Homotheties and Similarities a File of the Geometrikon Gallery by Paris Pamfilos

Homotheties and Similarities A file of the Geometrikon gallery by Paris Pamfilos

We shall not cease from exploration And the end of all our exploring Will be to arrive where we started And know the place for the first time.

T.S. Eliot, Little Gidding

Contents (Last update: 30‑05‑2021)

1 Homotheties 1 2 Homotheties and triangles 2

3 Homotheties with different centers 4 4 Homotheties and translations 5 5 Representation and group properties of homotheties 6

6 Similarities, general definitions 7

7 Similarities defined by two segments 9

8 Similarities and orientation 13 9 Similarities and triangles 14

10 Triangles varying by similarity 16

11 Relative position in similar figures 18

12 Polygons on the sides of a triangle 19

13 Representation and group properties of similarities 24

14 Logarithmic spiral and pursuit curves 24

1 Homotheties

Homotheties of the plane and their generalization, the “similarities”, to be discussed be‑ low, are transformations which generalize those of “isometries” or “congruences” of the plane (see file Isometries). Given a number 휅 ≠ 0 and point 푂 of the plane, we call “homothety” of “center” 푂 and “ratio” 휅 the transformation which corresponds: a) to point 푂, itself, b) to every point 푋 ≠ 푂 the point 푋′ on the line 푂푋, such that the following signed ratio relation holds: 푂푋′ = 휅. 푂푋 A direct consequence of the definition is, that for every point 푂 the homothety of center 2 Homotheties and triangles 2

O X X'

Figure 1: Homothety

푂 and ratio 휅 = 1 is the identity transformation. Often, when the ratio is 휅 < 0 we say that the transformation is an “antihomothety”. Its characteristic is that point 푂 is between 푋 and 푋′.

Theorem 1. The composition of two homotheties with center 푂 and ratios 휅 and 휆 is a homothety of center 푂 and ratio 휅 ⋅ 휆.

Proof. Obvious consequence of the definition. If 푓 and 푔 are the two homotheties with the same center 푂 and ratios respectively 휅 and 휆, then, for every point 푋, points 푌 = 푓 (푋), 푍 = 푔(푌) and 푂 will be four points on the same line and will satisfy,

푂푌 푂푍 푂푍 푂푍 푂푌 = 휅, = 휆 ⇒ = ⋅ = 휆 ⋅ 휅 . 푂푋 푂푌 푂푋 푂푌 푂푋

Corollary 1. The inverse transformation of a homothety 푓 , of center 푂 and ratio 휅, is the homothety 1 with the same center and ratio 휅 . Remark 1. The homothety is a special transformation closely connected with Thales’ the‑ orem and the “similarity of triangles”, i.e “triangles which have equal corresponding angles” 퐿푒푓 푡푟푖푔ℎ푡푎푟푟표푤 “triangles which have proportional corresponding sides”.

Two triangles, and more general two shapes {Σ, Σ′} are called “homothetic” when there exists a homothety 푓 mapping one onto the other 푓 (Σ) = Σ′.

Exercise 1. Find all homotheties that transform a given point 푋 to another point 푋′ ≠ 푋.

Homothetic triangles are particular cases of “similar triangles” and are the key to in‑ vestigate properties of more general homothetic shapes.

2 Homotheties and triangles

Theorem 2. A homothety 푓 with center 푂 maps a triangle 푂푋푌 to a similar triangle 푂푋′푌′ with {푋′ = 푓 (푋), 푌′ = 푓 (푌)}, the sides {푋푌, 푋′푌′} being parallel.

Proof. A simple application of Thales’ theorem.

Corollary 1. A homothety maps a line 휀 to a parallel line 휀′.

Exercise 2. Find all homotheties that transform a given line 휀 to a given parallel to it 휀′. Distin‑ guish the cases {휀 ≠ 휀′ , 휀 = 휀′}.

Theorem 3. A homothety maps a triangle 퐴퐵퐶 to a similar triangle 퐴′퐵′퐶′.

Proof. Application of the previous corollary and Thales’ theorem.

Corollary 2. A homothety 푓 preserves the angles and multiplies the distances between points with its ratio. In other words, for every pair of points 푋, 푌 and their images 푋′ = 푓 (푋), 푌′ = 푓 (푌) holds |푋′푌′| = 휅|푋푌| and for every three points the respective angles are preserved 푌′̂푋′푍′ = 푌푋푍̂ . 2 Homotheties and triangles 3

Exercise 3. Show that there is no genuine homothety with ratio 푘 ≠ 1 mapping a triangle to itself.

Theorem 4. Two triangles 퐴퐵퐶 and 퐴′퐵′C′, which have their corresponding sides parallel are similar.

C C 1 C' Α Β Β Β 2 1 Α' Β' C 2 Figure 2: Triangles with corresponding sides parallel

Proof. Translate triangle 퐴′퐵′C′ and place it in such a way, that the vertices 퐴 and 퐴′ coincide and the lines of their sides 퐴퐵, 퐴C coincide respectively with 퐴′퐵′ and 퐴′C′ (See Figure 2). The translated triangle will take the position 퐴퐵1C1 or 퐴퐵2C2, with its third side parallel to 퐵C. Therefore, it will be similar to 퐴퐵퐶, while it is also congruent to the initial 퐴′퐵′C′.

Theorem 5. For two triangles 퐴퐵퐶 and 퐴′퐵′C′, which have parallel corresponding sides, the lines 퐴퐴′, 퐵퐵′ and CC′, which join the vertices with the corresponding equal angles, either pass through a common point and the triangles are homothetic, or are parallel and the triangles are congruent.

A' A

A B O B' C' B' C B O C' C A'

Figure 3: Homothetic triangles

Proof. Let 푂 be the intersection point of 퐴퐴′ and 퐵퐵′. We will show that CC′ also passes |퐴퐵| |푂퐴| |푂퐵| through point 푂. According to Thales, we have equal ratios |퐴′퐵′| = |푂퐴′| = |푂퐵′| =휅. Con‑ C C″ |푂C| ′ ′C″ sider therefore on 푂 point with |푂C″| =휅. The created triangle 퐴 퐵 has sides pro‑ portional to those of 퐴퐵퐶, therefore it is similar to it and consequently has the same angles. It follows, that 퐴′퐵′C′ and 퐴′퐵′C″ have 퐴′퐵′ in common and same angles at 퐴′ and 퐵′, therefore they coincide and C′ = C″, in other words, 푂C passes through C′ too. This reasoning shows also that, if the two lines 퐴퐴′ and 퐵퐵′ do not intersect, that is if they are parallel, then the third line will also be necessarily parallel to them and 퐴퐵퐵′퐴′, 퐵CC′퐵′ and 퐴CC′퐴′ will be parallelograms, therefore the triangles will have correspond‑ ing sides equal. 3 Homotheties with different centers 4

3 Homotheties with different centers

Theorem 6. The composition of two homotheties 푓 and 푔 with different centers 푂 and 푃 and ratios respectively 휅 and 휆, with 휅 ⋅ 휆 ≠ 1, is a homothety with center 푇 on the line 푂푃 and ratio equal to 휅 ⋅ 휆.

Proof. The proof is an interesting application of Menelaus’ theorem (see file Menelaus’ theorem). Let 푋 be an arbitrary point and 푌 = 푓 (푋), 푍 = 푔(푌). This defines the triangle 푂푌푃 and the points 푋, 푍 are contained in its sides 푂푌 and 푌푃 respectively. Let 푇 be the intersection point of 푍푋 with 푂푃. Applying Menelaus’ theorem we have,

Z Y X T P O

Figure 4: Composition of homotheties with 휅휆 ≠ 1

푋푂 푍푌 푇푃 푇푃 푋푌 푍푃 ⋅ ⋅ = 1 ⇒ = ⋅ . 푋푌 푍푃 푇푂 푇푂 푋푂 푍푌 However, for the oriented line segments holds

푋푌 푋푂 + 푂푌 푂푌 푋푌 = 푋푂 + 푂푌 ⇒ = = 1 + = 1 − 휅, 푋푂 푋푂 푋푂 푍푌 푍푃 + 푃푌 푃푌 1 푍푌 = 푍푃 + 푃푌 ⇒ = = 1 + = 1 − ⇒ 푍푃 푍푃 푍푃 휆 푇푃 푋푌 푍푃 1 휆 ⋅ (1 − 휅) = ⋅ = (1 − 휅) ⋅ ⎜⎛ ⎟⎞ = . 푇푂 푋푂 푍푌 1 휆 − 1 ⎝1 − 휆 ⎠ The last formula shows, that the position of 푇 on the line 푂푃 is fixed and independent 푇푍 of 푋. In addition, the ratio 휇 = 푇푋 is calculated, by applying Menelaus’ theorem to the triangle 푂푋푇, this time with 푃푌 as secant:

푃푇 푍푋 푌푂 ⋅ ⋅ = 1 ⇒ 푃푂 푍푇 푌푋 푍푋 푌푋 푃푂 = ⋅ ⇔ 푍푇 푌푂 푃푇 푍푇 + 푇푋 푌푂 + 푂푋 푃푇 + 푇푂 = ⋅ ⇔ 푍푇 푌푂 푃푇 1 푂푋 푇푂 1 − = (1 + ) (1 + ) ⇔ 휇 푌푂 푃푇 1 1 휆 − 1 1 − = (1 − ) (1 − ) ⇔ 휇 휅 휆(1 − 휅) 휇 = 휅휆 .

Theorem 7. The composition of two homotheties 푓 and 푔 with different centers 푂 and 푃 respec‑ tively and ratios 휅 and 휆 with 휅 ⋅ 휆 = 1 is a translation by interval parallel to 푂푃. 4 Homotheties and translations 5

Z X

P O

Figure 5: Composition of homotheties with 휅휆 = 1

Proof. Let 푋 be an arbitrary point and 푌 = 푓 (푋), 푍 = 푔(푌). This defines the triangle 푂푌푃 and the points 푋, 푍 are contained in its sides 푂푌 and 푌푃 respectively. According to the hypothesis 푌푋 푌푂 + 푂푋 1 푌푍 푌푃 + 푃푍 푃푍 1 = = 1 − , = = 1 − = 1 − 휆 = 1 − . 푌푂 푌푂 휅 푌푃 푌푃 푌푃 휅 The equality of the ratios shows, that the line segment 푋푍 is parallel to 푂푃. From the similarity of triangles 푌푂푃 and 푌푋푍, follows that

푋푍 = (1 − 휆)푂푃, therefore 푋푍 has fixed length and direction.

4 Homotheties and translations

Theorem 8. The composition 푔 ∘ 푓 of a homothety and a translation 푔 is a homothety. Proof. Assume that the homothety has center 푂 and ratio 휅 and the translation is defined by the fixed, oriented line segment 퐴퐵. Let also 푋 ≠ 푂 be arbitrary and 푌 = 푓 (푋), 푍 = 푔(푌). Assume finally that 푃 is the intersection of the line 푋푍 and the line 휀 is the parallel

Y A B Z X

ε P O

Figure 6: Compositions of homotheties and a translation to 퐴퐵 from 푂 (See Figure 6). From the similarity of the triangles 푋푌푍 and 푂푋푃 follows that 푂푃 푂푃 푂푋 푂푋 1 1 1 = = = = = ⇒ 푂푃 = 퐴퐵. 퐴퐵 푌푍 푌푋 푌푂 + 푂푋 푌푂+푂푋 1 − 휅 1 − 휅 푂푋 It follows that the position of 푃 on 휀 is fixed and independent of 푋. Also for the ratio, 푃푍 푂푌 = = 휅. 푃푋 푂푋 Therefore the composition 푔 ∘ 푓 is a homothety of center 푃 and ratio 휅.

Exercise 4. Show that the composition 푔 ∘ 푓 of a translation 푓 and a homothety 푔 is a homothety. Remark 2. The last theorems and the exercise show that homotheties and translations build a closed, as we say, set of transformations with respect to composition. We saw something similar also for rotations and translations (see file Isometries). 5 Representation and group properties of homotheties 6

Χ Ο X1 a ε Ο' Χ' = g(X) X 2 b Χ'' = h(X)

Figure 7: Composition of a homothety and a reflection

Exercise 5. Consider a line 휀 and a point 푂 at distance 푎 from it. Let 푓1 be the homothety with center at 푂 and ratio 푘 and 푓2 the reflection in 휀. Show that the compositions 푔 = 푓2 ∘ 푓1 and ℎ = 푓1 ∘ 푓2 differ by a translation. In other words, for every point 푋 of the plane it is valid ℎ(푋) − 푔(푋) = 푏, where 푏 is a line segment of length |2푎(1 − 푘)| and direction orthogonal to 휀 (See Figure 7).

Exercise 6. Given two circles with different radii, show that there exist homotheties which map one to the other. How many are there? What are their centers and ratios?

Exercise 7. Given two circles 휅 and 휆, draw a line intersecting them, which forms chords 퐴퐵, CD, having given lengths ([2, 푝. 21]).

Exercise 8. Show that a shape , with more than one points, for which there is a homothety 푓 , different from the identity, leaving Σ invariant (푓 (Σ) = Σ), extends to infinity. Find a shape example with this property.

Hint: If 푓 leaves invariant, then also the inverse homothety 푔 = 푓 −1 will leave it invariant. 1 If {푂, 푘} is the center and the ratio of 푓 , then {푂, 푘 } will be respectively the center and ratio of the inverse homothety. Thus, we can assume 푘 > 1. Then if 푋 is an arbitrary point of Σ, the 푋′ = 푓 (푋) will satisfy |푂푋′| = 푘|푂푋|. Repeating this procedure we find 푋″ = 푓 (푋′), with |푂푋″| = 푘2|푂푋| and after 푛 similar steps, we find points 푋(푛) = 푓 (푋(푛−1)), with |푂푋(푛)| = 푘푛|푂푋|. A shape example with the aforementioned property is a set of lines through a fixed point 푂.

5 Representation and group properties of homotheties

Fixing a cartesian coordinate system, the homothety with center 푂 and ratio 푘 is repre‑ sented using vectors by :

푌 = 푂 + 푘 ⋅ (푋 − 푂) ⇔ {푦1 = 표1 + 푘(푥1 − 표1) , 푦2 = 표2 + 푘(푥2 − 표2)}. (1)

Using matrices, this is equivalent to :

푦1 푘 0 (1 − 푘)표1 푥1 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜푦2⎟ = ⎜0 푘 (1 − 푘)표2⎟ ⋅ ⎜푥2⎟ . (2) ⎝ 1 ⎠ ⎝0 0 1 ⎠ ⎝ 1 ⎠

The product of two such matrices is of the same form :

′ ′ ′ ″ ″ ″ 푘 0 (1 − 푘)표1 푘 0 (1 − 푘 )표1 푘 0 (1 − 푘 )표1 ⎜⎛ ⎟⎞ ⎜⎛ ′ ′ ′ ⎟⎞ ⎜⎛ ″ ″ ″⎟⎞ ⎜0 푘 (1 − 푘)표2⎟ ⋅ ⎜ 0 푘 (1 − 푘 )표2⎟ = ⎜ 0 푘 (1 − 푘 )표2⎟ , (3) ⎝0 0 1 ⎠ ⎝ 0 0 1 ⎠ ⎝ 0 0 1 ⎠ 6 Similarities, general definitions 7

푘(1 − 푘′) 1 − 푘 with 푘″ = 푘 ⋅ 푘′ and 푂″ = 푂 + 푂′ . 1 − 푘푘′ 1 − 푘푘′ Equation (2) shows that the general homothetic transformation is a product of a homoth‑ ety centered at the origin and a translation with corresponding matrix representations:

푘 0 0 1 0 (1 − 푘)표1 ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ ⎜0 푘 0⎟ and ⎜0 1 (1 − 푘)표2⎟ . ⎝0 0 1⎠ ⎝0 0 1 ⎠

In the language of “groups” the preceding discussion and matrix representations reflect the following properties:

1. The set of “homotheties + translations” constitutes a group 퐺.

2. The homotheties with a given fixed center 푂 constitute a subgroup 퐺푂 of 퐺. 3. The set of translations constitute also a subgroup 푇 of 퐺.

6 Similarities, general definitions

The first encounter with the notion of similarity is perhaps thatof “similar triangles”, i.e. “triangles having equal respective angles” ⇔ “triangles having proportional respective sides”. The material discussed here handles this notion and its generalizations for more general shapes, like polygons, from the viewpoint of transformations of the plane. “Similarity” is called a transformation 푓 of the plane, which multiplies the distances of points with a constant 휅 > 0, which is called “ratio” or “scale” of the similarity. By definition then, for every pair of points {푋 , 푌} a similarity corresponds points

푋′ = 푓 (푋) , 푌′ = 푓 (푌) , which satisfy |푋′푌′| = 휅 ⋅ |푋푌| .

This general definition includes the “isometries” or “congruences”, for which 휅 = 1, and the homotheties. Simillarities not coincident with isometries, in other words, similarities for which 휅 ≠ 1 are called “proper” similarities. As we will see further down (Theorem 15), proper similarities are divided into two categories: “direct similarities” or “rotational similarities” and “antisimilarities” or “reflective similarities” [3, 푝. 217].

A direct similarity or rotational similarity is defined as a composition 푔 ∘ 푓 of a rotation 푓 and a homothety 푔, which shares the same center with 푓 . The rotation angle of 푓 is called “angle of similarity”. An antisimilarity is defined as a composition 푔 ∘ 푓 of a reflection 푓 and a homothety 푔 with center on the axis of the reflection 푓 . The axis of 푓 is called “axis of antisimilarity”.

Remark 3. In both categories therefore there exists a point, the center 푂 of the homothety 푔 which is fixed under the transformation. Obviously, proper similarities cannot have also a second fixed point 푇 different from 푂. For if they had, then for the two points and their images 푂′ = 푓 (푂) = 푂, 푇′ = 푓 (푇) = 푇 would hold |푂푇| = |푂′푇′|, while a proper similarity requires |푂′푇′| = 휅|푂푇| with 휅 ≠ 1. This unique fixed point is called “center” of the proper similarity.

The order of the transformations, which participate in the definition of a proper sim‑ ilarity, is irrelevant because of the following theorem. 6 Similarities, general definitions 8

Ζ Υ

ω Ο Χ Υ'

Figure 8: Commutativity of rotation and concentric homothety

Theorem 9. The two transformations, which participate in the definition of a proper similarity, commute (푔 ∘ 푓 = 푓 ∘ 푔).

Proof. Let us see the proof for the direct similarities, which are compositions 푔 ∘ 푓 of rota‑ tions 푓 and homotheties 푔 (See Figure 8). The proof for antisimilarities is similar. For the proof then, it suffices to observe the orbit of an arbitrary point 푋 under the application of the two transformations. According to 푔 ∘ 푓 , we first rotate 푋, about the center 푂 of the rotation, to 푌 and next we take the homothetic 푍 of 푌. It holds therefore (푋푂푌) = 휔 and 푂푍 푂푌 = 휅, where 휔 is the angle of rotation of 푓 and 휅 the homothety ratio of 푔. According to 푓 ∘ 푔, we first take the homothetic 푌′ of 푋 and next we rotate 푌′ by 휔. It is obvious that the two processes give the same final result, which is the point 푍.

Y Χ'

ω O X

Figure 9: Triangles 푂푋푌 for direct similarities

Theorem 10. For every direct similarity 푓 with center 푂 and rotation angle, which is not a mul‑ tiple of 휋, the triangles 푂푋푌 with 푌 = 푓 (푋), which result for the different positions of 푋 on the plane, are similar. Proof. Direct consequence of the definition, according to which 푋푂푌̂ is the rotation angle |푂푌| 휔 and the ratio |푂푋| is the ratio 휅 of the similarity,

(Ι) (ΙΙ) κ Υ ε' Χ' X O ε Α ε O Χ Υ

Figure 10: Antisimilarity ... and Apollonian circles

Theorem 11. For every antisimilarity 푓 with center 푂 and axis 휀 and every point 푋 of the plane, for which points 푂, 푋, 푌 = 푓 (푋) are not collinear, the angles 푋푂푌̂ have the same bisectors, which coincide with 휀 and its orthogonal 휀′ at 푂. Points 푋 of 휀 and 휀′ are the only points for which 푂, 푋, 푌 are collinear. 7 Similarities defined by two segments 9

Proof. Direct consequence of the definition, according to which the lines 푂푋, 푂푌 are al‑ ways symmetric relative to 휀 (See Figure 10‑I).

Exercise 9. Show that, if 푓 is an anitsimilarity with center at 푂 and axis 휀, then for every point |푂푌| |퐴푌| 푋 different from 푂 and its image 푌 = 푓 (푋), holds |푘| = |푂푋| = |퐴푋| , where 푘 is the similarity ratio and 퐴 is the intersection point of line 푋푌 with line 휀. Conclude that the Apollonian circle 휅 of the segment 푋푌 for the ratio |푘| (see file Apollonian circles), passes through points {푂, 퐴} and points {푋, 푌} are inverse with respect to 휅 (See Figure 10‑II).

B' ε

A B

Figure 11: {퐴퐵, 퐴′퐵′} equally inclined to the axis 휀

Exercise 10. Show that, if 푓 is an anitsimilarity with center at 푂 and axis 휀, then for any two points {퐴, 퐵} and their images {퐴′, 퐵′} the segments {퐴퐵, 퐴′퐵′} are equally inclined to the axis, i.e. a bisector of their angle is parallel to 휀 (see figure 11).

Exercise 11. Show that for every triple of non collinear points 푋, 푌, 푍 and their images 푋′, 푌′, 푍′ through a similarity, triangles 푋푌푍 and 푋′푌′푍′ are similar.

Exercise 12. Show that a direct similarity maps a triangle 퐴퐵C to a similar triangle 퐴′퐵′C′, which is also similarly oriented to 퐴퐵C. An antisimilarity reverses the orientation of the triangles.

Exercise 13. Show that a similarity maps a line 휀 to a line 휀′ and a circle 휅 to a circle 휅′.

Exercise 14. Show that two similarities 푓 , 푔, which are coincident at two different points 퐴 and 퐵, they are coincident at every point of the line 퐴퐵. Conclude then, that the composition of the transformations 푔−1 ∘ 푓 is either the identity transformation or a reflection.

Exercise 15. Show that two similarities 푓 , 푔, which are coincident at three non collinear points, they are coincident at every point of the plane.

7 Similarities defined by two segments

A basic property of similarities is expressed with the following theorem:

Theorem 12. For two line segments 퐴퐵 and 퐴′퐵′ of the plane, of different length, there exists a unique direct similarity which maps 퐴 to 퐴′ and 퐵 to 퐵′, consequently mapping 퐴퐵 to 퐴′퐵′. 7 Similarities defined by two segments 10

Β Α'

Α Β'

Figure 12: Similarity from two line segments

Proof. Leaving the special cases for the end, let us assume that the two segments are in general position and the lines they define intersect at a point 푇. This defines two circles (퐴퐴′푇) and (퐵퐵′푇) which intersect not only at 푇 but also at a second point 푂. The quadri‑ laterals 푇퐵푂퐵′ and 푇퐴푂퐴′ are inscriptible in circle, therefore their angles at 푂 are equal as supplementary to the angle at 푇. This shows that (퐴푂퐴′) = (퐵푂퐵′) and defines the rotation 푓 of the similarity. From this property follows that the angles of triangles 퐴′퐵′푂 and 퐴퐵푂 at 푂 are equal as are their angles at 퐴′ and 퐴 (as internal and opposite external in quadrilateral 퐴푂퐴′푇). It follows that the two triangles are similar and the similarity |퐴′퐵′| ratio is 휅 = |퐴퐵| . The wanted similarity then is the composition of the rotation 푓 and the similarity with ratio 휅 and center 푂.

Α' Β' A A Ο Ο B Β' B Α'

Figure 13: Similarity from two parallel line segments

In the special case where 푇 does not exist, that is when 퐴퐵 and 퐴′퐵′ are parallel and not collinear, then 푂 is the intersection point of 퐴퐴′ and 퐵퐵′. If 퐴퐵 and 퐴′퐵′ are equally |퐴′퐵′| oriented, then the wanted similarity is the homothety with center 푂 and ratio 휅 = |퐴퐵| . If 퐴퐵 and 퐴′퐵′ are inversely oriented, then the wanted similarity is the composition of the rotation 푓 by 휋 about 푂 (which coincides with point symmetry relative to 푂) and the |퐴′퐵′| ′ ′ homothety with ratio 휅 = |퐴퐵| relative to 푂. The reasoning for collinear 퐴퐵 and 퐴 퐵 is similar, but I leave this case as an exercise. The uniqueness of this similarity follows from the fact that the arguments can be re‑ versed. If 푂 is the center of a similarity, which maps 퐴퐵 to 퐴′퐵′, then for the angles, (퐴푂퐴′) = (퐵푂퐵′) and further the triangles 퐴푂퐵 and 퐴′푂퐵′ will be similar. This however means that the quadrilaterals 퐴푂퐴′푇 and 퐵푂퐵′푇 are inscriptible in circles and 푂 is the intersection point of the circles (퐴푇퐴′) and (퐵푇퐵′), as in the previous case.

Theorem 13. For two line segments 퐴퐵 and 퐴′퐵′ of the plane, of different length, there exists a unique antisimilarity, which maps 퐴 to 퐴′ and 퐵 to 퐵′, consequently mapping 퐴퐵 to 퐴′퐵′.

|퐴′퐵′| Proof. We want an antisimilarity whose ratio 휅 = |퐴퐵| we know. Therefore it suffices to find its center 푂′. This point will be on the bisectors of the angles 퐴푂̂′퐴′ and 퐵푂퐵̂ ′ (the‑ orem 11). These bisectors will intersect the respective sides 퐴퐴′ and 퐵퐵′ of the triangles 7 Similarities defined by two segments 11

O B'

A O'

A' k B k A

Figure 14: Antisimilarity from two line segments

퐴푂퐴′ and 퐵푂퐵′ at points which divide them in ratio 휅. Therefore 푂′ will be contained in the two Apollonian circles 푘퐴 and 푘퐵, which are respectively the loci of the points which divide segments 퐴퐴′ and 퐵퐵′ in ratio 휅. Consequently it will coincide with an intersec‑ tion point of these circles. A similar property will be valid also for the center 푂 of the direct similarity, which is guaranteed by the previous theorem. Therefore this, too will be contained in the intersection of 푘퐴 and 푘퐵. Consequently the two circles will intersect. From the equality of ratios |푂퐴| |푂퐵| |푂′퐴| |푂′퐵| |퐴퐵| = = = = , |푂퐴′| |푂퐵′| |푂′퐴′| |푂′퐵′| |퐴′퐵′| it follows that triangles 푂′퐴퐵 and 푂′퐴′퐵′ are similar and 푂퐴퐵, 푂퐴′퐵′ are also equal. In the case where the two circles intersect at exactly two points (See Figure 14), it is impossible for both pairs of similar triangles to consist of similarly oriented triangles. This, because otherwise we would have two direct similarities with centers at 푂 and 푂′, something which is excluded by the previous theorem. Therefore one of the two pairs will consist of reversely oriented triangles and consequently one of the two will be an antisimilarity and the other a direct similarity (see remark 4).

Δ' Β' Β Δ Γ' Α Γ Α'

Figure 15: Coincidence of centers of similarity and antisimilarity

If the two points 푂 and 푂′ coincide then 퐴퐴′ and 퐵퐵′ must be parallel (See Figure 15). Indeed, then, the bisectors of the angles 퐴푂퐴′ and 퐵푂퐵′ will coincide and the two circles ′ ′ 푘퐴 and 푘퐵 will be tangent at 푂. However the lines 퐴퐴 and 퐵퐵 contain the diametrically opposite pairs of points C, C′ and D, D′ respectively, which are defined by the mutually orthogonal bisectors which pass through 푂. Because of the circle tangency at 푂, the di‑ ameters CC′ and DD′, which are excised by the two orthogonal lines on the circles are parallel, something which proves the claim. In this case the direct similarity has rotation angle 퐴푂퐴̂ ′ and the antisimilarity has axis line CD. In the special case, in which the lines 퐴퐵 and 퐴′퐵′ are parallel, the two Apollonian circles pass through the intersection point 푂 of 퐴퐴′ and 퐵퐵′, which is a homothety center 7 Similarities defined by two segments 12

Α'

Β' Β Ο Ο'

Figure 16: Centers of similarity and antisimilarity when 퐴퐵||퐴′퐵′ and, consequently, the center of a direct similarity between 퐴퐴′ and 퐵퐵′. The antisimilar‑ ity center coincides in this case also with the other intersection point 푂′ of the two circles (See Figure 16).

Exercise 16. Complete the proof of the last two theorems, by examining the case where 퐴퐵 and 퐴′퐵′ are on the same line.

T B'

B P2

P1 S O λ 2 1 A

λ2 A' D1 O1

κ2 Q κ1

Figure 17: The line 푄푇 and the right angle 푄퐷̂1퐷2

Theorem 14. Let {퐷1, 퐷2} be the similarity centers of the direct and indirect(antisimilarity) map‑ ping the segment 퐴퐵 onto 퐴′퐵′ (see figure 17). Let also 푄 denote the intersection 퐴퐵 ∩ 퐴′퐵′ ′ ′ and consider the circles {휅1 = (퐵푄퐵 ), 휅2 = (퐴푄푄 )}, whose second intersection defines 퐷1. The ′ ′ second intersection point 퐷2 ≠ 퐷1 of the Apollonian circles {휆1, 휆2} of the segments {퐴퐴 , 퐵퐵 } w.r.t. the ratio 푟 = 퐴퐵/퐴′퐵′ defines the center of the antisimilarity mapping 퐴퐵 onto 퐴′퐵′. Let also {푆 = 휅1 ∩ 휆1, 푇 = 휅2 ∩ 휆2}. The following are valid properties. 1. Triangles {퐴푆퐴′, 퐵푇퐵′} are similar. 2. Points {푇, 푆, 푄} are collinear.

3. 퐷2 is collinear with {푇, 푆, 푄}.

4. The angle 퐷2̂퐷1푄 is right. ′ ′ 5. The lines {푄퐷1, 푄퐷2} are harmonic conjugate w.r.t. {퐴퐵, 퐴 퐵 }. 8 Similarities and orientation 13

Proof. Nr‑1 is valid because the ratios 푆퐴/푆퐴′ = 푇퐵/푇퐵′ = 푟 and the angles are equal: 푆̂ = 푇̂ = 휋 − 푄.̂ Nr‑2 is valid because 퐴푄푆̂ = 퐴퐴̂′푆 = 퐵퐵̂′푇 = 퐵푄푇.̂ Nr‑3 is valid because the line 푇푆 is characterized by the ratio of distances of its points ′ ′ from the segments 푑(푋, 퐴퐵)/푑(푋, 퐴 퐵 ) = 푟, satisfied by {푆, 푇}. But 퐷2 satisfies also this condition, hence belongs to that line. Nr‑4 follows by an angle chasing argument. 퐷2̂퐷1푂2 = 퐷1̂푇퐷2 because 퐷1푂2 is tan‑ ′ gent to 휆2 at 퐷1. This follows from the fact that 휅2 passing through {퐵, 퐵 }, which are in‑ ̂ 1 ̂ ̂ verse relative to 휆2 is orthogonal to 휆2. Also 푂2퐷1푄 = 2 (휋 − 퐷1푂2푄) = 휋/2 − 퐷1푇푄. Nr‑5 is a consequence of the characterization of their points to have ratio of distances from {퐴퐵, 퐴′퐵′}: 푑(푋, 퐴퐵)/푑(푋, 퐴′퐵′) = 푟.

Remark 4. The preceding theorem makes more precise the distinction between the centers {퐷1, 퐷2} respectively of the direct similarity and the antisimilarity mapping the segment ′ ′ 퐴퐵 onto 퐴 퐵 : In the right angled triangle 푄퐷1퐷2 the right angle is at 퐷1.

ν T 2 B'

B P2

M P1 O λ 2 1 A S

λ2 A' D1 O1

κ2 Q κ1

Figure 18: Collinear points {푂2, 푆, 푀}

Exercise 17. With the notation and conventions of the preceding theorem show that:

1. The triangle 퐷1푂2푃2 is similar to △퐷2퐷1푄 (see figure 18).

2. The line 푂2푆 passes through the second intersection point 푀 of the two circles 휆1 and 푃2퐷1푂2 .

Hint: For nr‑2 consider 푀 as second intersection of the circle 휈2 = (푃2퐷1푂2) with line 푂2푆 and show that 푀 ∈ 휆2. For this it suffices to show that 퐷̂1푀푆 = 퐷̂1퐷2푆.

8 Similarities and orientation

Theorem 15. Every proper similarity is a direct similarity, if it preserves the orientation of tri‑ angles and an antisimilarity if it reverses the orientation of triangles. 9 Similarities and triangles 14

Proof. Indeed, let 푋, 푌 be two different points and 푋′ = 푓 (푋), 푌′ = 푓 (푌) their images by the similarity. Assume also that 푓 preserves the orientation of triangles and 푔 is the direct similarity, which maps 푋 to 푋′ and 푌 to 푌′ (Theorem 12). Then the two similarities 푓 and 푔 coincide on the entire line 푋푌 (Exercise 12). Let 푍 be a point not on the line 푋푌. The triangle 푋푌푍 maps by 푓 to the similar and similarly oriented (to 푋푌푍) triangle 푋′푌′푍′. The same happens with 푔. It also maps 푋푌푍 to a similar and similarly oriented triangle 푋′푌′푍″. Triangles 푋푌푍, 푋′푌′푍′, 푋′푌′푍″ are similar and similarly oriented, and the last two have 푋′푌′ in common. Therefore they either coincide or one is the mirror image of the other. The latter however cannot happen, because then the two triangles would have reverse orientation. Therefore the triangles coincide and consequently 푍′ = 푍″, in other words 푓 and 푔 are coincident on three non collinear points, therefore they are coincident everywhere and holds 푓 = 푔. The case where the transformation 푓 reverses the orientation of the triangles is proved similarly.

Corollary 2. Every proper similarity has exactly one fixed point.

Exercise 18. Determine the fixed point of a given proper similarity 푓 .

Hint: Use Theorem 12 for direct similarities and Theorem 13 for antisimilarities ([4, 푝. 74]).

9 Similarities and triangles

Exercise 19. Show that, for two similar but not congruent triangles 퐴퐵C and 퐴′퐵′C′, there exists a unique proper similarity which maps 퐴퐵C to 퐴′퐵′C′.

Hint: Use the similarity (direct or antisimilarity) which maps 퐴퐵 to 퐴′퐵′.

The next two exercises show, that in the definition of the proper similarity it is not neces‑ sary to restrict ourselves to homotheties and rotations (resp. reflections) with coincident centers (resp. with homothety center on the the axis of the reflection). Even if the cen‑ ters are different (resp. the center is not on the axis of reflection), the composition ofa homothety and a rotation (resp. reflection) is a proper similarity.

Exercise 20. Show that the composition 푔 ∘푓 of a homothety 푓 with center 푂 and a rotation 푔 with center 푃 ≠ 푂 is a direct similarity with rotation angle that of 푔, ratio that of 푓 and center which is determined by 푓 and 푔. Show that the same happens also for the composition 푓 ∘ 푔.

Z Y Z (I) (II) X X Y B B M M A A

O P O P

Figure 19: Composition of homothety and rotation 9 Similarities and triangles 15

Hint: Let 휅 be the ratio of the homothety 푓 and 휔 be the angle of the rotation 푔. There exists an isosceles triangle 푃퐴퐵, with vertex at the center 푃 of the rotation, whose two other vertices 퐴, 퐵 are centers of the similarities 푔 ∘ 푓 and 푓 ∘ 푔 ((I) and (II) in figure 19, respectively). This triangle can be constructed using two characteristic properties it has: a) an apical angle equal to 휔 and b) 퐵 = 푓 (퐴). Indeed, if such a triangle exists, then 퐵 will see the line segment 푂푃 under the angle 휋−휔 휋+휔 2 and 퐴 will see 푂푃 under the angle 2 . Both of the latter if 휅 > 1. If 휅 < 1 the roles of 퐴 and 퐵 must be reversed. Let us then assume that 휅 > 1 and that 푓 (퐴) = 퐵. Point 퐵 is 휋−휔 on the intersection of the arc of the points which see 푂푃 under angle 2 and of the arc which results through the homothety 푓 from the arc of points which see 푂푃 under angle 휋+휔 2 . Consequently point 퐵 is constructible and from it the isosceles 푃퐴퐵 with angle 휔 at 푃 is also constructible. Then 푔(푓 (퐴)) = 푔(퐵) = 퐴, therefore point 퐴 is a fixed point of ℎ = 푔 ∘ 푓 . Let 푋 an arbitrary point, 푌 = 푓 (푋) and 푍 = 푔(푌). The angle (푋퐴푍) = 휔. Indeed, the triangles 푃퐴푍 and 푃퐵푌 are congruent, because they have |푃퐴| = |푃퐵| by hypothesis, |푃푌| = |푃푍|, since point 푍 results from 푌 through a rotation about 푃 and the angles 퐴푃푍̂ , ̂ ̂ |푂퐵| 퐵푃푌 are equal since both added to 푍푃퐵 give 휔. Also, because of the similarity, 휅 = |푂퐴| = |푂푌| |퐴푍| |푂푋| , therefore 퐴푋 and 퐵푌 are parallel and |퐵푌| = |퐴푍|. Therefore |퐴푋| = 휅 and the angle between the lines 퐴푍 and 퐴푋 is equal to the angle between 퐴푍 and 퐵푌, which is 휔. Consequently, the correspondence 푍 = 푔(푓 (푋)) coincides with the composition 푔′ ∘ 푓 ′, where 푓 ′ is the rotation about 퐴 by 휔 and 푔′ is the homothety relative to 퐴 with ratio 휅. We have then 푔∘푓 = 푔′ ∘푓 ′ and the second composition satisfies the definition of the direct rotation. The proof of the claim for the other ordering of the composition, that is 푓 ∘ 푔 (corre‑ sponding to case (II) in figure 19) is similar.

Exercise 21. Show that the composition 푔 ∘ 푓 of a homothety 푓 with center 푂 and a reflection 푔 with axis 휀, which does not contain 푂, is an antisimilarity with axis a line 휀′, parallel to 휀 and center the projection 푃 of 푂 on 휀′. Show that the same happens also for the composition 푓 ∘ 푔.

(I) X' (II) ε' ε X ε O X O Y ε'

P Z Y

M Z M P Y'

Figure 20: Composition of homothety and reflection

Hint: The key role here is played by the circle with center the projection 푀 of 푂 on the 휅−1 axis of 푔 and radius 푟 = 휅+1 |푂푀|. The intersection point 푃 of this circle with 푂푀, which is contained between points 푂 and 푀 is proven to be a fixed point of 푔 ∘ 푓 . Its diametrically opposite is proven to be a fixed point of 푓 ∘ 푔 (cases (I) and (II) respectively in figure 20). The rest follows easily from the figures, in which 푋 is an arbitrary point of the plane, 푌 = 푓 (푋) (resp. 푌 = 푔(푋)) and 푍 = 푔(푌) (resp. 푍 = 푓 (푌)). In the first case 푔 ∘ 푓 = 푔′ ∘ 푓 ′, where 푓 ′ is the homothety with center 푃 and ratio equal to the ratio 휅 of 푓 and 푔′ is the 10 Triangles varying by similarity 16 reflection relative to the line 휀′, which is parallel to 휀 and passes through point 푃. In the second case 푓 ∘ 푔 = 푓 ″ ∘ 푔″, where 푔″ is the reflection relative to 휀′, which is parallel to 휀 and passes through point 푃 and 푓 ″ is the homothety with center 푃 and ratio equal to 휅. The figures show the trajectory of the arbitrary point 푋 under the application of these new transformations. In the first case point 푋 maps by 푓 ′ to 푋′, which next, by 푔′ maps to 푍. In the second case point 푋 maps by 푔″ to 푌′, which, by 푓 ″ maps to 푍′.

Exercise 22. Show that the composition of two direct similarities is a direct similarity with rota‑ tion the sum of the rotations and ratio the product of the ratios if their angles sum up to 휔 + 휔′ ≠ 2푘휋 and the ratios 휅 and 휅′ satisfy 휅 ⋅ 휅′ ≠ 1, otherwise it is a translation.

Hint: Combination of the two previous exercises and of the fact, that the product of rota‑ tions is a rotation (see file Isometries). Write the two similarities in their “normal” form, as compositions of homotheties and rotations with the same center: 푔 ∘ 푓 and 푔′ ∘ 푓 ′. Then their composition would be (푔′ ∘푓 ′)∘(푔∘푓 ) = (푔′ ∘푓 ′)∘(푓 ∘푔) = 푔′ ∘(푓 ′ ∘푓 )∘푔. Consider next the aforementioned fact about the composition of two rotations, which is also a rotation, applied to 푓 ′ ∘ 푓 and subsequently to the resulting rotation or translation ℎ apply Exercise 20 or Theorem 8, etc ([9, 푝. 42, II]).

Exercise 23. State and prove an exercise similar to the previous one for the composition of two antisimilarities and the composition of an antismilarity and a direct similarity.

As it happens with isometries and congruence, so it happens also with similarity trans‑ formations, which are at the root of a general definition of the similarity for plane shapes: Two shapes 푆, 푆′ of the plane are called “similar”, when there exists a similarity 푓 which maps one to the other (푓 (푆) = 푆′).

10 Triangles varying by similarity

Theorem 16. The triangle 퐴퐵퐶 varies in such a way, that the measure of its angles remain fixed, its vertex at 퐴 also remains fixed and its vertex 퐵 moves along a fixed line 휀. Then, its third vertex C moves along another fixed line 휁, which forms with 휀 an angle equal to 퐵퐴̂C.

α C' γ γ β C

Α α

β β ε

Ο Β Β'

Figure 21: Variable triangle with fixed angles

Proof. Let 퐴퐵퐶 be one triangle with the aforementioned properties (See Figure 21). Con‑ sider the circumscribed circle of this triangle and the second intersection point 푂 of this circle (different from 퐵) with line 휀. Because 푂 sees 퐵C under the same angle as 퐴, the 11 Relative position in similar figures 17 angle at 푂 will be fixed and equal to the triangle angle 훼 (the angles remain fixed, only the dimensions and the position of the triangle change). Because the quadrilateral 퐴푂퐵C is inscribed in the circle, the angle 퐴푂퐵̂ , which is opposite to 훾, will be its supplemen‑ tary, something which completely determines the position of point 푂. Consequently, C is contained in the line, which passes through point 푂 and forms angle 훼 with 휀.

(Ι) (ΙΙ) C Β' β Β γ κ Α α C γ

α β ε Ο Α Α' Β ε

Figure 22: Vertices on lines Vertices on line and circle

Exercise 24. Construct a triangle 퐴퐵C with given angles, whose vertices {퐴, 퐵, 퐶} are contained respectively in three lines {훼 , 훽 , 훾}.

Hint: Consider an arbitrary point C of 훾 and triangles C퐴′퐵′ with angles equal respectively to the given and the vertex 퐴′ on line 훼 (See Figure 22‑I). Then (Theorem 16) vertex 퐵′ is contained always in a fixed line 휀. Consider the intersection point 퐵 of 휀 and of 훽 and define the triangle 퐴퐵C.

Exercise 25. Construct a triangle 퐴퐵C with given angles, whose vertex at 퐴 is a fixed point, the vertex at 퐵 is contained in a line 휀 and the vertex C is contained in a circle 휅 (See Figure 22‑II).

ε Δ Γ Ζ' (I) (II) Ρ E ζ Ρ Ζ

E' ε Ε Ε' Z' ABΗ Z

Figure 23: Equilateral inscribed in square Equilateral between parallels

Hint: Consider all equilateral triangles 푃푍′퐸′, with 푍′ on line 퐴퐵. According to Theorem 16, the other vertex 퐸′ of all these triangles varies on a definite line 휀 (See Figure 23‑I). One intersection point 퐸 of this line and the square, different from the intersection point 퐻 of 휀 with 퐴퐵, defines the base of the wanted triangle 푃푍퐸.

Exercise 26. Construct an equilateral triangle, which has one vertex at a given point and the other two vertices on given parallel lines (See Figure 23‑II). Also construct an equilateral triangle which has its three vertices on three given parallel lines. 11 Relative position in similar figures 18

Ν Α Ζ' Α' Β Ζ Ν' Ε' Μ Μ' D' Ε C Β' D C'

Figure 24: 푀 and 푀′ have the same relative position with respect to the polygon

11 Relative position in similar figures

Given two similar polygons 푝 = 퐴퐵CD... and 푝′ = 퐴′퐵′C′D′..., we say that the points 푀 and 푀′ have respectively the same “relative position” with respect to the polygons, when all the triangles which are formed by connecting 푀 and 푀′ with corresponding vertices of the polygons are respectively similar. In figure 24 points 푀 and 푀′ have the same relative |푀퐴| |푀퐵| position with respect to the two similar polygons. The ratios |푀′퐴′| = |푀′퐵′| = ... are all equal to the similarity ratio of the two polygons. It follows directly from the definition, that the distances between two points with the same relative positions have themselves |푀푁| ratio |푀′푁′| equal to the ratio of similarity of the two polygons. More generally, it can be proved easily, that if the points 푋, 푌, 푍, ... and 푋′, 푌′, 푍′, ... have respectively the same relative position with respect to the similar polygons 푝 and 푝′, then the polygons 푋푌푍... and 푋′푌′푍′... are similar and their similarity ratio is equal to the similarity ratio of 푝, 푝′. Some special points in 푝, 푝′ which have the same relative position, are corresponding vertices, the middles of corresponding sides, etc. Remark 5. Two similar polygons {푝, 푝′} define a similarity 푓 which maps one to the other: 푓 (푝) = 푝′. Thus, to say that {푋, 푋′} have the same relative position with respect to {푝, 푝′} is the same with saying that they correspond under 푓 ∶ 푓 (푋) = 푋′. Often in applications we consider polygons 푝 = 퐴퐵CD... which vary, remain however similar to a fixed polygon. We say then that the polygon varies “by similarity”. Points which are co‑varied with such a polygon, retaining however their relative position with respect to the similar polygons, we say that they remain “similarly invariant”. In figure 24 polygon 푝′ = 퐴′퐵′C′... results by similarity from 푝 = 퐴퐵C... and, following this change, points 푀, 푁 remain similarly invariant relative to the changing polygon, taking the same relative positions 푀′, 푁′ with respect to the similar polygon 푝′. Such a variation by simi‑ larity we met already in Theorem 16, which can be re‑expressed as follows: If the triangle 퐴퐵C varies by similarity, so that one of its vertices remains fixed, while another moves on a line, then the third vertex as well moves on a line. The next two theorems generalize this property. Theorem 17. Suppose that the polygon 푝 = 퐴퐵CD... varies by similarity and in such a way, that the point 푀 of its plane retains its position fixed, not only its relative position with respect to 푝, but also its absolute position on the plane. Suppose further that, under this variation, the similarly invariant point 푋 of the polygon moves on a line 휀푋, then every other similarly invariant point 푌 of 푝 will move on a line 휀푌.

Proof. This follows directly by applying Theorem 16 to triangle 푀푋푌, which, according to the previous comments, varies but remains similar to itself. Let us note that the polygon’s vertices are special points which remain similarly invariant, therefore they too will draw 12 Polygons on the sides of a triangle 19

ε Υ Ε' Α ε 0 Ε Χ 0 Α D' Υ Μ Ε 0 0 D Μ 0 Α' Υ Β Χ Χ' 0 0 Υ' D Β Χ C ε 0 Γ C' Β' C

Figure 25: Variation by similarity with 푀’s position relatively and absolutely fixed

lines. In figure 25 the line 휀C is shown, which the vertex C of the polygon slides on. In the same figure can also be seen a similar polygon 푝0 = 퐴0퐵0C0... to 푝, on which the corresponding points 푀0, 푋0, 푌0 can be distinguished. These points have in 푝0 the same relative position with that of 푀, 푋, 푌 in 푝.

Theorem 18. Suppose that the polygon 푝 = 퐴퐵CD... varies by similarity and in such a way, that the point of its plane 푀 retains its position fixed, not only its relative position with respect to 푝, but also its absolute position on the plane. Suppose also that, during this variation, the similarly invariant point 푋 of the polygon moves on a circle 휅푋, then every other similarly invariant point 푌 of 푝 will be moving on a circle 휅푌.

κ Χ Α Μ 0 Ο κ Υ Ε Μ Β Β 0 0 Χ Χ Ο' 0 Υ Ε Υ 0 0

C D 0 D C 0

Figure 26: Variation by similarity with the position of 푀 relatively and absolutely fixed II

Proof. During the variation of the polygon 푝, the triangle 푀푋푌, whose vertices remain similarly invariant, will remain similar to itself. The angles of this triangle will remain fixed, point 푀 will remain fixed and 푋 will move on a fixed circle 휅푋(푂, 휌) (See Figure 26). We consider the triangle 푀푂푋 and we construct its similar 푀푂′푌, such that the angle 푌푀푂̂ ′ is equal to 푋푀푂̂ and 푂̂′푌푀 is equal to 푂푋푀̂ . Triangles 푀푂푂′ and 푀푋푌 have then their sides at 푀 proportional and their angles at 푀 equal. Consequently the triangles are similar. Because 푂푀 is fixed, it follows that 푀푂′ is also fixed, consequently point 푂′ will be fixed. From the similarity of triangles 푂푋푀, 푂′푌푀 it follows that 휌′ = |푂′푌| will ′ ′ also be fixed, therefore 푌 will be moving on the circle 휅푌(푂 , 휌 ). Figure 26 also shows a polygon 푝0 = 퐴0퐵0C0..., similar to 푝, on which the corresponding points 푀0, 푋0, 푌0 are shown. These points have on 푝0 the same relative position with that of 푀, 푋, 푌 on 푝.

12 Polygons on the sides of a triangle

Theorem 19. The ratio of areas of two similar polygons Π and Π′ is 휖(Π) = 휅2, 휖(Π′) 12 Polygons on the sides of a triangle 20 where 휅 is the similarity ratio of the two polygons. Proof. From one vertex in Π and its corresponding in Π′ draw the diagonals and divide Π and respectively Π′ into triangles respectively similar to the previous. The areas of the polygons are written as a sum of the areas of these triangles 휖(Π) = 휖(푡1) + 휖(푡2) + ... ′ ′ ′ and respectively 휖(Π ) = 휖(푡1) + 휖(푡2) + … . The corresponding triangles are similar ′ 2 ′ 2 therefore휖(푡1) = 휅 휖(푡1), 휖(푡2) = 휅 휖(푡2), ... and the assertion follows using these rela‑ tions in the previous equations for areas:

′ ′ ′ 휖(Π ) = 휖(푡1) + 휖(푡2) + ... 2 2 = 휅 휖(푡1) + 휅 휖(푡2) + ... 2 = 휅 ⋅ (휖(푡1) + 휖(푡2) + ...) = 휅2휖(Π),

Remark 6. The preceding relation, between areas of similar polygons, leads to another form of the Pythagorean theorem in which, instead of squares on the sides of the right triangle, we construct on the sides polygons similar to a given polygon.

Π 2 Π B C b 1 a D c Π E Π A 3

Figure 27: Generalized theorem of Pythagoras

Theorem 20. Given is a polygon Π = 퐴퐵CD... and a right triangle. On the sides of the right triangle are constructed polygons Π1 , Π2 , Π3 similar to Π , such that the sides of the right triangle 푎, 푏, 푐 are homologous to the side 퐴퐵 of Π (See Figure 27). Then the sum of the areas of the polygons on the orthogonal sides is equal to the area of the polygon on the hypotenuse

휖(Π1) + 휖(Π2) = 휖(Π3). Proof. Let |퐴퐵| = 푑 be the length of the side 퐴퐵 of Π . According to Theorem 19, the ratios of areas will be 휖(Π ) 푎2 휖(Π ) 푏2 휖(Π ) 푐2 1 = , 2 = , 3 = . 휖(Π) 푑2 휖(Π) 푑2 휖(Π) 푑2 The claim follows directly by solving for 푎2, 푏2, 푐2 the previous relations and replacing in the theorem of Pythagoras: 푎2 + 푏2 = 푐2.

Exercise 27. Construct a square D퐸푍퐻, inscribed in the triangle 퐴퐵퐶 with its side D퐸 on 퐵C. Hint: Let us suppose that the square has been constructed (See Figure 28‑I). We extend 퐴D and 퐴퐸 until they meet the orthogonals of 퐵C at 퐵 and C, respectively, at points 퐾 and 퐼. (퐴퐻D, 퐴퐵퐾) and (퐴푍퐸, 퐴C퐼) are pairs of similar triangles, consequently |퐻D| |퐴퐻| |푍퐸| |퐴푍| = , = . |퐵퐾| |퐴퐵| |C퐼| |퐴C| 12 Polygons on the sides of a triangle 21

A A (I) (II) I K H Z Λ Θ C B N M B C DE D H O

E Z

Κ I H

Figure 28: Square inscribed in triangle Hexagon inscribed in triangle

Because 퐻푍 and 퐵C are parallel, the ratios on the right sides of the equalities are equal, therefore the ratios on the left sides will be equal. This implies that |퐵퐾| = |C퐼|, therefore 퐵C퐼퐾 is a rectangle and consequently the triangles 퐴D퐸 and 퐴퐾퐼 are similar. Because |D퐸| |퐴D| |퐴퐻| C |퐾퐼| = |퐴퐾| = |퐴퐵| , it follows that 퐾퐼 and 퐵퐾 are equal, therefore 퐵 퐼퐾 is a square. This square can be constructed directly and by drawing 퐴퐾 and 퐴퐼 we define D퐸 on 퐵C and from this D퐸푍퐻, which is proved being a square with similar reasoning. Exercise 28. Construct an equilateral hexagon with three sides on the sides of triangle 퐴퐵퐶 and three parallel to them. Hint: Let Θ퐼퐾Λ푀푁 be the wanted hexagon (See Figure 28‑II). Construct the similar to it on side 퐵C. For this, define equal segments on the extensions of 퐴퐵 and 퐴C respectively: |퐵퐻| = |퐵C| = |CD| and consider the middle 푂 of 퐻D. Next consider the symmetric 퐸, 푍 relative to 푂 of 퐵, C respectively. The first hexagon is equilateral by assumption and the second by construction. Also the two hexagons have corresponding sides parallel, hence they are also homothetic. The three points 퐴, 푁 and 푍 are collinear, as well as the three points 퐴, 푀 and 퐸. Triangles 퐼Θ푁 and 퐵퐻푍 are isosceli with equal apical angles Θ and 퐻. Therefore 퐼푁 and 퐵푍 are parallel and their ratio is |퐼푁| |Θ퐼| |퐼퐾| = = . |퐵푍| |퐻퐵| |퐵C| However 퐴퐼퐾 and 퐴퐵퐶 are similar, consequently |퐼퐾| |퐴퐼| = = 휅. |퐵C| |퐴퐵| The two hexagons, Θ퐼퐾Λ푀푁 and 퐻퐵CD퐸푍 are homothetic relative to 퐴 with homothetic ratio 휅. The equilateral hexagon 퐻퐵CD퐸푍 can be constructed directly and, by drawing 퐴푍 and 퐴퐸, we define the points 푁 and 푀 on 퐵C. From there on the construction of Θ퐼퐾Λ푀푁 is easy, through parallels to the sides of the triangle 퐴퐵퐶. Exercise 29. Show that the equilateral hexagon of the preceding exercise is unique, consequently the three hexagons which result from the previous construction, but starting from a different tri‑ angle side, coincide. Hint: From the previous exercise, it follows that for every hexagon like Θ퐼퐾Λ푀푁 (See Figure 28‑II), its vertex 푁 is contained in a line 퐴푍 which is dependent only from the 12 Polygons on the sides of a triangle 22 triangle 퐴퐵C. This line intersects 퐵C at exactly one point, 푁, and consequently there is only one triangle 퐼Θ푁 defined and similar to 퐵퐻푍 with its vertex 푁 on 퐵C. From the uniqueness of 푁 follows that of Θ , next of 퐼 etc. The second part of the exercise is a direct logical consequence of its first part.

M N Ι Α Κ K Λ (Ι) Ξ (ΙΙ) A Ρ Π Ν I Ξ T C Π Θ B O Τ Ο D D Β Σ Ρ H Σ Λ Μ Θ Η Z E C

Figure 29: Escribed equilateral hexagons Circumscribed parallelogram

Exercise 30. Construct the three escribed equilateral hexagons on the sides of triangle 퐴퐵퐶 (See Figure 29‑I). Show that the lines which join the centers of symmetry Π, 푃 and Σ of these hexagons pass through corresponding vertices of the triangle 퐴퐵퐶. Hint: Show first that the halves of two such polygons, like for example the quadrilaterals 퐶퐷퐻퐵 and 퐶퐴푀푂 are similar. Exercise 31. From the vertices of the convex quadrilateral 퐴퐵퐶퐷 draw parallels to the diagonals, which do not contain them. This forms a parallelogram 퐻ϴ퐼퐾 (See Figure 29‑II). Show that the point Π of the intersection of its diagonals, the intersection point Σ of the diagonals of 퐴퐵퐶퐷 as well as the middles of its diagonals, define the vertices of another parallelogram. Also show that the parallelogram Λ푀푁Ξ of the middles of the sides of 퐴퐵퐶퐷 is homothetic to 퐻ϴ퐼퐾 and find the ratio and the center of the homothety.

Ν Ι

Μ Π Η Ο Κ

Β D Λ Ρ Θ Ε Ξ

Figure 30: Homothetic to quadrilateral

Exercise 32. In the convex quadrilateral 퐴퐵퐶퐷 we define the centroids 퐻, ϴ, 퐾, 푀 of the triangles 퐴퐵퐶, 퐵퐶퐷, 퐶퐷퐴, 퐷퐴퐵, respectively. Show that 퐻ϴ퐾푀 is homothetic to 퐴퐵퐶퐷 with homothety 12 Polygons on the sides of a triangle 23

1 ratio 3 and homothety center the common middle 푂 of the line segments which join the middles of its opposite sides (See Figure 30).

Hint: First show that the triangles like 퐵푁퐸 and 퐾푃Π are homothetic relative to the ho‑ mothety with center 푂 and ratio 3. Points 푃, Π are respectively the middles of 퐾ϴ and 퐾푀. Next exercise gives an application of similarity, which produces a relatively simple solution to a complex problem. The problem is related to the construction of triangles on the sides of a given triangle. Given a triangle 퐴퐵C and three other triangles 휏1, 휏2, 휏3, we choose a side on each one of the three last triangles. Next we construct on the sides of triangle 퐴퐵C externally lying triangles 퐵DC, C퐸퐴, 퐴푍퐵 respectively similar to 휏1, 휏2, 휏3, on 퐴퐵C, so that the respectively similar to the selected sides of the three triangles coincide with the sides of the triangle ([5, 푝. 141]). I call such a construction briefly: a welding by similarity of 휏1, 휏2, 휏3 onto 퐴퐵C. The resulting points {퐷, 퐸, 푍} I call vertices of the welding (See Figure 31).

Ζ Ε Α Z X τ f g 3 τ h 2 Y Β C D Ω

τ1

Figure 31: Triangles on the sides of 퐴퐵C

Exercise 33. Construct the triangle 퐴퐵C from the triangles 휏1, 휏2, 휏3 and the respective vertices D, 퐸, 푍 of a welding by similarity on 퐴퐵C.

Hint: From the given triangles and points D, 퐸, 푍 there are defined respectively three |푍퐵| similarities 푓 , 푔, ℎ. Similrarity 푓 has center 푍, angle 휔1 = (퐴푍퐵) and ratio 휅1 = |푍퐴| . D DC |DC| Similarity 푔 has center , angle 휔2 = (퐵 ) and ratio 휅2 = |D퐵| . Finally, the similarity C |퐸퐴| ℎ has center 퐸, angle 휔3 = ( 퐸퐴) and ratio 휅3 = |퐸C| . The angles and the ratios of the similarities are determined completely from the given triangles 휏1, 휏2, 휏3. We then observe that point 퐴 satisfies

푓 (퐴) = 퐵, 푔(퐵) = C, ℎ(C) = 퐴 ⇒ (ℎ ∘ 푔 ∘ 푓 )(퐴) = 퐴.

Point 퐴 therefore coincides with the unique fixed point of the composed similarity ℎ∘푔∘푓 . Consequently point 퐴 is determined from the givens (even if its actual construction is somewhat involved). As soon as 퐴 is determined, the rest of the vertices of the wanted triangle are constructed by applying the similarities: 퐵 = 푓 (퐴) and C = 푔(퐵).

The preceding exercise includes many interesting special cases, which offer themselves for further study, for example, when the three triangles 휏1, 휏2, 휏3 coincide or have a more special form (isosceles, equilateral) or when for the corresponding ratios holds 푘1푘2푘3 = 1. 13 Representation and group properties of similarities 24

13 Representation and group properties of similarities

Fixing a cartesian coordinate system, a rotation about the origin and a reflection on a line through the origin is described respectively by the matrices 푅휙 of the form:

cos(휙) − sin(휙) cos(휙) sin(휙) ( ) and ( ). sin(휙) cos(휙) sin(휙) − cos(휙)

The reflective behavior of the second matrix is seen by applying it to the vectors

푎 = (cos(휙/2), sin(휙/2))푡 and 푏 = (− sin(휙/2), cos(휙/2))푡 , of which, the first maps to itself and the second, which is orthogonal to the first, maps to its negative. This identifies the line of the first vector {푡푎, 푡 ∈ ℝ} with the axis 휀 of the reflection. The general similarity, according to the definition, is described by thevector equation:

′ ″ ′ 푋 = 푂 + 푅휙(푋 − 푂) multiplied by 푋 = 푂 + 푘(푋 − 푂) .

This in matrix notation is expressed through the product of matrices

푘 0 (1 − 푘)표1 푢 푣 표1 − (푢표1 + 푣표2) ⎜⎛ ⎟⎞ ⎜⎛ ⎟⎞ 퐻 ⋅ 푅 with 퐻 = ⎜0 푘 (1 − 푘)표2⎟ and 푅 = ⎜푤 푧 표2 − (푤표1 + 푧표2)⎟ , ⎝0 0 1 ⎠ ⎝0 0 1 ⎠ where the constants represent the matrix:

푘푢 푘푣 (1 − 푘푢)표 − 푘푣표 푢 푣 ⎛ 1 2 ⎞ 푅 = ( ) ⇒ 퐻 ⋅ 푅 = ⎜푘푤 푘푧 −푘푤표 + (1 − 푘푧)표 ⎟ . 휙 푤 푧 ⎜ 1 2⎟ ⎝ 0 0 1 ⎠

In the language of “groups” the preceding discussion and matrix representations reflect the following properties:

1. The set of “similarities” constitutes a group 퐺.

2. The “direct” similarities constitute a subgroup 퐺+ of 퐺 . They are characterized by the sign of the determinant of the matrix representing the similarity, which is positive.

3. The “antisimilarities” constitute a “coset” 퐺− in 퐺, so that 퐺 = 퐺+ ∪ 퐺−. The an‑ tisimilarities are characterized by the sign of the determinant of their matrix repre‑ sentation, which is negative.

4. The homotheties constitute a subgroup 퐻 of 퐺+.

14 Logarithmic spiral and pursuit curves

Exercise 34. Starting from a “golden” rectangle 퐾푃ϴ퐼 and successively subtracting the squares of its small sides, we construct a sequence of other, pairwise similar rectangles (See Figure 32). Show that each one of them, for example 푀ϴ퐻Λ, results from its previous 푍퐼ϴ퐻 through a similarity 휋 √5−1 푓 with center the intersection point 푂 of 퐾ϴ, 퐻퐼, rotation angle 2 and ratio 푥 = 2 . 14 Logarithmic spiral and pursuit curves 25

Κ Ζ Ι

ΛΜ O Π Ξ

Ρ ΗΘΝ

Figure 32: Golden section rectangles and logarithmic spiral

Hint: Simple use of the definitions and the properties of the golden section .

Figure 32, shows a curve, called a logarithmic spiral, which passes through a vertex of the initial rectangle 푃, as well as its successive positions 푍 = 푓 (푃), 푀 = 푓 (푍), 푁 = 푓 (푀), = 푓 (푁), ..., which result by applying repeatedly the similarity 푓 ([1, 푝. 227]). Similar sequences of points and logarithmic spirals containing them result by starting from any point 푃 ≠ 푂 and taking the successive 푓 (푃), 푓 2(푃), 푓 3(푃), ..., etc.

Figure 33: Logarithmic spirals as pursuit curves

This kind of curve shows up also as a “pursuit trajectory” in “pursuit problems” , like that with 4 bugs initially placed at the vertices of a square. The bugs start pursuing each other, moving at a constant speed. Each time their positions are at the vertices of a square, which gradually shrinks and simultaneously rotates, until they all meet at the center of the square. Figure 33, on the left, shows the positions of the bugs at different moments in time and the corresponding square defined by their positions. The same figure, onthe BIBLIOGRAPHY 26 right, shows the corresponding curves for three bugs, which start at the vertices of an equilateral triangle ([8, 푝. 136], [6, 푝. 203], [7, 푝. 109]).

Exercise 35. On the sides {퐴퐵, C퐴} of the triangle 퐴퐵C we take respectively points {퐸, D} such that {퐴퐸 = 푥 ⋅ 퐴퐵, CD = 푥 ⋅ C퐴}. Show that the segment D퐸 becomes minimal, when it is or‑ thogonal to the median 퐴푀 of the triangle. Compute the minimal length of D퐸 and the value of 푥 for which this is obtained.

Hint: Draw CD′ parallel, equal and equal oriented to D퐸. Show that D′ lies on the me‑ dian 퐴푀 of the triangle. The minimal D퐸 is the altitude from C of 퐴C퐴′, where 퐴′ the symmetric of 퐴 relative to 푀.

Bibliography

[1] J. M. Aarts. Plane and Solid Geometry. Springer Verlag, Berlin, 2008.

[2] August Adler. Theorie der Geometrischen Konstruktionen. Goeschensche Verlagshand‑ lung, Leipzig, 1906.

[3] William Barker and Roger Howe. Continuous Symmetry, From Euclid to Klein. American Mathematical Society, 2007.

[4] H Coxeter. Introduction to Geometry. John Wiley and Sons Inc., New York, 1961.

[5] Roland Deaux. Introduction to the Geometry of Complex Numbers. Dover, New York, 1956.

[6] L. Graham. Ingenious Mathematical Problems and Methods. Dover Publications, New York, 1959.

[7] Paul J. Nahin. Chases and Escapes, The Mathematics of Pursuit and Evasion. Princeton University Press, Princeton, 2007.

[8] Hugo Steinhaus. Mathematical Snapshots. Oxford University Press, 3rd Edition, Ox‑ ford, 1983.

[9] I. M. Yaglom. Geometric Transformations I, II, III, IV. The Mathematical Association of America, 1962.

Related topics 1. Isometries 2. Menelaus’ theorem