” like the small one on the left of Figure 1 are not allowed. With the notation , note that there are -tuples for which does not exist. However, similarly to Czédli and Kunos [5] and Czédli [3], it follows from continuity that, for ,

 (1.1) if n≥3 or δ<2π, and all the ratios di/dj are sufficiently close to 1, then Fn\textupcd(δ,d1,…,dn) exists and it is unique.

### Constructibility

In this paper, constructibility is always understood as the classical geometric constructibility with straightedge and compass. (We prefer the word “straightedge” to “ruler”, because it describes the permitted usage better.) Due to the usual coordinate system of the plane, we can assume that a constructibility problem is always a task of constructing a real number from a sequence of real numbers. Geometrically, this means that we are given the points , , …, in the plane and we want to construct the point . Angles are also given by real numbers. Whenever we say that the central angle is given, this means that the real number is given. From the perspective of constructibility, any other usual way of giving is equivalent to giving , that is the point . The advantage of using over, say, is that uniquely determines . As opposed to the constructibility of a concrete point from other concrete points, the concept of constructibility in general is more involved; the reader may want to but need not see Czédli [3] and Czédli and Kunos [5] for a rigorous definition. The reader of this paper may safely assume that “constructible in general” means “constructible for all meaningful data”.

### Our results

Our first target is to decide whether the -fan can be constructed from in general. We are going to prove the following.

###### Theorem 1.1.

The -fan is geometrically constructible in general from if and only if . Furthermore, if , then there exist rational numbers , , …, such that holds and the -fan exists, but this -fan cannot be constructed from .

For many values of , the inequality above can easily be strengthened to the equality . For example, for and , where , even the -fan cannot be constructed; this follows trivially from the Gauss–Wantzel Theorem, [7], from which we know that the regular nonagon (also known as -gon) cannot be constructed. Note that this easy argument is not applicable if is a power of 2. Note also that the constructibility from rational parameters is equivalent to the constructibility from , that is, from the points and .

In view of earlier results where was fixed as or , it is reasonable to consider the problem also for the case where is fixed and only the side lengths are “general”. In particular, if is a constructible angle like , or , then we can consider it only information rather than a part of the data. As a preparation for our second theorem, we introduce the following notation for and :

 N\textupnum(δ) xxxialso exists and it is uniquely determined, % but it is xxxinot constructible from ⟨cos(δ/2),d1,…,dn⟩}, and N\textupan(m) :={δ∈(0,2π]:m∈N\textupnum(δ)}.

The superscripts above come for “numbers” and “angles”, respectively, while “” comes from the prefix “non” in “non-constructible”. As usual, stands for the open interval of real numbers. Now, we are in the position to formulate our second statement.

###### Theorem 1.2.

The following six assertions hold.

1. . In particular, .

2. . In particular, .

3. For every such that is transcendental, we have that and, in particular, .

4. For every , .

5. For , but .

6. For , let . With this notation, whenever belongs to , then .

###### Remark 1.3.

Note that if and , then for all -tuples such that the -fan exists, this -fan is constructible from . However, this is not true for since and do not make sense; the first is not determined uniquely while the second does not exist.

It is a surprising gap in 1.2(i) that does not belong to . The redundancy in the theorem focuses our attention to . However, we do not have a satisfactory description of . Note that it follows from 1.2(vi) that

 {jπ4:4≠j∈{1,2,…,7}}∪{kπ6:6≠k∈{1,2,…,11}}⊆N\textupan(3).

We could let and run up to 8 and 12, respectively, but the inclusion above for and follows from 1.2(i), not from 1.2(vi).

The -fan determined by its central angle and its side lengths , see Figure 1, will be denoted by ; the subscript comes from “side lengths”. Due to the Limit Theorem from Czédli and Kunos [5], the constructibility problem for is easier than that for . This fact and space considerations explain that the present paper contains only the following result on side lengths.

###### Proposition 1.4.

For , the -fan is constructible in general from if and only if .

###### Remark 1.5.

For a fixed , the situation can be different. We know from school and Czédli and Kunos [5] or Screiber [6] that and can be constructed from and , respectively, in general. On the other hand, we know from Czédli [3, Theorem 1.1(v)] that exists but it cannot be constructed from its side lengths.

### Outline

The rest of the paper is devoted to the proofs of our theorems and also to some additional statements that make these theorems a bit stronger by tailoring special conditions on possible data determining non-constructible -fans. Section 2 lists some well-known concepts, notations, and facts from algebra for later reference; readers familiar with irreducible polynomials and field extensions may skip most parts of this section. Section 3 contains the above-mentioned additional statements as propositions, and it contains almost all the proofs of the paper.

## 2. A short overview of the algebraic background

A polynomial is primitive if the greatest common divisor of its coefficients is 1. The following well-known statement is due to C. F. Gauss; we cite parts (i) and (iii) from Cameron [1, Theorem 2.16 (page 90) and Proposition 7.24 (page 260)], while (ii) follows from (iii).

###### Lemma 2.1.

If is a unique factorization domain with field of fractions , then

1. the polynomial ring is also a unique factorization domain,

2. if a polynomial is irreducible in , then it is also irreducible in , and

3. a primitive polynomial is irreducible in iff it is irreducible in .

For the ring of integers and , the field of fractions of is , the field of rational -variable functions over . Note that for , we say that these numbers are algebraically independent over if the map from to extends to a field embedding to . For , this means that is a transcendental number (over ). The field generated by is denoted by ; it is isomorphic to provided that are algebraically independent over . We often write instead of , even if is not transcendental.

Given a unique factorization domain with field of fractions , the polynomial rings , , and are well known to be isomorphic. This fact allows us to write and instead of . That is, , , and are essentially the same polynomials but we put an emphasis on and . Therefore, the following convention applies in the paper:

 (2.1) no matter which of f(x,y)∈R[x,y], fx(y)∈R[x][y], and fy(x)∈R[y][x] is given first, we can also use the other two.

Note that in many cases but not always, and will be and . The degree of a polynomial will be denoted by or .

The following statement is well known and usually taught for MSc students; see, for example, Czédli and Szendrei [4, Theorem V.3.6]; see also the list of references right before Czédli and Kunos [5, Proposition 3.1].

###### Proposition 2.2.

Let . If there exists an irreducible polynomial

 h(x)∈Q(c1,…,ct)[x]

such that and is not a power of , then is not constructible from (or, equivalently and according to the present terminology, is not constructible from .

The following statement is also well known, and it is even trivial for fields rather than unique factorization domains; having no reference at hand, we are going to give a proof.

###### Lemma 2.3.

Let be a unique factorization domain with field of fractions . Let be a primitive quadratic polynomial. If its discriminant, , is not a square in , then is irreducible in and, consequently, also in .

###### Proof.

Suppose to the contrary that is reducible. Since it is primitive, it cannot have a nontrivial divisor of degree 0. Hence, there are such that . Comparing the leading coefficients, . Since is a root of , the well-known formula gives that

 −b1a1=−b±√D2a,

After multiplying by , we obtain that . Therefore, is a square of . This contradicts our hypothesis and proves the lemma. ∎

## 3. Proofs and propositions

###### Proposition 3.1.

If is an even integer, then for every real number belonging to the open interval , there exists a rational number such that the -fan exists but it cannot be constructed from .

###### Proof.

The case has been settled in Czédli [3]; see Cases 3 and 4 in page 68 there and note that our corresponds to in [3] and and are equivalent data from the perspective of geometric constructibility. Hence, we can assume that . We denote by ; it belongs to the open interval and it is distinct from 0. The smallest subfield of that includes is denoted by . We now from (1.1) that if is sufficiently close to 1, then exists. This fact and the Rational Parameter Theorem of Czédli and Kunos [5, Theorem 11.1] yield that it suffices to show that is not constructible for those in a small neighborhood of 1 that are transcendental over . Since is isomorphic to the field of rational functions over for these transcendental , we can treat later as an indeterminate . Note that this paragraph, that is the first paragraph of the present proof, would also be appropriate for ; this fact will be needed only in another proof of the paper.

Let ; it is an odd number and . As always in this paper, denotes the radius of the circular arc. We let . As Figure 1 approximately shows, for the “half angles” and , we have that

 (3.1) cosα=u,sinα=√1−u2,cosβ=cu,sinβ=√1−c2u2.

Since we work with half angles, , whereby . Using the well-known formula for the cosine of a difference, we obtain that

 (3.2) cos(δ/2−β) =cos(δ/2)cosβ+sin(δ/2)sinβ =pcu+√1−p2⋅√1−c2u2.

We also need the following well-known equality, which we combine with (3.1):

 (3.3) cos(kα) =k∑2∣j=0(−1)j/2(kj)(cosα)k−j⋅(sinα)j =k∑2∣j=0(−1)j/2(kj)uk−j⋅(1−u2)j/2=:g(k)\textupcos(u).

Note that is a polynomial over since above runs through even numbers. Since the coefficient of is

 (3.4) k∑2∣j=0(−1)j/2(kj)(−1)j/2=k∑2∣j=0(kj)=2k−1≠0,

we conclude that

 (3.5) the leading coefficient of g(k)\textupcos(u) is a positive integer and the degree of u in g(k)\textupcos(u) is k.

Since , (3.2) and (3.3) give the same real number. After rearranging the equality of (3.2) and (3.3) and squaring,

 (3.6) (g(k)\textupcos(u)−pcu)2=(1−p2)(1−c2u2).

This encourages us to consider the polynomial

 (3.7) f[\textupe,k](x,y)=(g(k)\textupcos(x)−pyx)2−(1−p2)(1−y2x2)∈Q(p)[x,y],

which is obtained from (3.6) by substituting and rearranging. The superscript of reminds us to “even” and . Since and it is odd, the degree of in is by (3.5), whence is not a power of 2. Note that remains the same if we replace by , since is transcendental over . Therefore, Proposition 2.2 will imply the non-constructibility of and that of our polygon as soon as we show that is an irreducible polynomial. Let be the canonical isomorphism that acts identically on and maps to . This extends to an isomorphism with the property in the usual way. It suffices to show that is irreducible in . But and is the field of fractions of . Thus, by Lemma 2.1, it suffices to show that is irreducible in . So, in the rest of the proof, we deal only with the irreducibility of the polynomial .

Rearranging (3.7) according to the powers of , we obtain that

 (3.8) f[\textupe,k]x(y) =(p2x2+(1−p2)x2)⋅y2−2xpg(k)\textupcos(x)⋅y +(g(k)\textupcos(x)2−(1−p2)) =x2⋅y2−2pxg(k)\textupcos(x)⋅y+(g(k)\textupcos(x)2+p2−1)∈Q(p)[x][y].

Since , we have that , whence is not an integer. Thus, since , the constant term in is nonzero. In , which is a unique factorization domain, is an irreducible element. The above-mentioned nonzero term guarantees that does not divide . Thus, is a primitive polynomial over , and we are going to apply Lemma 2.3. To do so, we compute the discriminant of as follows:

 (3.9) D[\textupe,k]x :=4p2x2g(k)\textupcos(x)2−4x2(g(k)\textupcos(x)2+p2−1) :=4(p2−1)x2⋅(g(k)\textupcos(x)2−1).

Since , it follows from (3.5) that

 (3.10) D[\textupe,k]t tends to −∞ as t∈Q(p) tends to ∞.

Now if was of the form for some , then we would have that for all and (3.10) would be impossible. Hence, is not a square in and Lemma 2.3 yields the irreducibility of , as required. This completes the proof of Proposition 3.1. ∎

Remark 1.3 explains why we consider rather than in the following statement.

###### Proposition 3.2.

If , then for every real number , the -fan can be constructed from in general.

###### Proof.

We assume that since the case is trivial by elementary geometrical considerations. We can also assume that the scale is chosen so that . Let . It is clear by (3.1) and (3.3) that . Substituting this into (3.6), an easy calculation leads to

 (3.11) (c2−2pc+1)u2+p2−1=0.

Since , is distinct from . Hence, (3.11) gives that the coefficient of is nonzero. Thus, is the root of a quadratic polynomial over the field , whereby it is constructible. So are and our 2-fan. ∎

###### Proposition 3.3.

If is an odd integer, then for every real number belonging to the open interval , there exists a rational number such that

1. if , then the -fan exists but it cannot be constructed from , and

2. if , then the -fan exists but it cannot be constructed from .

###### Proof.

First, we deal with (i). Let ; note that is odd and . The first paragraph of the proof of Proposition 3.1 for and (3.1) will be used. In particular, is assumed to be transcendental, whence so is . Since

 (3.12) cos(δ/2−2β) =cos(δ/2)cos(2β)+sin(δ/2)sin(2β) =p(2cos2(β)−1)+√1−p2⋅2sinβ⋅cosβ =p(2c2u2−1)+2cu⋅√1−p2⋅√1−c2u2

and gives that , (3.3) and (3.12) give the same value. Rearranging the equality of these two values and squaring, we have that

 (3.13) (g(k)\textupcos(u)−p(2c2u2−1))2=4c2u2(1−p2)(1−c2u2).

Since and are mutually constructible from each other, we can assume that rather than is given. Rearranging (3.13) and substituting for , we obtain that is a root (in ) of the following polynomial

 (3.14) f[\textupo,k]y(x) =f[\textupo,k](x,y)=f[\textupo,k]x(y) =(g(k)\textupcos(x)−p(2yx2−1))2−4yx2(1−p2)(1−yx2) =4x4⋅y2−(4px2g(k)\textupcos(x)+4x2)⋅y+(p+g(k)\textupcos(x))2.

Observe that since . Thus, is not a power of since is odd. Hence, by the same reason as in the paragraph right after (3.7), it suffices to show that the quadratic polynomial is irreducible in . The assumption gives that . Since is even in (3.3) but now is odd, the constant term in is 0. So the constant term of is . Hence, , which is an irreducible element in the unique factorization domain and the only prime divisor of , does not divide . Thus, the quadratic polynomial , see the last line of (3.14), is primitive. Its discriminant is

 (3.15) D[\textupo,k]x =(4px2g(k)\textupcos(x)+4x2)2−16x4⋅(p+g(k)\textupcos(x))2 :=16x4(p2−1)⋅(g(k)\textupcos(x)2−1)∈Q(p)[x],

which tends to as tends to . This leads to non-constructibility in the same way as (3.10) did.

Case (ii) needs an entirely different approach, which has already be given in Case 4 in pages 68–69 of Czédli [3]; take into account that our corresponds to in [3] and our and the in [3] are equivalent for constructibility. ∎

###### Proposition 3.4.

There exist rational numbers , , and such that with the angle , the -fan exists but it cannot be constructed from . Also, for every such that is transcendental, there are rational numbers and such that exists but it cannot be constructed from .

###### Proof.

Like for the special values of considered in Czédli [3] and Czédli and Kunos [5], the 3-fan depends continuously on its parameters. We will soon see from (3.16) that the corresponding dependence on is polynomial, whereby it remains polynomial even after fixing some parameters and letting only the rest remain indeterminates. Hence, a repeated use of the Rational Parameter Theorem of Czédli and Kunos [5, Theorem 11.1] shows that it suffices to prove that cannot be constructed in general from its parameters . Hence, we can treat , and as algebraically independent numbers over , whereby we can consider them indeterminates , , and , respectively. Note that although the rest of this proof is conceptually easy and it is hopefully readable without computers, the real verification has been done by computer algebra; reference will be given later.

We denote the half-angles corresponding to the sides at distances , and 1 of our 3-fan by , and , respectively. For convenience, we let . Then we have that , whereby

 0 =h1:=cos2(α+β)−cos2(δ′−γ) =(cosαcosβ−sinαsinβ)2−(cosδ′cosγ+sinδ′sinγ)2 =cos2αcos2β+sin2αsin2β−cos2δ′cos2γ−sin2δ′sin2γ−s1,

where . Our purpose is to get rid of the sines in that are raised to odd exponents. Note that neither , nor its square has sines with odd exponents. Since , so is

 h2 :=(h1+s1)2−s21 :=(h1+s1)2−4cos2αcos2βsin2αsin2β−4cos2δ′cos2γsin2δ′sin2γ−s2,

where . Clearly, neither , nor its square has sines with odd exponents. Finally, since , so is

 h3:=(h2+s2)2−s22.

Now we are in the position that after expanding , all the sines are raised to even exponents. Hence, after substituting , …, for , …, in , we obtain a quaternary polynomial over such that

 0=h3=h4(cosα,cosδ,cosγ,cosδ′).

As we did this before, see Figure 1 with different notation, , , and , while . Substituting these equalities into , we obtain a nonzero quaternary polynomial over such that

 0=h4(cosα,cosδ,cosγ,cosδ′)=h