Equisectional equivalence of triangles

# Equisectional equivalence of triangles

## Abstract

We study equivalence relation of the set of triangles generated by similarity and operation on a triangle to get a new one by joining division points of three edges with the same ratio. Using the moduli space of similarity classes of triangles introduced by Nakamura and Oguiso, we give characterization of equivalent triangles in terms of circles of Apollonius (or hyperbolic pencil of circles) and properties of special equivalent triangles. We also study rationality problem and constructibility problem.

Key words and phrases. Triangle, moduli space, circle of Apollonius, hyperbolic pencil of circles.

2010 Mathematics Subject Classification: 51M04.

## 1 Introduction

We study elementary geometric operations on triangles defined as follows. Let be a triangle, and be a real number. Let , and be division points of the edges , and by respectively, namely,

 A′=qB+(1−q)C,B′=qC+(1−q)A,C′=qA+(1−q)B.

Let ( or ) be the intersection of the lines and ( and or and respectively). Define equisection operators and , where can be defined when , by

 Tq(△ABC)=△A′B′C′ and % Sq(△ABC)=△A′′B′′C′′.

The operators have been studied in articles such as [CD, D, NO, S], et. al.

In this note we study the equivalence relation (denoted by ) of the set of triangles (denoted by ) generated by similarity and , which we shall call equisectional equivalence. The equivalence relation generated by similarity and shall be called rational equisectional equivalence and denoted by . We say two triangles and are equisectionally equivalent (or rational equisectionally equivalent) if (or respectively). We remark that we use the term “similarity” as the equivalence under orientatipon preserving homothetic transformation in this article. We say two triangles are reversely similar if they are equivalent under orientation reversing homothetic transformation.

Nakamura and Oguiso introduced the moduli space of similarity classes of triangles in [NO], which is a strong tool for the study of and . Using their results (explained in Section 3), we give (projective) geometric characterization of equisectionally equivalent triangles. Namely, two triangles with a common base, say , with the third vertices, say and , in the same side of the base are equisectionally equivalent if and only if and are on the same circle of Apollonius with foci being two vertices (denoted by and ) of regular triangles with the common base . Therefore, each equisectional equivalence class with a given base corresponds to a circle of Apollonius with foci and . It is an element of a hyperbolic pencil of circles defined by and from a projective geometric viewpoint.

We then study properties of triangles of the following three special types, right triangles, isosceles triangles, and trianges with sides in arithmetic progression (which shall be denoted by SAP triangles), that appear in the same equisectional equivalence class. There are (at most) two similarity classes of such triangles for each type, which are reversely similar in the case of right or SAP triangles, or the base angles of which satisfy in the case of isosceles triangles. For each type we explicitly give the ratio such that maps one to the other in the same equisectional equivalence class, which implies that a pair of triangles and of one of the above special types with rational edges satisfies if and only if .

We finally study compass and straightedge constructibility of for a given pair of triangles.

## 2 The statement of the main results

###### Definition 1

Let be a triangle. Let be a half plane containing with boundary the line , and and be two points () such that and are regular triangles. Define () and by

where means the area of .

We remark that both and are independent of the choice of the base of the triangle. A locus of points such that is a given positive constant is a circle, called a circle of Apollonius with foci and . Put

 CA={X:|XD|/∣∣X¯¯¯¯¯D∣∣=α(Δ)}.

Note that when is a regular triangle. The quantity takes the value if and only if is a regular triangle, and approaches as becomes thinner and thinner. In that sense, it can be considered as measuring how far a triangle is from a regular triangle.

###### Theorem 2

Given two triangles and . Let be a point in such that is similar to . Then the following conditions are equivalent:

1. is equisectionally equivalent to .

2. , in other words, is on the circle of Apollonius with foci and that passes through .

3. .

4. Let and be points in such that and are similar to in such a way that each vertex of or corresponds to a vertex of in the same sequential order through the similarity (Figure 4). Then is on the circle that passes through , and . When is a regular triangle we agree that the circle through , and consists of a single point.

The set of circles of Apollonius with foci and is called a hyperbolic pencil of circles defined by and (or a Poncelet pencil with limit points (or Poncelet points) and ). It consists of circles that are orthogonal to any circle passing through and (Figure 5 left). A set of circles through and is called an elliptic pencil (or a pencil of circles with base points).

Let be the set of similarity classes of triangles and denote the similarity class of a triangle . Nakamura and Oguiso’s result implies that the sets of similarity classes of equisectionally equivalent triangles form a codimension foliation of with a unique singularity that corresponds to regular triangles. We study the intersection of each leaf and another codimension subspace of which is the set of similarity classes of one of the following three special triangles, isosceles triangles, right triangles, and SAP triangles (i.e., triangles with sides in arithmetic progression) (the reader is referred to [MOS] for the properties of SAP triangles).

###### Corollary 3

Two right triangles are equisectionally equivalent if and only if they are either similar or reversely similar. Any triangle is equisectionally equivalent to a right triangle if and only if .

###### Corollary 4

Two SAP triangles are equisectionally equivalent if and only if they are either similar or reversely similar. Any trianle is equisectionally equivalent to such a triangle.

###### Proposition 5

Two isosceles triangles and are equisectionally equivalent if and only if either they are similar or the base angles satisfy . When , if and only if or . Any triangle is equisectionally equivalent to an isosceles triangle.

Corollaries 3 and 4 imply the importance of an equisectional operator that maps a similarity class of a given triangle to that of its mirror image.

###### Lemma 6

Let denote a triangle with side lengths in the anti-clockwise order. Then if and only if

 q=b2−1a2+b2−2,1−a2b2+1−2a2, % or a2−b21+a2−2b2.
###### Theorem 7

Suppose and are isosceles (or right or SAP) triangles such that all the sides are rational numbers. If then is also rational, namely, if and only if .

## 3 The moduli space by Nakamura and Oguiso

Nakamura and Oguiso gave a bijection between and the open unit disc in , and showed that the set of equisection operators acts on as rotations.

Let us first introduce the result of Nakamura-Oguiso [NO]. We work in the complex plane . Let be the upper half plane , and be the open unit discs in variables and respectively. Put . Define , and by

 λ(z)=11−z,μ(Z)=ρ2Z,f(z)=ρ2−ρzρ2+z=−ρz−ρz−ρ−1,g(Z)=Z3. (1)

Then we have the following commutative diagram:

 Hf−−−−→Bg−−−−→Dλ⏐⏐↓⏐⏐↓μ∥∥Hf−−−−→Bg−−−−→D

Let us fix the base of a triangle to be so that a triangle can be identified by the vertex . Suppose is similar to the triangle . Since the three choices of the base, , or corresponds to , or , there is a bijection ([NO]) given by

 φ([△z01])=g∘f(z)=(z−ρz−ρ−1)3. (2)

Let us call the Nakamura-Oguiso moduli space of the similarity classes of triangles.

From the construction of the moduli space and the property of linear fractional transformations, it follows that the set of isosceles triangles is expressed by a real axis in (explained in Section 4), and reversely similar triangles by complex conjugate numbers, as was pointed out in [NO].

The equisection operators and the equisectional equivalence relation on , denoted by the same symbols, can be naturally induced from those on . We shall express the operators on given by and simply by and respectively. Since our and are equal to and in [NO] respectively, Theorem 1 of [NO] implies

###### Theorem 8

([NO]) The operator acts on as a rotation by angle

 6arg((1−2q)ρ−(1−q))=6arg(−1+(1−2q)√3i), (3)

and by angle .

Theorem 8 implies that is in fact an equivalence relation, that if and only if there is a real numbers such that is similar to , and that the equivalence relation generated by similarity and is identical with the equisectional equivalence.

The following corollaries can be obtained by simple computation.

###### Corollary 9

([NO])

1. if and only if , or .

2. and if and only if or .

###### Corollary 10

if and only if

 q′=q,q′=2q−13q−1(q≠13), or q′=q−13q−2(q≠23).
###### Corollary 11

([NO]) if and only if

 q′=1−q,q′=2q−13q−2(q≠23), or q′=q3q−1(q≠13).
###### Remark 12

The six functions of which appear in the right hand sides in Corollaries 10 and 11 form a non-abelian group with the operation of composition, which is isomorphic to the full permutation group of three elements.

###### Corollary 13

Given real numbers and . if and only if

 q′′=3qq′−2(q+q′)+16qq′−3(q+q′)+1,q′′=−q+q′−13qq′−3(q+q′)+2, or q′′=3qq′−(q+q′)3qq′−1.
###### Proof.

The formula (3) implies that is the rotation by angle

 6arg[(−1+(q+q′))ρ+(q+q′)−3qq′].

Since

 6arg(Aρ+B)≡6arg[(1−2A+BA+2B)ρ−(1−A+BA+2B)](modulo 2π),

substitution and implies that if we put

 q′′=A+BA+2B=3qq′−2(q+q′)+16qq′−3(q+q′)+1,

then . The other two values for can be obtained by Corollary 10. ∎

Corollaries 9 (1), 11, and 13 show that the rational equisectional equivalence is in fact an equivalence relation.

## 4 Proofs of the main results

In what follows, we restrict ourselves to the case of non-regular triangles.

Proof of Theorem 2.   Theorem 8 shows that a set of equisectionally equivalent triangles corresponds to a circle with center in . Therefore, the formula (2) implies

 [△z01]∼[△z′01]⟺∣∣∣(z−ρz−ρ−1)3∣∣∣=∣∣ ∣∣(z′−ρz′−ρ−1)3∣∣ ∣∣⟺|z−ρ||z−ρ−1|=|z′−ρ||z′−ρ−1|,

which proves the equivalence between (1) and (2).

Since , , and are similar, the above argument implies that a circle of Apollonius with foci and that passes through also passes through and , which proves the equivalence of (1) and (4).

The equivalence between (2) and (3) follows from computation. Since

 |z−ρ|2|z−ρ−1|2=1−2√3Imz|z−ρ−1|2, (4)
 |z−ρ||z−ρ−1|=|z′−ρ||z′−ρ−1| if and only if Imz|z−ρ−1|2=Imz′|z′−ρ−1|2,

which, translated to a scale-invariant statement, is equivalent to (3). We remark that the formula (4) implies .

Let us give projective geometric explanation of the equivalence between (1) and (2). The set of lines through , which are considered as circles through and , is an elliptic pencil of circles defined by and , and the set of concentric circles with center is a hyperbolic pencil of circles defined by and . They are mutually orthogonal. A linear fractional transformation is a conformal map (i.e., it preserves the angles) that maps circles (which include lines that can be considered as circles through ) to circles, and hence it maps an elliptic pencil (or a hyperbolic pencil) of circles defined by a pair of points to an elliptic pencil (or a hyperbolic pencil) defined by a pair of corresponding points.

Since is a linear fractional transformation which maps and to and , it maps the set of lines through to an elliptic pencil consisting of the circles through and , and the set of concentric circles with center to a hyperbolic pencil defined by and , which consists of circles of Apollonius with foci and (Figure 5). Now it follows from the construction of Nakamura-Oguiso moduli space that the vertices of equisectionally equivalent triangles form a circle of Apollonius with foci and , which proves the equivalence between (1) and (2).

In what follows, it sometimes makes things easier to work in a a fundamental domain of ,

 Ω={z∈H:|z|<1,|z−1|≤1}∪{ρ}, (5)

which corresponds to studying triangles with the longest edge (one of the longest edges) being fixed to .

Let us explain why the isosceles triangles corresponds to in the real axis in the Nakamura-Oguiso moduli space .

###### Definition 14

Let be a circle through three points , and . When one of the three points is , is a line. We assume that is oriented by the cyclic order of , and .

In , a vertex of an isosceles triangle lies either on the line or on the circle . Since is a linear fractional transformation with

 f:ρ↦0,0↦1,1↦ρ2,ρ−1↦∞, and ∞↦−ρ,

it maps a circle () to the real axis (), another circle () to a line joining and (), a line () to a line segment joining and (), the real axis () to the unit circle (), and to one third of the open unit disc . Therefore, the images of of and are and respectively.

Proof of Proposition 5.   We work in the Nakamura-Oguis moduli space . The isosceles triangles correspond to the real axis in . Each circle with center , which corresponds to a set of equisectionally equivalent non-regular triangles, intersects the real axis in two points. It proves the third statement. It also proves that if satisfies with two isosceles triangles and then either or is a rotation by on .

Theorem 8 shows that is a rotation by angle for some if and only if or , which proves the second statement.

The equation follows directly from the fact that as is illustrated in Figure 6.

Proof of Corollary 3.   In the fundamental domain , a vertex of a right triangle lies on the upper half hemi-circle , which intersects any circle of Apollonius in at most two points, which are symmetric in the line . The extremal value of is given by a right isosceles triangle.

Proof of Corollary 4.   In the fundamental domain , a non-regular triangle such that the ratio of the edge lengths is corresponds to

 P±(d)=(1±(2d−3d2)2,√32√(1−d)(1−d2)(1−3d))(0

Let be a curve . We show that each of , say , intersects any circle of Apollonius with foci and , , in exactly one point.

Firstly, since is an open curve joining and a point on the real axis, it must intersect . Secondly, if we put 1, then , and hence . Let , then, as is an increasing function and a decreasing function of , we have on , whereas can be expressed as a graph of an increasing function. Therefore intersects in at most one point.

We remark that the statement can also be proved by computation showing that is a monotonely increasing function of with and .

Proof of Lemma 6.   Suppose is expressed by a complex number in the fundamental domain . Since and , and satisfy

 x=12(a2−b2+1),y2=14(b2−(a−1)2)((a+1)2−b2). (6)

Let and . Then

 tanθ=Im(−ρz−ρz−ρ−1)Re(−ρz−ρz−ρ−1)=√3x2+y2−2xx2+y2+2x−2.

Substitution of (6) gives

 tanθ=√3b2−12a2−b2−1.

Since is a mirror image of , we have , and since , we have

 argφ(zab)=argφ(zba)+6θmodulo 2π.

On the other hand, acts on as a rotation by , where . Suppose

 −(1−2q)√3=√3b2−12a2−b2−1. (7)

It means , which implies modulo and hence modulo , which implies , i.e., .

The equation (7) gives

 q=1−a2b2+1−2a2.

The other two values follow from the above by Corollary 10.

Proof of Theorem 7.   When the conclusion follows from Corollary 9. When the conclusion follows from Proposition 5 for isosceles triangles and from Corollaries 3, 4 and Lemma 6 for right and SAP triangles.

## 5 Compass and straightedge constructibility

###### Proposition 15

Given two triangles and . Whether is equisectionally equivalent to or not can be determined using a straightedge and compass, and if the answer is affirmative, a real number such that is compass-and-straightedge constructible.

###### Proof.

One can construct the following with a compass and straightedge in the following order:

1. Two points and () such that both and are regular triangles.

2. A vertex in such that .

3. An oriented circle and another oriented circle (see Definition 14).

4. A circle of Apollonius with foci and that passes through , since the center is the intersection of a bisector of the edge and a tangent line of at point .

5. A decision whether or not, since the answer is affirmative if and only if .

Assume in what follows.

1. The signed angle () at point from the oriented circle to .

2. At least one such that

 θ2+nπ3∈(π2,3π2) modulo 2π.
3. The value which is given by

 q=12[1+1√3tan(θ2+nπ3)], (8)

where is given by the proceeding step.

We explain the process (6),(7),(8). By working in the fundamental domain , we may assume that , and . By formulae (2) and (3), we want a real number such that

 6arg(−1+(1−2q)√3i)=3(arg(z′−ρz′−ρ−1)−arg(z−ρz−ρ−1))modulo2π. (9)

Since the right hand side divided by is equal to the singed angle at from the oriented line to the oriented line , and a linear fractional transformation maps , and to , and respectively, the signed angle is equal to the signed angle at point from the oriented circle to the oriented circle .

If is given by (8), then

 −1+(1−2q)√3i=−1−itan(θ2+nπ3),

and therefore, (modulo ), which means (9). ∎

Jun O’Hara

Department of Mathematics and Informatics,Faculty of Science,

Chiba University

1-33 Yayoi-cho, Inage, Chiba, 263-8522, JAPAN.

E-mail: ohara@math.s.chiba-u.ac.jp

### Footnotes

1. The image of (or ) of is the intersection of and the upper half plane (or the lower half plane respectively).

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