PHD THESIS

ISOPTIC CURVES AND SURFACES

Csima Géza

Supervisor: Szirmai Jenő associate professor BUTE Mathematical Institute, Department of Geometry

BUTE 2017 Contents

0.1 Introduction ...... V

1 Isoptic curves on the Euclidean plane 1 1.1 Isoptics of closed strictly convex curves ...... 1 1.1.1 Example ...... 3 1.2 Isoptic curves of Euclidean conic sections ...... 4 1.2.1 Ellipse ...... 5 1.2.2 Hyperbola ...... 7 1.2.3 Parabola ...... 10 1.3 Isoptic curves to finite point sets ...... 12 1.4 Inverse problem ...... 15

2 Isoptic surfaces in the Euclidean space 18 2.1 Isoptic hypersurfaces in En ...... 18 2.1.1 Isoptic hypersurface of (n−1)−dimensional compact hypersur- faces in En ...... 19 2.1.2 Isoptic surface of rectangles ...... 20 2.1.3 Isoptic surface of ellipsoids ...... 23 2.2 Isoptic surfaces of polyhedra ...... 25 2.2.1 Computations of the isoptic surface of a given regular tetrahedron 27 2.2.2 Some isoptic surfaces to Platonic and Archimedean polyhedra 29

3 Isoptic curves of conics in non-Euclidean constant curvature geome- tries 31 3.1 Projective model ...... 31 3.1.1 Example: Isoptic curve of the line segment on the hyperbolic and elliptic plane ...... 32 3.1.2 The general method ...... 33 3.2 Elliptic conics and their isoptics ...... 34 3.2.1 Equation of the elliptic ellipse and hyperbola ...... 34 3.2.2 Equation of the elliptic parabola ...... 36

II 3.2.3 Isoptic curves of elliptic ellipse and hyperbola ...... 36 3.2.4 Isoptic curves of elliptic parabola ...... 37 3.3 Hyperbolic conic sections and their isoptics ...... 39 3.3.1 Generalized angle of straight lines ...... 39 3.3.2 Classification of generalized conic sections on the hyperbolic plane in dual pairs ...... 40 3.3.3 Isoptic curves of generalized hyperbolic conic sections . . . . . 45

4 On the geometry of SLg2R 52

4.1 The projective model for SLg2R ...... 52 4.2 Geodesic and translation curves ...... 55 4.2.1 Geodesic curves ...... 55 4.2.2 Translation curves ...... 57 4.3 Geodesic and translation triangles ...... 58 4.3.1 Geodesic triangles ...... 58 4.3.2 Translation triangles ...... 64 4.4 Isoptic surface of a line segment with translation curves ...... 67

III Acknowledgment

There are many people, who helped me in the writing of this thesis. I owe my super- visor, Jenő Szirmai gratitude for his professional help, guidance, useful suggestions and trust. Likewise I would like to thank professor Emil Molnár for his support and useful remarks. Many thanks to G. Horváth Ákos for his teaching and the opportu- nity to finish my Ph.D program at the Department of Geometry. I would also like to thank to all members of the department for their support. I would like to express my gratitude towards the Doctoral School of Mathematics and Computer Science for offering my fellowship providing me with the opportunity to get my degree. I owe a great deal of gratitude to my family, especially to my parents who sup- ported me throughout my education. I am also very grateful to my beloved Anna- maria, who supported, encouraged and helped me through every hardness in the past decade. I would like to thank to my sister Zsuzsanna and my brother Péter for show- ing me direction and to my grandparents, who gave us the possibility for a better life. Last, but not least, I would like to express my gratitude to all my teachers who spent time and energy on me. I would like to highlight my high school math teachers István Pécsi and Dénes Veres and my impeccable class master Béla Lapu.

Köszönetnyilvánítás

Dolgozatom megírásához rengetegen segítettek hozzá. Szakmai segítségéért, irány- mutatásáért, hasznos észrevételeiért és a bennem való bizalmáért hálával tartozom témavezetőm, Szirmai Jenő felé. Hasonlóképp szeretném megköszönni Molnár Emil professzornak a sok hasznos megjegyzését és támogatását. Köszönet illeti G. Horváth Ákost tanításaiért és a lehetőségért, hogy a Geometria Tanszéken fejezzem be dok- tori képzésemet. Szeretném megköszönni a tanszék összes tagjának is a támogatást. Szeretném kifejezni hálámat a Matematika- és Számítástudományok Doktori Iskola felé, akik ösztöndíjjukkal lehetőséget biztosítottak a fokozatszerzésre. Nagyon nagy hálával tartozom családomnak, főként szüleimnek, akik tanulmánya- im teljes időtartama alatt támogattak. Ugyancsak nagyon hálás vagyok szeretett páromnak, Annamárinak, aki támogatott, bátorított és átsegített minden nehézségen az elmúlt évtizedben. Szeretném megköszönni nővéremnek Zsuzsannának és bátyám- nak Péternek, hogy irányt mutattak és nagyszüleimnek, akik egy jobb élet lehetőségét adták nekünk. Végül, de nem utolsó sorban szeretnék köszönetet mondani minden tanáromnak, akik időt és energiát fordítottak rám. Szeretném kiemelni középiskolai matematika tanáraimat, Pécsi Istvánt és Veres Dénesnek és utólérhetetlen osztályfő- nökömet, Lapu Bélát.

IV 0.1 Introduction

The topic of my thesis concerns isoptic curves and surfaces in Euclidean and non- Euclidean geometries. This is the problem of generalizing the inscribed angle theorem. The question is quite simple: "What is the locus of points where a given object subtends a given angle?" It is well known that the locus of points in the plane from a segment subtends a given angle α (0 < α < π) is the union of two arcs except for π the endpoints with the segment as common chord. If this α is equal to 2 then we get the Thales circle. It is also known, that for an ellipse, any point on the circle around √ the center of the ellipse with radius a2 + b2 has this property, that the tangent lines drawn from this point to the ellipse form a right angle. For the parabola, only the points on its directrix satisfy this property. Although the name "isoptic curve" was suggested by Taylor in 1884 ([62]), refer- ence to former results can be found in [63]. In the obscure history of isoptic curves, we can find the names of la Hire (cycloids 1704) and Chasles (conics and epitrochoids 1837) among the contributors of the subject. A very interesting table of isoptic and orthoptic curves is introduced in [63], unfortunately without any exact reference of its source. However, recent works are available on the topic, which shows its timeliness. In [6] and [7], the Euclidean isoptic curves of closed strictly convex curves are studied using their support function. Papers [34, 68, 69] deal with Euclidean curves having a circle or an ellipse for an isoptic curve. Further curves appearing as isoptic curves are well studied in Euclidean plane geometry E2, see e.g. [36, 67]. Isoptic curves of conic sections have been studied in [25] and [55]. There are results for Bezier curves by Kunkli et al. as well, see [32]. Many papers focus on the properties of isoptics, e.g. [37, 38, 39], and the references therein. There are some generalizations of the isoptics as well e.g. equioptic curves in [52] by Odehnal or secantopics in [49, 57] by Skrzypiec. I encountered this question along with my supervisor during the third semester of my BSc studies within the framework of Independent Research Task I. The simplic- ity of the proposition encouraged me to continue this research alongside my studies even after the semester ended. On the recommendation of my supervisor, I began to study the topic in non-Euclidean geometries, especially in the Bolyai-Lobachevsky hyperbolic geometry. I did not find any existing result in this direction, despite the fact that the examination of problems in non-Euclidean geometries is crucial to un- derstand and describe both our small-scale and astronomical world. Simultaneously, efforts were made to expand the examination to higher dimensional spaces. It is pos- sible to examine the question in non-Euclidean spaces and we obtained some results in SL^2R geometry, but further reaching results of this research is beyond this thesis.

V This research topic was interesting to me due to the many applications. In Eu- clidean space E3 our investigation concerns sight. Therefore the results can be used in any area which is concerned with the quality of visibility. Planar investigations are also related to several areas of the mathematics. Isoptic curves are related to the discipline of shape recognition from X-ray images called geometric tomography (see [21]). Furthermore, recent results in the topic were motivated by the packing problem (see [64] Remark 4.12). It seems natural that every result in the Euclidean geometry can be applied to computer graphics. Perhaps one day isoptic curves will be implemented in CAD systems as well. In the first chapter we examine isoptic curves on the Euclidean plane. First we give an overview of some preliminary results concerning strictly convex closed and differentiable curves. The isoptic curves of conic sections have been well known for a long time [5], but we give a new approach using constructional procedures for their study. Next investigating isoptic curves of polygons, we find an algorithm to determine the isoptic curve for any finite point set. Finally, we examine the inverse question investigated by Kurusa in [34], does the isoptic of a convex body determines the convex body itself? In the second part of the thesis, we consider the Euclidean space E3. There are several possibilities of how to generalize the definition of isoptic curves to that of the isoptic surface, here we will choose the notion of solid angle. The disadvantage of this approach is that restricting the isoptic surface to any plane will not necessarily result in the isoptic curve of the restricted curve. However, some results are connected to with the detector placement problem, and the radiation level problem in nuclear physics ([24]). We will close the chapter with the discussion of isoptic surfaces of some Platonic and Archimedean solids. In the third chapter we again return to the plane, but this time the geometry is no longer Euclidean. We will consider the elliptic and the extended hyperbolic plane as embedded into three dimensional projective space P3. This will be useful for computations as real coordinates can be assigned to each point and straight line. The main subject of our examination will be the conic sections which are quite interesting both in the elliptic and hyperbolic plane. The literature on the classification of hyperbolic conic section is huge ([18, 19, 41]). To determine isoptic curves it will be inevitable to clarify the definition of the generalized angle and distance. Naturally, we give the equation of the isoptic curve to every type of conic and visualize them as well. In the last chapter, we get acquainted with one of the eight Thurston geometries, that of SL^2R [62]. We introduce the geodesic and translation curves that are different in this geometry, and we give some interesting results concerning the sum of interior

VI angles in triangles. The isoptic surfaces in this geometry is a brand new direction in our research. However, even the isoptic surface of a line segment is quite interesting due to the twisted nature of this geometry. The Wolfram Mathematica software package was utilized to create all the figures and perform all the necessary computations presented in this thesis.

VII Chapter 1

Isoptic curves on the Euclidean plane

1.1 Isoptics of closed strictly convex curves

In this chapter we work in the Euclidean plane E2. Let us introduce the following definition:

Definition 1.1.1 ([63]) The locus of the intersection of tangents to a curve (or curves) meeting at a constant angle α (0 < α < π) is the α−isoptic of the given curve (or curves). The isoptic curve with right angle called orthoptic curve.

Figure 1.1

In order to conduct further investigations on isoptics we need to summarize some preliminary results on the support function.

1 Definition 1.1.2 Let C be a closed, strictly convex curve which surrounds the ori- gin. Let p(t) where t ∈ [0, 2π[ be the distance from 0 to the support line of C being perpendicular to the vector eit. The function p is called a support function of C.

It is well-known [2] that the support function of a planar, closed, strictly convex curve C is differentiable. For now we would like to express the isoptic of C using the support function. We claim the following lemma omitting the proof which can be found for example in [66].

Lemma 1.1.3 ([66]) If f(x, y, t) = 0 is a family of straight lines, then the equation of the envelope of these lines can be obtained by eliminating the variable t from the d two equations f(x, y, t) = 0 and dt f(x, y, t) = 0.

This is used in [66] to prove the following theorem.

Theorem 1.1.4 ([66]) Given a planar, closed, strictly convex curve C in polar coor- dinates with the radius z a function of angle t, where t ∈ [0, 2π). Then the following equation holds z(t) = p(t)eit +p ˙(t)ieit.

The corollary of this theorem is that we may use this parametrization to determine the isoptic curve of C. The angle of p(t) and p(t+π−α) is α, since the p(t), p(t+π−α) and their support lines determine a cyclic quadrilateral (see Figure 1.2). Our goal is to determine the intersection of these tangent lines which is the fourth vertex opposite the origin. A proof can be found in [6], but we give another one using coordinate geometry and Wolfram Mathematica.

Theorem 1.1.5 ([6]) Let C be a plane, closed, strictly convex curve and suppose that the origin is in the interior of C. Let p(t), t ∈ [0, 2π] be the support function of C. Then the α-isoptic curve of C has the form 1 ! z (t) = p(t)eit + −p(t) cot(π − α) + p(t + π − α) ieit. α sin(π − α)

Proof[CsG] Let us determine the intersection of the support lines of p(t) and p(t + π − α) using analytic geometry. This is the fourth vertex of the cyclic quadrilateral. Using the normal form of the tangent lines:  cos(t)x + sin(t)y = p(t)  (1.1) cos(t + π − α)x + sin(t + π − α)y = p(t + π − α) 

Solving the (1.1) system of equation be Wolfram Mathematica obtain:

2 Figure 1.2 x -> p[t] Cos[t] - (p[t+π-α] + p[t] Cos[α]]) Csc[α]] Sin[t], y -> p[t+π-α] Cos[t] Csc[α]] + p[t] (Cos[t] Cot[α]] + Sin[t])

If we multiply y by i, add the result to x and perform the following simplification

FullSimplify[ ExpToTrig[ TrigToExp[ p[t] Cos[t] - (p[t+π-α] + p[t] Cos[α]]) Csc[α]] Sin[t] + i (p[t+π-α] Cos[t] Csc[α]] + p[t] (Cos[t] Cot[α]] + Sin[t]))]/eit]] we obtain: p[t] + i p[t] Cot[α]] + i p[t+π-α] Csc[α].

1 Using the fact that cot(α) = − cot(π − α) and csc(α) = sin(π−α) the theorem follows. 

Remark 1.1.6 Further properties of isoptic curves can be found in [39], [37], [59], [57] and [7].

1.1.1 Example

A typical example may be an ellipse with semi-axes a and b such that they coincide with x and y coordinate axes respectively i.e. we have a central ellipse. In this case

3 we have a well-known simple parametrization: (a cos(t), b sin(t)), t ∈ [0, 2π). Now we calculate the support function of the ellipse and use Theorem 1.1.5 to determine the isoptic curve. It is important to note that the tangent does not belong to the value t but a function of t, we denote it ϕ(t). Using the derivative of the parametrization we obtain the equation of the tangent line:

b cos(ϕ(t))x + a sin(ϕ(t))y = ab (1.2)

According to the well-known distance formula between a point and a line: ab p(t) = q (1.3) a2 sin2(ϕ(t)) + b2 cos2(ϕ(t))

Now we determine ϕ(t). Notice that the tangent line corresponding to ϕ(t) is perpendicular to (cos(t), sin(t)). According to (1.2) we have the following equation:

−a cos(t) sin(ϕ(t)) + b sin(t) cos(ϕ(t)) = 0 (1.4)

Solving equation (1.4) we have divided with trigonometric functions therefore we kπ consider the t = 2 , k = 0, 1, 2, 3 cases separately. The tangent line will be parallel with the x or y axis therefore the support function will be a, b, a, b respectively. Now, excluding these cases we can divide equation (1.4) with cos(t) and cos(ϕ(t)). Applying the arctan function we obtain:

b ! ϕ(t) = arctan tan(t) a

Substituting this into (1.3), we get the support function:

q p(t) = a2 cos2(t) + b2 sin2(t) (1.5)

Finally we can apply Theorem 1.1.5 with some simplification to obtain: q 2 2 2 2 it zα(t) = a cos (t) + b sin (t)e + q 1 q ! + cot(α) a2 cos2(t) + b2 sin2(t) + a2 cos2(t − α) + b2 sin2(t − α) ieit sin(α) (1.6)

1.2 Isoptic curves of Euclidean conic sections

In this section, we determine the isoptic curves of Euclidean conic sections. There are a lot of possibilities to give the equations of the isoptics (see e.g [36]). However we give here a new approach using the construction method of the tangent lines from an outer point (see e.g in [58]). This procedure requires less computation.

4 1.2.1 Ellipse

The isoptic curve of an ellipse has already been discussed, however we only gave a parametric representation. Now, our goal is to give an implicit equation for the isoptic curve. In the next subsection, we recall the construction method of the tangent lines from an outer point, then we apply it during our analytic calculations. Finally we determine the isoptic curve of a given ellipse using vector algebra.

1.2.1.1 Construction method

Let us suppose that we have an ellipse given by its foci F1, F2 and main axis. More- over, let K be an external point with respect to the ellipse. We can choose a foci freely e.g. F2. Now we draw a circle with radius 2a around F2. It is easy to see that the reflection of F1 on any tangent line is on this circle because every tangent to an ellipse forms equal angles with the focal radii of its point of tangency. Let us denote the reflection points of F1 over the tangent lines through K by E1 and E2 (see Figure 1.3).

Figure 1.3: Ellipse: Construction of tangents passing trough an external point

Now, if we want to ensure that the tangents overpass the external point K then the distance between E1, E2 and K must be equal to the distance between F1 and

K. Thus E1 and E2 are on the circle with center K and radius d(K,F1). Hence E1,2 is on the first circle as well, thus they can be obtained intersecting the circles. To draw the tangent lines we only have to take the perpendicular bisectors of E1F1 and

E2F1. It is important to remark that we do not have to determine the points of tangency, we can find the equation of the tangents directly making our calculation faster.

5 1.2.1.2 Calculations of tangent lines

Let us suppose that we have a canonical ellipse (see [28]) with semi-axis a and b i.e. they coincide with the x and y coordinate axis respectively. We can also assume that + a > b ∈ R and K(x0, y0) is an external point. It is well-know that the foci with √ √ 2 2 2 2 these assumptions have the coordinates: F1(− a − b , 0) and F2( a − b , 0). The next step of the construction above is to determine the equations of the mentioned circles. Thus we need the distance between K and F1:

r √ 2 r √ 2   2 2 2  2 2 2 d(K,F1) = x0 − − a − b + y0 = x0 + a − b + y0

Now, we are able to give the equations of the two circles:

 √ 2  x − a2 − b2 + y2 = 4a2  √ 2 (1.7) 2 2  2 2  2 (x − x0) + (y − y0) = a − b + x0 + y0 

Solving (1.7) results in two solutions denoted by E1 and E2. The normal vectors −−−→ −−−→ of the tangent lines passing trough K are E1F1 and E2F1. Using the point K and the normals, we are able to determine the equations of the tangent lines but it is unnecessary, because the vectors are sufficient to determine the angle.

1.2.1.3 Isoptic of the ellipse

We use the fact that: D−−−→ −−−→E F1E1, F1E2 cos(π − α) = −−−→ −−−→ (1.8)

F1E1 F1E2 After some simplification we obtain:

D−−−→ −−−→E F1E1, F1E2 a2 + b2 − x2 − y2 −−−→ −−−→ = q 2 2 2 2 2 2 2 2 4 F1E1 F1E2 (−a + b + x ) + 2y (a − b + x ) + y Summarizing the results above we can state:

Theorem 1.2.1 Given a canonical ellipse with semi-axis a and b in a coordinate system on the Euclidean plan E2. The α-isoptic curve (0 < α < π) of the considered ellipse has the equation

x2 + y2 − a2 − b2 cos α = q . (−a2 + b2 + x2)2 + 2y2 (a2 − b2 + x2) + y4

Figures 1.4 and 1.5 show the isoptic curve (blue) of the ellipse (red) for some values of α.

6 15 10

10

5

5

0 0

-5

-5

-10

-15 -10 -15 -10 -5 0 5 10 15 -10 -5 0 5 10

Figure 1.4: a = 4,b = 1, 5, α = π/6, a = 5, b = 3, α = π/3

6

4 5

2

0 0

-2

-5 -4

-6 -5 0 5 -6 -4 -2 0 2 4 6

Figure 1.5: a = 5, b = 3, α = π/2, a = 5, b = 3, α = 2π/3

1.2.2 Hyperbola

The next target of our investigation of isoptic curves of Euclidean conic sections is the hyperbola, because the method and the calculation is very similar to the ellipse. The external points of a hyperbola lie between its branches. From these points two tangent lines can be drawn unless the point and at least on of the asymptotes are incident. In this case according to the projective approach we can assume that an asymptote is also a tangent line with an ideal point of tangency.

1.2.2.1 Construction method

The procedure is almost the same as before with the ellipse. We draw a circle around one of the foci with radius 2a, furthermore another circle around the external point

7 K trough the other foci. The resulting intersections will be crucial to determine the tangent lines, hence they will be the corresponding perpendicular bisectors (see Figure 1.6). The exact procedure is detailed in the case of ellipse.

Figure 1.6: Hyperbola: Construction of tangents passing trough an external point

1.2.2.2 Calculations of tangent lines

As before, we can assume here also that the hyperbola is in canonical position with + semi-axis a and b, where a, b ∈ R and let K(x0, y0) be an external point. The coordi- √ √ 2 2 2 2 nates of the foci can be calculated here as well: F1(− a + b , 0) and F2( a + b , 0). The next step is to determine the equations of the circles, but we have to calculate the distance between K and F1 first.

r √ 2 r √ 2   2 2 2  2 2 2 d(K,F1) = x0 − − a + b + y0 = x0 + a + b + y0 Now, the equations of the circles are:

 √ 2  x − a2 + b2 + y2 = 4a2  √ 2 (1.9) 2 2  2 2  2 (x − x0) + (y − y0) = a + b + x0 + y0 

Solving the system of equations (1.9) we get the coordinates of E1 and E2 (see Figure 1.6). Reasoning similar as in the previous section we do not need the equation of the tangent lines to determine the isoptic curve, only their normals are necessary.

1.2.2.3 Isoptic of the hyperbola

To determine the isoptic curves of a hyperbola we do not need any new idea only the computations change slightly therefore we can claim without further explanation:

8 Theorem 1.2.2 Given a canonical hyperbola with semi-axis a and b in a coordinate system on the Euclidean plane E2. The α and π − α-isoptic curves (0 < α < π) of the hyperbola considered has the equation (−a2 + b2 + x2 + y2)2 cos2 α = . (a2 + b2 − x2)2 + 2y2 (a2 + b2 + x2) + y4

Remark 1.2.3 In some papers e.g [52] the equations of isoptic curves of the ellipse and hyperbola arise on the square to get not only the α-isoptic but the π − α-isoptic as well. For the ellipse, this is not essential. In the case of a hyperbola, the two asymptotes split the space into four domains, two of them contain a hyperbola branch (focal domains), the other two are empty. Let K be an outer point of the hyperbola. If K is in a focal domain, then the tangent lines are tangent to the same branch of the hyperbola, else they touch both branches. In these cases, the isoptic angles are complementary to each other, i.e. they sum up to π. Therefore, we take the square of the equation and thus obtain both types of isoptic curves.

15 10

10

5

5

0 0

-5

-5

-10

-15 -10 -15 -10 -5 0 5 10 15 -10 -5 0 5 10

Figure 1.7: a = 4, b = 1, 5, α = π/6, a = 5, b = 3, α = π/3

Figures 1.7 and 1.8 show the isoptic curve (blue) of the hyperbola (red) for some values of α.

Remark 1.2.4 It is easy to see that the origin is a "saddle point" with respect to the angle of the tangent lines. From any points inside focal domains, the angle of the tangents are greater, and from the other domains they are less then from the origin. Therefore, the isoptic curves do not exists in the interval b2 − a2 ! a2 − b2 !! arccos , arccos b2 + a2 a2 + b2 if the condition b > a holds (see the right Figure 1.8).

9 10 10

5 5

0 0

-5 -5

-10 -10 -10 -5 0 5 10 -10 -5 0 5 10

Figure 1.8: a = 5, b = 3, α = π/2, a = 1, 5, b = 3, α = π/12

1.2.3 Parabola

We will follow the procedure described above to determine the isoptic curve of a parabola. Namely, we follow the construction method analytically then fix the angle of the tangent lines. It is also important to mention that in this case the external points are those which are closer to the directrix than to the foci. Similarly to other conic sections, two tangent lines can be drawn to the parabola from every external point.

1.2.3.1 Construction method

Let us suppose that a parabola is given by its foci F and directrix d. It is well-known that a tangent to the parabola at any point bisects the angle between the line joining the point to the foci, and the line trough the point perpendicular to the directrix. Thus the reflections of the foci respected to the tangents fit on the directrix therefore these E1,2 points can be obtained intersecting the circle around K with radius d(K,F ) and the directrix. The tangents will be the perpendicular bisectors of E1F and E2F .

1.2.3.2 Calculations of tangent lines

In the previous cases canonical position was suitable but now we may choose axial position which means that the directrix coincides with the line y = 0 and the foci sits on the y axis with F (0, p) coordinates. With these assumptions, our calculation will be easier and the solutions will not be so complicated as before.

First we suppose that K(x0, y0) is an external point of the given parabola and we need the distance between K and F to find the equation of the circle:

10 Figure 1.9: Parabola: Construction of tangents passing trough an external point

2 2 d(K,F ) = (p − y0) + x0. From which the equation of the circle:

2 2 2 2 (x − x0) + (y − y0) = (p − y0) + x0.

The intersection of this with the straight line y = 0 gives:

q 2 2 E1 : x1 = x0 − p − 2py0 + x0 y1 = 0, q 2 2 E2 : x2 = x0 + p − 2py0 + x0 y2 = 0. To find the tangent lines we need their normals:

 q  −−→ 2 2 FE1 = x0 − p − 2py0 + x0 , −p ,

 q  −−→ 2 2 FE2 = x0 + p − 2py0 + x0 , −p . Finally the point K, and the normals determine the tangents. However the equa- tions are not necessary in order to compute the isoptic curve.

1.2.3.3 Isoptic of the parabola

As we claimed before the cosine of the tangents’ supplementary angle can be fixed by using the scalar product of the normals. Let us use the following coordinates: −−→ −−→ FE1 = (x1, y1) and FE2 = (x2, y2). We have to calculate the fractional value of (1.8). We determine the numerator and the denominator separately:

x1y2 + x2y2 = 2py0

11 q q 2 2 2 2 2 2 2 (x1 + y1)(x2 + y2) = 4p ((p − y0) + x0) Performing the division

x1y2 + x2y2 2py0 q = q . 2 2 2 2 2 2 2 (x1 + y1)(x2 + y2) 4p ((p − y0) + x0)

Simplifying the results above and using x = x0 and y = y0 we obtain the following theorem:

Theorem 1.2.5 Given an axial parabola with its foci F (0, p) in a coordinate system on the Euclidean plan E2. The α-isoptic curves (0 < α < π) of the considered parabola has the equation

y cos α = −q . (p − y)2 + x2

4 4

2 2

0 0

-2 -2

-4 -4

-4 -2 0 2 4 -4 -2 0 2 4

Figure 1.10: p = 1/2, α = π/6, p = 1/2, α = π/3

Remark 1.2.6 A quadratic Beziér curve can be considered to be a parabolic arc. For isoptic curves to Beziér curves with higher degree see [32].

1.3 Isoptic curves to finite point sets

In this Section, we consider isoptic curves of finite point sets on the Euclidean plane. In the Section 1.1 we used the fact that the given curve C was parametric. Even the first derivative of the parametrization was utilized. Construction and analytic methods were applied to determine the isoptic curve of conic sections. Now, neither of these possibilities are available, since there is no curve at all. We also have to clarify, what the isoptic curve of a point set is.

12 4 4

2 2

0 0

-2 -2

-4 -4

-4 -2 0 2 4 -4 -2 0 2 4

Figure 1.11: p = 1/2, α = π/2, p = 1/2, α = 2π/3

Definition 1.3.1 A given finite point set P is seen under α (0 < α < π) from P if the smallest angle with vertex P which contains all points in P is α.

Definition 1.3.2 The isoptic curve of an arbitrarily given finite point set P is the locus of points P where P is seen under a given fixed angle α (0 < α < π). The isoptic curve with a right angle is called orthoptic curve.

Now, we can easily see from the definition that:

Corollary 1.3.3 To determine the isoptic curve of a finite point set it is sufficient to determine the isoptic curve of its convex hull.

We may suppose that the points are in general position, i.e no three points are collinear. There are numerous algorithms to determine the convex hull of finitely many points (see [1]), therefore we may assume that we have n points (Xi where i ∈ 1, . . . , n) in clockwise order. Let C be a closed curve, obtained by drawing every

XiXi+1 segment (if i = n, then Xi+1 = X1). The resulting curve will be a closed simple n-gon, and the isoptic curve of the given finite point set P will be the isoptic curve of C. Despite our first approach being useful to visualize the isoptic curve, it will not appropriate for giving a formula or a parametrization of the curve, however the idea is very simple and useful. A tangent line of a polygon may be of two types, it is either an edge line or it passes through exactly one vertex. When both support lines pass through a vertex then we can assign a diagonal. This segment does not change until at least one of the tangents changes its point of tangency. Therefore the isoptic curve of the polygon will be the union of finitely many circular arcs.

13 Lemma 1.3.4 Given an angle α (0 < α < π) and an n-gon with X1, X2,... , Xn vertices. We consider all segments formed by XiXj. Let Ω be the union of the discs each one of which corresponds to a circular arc where a segment XiXj subtends α. Then the isoptic curve of the n-gon will be ∂Ω.

Proof.[CsG] Let X0 and Y0 be points inside the n-gon and P be a point on ∂Ω. It needs to be shown that X0PY0∠ ≤ α. 0 0 Let the straight line X0Y0 intersect the polygon in X and Y . Then it is obvious 0 0 that X PY ∠ ≥ X0PY0∠.

If X0, P and Y 0 are collinear, then they form zero angle. Otherwise, the n-gon has 0 at least one vertex X on the opposite side of Y0 with respect to the line PX . This holds because PX0 intersects another edge of the polygon due to the Pasch-axiom. 0 The point Y (∃j 6= i : Y = Xj) can be determined similarly with PY . It is easy to 0 0 see that XPY ∠ ≥ X PY ∠ ≥ X0PY0∠ and XY is a diagonal of the n-gon therefore α ≥ XPY ∠.  Remark 1.3.5 The statement that ’...the isoptic curve of a convex polygon is the union of circular arcs’ can be found in [34] without any proof, only the inscribed angle theorem is mentioned as justification.

Now, our goal is to give an implicit equation for ∂Ω. It is supposed that the polygon is closed, convex and it has n vertices X1, X2, ...Xn in clockwise order. If we want to determine the subtended angle of a line segment from a planar point P then we project it to a unit circle around P and compute its arc length. Only one question remains, that of which segment should be projected with respect to P ?A segment XiXi+1 is visible from P if and only if the line XiXi+1 separates P and the polygon. This will be detailed in the following chapter.

14 Notice that if our n-gon is convex then its projection will cover the circular arc exactly twice. Therefore if a point P is on the isoptic curve of the polygon, then the sum of all XiPXi+1∠ (i = 1 . . . n and Xn+1 = X1) have to be 2α. All XiPXi+1∠ (i = 1 . . . n and Xn+1 = X1) angles can be derived from the scalar product:  D−−→ −−→ E  PX , PX  i i+1  XiPXi+1 = arccos   . ∠ rD−−→ −−→ ED−−→ −−→ E PXi, PXi PXi+1, PXi+1 Now we have the following theorem:

Theorem 1.3.6 (CsG) Given a convex n-gon with the vertices X1, X2, ...Xn T (Xi = (xi, yi) ) in clockwise order and α, (0 < α < π). Then the α-isoptic curve of the n-gon considered has the equation:

 D−−→ −−→ E  n PX , PX X  i i+1  2α = arccos   (1.10) rD−−→ −−→ ED−−→ −−→ E i=1 PXi, PXi PXi+1, PXi+1

T where Xn+1 = X1 and P = (x, y) .

Remark 1.3.7 The isoptic curve of the triangle has already been determined in [38].

1.4 Inverse problem

In the previous sections we investigated isoptic curves of a given curve or point set. Now, we consider its inverse problem. Let K be a convex body in the Euclidean plane α α and CK be its isoptic curve. The question arises whether the isoptic CK of a convex body K determines the convex body itself? The first important work on this topic was by Green [23] in 1950. The investigation was for circles, specifically ’. . . the fact α that K subtends a constant angle on CK implies that K is such a circle.’ The answer is quite surprising, since:

α Theorem 1.4.1 ([23]) If 1 − π is irrational or rational with even numerator in its α α lowest terms and CK is a circle, then K can only be a circle of radius sin 2 concentric α with CK . Otherwise infinitely many non-circular K exist.

α Proof(outline if CK is the unit circle). The main idea of the proof is that the support α function of K is not only 2π periodic, but 2(π −α) = 2π(1− π ) periodic as well. And it is well known that every linear combination of two period are also a period. α If 1− π is irrational, then the periods are dense on the interval [0, 2π], which implies α that the support function of K is constant. It can be proved that π(1− π ) is a period α if 1 − π is rational with even numerator in its lowest terms. Then the permanence

15 α m of the support function occurs. The only remaining case is when 1 − π = n π is rational with odd numerator in its lowest terms. Then if  > 0 is sufficiently small,  πm  the support function p(Θ) = cos 2n +  sin(nΘ) satisfies the required conditions. The following figure shows the existence of non-circular K convex body with circular isoptic:

1.0

0.5

0.0

-0.5

-1.0

-1.0 -0.5 0.0 0.5 1.0

 3  Figure 1.12: p(Θ) = cos 8 π + 0.0232 sin(4Θ)

Of course the isoptic of this curve is circle only for a fixed angle. Klamkin in 1988 [31] conjectured that if K has two concentric circular isoptic then K itself has to also be a circle. In 1990 Nitsche proved this conjecture ([51]). Some recent results α consider the relation between the convex body K and its isoptic CK :

Theorem 1.4.2 ([20]) Let K be a convex body in the plane and let α ∈ (0, π). If α CK is homothetic to ∂K then K is a Euclidean disc.

Theorem 1.4.3 ([20]) Let K be a convex figure in the plane and α ∈ (0, π). Then

α 1 t(CK ) ≥ 2 t(K),   α  sin 2 with equality if and only if K is a Euclidean disc.

Uniqueness has been investigated not only for circular isoptics, but for general star shaped continuous curves as well. In 2012 Kurusa proved the following theorem:

α Theorem 1.4.4 ( [34]) Let 1 − π be an irrational or a rational number with even α α numerator in its lowest terms. If CK1 ≡ CK2 for convex bodies K1 and K2, then

K1 ≡ K2.

16 Obviously, we can not cover the remaining case so easily, since Green proved that isoptic circles do not have unique convex bodies. But we have positive answer for α uniqueness if CK is given and K is a polygon or a convex body rotationally symmetric α with angle α such that 1− π is irrational or rational with even numerator in its lowest terms. Since one isoptic is clearly not enough to reconstruct the convex body, the follow- ing open question arises after the work of Klamkin and Nitsche on circles: "How many isoptics of a convex body are necessary to permit its exact reconstruc- tion?" [34]

17 Chapter 2

Isoptic surfaces in the Euclidean space

In this chapter we consider isoptic surfaces, in the first section we generalize the definition to Euclidean spaces of arbitrary dimension n (n ≥ 3). We will not deter- mine isoptic surfaces generally, but their geometrical properties through numerical computations. Then an intermediate case will be considered, where we investigate the isoptic hypersurface for any compact domains which is a subset of a hyperplane. This results will be applied to determine isoptic surfaces of rectangles. Finally, we generalize the algorithm described in the Section 1.3. to adapt it for polytopes and apply it to various Platonic and Archimedean polyhedra.

2.1 Isoptic hypersurfaces in En

First, we generalize the notion of angle to Euclidean space E3. From this statement it is possible to generalize it to Euclidean spaces of any dimension n (n ≥ 3). The notion of solid angle is well known and widely studied in the literature (see [22]).

3 Definition 2.1.1 The solid angle ΩS(P ) subtended by a surface S in E is defined as the surface area of the projection of S onto the unit sphere around P .

In the International System of Units (SI), a solid angle is expressed in a dimen- sionless unit called a steradian (symbol: sr)1, e.g. the solid angle subtended by the whole Euclidean space E3 is equal to 4π steradians. Moreover, this notion has sev- eral important applications in physics, (in particular in astrophysics, radiometry or photometry) (see [4]) computational geometry (see [29]) and we can easily generalize it to any dimension. Some results for higher dimensional solid angles can be found in [53]. The Definition 1.1.1 can also be generalized:

1https://en.wikipedia.org/wiki/Solid_angle

18 α n Definition 2.1.2 ([11]) The isoptic hypersurface HD in E (n ≥ 3) of an arbi- trary 2 ≤ d ≤ n dimensional compact set D is the locus of points P where the d−dimensional measure of the projection of D onto the unit (n − 1)-sphere around P n−1 π 2 is a given fixed value α (0 < α < n−1 )) (see Figure 2.1). Γ( 2 )

Remark 2.1.3 Another approach can be find in [50], with numerous possibilities for application, see e.g. [33] Section 3.

Figure 2.1: Projection of a compact domain D onto the unit sphere in E3

2.1.1 Isoptic hypersurface of (n − 1)−dimensional compact hypersurfaces in En

Now, we consider a compact (n − 1) dimensional hypersurface D (n ≥ 3) lying in a hyperplane of En, given by a usual parametric system. Applying an appropriate transformation, we may assume parametrization of D has the form:   fe1(x1, x2, . . . , xn−1)      fe2(x1, x2, . . . , xn−1)     .  φe(x , x , . . . , x ) =  .  , (2.1) 1 2 n−1  .      fen−1(x1, x2, . . . , xn−1)   0

0 0 0 where xi ∈ [ai, bi],(ai, bi ∈ R), (i = 1, . . . , n − 1). Take a point P (x1, x2, . . . , xn) = 0 0 P (x ) assuming xn > 0 . Projecting D onto the unit sphere with center P , we have the following parametrization:

19     f1(x1, x2, . . . , xn−1) f1(x)         f2(x1, x2, . . . , xn−1) f2(x)     φ(x1, x2, . . . , xn−1) =  .  =  .  , (2.2)  .   .   .   .      fn(x1, x2, . . . , xn−1) fn(x) where if i 6= n

0 fei(x1, . . . , xn−1) − x1 fi(x) = q 0 2 0 2 0 2 (fe1(x1, . . . , xn−1) − x1) + ··· + (fen−1(x1, . . . , xn−1) − xn−1) + (xn) otherwise

0 −xn fn(x) = q 0 2 0 2 0 2 (fe1(x1, . . . , xn−1) − x1) + ··· + (fen−1(x1, . . . , xn−1) − xn−1) + (xn) It is well known, that the surface area can be calculated using the formula bellow:

Z b1 Z b2 Z bn−1 √ 0 0 0 S(x1, x2, . . . , xn) = ··· det G dxn−1 dxn−2 ... dx1 (2.3) a1 a2 an−1 where  T   ∂f1 ∂f1 ... ∂f1 ∂f1 ∂f1 ... ∂f1  ∂x1 ∂x2 ∂xn−1   ∂x1 ∂x2 ∂xn−1   ∂f2 ∂f1 ∂f1   ∂f2 ∂f1 ∂f1   ...   ...  T  ∂x1 ∂x2 ∂xn−1   ∂x1 ∂x2 ∂xn−1  G = J J =  . . . .   . . . .  .  . . .. .   . . .. .   . . .   . . .      ∂fn−1 ∂fn−1 ... ∂fn−1 ∂fn−1 ∂fn−1 ... ∂fn−1 ∂x1 ∂x2 ∂xn−1 ∂x1 ∂x2 ∂xn−1

α The isoptic hypersurface HD by Definition 2.1.2 is the following:

 n−1  α n 0 n 0 0 0 o π 2 HD = x ∈ E |α = S(x1, x2, . . . , xn) , where α ∈ 0 , n−1  . (2.4) Γ( 2 )

In the general case, the isoptic hypersurfaces can only be determined through numerical computations. In the next subsections we show an application of the procedure above.

2.1.2 Isoptic surface of rectangles

Suppose that n = 3 and D ⊂ E2 is a rectangle lying in the [x, y] plane in a given Cartesian coordinate system. Moreover, we may assume, that it is given by the parametrization     ϕ(x, y) x         φe(x, y) = ϑ(x, y) = y , (2.5)     0 0

20 where x ∈ [−a, a] and y ∈ [−b, b](a, b ∈ R+). Following the algorithm described in the previous section, we project the surface onto a unit sphere around P (x0, y0, z0) we obtain the following parametrization:

    √ ϕ(x,y)−x0 √ x−x0 (ϕ(x,y)−x )2+(ϑ(x,y)−y )2+z2 (x−x )2+(y−y )2+z2  0 0 0   0 0 0      φ(x, y) = √ ϑ(x,y)−y0  = √ y−y0  . (2.6)  (ϕ(x,y)−x )2+(ϑ(x,y)−y )2+z2   (x−x )2+(y−y )2+z2   0 0 0   0 0 0  √ −z0  √ −z0  2 2 2 2 2 2 (ϕ(x,y)−x0) +(ϑ(x,y)−y0) +z0 (x−x0) +(y−y0) +z0

Remark 2.1.4 It is clear, that the computations are similar if D is a normal domain. The difference appears only on the boundaries of the integrals.

Now, we need the partial derivatives to calculate the spherical surface area (see (2.3)):

 2 2    (y−y0) +z0 −(x−x0)(y−y0) (x−x )2+(y−y )2+z2 3/2 (x−x )2+(y−y )2+z2 3/2 ( 0 0 0 )  ( 0 0 0 )     (x−x )2+z2   −(x−x0)(y−y0)   0 0  φx(x, y) =  (x−x )2+(y−y )2+z2 3/2  , φy(x, y) =  (x−x )2+(y−y )2+z2 3/2  ( 0 0 0 )  ( 0 0 0 )   z0(x−x0)   z0(y−y0)  2 2 2 3/2 2 2 2 3/2 ((x−x0) +(y−y0) +z0 ) ((x−x0) +(y−y0) +z0 ) Taking the cross product of the vectors above, we obtain:

 z0(x0−x)  (x−x )2+(y−y )2+z2 2 ( 0 0 0 )     z0(y0−y)  φx(x, y) × φy(x, y) = 2 2 2 2 ((x−x0) +(y−y0) +z0 )   2   z0  2 2 2 2 ((x−x0) +(y−y0) +z0 )

Now we can substitute |φx(x, y) × φy(x, y)| in formula 2.3:

+a +b Z Z |z0| S(x0, y0, z0) = 3 dy dx = (2.7) −a −b 2 2 2 2 ((x − x0) + (y − y0) + z0)     = arctan √ (a−x0)(b−y0) + arctan √ (a+x0)(b−y0) + 2 2 2 2 2 2 z0 (a−x0) +(b−y0) +z0 z0 (a+x0) +(b−y0) +z0     + arctan √ (a−x0)(b+y0) + arctan √ (a+x0)(b+y0) 2 2 2 2 2 2 z0 (a−x0) +(b+y0) +z0 z0 (a+x0) +(b+y0) +z0 Remark 2.1.5 It is easy to see, if a → ∞ and b → ∞, then the angle tends to

2π for every z0. The normalized cross product of the two partial derivatives can be interpreted as a weight function on this elliptic plane. Now, if we have a domain on the plane, we can integrate this function over the domain to obtain the angle. But the symbolic integral for a given domain almost never works, so in this case, it is suggested again, to use numerical approach.

Using the results above, we can claim the following theorem, given by Gotoh and Yagi in [24] and Csima in [11] independently:

21 Theorem 2.1.6 ([11, 24]) Given a rectangle D ⊂ E2 lying in the [x, y] plane in a given Cartesian coordinate system. Moreover, we can assume that it is centered at the origin with edges (2a, 2b). Then the isoptic surface for a given angle α (0 < α < 2π) is determined by the following equation:     α = arctan √ (a−x)(b−y) + arctan √ (a+x)(b−y) + z (a−x)2+(b−y)2+z2 z (a−x)2+(b−y)2+z2     + arctan √ (a−x)(b+y) + arctan √ (a+x)(b+y) . z (a−x)2+(b−y)2+z2 z (a−x)2+(b−y)2+z2

The following figures show the isoptic surface of the rectangle:

π Figure 2.2: 2a = 9, 2b = 13, α = 2 2a = 9, 2b = 13, α = π

π π Figure 2.3: 2a = 11, 2b = 5, α = 12 2a = 7, 2b = 11, α = 6

Remark 2.1.7 The figures shows us, that this topic has numerous application pos- sibilities, for example designing stadiums, theaters or cinemas (see Figure 2.4). It can be interesting, to have a stadium, with the property that from every seat on the grandstand, the field can be seen under the same angle. Designing a lecture hall, it is important, that the screen or the blackboard is clearly visible from every seat. In this case, an isoptic lecture hall is not feasible, but the hall can be optimized (see Figure 2.5).

22 Figure 2.4: MetLife Stadium: http://www.bonjovi.pl/forum/topics58/25-27072013- east-rutherford-vt3278.htm

Figure 2.5: http://indogenius.com/2013/05/finding-your-best-fit-university/

2.1.3 Isoptic surface of ellipsoids

Now, we extend the method introduced in the previous section. We use a standard analytic approach to determine the isoptic surface of an ellipsoid, but only a numerical solution can be derived due to its computational complexity. It is well known that the ellipsoid has the following parametrization:

23   a cos u cos v     φe(u, v) = b cos u sin v  , (2.8)   c sin u π π + where u ∈ [− 2 , 2 ], v ∈ [0, 2π] and a, b, c ∈ R . Projecting this onto the unit sphere around P (x0, y0, z0) we obtain (see (2.6)):

  √ a cos u cos v−x0 2 2 2  (a cos u cos v−x0) +(b cos u sin v−y0) +(c sin u−z0)    √ b cos u sin v−y0  φ(u, v) =  2 2 2  . (2.9)  (a cos u cos v−x0) +(b cos u sin v−y0) +(c sin u−z0)  √ c sin u−z0  2 2 2 (a cos u cos v−x0) +(b cos u sin v−y0) +(c sin u−z0) Now, we can determine the isoptic surface of the ellipsoid (see (2.3), (2.4)):

π Z 2π Z 2 2α = |φu(u, v) × φv(u, v)| du dv. (2.10) π 0 − 2

π Figure 2.6: Numerical isoptic surface for ellipsoid with a = 8, b = 4, c = 1, α = 4

This integral can be computed only numerically therefore we only the figure of the solution without any implicit equation (see Figure 2.6).

24 2.2 Isoptic surfaces of polyhedra

Now we discuss the algorithm developed to determine the isoptic surface of a given polyhedron.

1. We assume that an arbitrary polyhedron P is given by the usual data structure.

This consists of the list of facets FP with the set of vertices Vi in clockwise order. Each facet can be embedded into a plane. It is well known, that if a ∈ R3 \{0} and b ∈ R then {x ∈ R3|aT x = b} is a plane and {x ∈ R3|aT x ≤ b} define a half space. Every polyhedron is the intersection of finitely many half spaces. Therefore an arbitrary polyhedron can also be given by a system of inequalities Ax ≤ b where A ∈ Rm×3 (4 ≤ m ∈ N), x ∈ R3 and b ∈ Rm.

2. For an arbitrary point P ∈ E3 with coordinates p we have to decide, which th i facets of P ’can be seen’ from it. Let us denote the i facet of P by FP (i = 1, . . . , m) and by ai the vector defined by the ith row of the matrix A i which characterize the facet FP . Since the polyhedron P is given by the system of inequalities Ax ≤ b, where i each inequality a x ≤ bi (i ∈ {1, 2 . . . , m}) is assigned to a certain facet, there- i i fore the facet FP is visible from P if an only if the inequality a p > bi holds.

i i Now, we define the characteristic function IP (x) for each facet FP :   1 aix > b i (x) = i IP i  0 a x ≤ bi.

3. Using the Definition 2.1.1, let Ωi(P ) := Ω i (P ) be the solid angle of the facet FP i FP regarding the point P .

To determine Ωi(P ), we use the methods of spherical geometry. Let us suppose i that FP contains ni vertices, Vij (xij ), (j = 1, . . . , ni) where the vertices are given in clockwise order. Projecting these vertices onto the unit sphere centered at

P we get a spherical ni-gon (see Figure 2.7), whose area can be calculated by the usual formula

Ωi(P ) = Θ − (ni − 2)π.

Here Θ is the sum of the angles τj of the spherical projection of the polygon i FP where the angles are measured in radians.

4. To obtain angles τj, we need to determine the angles between the two planes

containing the neighboring edges PV ij−1 , PV ij and PV ij , PV ij+1 . Thus for

25 i 3 Figure 2.7: Projection of a facet FP onto the unit sphere in E

j = 1, 2, . . . , n (i0 := ini and ini+1 := i1), we have: D−→ −→ −→ −→ E PV i × PV i , PV i × PV i  j−1 j j j+1  τj = π − arccos −→ −→ −→ −→ .   PV ij−1 × PV ij PV ij × PV ij+1 i 3 Finally, we get the solid angle function Ωi(x) of the facet FP for any x ∈ R : D−−→ −−→ −−→ −−→ E n i XV i × XV i , XV i × XV i X  j−1 j j j+1  Ωi(x) = 2π − arccos −−→ −−→ −−→ −−→ .   j=1 XV ij−1 × XV ij XV ij × XV ij+1 We can summarize our results in the following Theorem 2.2.1 ([13]) Consider a solid angle α (0 < α < 2π) and a convex polyhe- dron P given by its data structure and its set of inequalities. Then the isoptic surface of P can be determined by the equation m X i α = IP (x)Ωi(x). i=1 The results and the computations will be demonstrated in the following subsection through an example of the computation using the regular tetrahedron, along with some figures. Remark 2.2.2 1. The algorithm can easily be extended to non-closed directed sur- faces e.g. for subdivision surfaces.

2. If we have a convex polyhedron, then projecting its whole surface to the unit sphere, we obtain a double coverage (double solid angle) of the given polyhedron, therefore the algorithm can be changed i.e. it is not necessary to determine the visible facets. In this case the isoptic surfaces are determined by the following implicit equation: m 1 X α = Ωi(x). 2 i=1

26 2.2.1 Computations of the isoptic surface of a given regular tetrahedron

Following the steps of the algorithm described above, we will calculate the compound isoptic surface of a given regular tetrahedron T whose data structure is determined i by its vertices and facets where the facets FP are given by their clockwise ordered vertices:  s  2 1 1 1 1 ! A1 = 0, 0, − √  ,A2 = − √ , − , − √ , 3 2 6 2 3 2 2 6 1 1 1 ! 1 1 ! A3 = − √ , , − √ ,A4 = √ , 0, − √ , 2 3 2 2 6 3 2 6 n 1 o n 2 o FT , {A2,A3,A4} , FT , {A3,A2,A1} , n 3 o n 4 o FT , {A4,A1,A2} , FT , {A1,A4,A3} . This tetrahedron can also be given by its system of inequalities:

 √   √  0 0 −4 3   2  √ √  x  √       −8 6 0 4 3     3 2   √ √ √   y  ≤  √  .        4 6 −12 3 4 3     3 2   √ √ √  z  √  4 6 12 3 4 3 3 2

Since the final formula of the isoptic surface is too long to appear in print even for this simple example (the right side of the equation consist of twelve arccos function), 1 we choose i = 1 and derive only IP (x) and Ω1(x).  √ √ 1  1 −4 3z > 2 IP (x, y, z) = √ √ .  0 −4 3z ≤ 2

27  √ √  2 2  −24x + 4 3x(1 − 6y) − 12y − 24z − 4 6z + 3  Ω1(x, y, z) = 2π − arccos   −  r √ √   √ √  2 24x2 + 8 3x + 24z2 + 4 6z + 3 6x2 + 4 3x(3y − 1) + 18y2 − 12y + 24z2 + 4 6z + 3 " 1  √ √  − arccos √ −24x2 + 4 3x(6y + 1) + 12y − 24z2 − 4 6z + 3 2 3

r  √  √  √ √ √ √ √ √ √  2x2 72y2 + 120z2 + 20 6z − 1 − 16 3x3(6y + 1) + 4x 12 3y2 − 24 3yz2 − 12 2yz + 5 3y + 8 3z2 + 4 2z + 3 + ...

#  √   √  √ √ −1 ··· + 6y2 24z2 + 4 6z + 3 + 4y 24z2 + 4 6z + 3 + 192z4 + 48x4 + 64 6z3 + 80z2 + 8 6z + 3 −

" 1  √ √  − arccos √ 12x2 − 8 3x − 36y2 − 24z2 − 4 6z + 3 2 3 28

r √  √   √ √ √ √   √  12x4 − 16 3x3 + x2 −72y2 + 96z2 + 16 6z + 28 + 8x 6 3y2 − 8 3z2 − 4 2z − 3 + 108y4 + 12y2 24z2 + 4 6z − 1 + ...

√ √ −1# ··· + 192z4 + 64 6z3 + 80z2 + 8 6z + 3 (b) A part of the tetrahedral isoptic surface,

(a) The whole surface e.g. the surface φ1,2 is derived from the solid 3 4 angles of the facets FT and FT .

Figure 2.8: Isoptic surface of the regular tetrahedron for α = π/8

For the surface obtained, see Figure 2.8a, and Figure 2.8b showing a part of the tetrahedral isoptic surface for the solid angle α = π/8. The surface φ1,2 is derived by 3 4 the solid angles of the facets FT and FT . The isoptic surface φ1 is derived from the 2 3 4 solid angles of the facets FT , FT and FT and the isoptic surface φ2 is derived from 1 3 4 the solid angles of the facets FT , FT and FT .

2.2.2 Some isoptic surfaces to Platonic and Archimedean polyhedra

In the following, we apply our algorithm to some polyhedra. We note here that the algorithm provides the implicit equation of the compound isoptic surface related to a given polyhedron.

Remark 2.2.3 Despite the equation being obtainable in O(e) steps, where e is the number of edges, the rendering of these figures by Wolfram Mathematica takes 20–40 minutes. The implicit equation of the isoptic surface is so complicated that it seems difficult to draw further consequences from it.

29 (a) Isoptic surface of the cube for α = π/2. (b) Isoptic surface of the regular octahedron for α = π/7.

Figure 2.9: Isoptic surface of Platonic solids.

(a) Isoptic surface of the truncated cube for (b) Isoptic surface of the truncated octa- α = π. hedron α = 2π/3.

Figure 2.10: Isoptic surface of Archimedan solids.

30 Chapter 3

Isoptic curves of conics in non-Euclidean constant curvature geometries

3.1 Projective model

The following model (see [45] and [41]) is capable of uniting the 8 Thurston geometries ([62]). Let G be one of the constant curvature plane geometries. We use homogeneous 1 2 3 coordinates x = (x : x : x ) and u = (u1 : u2 : u3) in order to represent points X as well as straight lines u in projective space P3. Sometimes we write X = xR in order to express that the point X is determined by a certain vector x ∈ R3 and its non-trivial scalar multiples, similarly u = uR for lines. A point X = xR and a straight line u = uR are incident if x · uT = 0 with · denoting the usual matrix multiplication. Constant curvature plane geometries can be represented in projective space P3 using the bilinear form hx, yi = x1y1 + x2y2 + x3y3, (3.1) where  = 0, −1, +1, respectively, Euclidean, hyperbolic and elliptic geometries. Now we consider hyperbolic and elliptic planes, therefore  = ±1. Hyperbolic and elliptic distances and angles can be computed with the help of the bilinear form (3.1), see [45]. The distance d(X,Y ) of two proper points X = xR and Y = yR is given by hx, yi C(s) = q , (3.2) hx, xihy, yi where C(s) is the cosine function in elliptic geometry, and hyperbolic cosine functionin hyperbolic geometry.

31 Further, we find the angle α(u, v) enclosed by two proper straight lines u = uR and v = vR, respectively, with hu, vi cos α = q . (3.3) hu, uihv, vi

3.1.1 Example: Isoptic curve of the line segment on the hy- perbolic and elliptic plane

It is well known, that the α-isoptic curve of a line segment in the Euclidean plane contains of two circular arcs with central angle 2α symmetric to the straight line of the segment. In this subsection, we study the analogous question in the hyperbolic and elliptic planes together. Let G be either the hyperbolic H2 or the elliptic E 2 plane with  = −1 and  = 1 respectively. We use the usual projective models of the geometries above. Let two points A and B be given in the plane. Without loss of generality, we can assume that their homogeneous coordinates are A = (a : 0 : 1) and B = (−a : 0 : 1), T where (a ∈]0, 1]). We consider two straight lines u and v, where u = (u1 : u2 : 1) T passes trough points A and P , and v = (v1 : v2 : 1) passes through points B and P . By the incidence formula we get the following equations:   u1   1   A ∈ u ⇔ (a : 0 : 1) u2 = 0 ⇔ u1 = − ,   a 1 (3.4)   v1   1   B ∈ v ⇔ (−a : 0 : 1) v2 = 0 ⇔ v1 = .   a 1

  u1   a − x   P ∈ u ⇔ (x : y : 1) u2 = 0 ⇔ u2 = − , y 6= 0,   ya 1   (3.5) v1   a + x   P ∈ v ⇔ (x : y : 1) v2 = 0 ⇔ v2 = − , y 6= 0.   ya 1 The angle α between the above straight lines can be determined by the for- mula (3.3): ( + u v + u v ) cos(α) = 1 1 2 2 q 2 2 2 2 ( + u1 + u2)( + v1 + v2) with  = ±1 if G is elliptic or hyperbolic. Substituting coordinates from (3.4) and (3.5) into the above equation, we obtain:

32 Theorem 3.1.1 (CsG) Let us suppose that a line segment is given in the H2 hyper- bolic or E 2 elliptic plane by its endpoints A = (a : 0 : 1) and B = (−a : 0 : 1). Then for a given α(0 < α < π), the α-isoptic curve of AB in the hyperbolic and elliptic plane has an equation of the form:

1 a2−x2 ( − 2 + 2 2 ) cos(α) = a y a , (3.6) q 1 a−x 2 1 a+x 2 ( + a2 + ( ya ) )( + a2 + ( ya ) ) where  = ±1 if G is either the elliptic or the hyperbolic plane.

Remark 3.1.2

1. Choosing α = π/2 results in the orthoptic curve:

x2 y2 + = 1. (3.7) a2 a2 1−a2 This is an ellipse (without endpoints of the given segment) in the Euclidean sense, and it can be called the Thales curve. In the hyperbolic plane, if we increase the parameter a, then the Thales curve tends to a hypercycle (or an equidistant curve). That means the hypercycle is a special type of orthoptic curves with equation

x2 + 2y2 = 1.

2. In the hyperbolic plane, if a → 1, then equation (3.6) converges to the following equation  2 y x2 + = 1.   α  cos 2

3. In the elliptic plane, ignoring the condition a ∈]0, 1] if a → ∞, the right side of equation (3.6) converges to 1, therefore the full line in elliptic geometry has no isoptic curve at all.

3.1.2 The general method

The procedure above can be used to develop a more general method to determine the isoptic curves of a given hyperbolic or elliptic object. However we describe the method for conics. Let a hyperbolic or elliptic conic section C and one of its points P be given. Using the implicit function theorem and the equation of C, we can determine the equation of the tangent line in this point. After that, a system of equations for the coordinates of the tangent point from an exterior point K can be

33 1.0 3

2

0.5

1

0.0 -3 -2 -1 1 2 3

-1 -0.5

-2

-1.0 -3 -1.0 -0.5 0.0 0.5 1.0

Figure 3.1: The isoptic curve of the hyperbolic (left) and elliptic (right) line segment with parameters: hyperbolic: a = 0.4, α = π/6, elliptic: a = 0.8, α = 8π/18 given. This point has to satisfy the equation of the given curve and the tangent lines to this point have to contain K. This system can be solved for every K = (x0 : y0 : 1) outer point with respect to the parameters x0, y0. In the general case, the formulas of the solutions are complicated. Now, we have to follow the above method using the coordinates of the tangent points. The equation of the tangent lines from K can be determined by solving a system of equations for its coordinates. Finally, we have to fix the angle of the straight lines and we get the equation of the isoptic curve. But first of all, we need the conic equations.

3.2 Elliptic conics and their isoptics

In this section, we will discuss the equations of the conic sections and their isoptics in the elliptic plane E 2. We remark that in the elliptic plane the maximum distance between two points π is less than 2 ; therefore, in some cases the given curve cannot be seen under an arbitrarily small angle.

3.2.1 Equation of the elliptic ellipse and hyperbola

Definition 3.2.1 The elliptic ellipse is the locus of all points of the elliptic plane whose distances to two fixed points sum up to the same constant 2a.

Definition 3.2.2 The elliptic hyperbola is the locus of points where the absolute value of the difference of the distances from the two foci is a constant 2a.

We discuss the ellipse and the hyperbola together. We may suppose that the two foci are equidistant from the origin O, both located on the axis x with coordinates f1R = F1 = (f : 0 : 1) and f2R = F2 = (−f : 0 : 1) where 0 < f < π/2. Let

34 pR = P = (x : y : 1) ∈ E 2 a point of the conic section. The inequality 2a ≤ π must π hold since the distance between any two point is at most 2 . Using (3.2) we obtain the following equation:

    −1 hp, f1i −1 hp, f2i 1 cos q  + 2 cos q  = hp, pi hf1, f1i hp, pi hf2, f2i   −1 (1 + xf) = 2a ⇔ 1 cos q  = (1 + x2 + y2)(1 + f 2)   −1 (1 − xf) = 2a − 2 cos q  , (1 + x2 + y2)(1 + f 2) where 1,2 = ±1 and 1 + 2 ≥ 0. Applying cosine to both sides, we get the following equations: 1 + xf 1 − xf q = cos(2a)q (1+x2 +y2)(1+f 2) (1+x2 +y2)(1+f 2)    −1 1 − xf +2 sin(2a) sincos q  . (1+x2 +y2)(1+f 2) √ The next equation is obtained by applying the formula sin (cos−1(t)) = 1 − t2 and q by multiplying both sides by (1 + x2 + y2)(1 + f 2).

q 2 2 2 2 1 + xf = cos(2a)(1 − xf) + 2 sin(2a) (1 + x + y )(1 + f ) − (1 − xf) .

Now, if we simplify this equation and take its square, there is no  therein. Finally, we get the following equation:

x !2 y2 + 1 = 1. (3.8) tan(a) (1+f 2) cos2(a) − 1

If the distance between the two foci is less than 2a, it is an ellipse; if it is greater, then it is a hyperbola since cosine is monotonically decreasing on [0, π] and we have

hf , f i 1 − f 2 d(F ,F ) <> 2a ⇔ cos(2a) <> cos(d(F ,F )) = 1 2 = 1 2 1 2 q 1 + f 2 hf1, f1i hf2, f2i 2 1 ⇔ 2 cos2(a) − 1 <> − 1 ⇔ 1 <> 1 + f 2 cos2(a)(1 + f 2) 1 ⇔ 0 <> − 1. cos2(a)(1 + f 2)

Therefore, the elliptic ellipse and hyperbola are also an ellipse and a hyperbola in the model.

35 3.2.2 Equation of the elliptic parabola

Definition 3.2.3 An elliptic parabola is the set of points (X = (x : y : 1) ∈ E 2) in the elliptic plane that are equidistant to a proper point (the focus F ) and a proper line (the directrix e)(s = d(X; F ) = d(X; e)).

Without loss of generality, we can assume that the directrix (e) is the axis x and the coordinates of the focus point are F = (0 : p : 1). We remark that the coordinates of the foot point X0 of the perpendicular dropped form X to the x-axis are X0 = (x : 0 : 1). The distances d(X; F ) and d(X; e) = d(X; X0) can be computed by the formula (3.2): 1 + py 1 + x2 cos(s) = √ √ = √ √ . (3.9) 1 + x2 + y2 1 + p2 1 + x2 + y2 1 + x2 From (3.9) we obtain the equation of the elliptic parabola: (1 + py)2 −x2 + = 1. (3.10) 1 + p2

3.2.3 Isoptic curves of elliptic ellipse and hyperbola

Now, we will use the method described in 3.1.2 to determine the isoptic curves to the elliptic ellipses and hyperbolas. The first step is to determine the equation of the tangent lines (y 6= 0): x f 2 ! y0 = − 1 − . (3.11) y sin2(a)(1 + f 2) The equation (3.8) converges continuously to x2 = tan2(a), if y → 0. After that, we ˜ have to solve the following system of equations for Xi = (˜xi :y ˜i : 1), (i = 1, 2), where X = (x : y : 1) is a point in the model: x˜ f 2 ! − 1 − (x − x˜) +y ˜ = y y˜ sin2(a)(1 + f 2) (3.12) x˜ !2 y˜2 + 1 = 1. tan(a) (1+f 2) cos2(a) − 1 It is not so hard to determine the roots, but because of the complexity of the T T result, we omit it. Now we need u = (u1 : u2 : 1) and v = (v1 : v2 : 1) straight ˜ ˜ lines, fits on P , X1 and P , X2, respectively.

Using that (˜xi, y˜i) is known, we get the following system of equations:

u1x + u2y + 1 = 0 (3.13) u1x˜1 + u2y˜1 + 1 = 0,

v1x + v2y + 1 = 0 (3.14) v1x˜2 + v2y˜2 + 1 = 0.

36 Solving these systems, we can determine the tangent lines for all X = Rx outside of the conic section. The last step in the method of the previous section is to fix the angle. We summarize our results in the following theorem:

Theorem 3.2.4 ([12]) Let an elliptic ellipse or hyperbola be centered at the origin π of the projective model given by its semimajor axis a (a ∈ (0, 2 ) and its foci F1 =

(f : 0 : 1), F2 = (−f : 0 : 1), (0 < f < 1) such that 2a > d(F1,F2) or 2a < d(F1,F2) holds. The α-isoptic and (π − α)-isoptic curves (0 < α < π) of the considered ellipse or hyperbola in the elliptic plane have the equation

((1 + f 2) cos(2a)(x2 + y2 + 1) + f 2x2 − 1)2 cos2(α) = . (3.15) 2 (1 + f 2) y2 (f 2 + x2) + (f 2 − x2)2 + (1 + f 2)2 y4

Remark 3.2.5

1. The orthoptic curve of the elliptic ellipse and hyperbola is an ellipse with the following equation:

    1 + f 2 cos(2a) x2 + y2 + 1 + f 2x2 = 1.

2. The isoptic curve of the elliptic hyperbola exists if the following formula holds true 1 − (1 + f 2) cos(2a) !! cos α ≤ max , f 2 + (1 + f 2) cos(2a) f 2 π s 1 ! ! ∧ a ≥ ∨ f ≤ − 1 ∨ (α∈ / I) , 6 cos(2a)

where (1 + f 2) cos(2a) − 1! 1 − (1 + f 2) cos(2a)!! I = arccos , arccos . f 2 f 2

3.2.4 Isoptic curves of elliptic parabola

Using the same method, we can solve the following system of equations for (˜x, y˜):

x˜(1 + p2) (x − x˜) +y ˜ = y p +yp ˜ 2 (3.16) (1 + py)2 −x2 + = 1. 1 + p2

Finally, we have to solve the system of equations (3.13), (3.14). Now, the following theorem holds

37 (a) Ellipse: a = 1, f = 1; α = 2π/3

(b) Ellipse: a = 0.7, f = 0.8; α = π/3

(c) Hyperbola: a = 0.7, f = 1; α = π/2

(d) Hyperbola: a = 0.78, f = 1; α = 3π/4

Figure 3.2: Isoptic curve to elliptic ellipse and hyperbola

38 (a) p = 0.25; α = π/3 (b) p = 1.5, α = 2π/3

Figure 3.3: Isoptic curve to elliptic parabola

Theorem 3.2.6 ([12]) Let an elliptic parabola be given in the projective model by its focus F = (0 : p : 1) and its directrix e which coincides with an x-axis. The α-isoptic curve of this parabola (0 < α < π) in the elliptic plane has the equation y(py + 1) cos(α) = q . (3.17) (x2 + 1)((p2(x2 + 1) − 2py + y2 + x2)

Remark 3.2.7 The orthoptic curve of the elliptic parabola contains two straight lines 1 y = 0 and y = − p .

Remark 3.2.8 The figures of the isoptic curves also confirm the fact that in elliptic geometry there is only one class of conic sections. However, in the affine model of the projective plane used, these can be considered separately.

3.3 Hyperbolic conic sections and their isoptics

3.3.1 Generalized angle of straight lines

Having regard to the fact that the majority of the generalized conic sections have ideal and outer tangents as well, it is inevitable to introduce the generalized concept of the hyperbolic angle. In the extended hyperbolic plane there are three classes of lines by the number of common points with the absolute conic AC:

1. The straight line u = Ru is proper if card(u ∩ AC) = 2 ⇔ hu, ui > 0.

2. The straight line u = Ru is non-proper if card(u ∩ AC) < 2.

(a) If card(u ∩ AC) = 1 ⇔ hu, ui = 0 then u = Ru is called boundary straight line.

(b) If card(u ∩ AC) = 0 ⇔ hu, ui < 0 then u = Ru is called outer straight line.

We define the generalized angle between straight lines using the results of the papers [3], [26] and [65] in the projective model.

39 Definition 3.3.1 1. Suppose that u = Ru and v = Rv are both proper lines.

(a) If hu, uihv, vi − hu, vi2 > 0 then they intersect in a proper point (see Figure 3.4a) and their angle α(u, v) can be measured by

±hu, vi cos α = q . (3.18) hu, uihv, vi

(b) If hu, uihv, vi − hu, vi2 < 0 then they intersect in a non-proper point (see Figure 3.4b) and their angle is the length of their normal transverse and it can be calculated using the formula below:

±hu, vi cosh α = q . (3.19) hu, uihv, vi

(c) If hu, uihv, vi − hu, vi2 = 0 then they intersect in a boundary point and their angle is 0.

2. Suppose that u = Ru and v = Rv are both outer lines of H2(see Figure 3.4c). The angle of these lines will be the distance of their poles using the formula (3.19).

3. Suppose that u = Ru is a proper and v = Rv is an outer line (see Figure 3.4d). Their angle is defined as the distance of the pole of the outer line to the real line and can be computed by

±hu, vi sinh α = q . (3.20) −hu, uihv, vi

4. Suppose that at least one of the straight lines u = Ru and v = Rv is boundary line of H2 (see Figure 3.4e). If the other line fits the boundary point, the angle cannot be defined, otherwise it is infinite.

Remark 3.3.2 In the previous definition we fixed that except case 1. (a) we use real distance type values instead of complex angles which arise in other cases. The ± on the right sides are justifiable because we consider complementary angles i.e. α and π − α together.

3.3.2 Classification of generalized conic sections on the hy- perbolic plane in dual pairs

The literature of the hyperbolic conic section classification is very wide and it goes back until 1902 when it was first given by Liebmann (see [35]). We also note that

40 3 3 3

2 2 2

1 1 1

-3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3

-1 -1 -1

-2 -2 -2

-3 -3 -3 (a) Both u and v are real line, (b) Both u and v are real line, (c) Both u and v are outer proper intersection. outer intersection. line.

3

3

2

2

1

1

-3 -2 -1 1 2 3

-3 -2 -1 1 2 3

-1

-1

-2

-2

-3

-3 (d) Line u is outer but v is (e) Line u is boundary. proper.

Figure 3.4: Generalized angle of straight lines u = Ru and v = Rv there is a detailed theory of conic sections in the work of both Cooligde and Kagan(see [8] and [30]). There are numerous other works in this topic (see [17] and [54]), but in this section we will summarize and extend the results of K. Fladt (see [18] and [19]) about the generalized conic sections on the extended hyperbolic plane. Let us denote a point with x and a line with u. Then the absolute conic (AC) can be defined as a point conic with the xT ex = 0 quadratic form where e = diag {1, 1, −1} or due to the absolute polarity as line conic with uEuT = 0 where E = e−1 = diag {1, 1, −1}. Similarly to the Euclidean geometry we use the well-known quadratic form

T 1 1 2 2 3 3 2 3 1 3 1 2 x ax = a11x x + a22x x + a33x x + 2a23x x + 2a13x x + 2a12x x = 0 where det a 6= 0 for a non-degenerate point conic and

T 11 22 33 23 13 12 uAu = A u1u1 + A u2u2 + A u3u3 + 2A u2u3 + 2A u1u3 + 2A u1u2 = 0 where A = a−1 for the corresponding line conic defined by the tangent lines of the previous point conic. Using the polarity x = AuT and uT = ax follow since ux = 0.

41 Consider a one parameter conic family of our point conic with the (AC), defined by xT (a + ρe)x = 0.

Since the characteristic equation ∆(ρ) := det(a+ρe) is an odd degree polynomial, this conic pencil has at least one real degenerate element (ρ1), which consists of at 1 2 most two point sequences with holding lines p1 and p1 called asymptotes. Therefore we get a product

T 1 T 2 T 1 T 2 x (a + ρ1e)x = (p1x) (p1x) = x ((p1) p1)x = 0 with occasional complex coordinates of the asymptotes. Each of these two asymptotes has at most two common points with the (AC) and with the conic as well. Thus, the at most 4 common points with at most 3 pairs of asymptotes can be determined through complex coordinates and elements according to the at most 3 different eigenvalues

ρ1, ρ2 and ρ3. In complete analogy with the previous discussion in dual formulation we get that the one parameter conic family of a line conic with (AC) has at least one degener- ate element (σ1) which contains two line pencils at most with occasionally complex 1 1 holding points f1 and f2 called foci.

1 T 1 1 T 1 1 T T u(A + σ E)u = (uf1 )(uf2 ) = u(f1 (f2 ) )u = 0 For each focus at most two common tangent line can be drawn to AC and to our line conic. Therefore, at most four common tangent lines with at most three pairs of foci can be determined maybe with complex coordinates to the corresponding eigenvalues σ1, σ2 and σ3. Combining the previous discussions with [18] and [41] the classification of the conics on the extended hyperbolic plane can be obtained in dual pairs. First, our goal is to find an appropriate transformation, so that the resulted normalform characterizes the conic e. g. the straight line x1 = 0 is a symmetry axis of the conic section (a31 = a12 = 0). Therefore we take a rotation around the origin O(0, 0, 1)T and a translation parallel with x2 = 0. As it used before, the characteristic equation   a11 + ρ a12 a13     ∆(ρ) = det(a + ρe) = det  a21 a22 + ρ a23  = 0   a31 a32 a33 − ρ has at least one real root denoted by ρ1.

This is helpful to determine the exact transformation if the equalities ρ1 = ρ2 = ρ3 not hold. That case will be covered later. With this transformations we obtain the

42 normalform 1 1 2 2 2 3 3 3 ρ1x x + a22x x + 2a23x x + a33x x = 0. (3.21)

In the following we distinguish 3 different cases according to the other two roots:

1. Two different real roots Then the monom x2x3 can be eliminated from the equation above, by translat- ing the conic parallel with x1 = 0. The final form of the conic equation in this case, called central conic section:

1 1 2 2 3 3 ρ1x x + ρ2x x − ρ3x x = 0.

3 Because our conic is non-degenerate ρ3x 6= 0 follows and with the notations a = ρ1 and b = ρ2 our matrix can be transformed into a = diag {a, b, −1}, ρ3 ρ3 where a ≤ b can be assumed. The equation of the dual conic can be obtained −1 1 1 using the polarity E respected to (AC) by EAE = diag{ a , b , −1}. By the above considerations we can give an overview of the generalized central conics with representants:

Theorem 3.3.3 ([14]) If the conic section has the normalform ax2 + by2 = 1 then we get the following types of central conic sections (see Figure 3.5):

(a) Absolute conic: a = b = 1 (b) i. Circle: 1 < a = b ii. Circle enclosing the absolute: a = b < 1 (c) i. Hypercycle: 1 = a < b ii. Hypercycle enclosing the absolute: 0 < a < 1 = b (d) Hypercycle excluding the absolute: a < 0 < 1 = b (e) Concave hyperbola: 0 < a < 1 < b (f) i. Convex hyperbola: a < 0 < 1 < b ii. Hyperbola excluding the absolute: a < 0 < b < 1 (g) i. Ellipse: 1 < a < b ii. Ellipse enclosing the absolute: 0 < a < b < 1 (h) empty: a ≤ b ≤ 0

where either the conic and its dual pair lies in the same class or (i) and (ii) are 0 1 0 1 dual pairs with a = a and b = b .

43 2. Coinciding real roots

The last translation cannot be enforced but it can be proved that ρ2 = ρ3 =

a33−a22 2 follows. With some simplifications of the formulas in [18] we obtain the normalform of the so-called generalized parabolas.

Theorem 3.3.4 ([14]) The parabolas have the normalform ax2 + (b + 1)y2 − 2y = b − 1 and the following cases arise (see Figure 3.6):

(a) i. Horocycle: 0 < a = b ii. Horocycle enclosing the absolute: a = b < 0 (b) i. Elliptic parabola: 0 < b < a ii. Parabola enclosing the absolute: b < a < 0 (c) i. Two sided parabola: a < b < 0 ii. Concave hyperbolic parabola: 0 < a < b (d) i. Convex hyperbolic parabola: a < 0 < b ii. Parabola excluding the absolute: b < 0 < a

0 b2 0 where all (i) and (ii) are dual pairs with parameters a = − a and b = −b.

3. Two conjugate complex roots Then the last translation cannot be performed to eliminate the monom x2x3 but we can eliminate the monom x3x3 by an appropriate transformation described in [18]. Shifting to inhomogeneous coordinates and simplifying the coefficients we obtain:

Theorem 3.3.5 ([14]) The so-called semi-hyperbola has the normalform ax2 + 2by2 − 2y = 0 where |b| < 1 and its dual pair is projectively equivalent 0 1 0 with another semi-hyperbola with a = a and b = −b (see Figure 3.7).

4. Overviewing the above cases only one remains, when the conic has no symmetry

axis at all and ρ1 = ρ2 = ρ3. Ignoring further explanations we claim the following theorem:

Theorem 3.3.6 ([14]) If the conic has the normalform (1−x2 −y2)+2ay(x+ 1) = 0 where a > 0 then it is called osculating parabola. Its dual is also an osculating parabola by a convenient reflection (see Figure 3.8).

44 3.3.3 Isoptic curves of generalized hyperbolic conic sections

3.3.3.1 Isoptic curves of central conics

In this section we will extend the algorithm for determining the isoptic curves of conic sections described in [12] for generalized hyperbolic conic sections. We have to apply the generalized notion of angle, therefore the equation of the isoptic curve will not be given by a implicit formula but the isoptic curve consist of some piecewise continuous arcs that will be given by implicit equations. First we determine the equations of the tangent lines through a given external point P to a given conic section C. We will use not only the point conic but the corresponding line conic as well. The algorithm does not use the points of tangency but of course they can be determined from the tangents. This strategy provides a general simplification in the other cases as well. Let the external point P be given with homogeneous coordinates (x, y, 1)T . If a central conic is given by a, then the corresponding line conic is defined by A, where a−1 = A. Now, we know that P fits on the tangent lines u = Ru and v = Rv where u = (u1, u2, 1) and v = (v1, v2, 1) furthermore u and v satisfy the equation of the line conic.   u1x + u2y + 1 = 0  v1x + v2y + 1 = 0  2 2 2 2 u1 u2 v1 v2 a + b − 1 = 0  a + b − 1 = 0  Solving the above systems we obtain the coordinates of the straight lines u and v: √ √ 2 2 2 2 2 2 u = − ax+ aby (ax +by −1) v = −ax+ aby (ax +by −1) 1 √ax2+by2 1 √ax2+by2 (3.22) −by2+x aby2(ax2+by2−1) by2+x aby2(ax2+by2−1) u2 = ax2y+by3 v2 = − ax2y+by3 Of course, aby2 (ax2 + by2 − 1) ≥ 0 must hold otherwise P (x, y, 1)T is not an external point. We get the exact formula of the more parted isoptic curve related to the central conic sections (see Theorem 3.3.3) using the definition of the generalized angle (see Definition 3.3.1). The compound isoptic can be given by classifying the straight lines u = Ru and v = Rv according to their poles. We summarize our result in the following:

Theorem 3.3.7 ([14]) Let a central conic section be given by its equation ax2+by2 = 1 (see Theorem 3.3.3). Then the compound α-isoptic curve (0 < α < π) of the considered conic has the equation

(a ((b + 1)x2 − 1) + (a + 1)by2 − b)2 = 2 2 4 2 2 2 2 (a − 1) b y + 2(a − 1)b (b + a ((b − 1)x − 1)) y + (a(b − 1)x + a − b)

45  2  cosh (α), aby2 (ax2 + by2 − 1) ≥ 0 ∧    (1 − u2 − u2) (1 − v2 − v2) > 0 ∧  1 2 1 2   2 2  x + y > 1     = cos2(α), aby2 (ax2 + by2 − 1) ≥ 0 ∧   x2 + y2 < 1       2 2 2 2  sinh (α), aby (ax + by − 1) ≥ 0 ∧   2 2 2 2  (1 − u1 − u2) (1 − v1 − v2) < 0, wherein u1,2 and v1,2 are derived by (3.22).

In Figures 3.5 we visualize the isoptic curves of central conic sections. The fore- most figure shows, how different types of isoptics arise due to common tangents. We indicated the Cayley-Klein model circle with black, the conic with dashed line and the isoptic is shaded.

3.3.3.2 Isoptic curves of parabolas

We study the isoptic curves of the generalized parabolas given by their equations in Theorem 3.3.4. The algorithm described in the previous subsection can be repeated for further conics as well. The difference is only in the equation of the compound isoptic is because of the different conic equation.

The coordinates of the tangent line u = (u1, u2, 1) and v = (v1, v2, 1) can be given: √ 2 2 2 2 2 2 u = − ax (b+y−1)+y ab x (ax +b(y −1)+(y−1) ) 1 √ a(b−1)x3+b2xy2 ax2−b2y− ab2x2(ax2+b(y2−1)+(y−1)2) u2 = a(b−1)x2+b2y2 (3.23) √ 2 2 2 2 2 2 v = −ax (b+y−1)+y ab x (ax +b(y −1)+(y−1) ) 1 √ a(b−1)x3+b2xy2 ax2−b2y+ ab2x2(ax2+b(y2−1)+(y−1)2) v2 = a(b−1)x2+b2y2

Theorem 3.3.8 ([14]) Let a parabola be given by its equation ax2 +(b+1)y2 −2y = b − 1 (see Theorem 3.3.4). Then the compound α-isoptic curve (0 < α < π) of the considered conic has the equation

      2  2 2 2 2 2 2 2 4 a b 2x + y − 1 + (y − 1) + b y − 1 (y − 1) (y + 1) b −

     −1 2 2 2 2 2 2 2 2 2 2 −2a 2x + y + b(y + 1) − 1 b + a (y − 1) + b (y + 1) + 2b 2x + y − 1 =

46 (a) Domains for concave hyperbola with (b) Concave hyperbola: π notations of Definition 3.3.1 a = 0.3, b = 2, α = 2

(c) Hyperbola excluding the absolute: (d) Convex hyperbola: π 19π a = 0.5, b = −2, α = 3 a = 1.1, b = −1.5, α = 36

(e) Ellipse: (f) Ellipse enclosing the absolute: 7π π a = 2, b = 3, α = 18 a = 0.45, b = 0.8, α = 2

Figure 3.5: Isoptic curve to central conics

47  2  cosh (α), ab2x2 (ax2 + b (y2 − 1) + (y − 1)2) ≥ 0 ∧    (1 − u2 − u2) (1 − v2 − v2) > 0∧  1 2 1 2   2 2  x + y > 1     = cos2(α), ab2x2 (ax2 + b (y2 − 1) + (y − 1)2) ≥ 0 ∧   x2 + y2 < 1       2 2 2 2 2 2  sinh (α), ab x (ax + b (y − 1) + (y − 1) ) ≥ 0 ∧   2 2 2 2  (1 − u1 − u2) (1 − v1 − v2) < 0, wherein u1,2 and v1,2 are derived by (3.23).

The isoptic curves of the parabolas can be seen in Figure 3.6. The same convention has been used as on the previous figures.

3.3.3.3 Isoptic curves of semi-hyperbola

Theorem 3.3.9 ([14]) Let the semi-hyperbola be given by its equation ax2 + 2by2 − 2y = 0, where |b| < 1 (see Theorem 3.3.5). Then the compound α-isoptic curve (0 < α < π) of the considered conic has the equation

     2      2 2 2 4 2 2 2 2 2 2 2a b x + y − y + y − 1 y + 4a x + y b − 1 x + (by − 1) −

       −1 2 2 4 2 2 2 2 −4a y − 2x + y y + b y + x − 1 y + x − 2y + 1 =

 2  cosh (α), ax2 (a + 2y(by − 1)) ≥ 0∧    (1 − u2 − u2) (1 − v2 − v2) > 0∧  1 2 1 2   2 2  x + y > 1     = cos2(α), ax2 (a + 2y(by − 1)) ≥ 0∧   x2 + y2 < 1       2 2  sinh (α), ax (a + 2y(by − 1)) ≥ 0∧   2 2 2 2  (1 − u1 − u2) (1 − v1 − v2) < 0,

48 (a) Elliptic parabola: (b) Parabola enclosing the absolute: π 7π a = 2, b = 1.5, α = 3 a = −2.5, b = −5, α = 18

(c) Two sided parabola: (d) Concave hyperbolic parabola: π π a = −5, b = −2.7, α = 2 a = 1, b = 2, α = 2

(e) Convex hyperbolic parabola: (f) Parabola excluding the absolute: π π a = −2, b = 1.5, α = 3 a = 0.8, b = −0.4, α = 3

Figure 3.6: Isoptic curve to parabolas

49 wherein √ 2 u = − ax+ a(ax +2y(by−1)) 1 √ y ax2−y+x a(ax2+2y(by+1)) u2 = y2 √ 2 v = −ax+ a(ax +2y(by−1)) 1 √y ax2−y−x a(ax2+2y(by+1)) v2 = y2 . The isoptic curve of the semi-hyperbola can be seen in Figure 3.7.

π 8π (a) a = 1.4, b = 0.5, α = 4 (b) a = 1.4, b = 0.5, α = 18

Figure 3.7: Isoptic curve to semy-hyperbola

3.3.3.4 Isoptic curves of the osculating parabola

Theorem 3.3.10 ([14]) Let the osculating parabola be given by its equation (1 − x2 − y2)+ 2a(x+1)y = 0 (see Theorem 3.3.6). Then the compound α-isoptic curve (0 < α < π) of the considered conic has the equation (−2 (x2 + y2 − 1) + 2a(x + 1)y + a2(x + 1)2)2 = |a2(x + 1)3 (4(1 − x) + 4ay + a2(x + 1))|

 2  cosh (α), (x2 + y2 − 1 − 2a(x + 1)y) ≥ 0∧    (1 − u2 − u2) (1 − v2 − v2) > 0∧  1 2 1 2   2 2  x + y > 1     = cos2(α), (x2 + y2 − 1 − 2a(x + 1)y) ≥ 0∧   x2 + y2 < 1       2 2 2  sinh (α), (x + y − 1 − 2a(x + 1)y) ≥ 0∧   2 2 2 2  (1 − u1 − u2) (1 − v1 − v2) < 0,

50 wherein √ 2 2 2 u = −(1+ay)(x−ay)+ y (x +y −1−2a(x+1)y) 1 (x2+y2)−2axy+a2y√2 y2−ax(x+1)y+a2(x+1)y2+x y2(x2+y2−1−2a(x+1)y) u2 = − y((x2+y2)−2axy+a2y2) √ 2 2 2 v = − (1+ay)(x−ay)+ y (x +y −1−2a(x+1)y) 1 (x2+y2)−2axy+a2√y2 y2−ax(x+1)y+a2(x+1)y2−x y2(x2+y2−1−2a(x+1)y) v2 = − y((x2+y2)−2axy+a2y2) .

The Figure 3.8 shows some cases of the isoptic curve for the osculating parabola.

π 2π (a) a = 0.4, α = 3 (b) a = 0.4, α = 3

Figure 3.8: Isoptic curve to osculating parabola

Our method is suited for determining the isoptic curves to generalized conic sec- tions for all possible parameters. Moreover with this procedure above we may be able to determine the isoptics for other curves as well.

51 Chapter 4

On the geometry of SL^2R

The SL^2R space is the universal cover of SL2R. Recall that SL^2R is one of the eight Thurston geometries [56, 62] that can be derived from the 3-dimensional Lie group of all 2 × 2 real matrices with unit determinant. According to [40], the space of left invariant Riemannian metrics on the group SL^2R is 3-dimensional.

In Section 4.1 we describe the projective model of SL^2R and we shall use the standard Riemannian metric of SL^2R obtained by pull back transform applied to the infinitesimal arc-length-square at the origin. This coincides with the global projective metric belonging to the quadratic form (4.2). We also describe the isometry group of SL^2R. In Section 4.2 we give an overview not only of geodesics, but also of translation curves summarizing previous results, related to the projective model of

SL^2R suggested in [43, 60]. Our main results will be presented in Section 4.3, namely that the sum of the interior angles in a translation triangle must be greater than or equal to π. However in geodesic triangles this sum can be either smaller, greater, or equal to π. Finally, isoptic surface of the translation line segment will be considered in Section 4.4.

4.1 The projective model for SL^2R   d b Real 2 × 2 matrices   with the unit determinant ad − bc = 1 form a Lie c a transformation group by the standard product operation, taken to act on row matrices as point coordinates

  0 1 d b 0 1 0 1 0 1 (z , z )   = (z d + z c, z b + z a) = (w , w ) (4.1) c a with w1 b + (z1/z0) a b + za w = = = , w0 d + (z1/z0) c d + zc

52 z = z1/z0, on the complex projective line C∞. This group is a 3-dimensional mani- fold, because of its 3 independent real coordinates and with its usual neighborhood topology [56, 62]. In order to model the above structure in the projective sphere PS3 and in the projective space P3 (see [43]), we introduce the new projective coordinates (x0 : x1 : x2 : x3) where

a := x0 + x3, b := x1 + x2, c := −x1 + x2, d := x0 − x3.

Then it follows that

0 > bc − ad = −x0x0 − x1x1 + x2x2 + x3x3 (4.2) describes the interior of the one-sheeted hyperboloid solid H in the usual Euclidean coordinate simplex, with the origin E0(1 : 0 : 0 : 0) and the ideal points of the axis ∞ ∞ ∞ E1 (0 : 1 : 0 : 0), E2 (0 : 0 : 1 : 0), E3 (0 : 0 : 0 : 1). We consider the collineation 3 group G∗ that acts on the projective sphere PS and preserves a polarity, i.e. a scalar product of signature (− − ++). This group leaves the one-sheeted hyperboloid solid

H invariant. We have to choose an appropriate subgroup G of G∗ as isometry group, then the universal covering group and space Hf of H will be the hyperboloid model of SL^2R (see Fig. 4.1 and [43]).

Figure 4.1: The hyperboloid model.

Consider isometries given by matrices   cos φ sin φ 0 0     − sin φ cos φ 0 0  S(φ) =   , (4.3)    0 0 cos φ − sin φ   0 0 sin φ cos φ where φ ∈ [0, 2π). They constitute a one parameter subgroup which we denote by S(φ). The elements of S(φ) are the so-called fibre translations. We obtain a unique fibre line to each X(x0 : x1 : x2 : x3) ∈ Hf as the orbit by right action of S(φ) on X.

53 The coordinates of points, lying on the fibre line through X, can be expressed as the images of X by S(φ):

S(φ) (x0 : x1 : x2 : x3) −→ (x0 cos φ − x1 sin φ : x0 sin φ + x1 cos φ : (4.4) x2 cos φ + x3 sin φ : −x2 sin φ + x3 cos φ). In (4.3) and (4.4) we can see the 2π-periodicity by φ. It leads us to introduce the extension to φ ∈ R, as real parameter, to obtain the universal covers Hf and SL^2R, respectively, through the projective sphere PS3. The elements of the isometry group of SL2(R) (and so by the above extension the isometries of SL^2R) can be described j by the matrix (ai ) (see [43, 45])

 0 1 2 3  a0 a0 a0 a0    1 0 3 2 j ∓a0 ±a0 ±a0 ∓a0 (a ) =   , (4.5) i  0 1 2 3   a2 a2 a2 a2    1 0 3 2 ±a2 ∓a2 ∓a2 ±a2 where   −(a0)2 − (a1)2 + (a2)2 + (a3)2 = −1,  0 0 0 0   0 2 1 2 2 2 3 2  −(a2) − (a2) + (a2) + (a2) = 1,

 −a0a0 − a1a1 + a2a2 + a3a3 = 0,  0 2 0 2 0 2 0 2   0 1 1 0 2 3 3 2  −a0a2 + a0a2 − a0a2 + a0a2 = 0, and we allow positive proportionality, of course, with projective freedom.

We define the translation group GT , as a subgroup of the isometry group of

SL2(R), those isometries act transitively on the points of H and by the above exten- 0 1 2 sion on the points of Hf. GT maps the origin E0(1 : 0 : 0 : 0) onto X(x : x : x : x3) ∈ H. These isometries and their inverses (up to a positive determinant factor) can be given by     x0 x1 x2 x3 x0 −x1 −x2 −x3      1 0 3 2  1 0 3 2  −x x x −x  −1  x x −x x  T =   , T =   . (4.6)  2 3 0 1   2 3 0 1  x x x x  −x −x x −x      x3 −x2 −x1 x0 −x3 x2 x1 x0

∞ ∞ The horizontal intersection of the hyperboloid solid H with the plane E0E2 E3 provides the base plane of the model Hf = SL^2R. We introduce a so-called hyperboloid parametrization by [43] as follows   x0 = cosh r cos φ,    1  x = cosh r sin φ, (4.7)  x2 = sinh r cos (θ − φ),    3  x = sinh r sin (θ − φ),

54 where (r, θ) are the polar coordinates of the base plane, and φ is the fibre coordinate  + h π π i  r ∈ R , φ ∈ − 2 , 2 , θ ∈ [0, 2π) . We note that −x0x0 − x1x1 + x2x2 + x3x3 = − cosh2 r + sinh2 r = −1 < 0.

The inhomogeneous coordinates, which will play an important role in the E3-visua- lizations of SL^2R, are given by  x1  x := = tan φ,  0  x  x2 cos (θ − φ) y := 0 = tanh r , (4.8)  x cos φ  3  x sin (θ − φ)  z := = tanh r .  x0 cos φ 4.2 Geodesic and translation curves

The infinitesimal arc-length-square can be derived by the standard pull back method (see [42, 46, 61]). By the T −1-action presented by (4.6) on differentials (dx0 : dx1 : dx2 : dx3), we obtain (see [43]) the infinitesimal arc-length-square at any point of

SL^2R in coordinates (r, θ, φ): h i2 (ds)2 = (dr)2 + cosh2 r sinh2 r(dθ)2 + (dφ) + sinh2 r(dθ) .

Hence we obtain the symmetric metric tensor field gij on SL^2R by components:   1 0 0    2 2 2 2  gij := 0 sinh r(sinh r + cosh r) sinh r . (4.9)   0 sinh2 r 1 Similarly to the above computations we obtain the metric tensor for coordinates (x1, x2, x3):

 1+(x2)2+(x3)2 −x1x2−2x3 −x1x3+2x2  (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2  1 2 3 2   −x1x2−2x3 1+(x ) +(x ) x2x3  gij :=  (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2  . (4.10)   −x1x3+2x2 x2x3 1+(x1)2+(x2)2 (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2 (−1−(x1)2+(x2)2+(x3)2)2

4.2.1 Geodesic curves

The geodesic curves of SL^2R are generally defined as having locally minimal arc length between any two of their (sufficiently close) points.

By (4.9) the system of second order differential equations of the SL^2R geodesic curve is as follows (see [16]):  1h i  ˙ ˙ ˙ ˙  r¨ = sinh(2r) θ φ + sinh(4r) − sinh(2r) θ θ,  2  2r ˙ h i θ¨ = (3 cosh (2r) − 1)θ˙ + 2φ˙ , (4.11)  sinh (2r)   2  φ¨ = 2r ˙ tanh (r) [2 sinh (r) θ˙ + φ˙].

55 We can assume by homogeneity, that the starting point of a geodesic curve is the origin (1 : 0 : 0 : 0). Moreover,

   r(0) = 0,  r˙(0) = cos(α),     θ(0) = 0, and θ˙(0) = − sin(α),      φ(0) = 0,  φ˙(0) = sin(α) are the initial values in Table 4.1 for the solution of (4.11), and so the unit velocity will be achieved (see details in [16]). The solutions are parametrized by the arc-length s and the angle α from the initial condition: (r(s, α), θ(s, α), φ(s, α)).

Table 4.1: Geodesic curves

Direction Parametrization of a geodesic curve  √  π π r(s, α) = arsinh √cos α sinh(s cos 2α) − < α < cos 2α 4 4  √  θ(s, α) = −arctan √sin α tanh(s cos 2α) (H2 − like) cos 2α φ(s, α) = 2s sin α + θ(s, α)  √  2 π r(s, α) = arsinh 2 s α = ±  √  4 2 θ(s, α) = −arctan 2 s (light − like) √ φ(s, α) = 2s + θ(s, α) π π  √  − ≤ α < − r(s, α) = arsinh √ cos α sin(s − cos 2α) 2 4 − cos 2α π π  √  < α ≤ θ(s, α) = −arctan √ sin α tan(s − cos 2α) 4 2 − cos 2α (fibre − like) φ(s, α) = 2s sin α + θ(s, α)

The parametrization of a geodesic curve in the hyperboloid model with the geo- π graphical sphere coordinates (λ, α), as longitude and altitude, (−π < λ ≤ π, − 2 ≤ π α ≤ 2 ), and the arc-length parameter s ≥ 0, has been determined in [16]. The Euclidean coordinates X(s, λ, α), Y (s, λ, α), Z(s, λ, α) of the geodesic curves can be determined by substituting the results of Table 4.1 (see also [16]) into formula (4.8) as follows    X(s, λ, α) = tan (φ(s, α)),   tanh (r(s, α)) h i  Y (s, λ, α) = cos θ(s, α) − φ(s, α) + λ , cos (φ(s, α)) (4.12)   tanh (r(s, α)) h i  Z(s, λ, α) = sin θ(s, α) − φ(s, α) + λ .  cos (φ(s, α))

As is standard, the geodesic distance d(P,Q) between points P,Q ∈ SL^2R is defined as the arc length of the geodesic curve from P to Q.

56 4.2.2 Translation curves

We recall some basic facts about translation curves in SL^2R following [44, 45, 47, 48]. For any point X(x0 : x1 : x2 : x3) ∈ H (and later also for points in Hf) the translation map from the origin E0(1 : 0 : 0 : 0) to X is defined by the translation matrix T and its inverse presented in equation (4.6). Let us consider for a given vector (q : u : v : w) a curve C(t) = (x0(t): x1(t): 2 3 x (t): x (t)), t ≥ 0, in H starting at the origin: C(0) = E0(1 : 0 : 0 : 0) and such that

C˙(0) = (x ˙ 0(0) :x ˙ 1(0) :x ˙ 2(0) :x ˙ 3(0)) = (q : u : v : w), where C˙(t) = (x ˙ 0(t) :x ˙ 1(t) :x ˙ 2(t) :x ˙ 3(t)) is the tangent vector at any point of the curve. For t ≥ 0 there exists a matrix

  x0(t) x1(t) x2(t) x3(t)    1 0 3 2   −x (t) x (t) x (t) −x (t)  T (t) =    2 3 0 1   x (t) x (t) x (t) x (t)    x3(t) −x2(t) −x1(t) x0(t) which defines the translation from C(0) to C(t):

C(0) · T (t) = C(t), t > 0.

The t-parametrized family T (t) of translations is used in the following definition.

Definition 4.2.1 The curve C(t), t > 0, is said to be a translation curve if

˙ ˙ C(0) · T (t) = C(t), t > 0.

The solution, depending on (q : u : v : w) was determined in [44], where it splits into three cases. It was observed above that for any X(x0 : x1 : x2 : x3) ∈ Hf there is a suitable −1 transformation T , given by equation (4.6), which sends X to the origin E0 along a translation curve.

Definition 4.2.2 The translation distance ρ(E0,X) between the origin E0(1 : 0 : 0 : 0) and the point X(1 : x : y : z) is the length of the translation curve connecting them.

For a given translation curve C = C(t) the initial unit tangent vector (u, v, w) (in

Euclidean coordinates) at E0 can be presented as

u = sin α, v = cos α cos λ, w = cos α sin λ, (4.13)

57 π π for some − 2 6 α 6 2 and −π < λ 6 π. In Hf this vector is of length square −u2 + v2 + w2 = cos 2α. We always can assume that C is parametrized by the translation arc-length parameter t = s > 0. Then coordinates of a point X(x, y, z) of

C, such that the translation distance between E0 and X equals s, depend on (λ, α, s) as geographic coordinates according to the three cases considered above as follows (see Table 4.2).

Table 4.2: Translation curves

Direction Parametrization of a translation curve     π π x(s, α, λ) √ sin α − < α <   tanh(s cos 2α)   4 4     y(s, α, λ) = √  cos α cos λ  (H2 − like)   cos 2α   z(s, α, λ) cos α sin λ)     π x(s, α, λ) √ 1 α = ±   2s   4     y(s, α, λ) =  cos λ  (light − like)   2   z(s, α, λ) sin λ) π π − ≤ α < −     2 4 x(s, α, λ) √ sin α   tan(s − cos 2α)   π π     < α ≤ y(s, α, λ) = √  cos α cos λ  4 2   − cos 2α   (fibre − like) z(s, α, λ) cos α sin λ)

4.3 Geodesic and translation triangles

4.3.1 Geodesic triangles

We consider three points A1, A2, A3 in the projective model of SL^2R space. The geodesic segments ak between the points Ai and Aj (i < j, i, j, k ∈ {1, 2, 3}, k 6= i, j) are called sides of the geodesic triangle Tg with vertices A1, A2, A3. In Riemannian geometries the metric tensor (4.10) is used to define the angle θ between two geodesic curves. If their tangent vectors in their common point are u and v and gij are the components of the metric tensor then uig vj cos(θ) = ij (4.14) q i j i j u giju v gijv

It is clear by the above definition of the angles and by the metric tensor (4.10), that the angles are the same as the Euclidean ones at the origin of the projective model of SL^2R geometry. We note here that the angle of two intersecting geodesic curves depend on the orientation of the tangent vectors. We will consider the interior angles of the triangles

58 at the vertices denoted by ωi (i ∈ {1, 2, 3}).

4.3.1.1 Fibre-like right angled triangles

A geodesic triangle is called fibre-like if one of its edges lies on a fibre line. In this section we study the right angled fibre-like triangles. We can assume without loss of generality that the vertices A1, A2, A3 of a fibre-like right angled triangle (see Figure

4.2) Tg have the following coordinates:

2 3 A1 = (1 : 0 : 0 : 0),A2 = (1 : 0 : y : 0),A3 = (1 : x : 0 : 0) (4.15)

The geodesic segment A1A2 lies on the y axis, the geodesic segment A1A3 lies on the

Figure 4.2: Fibre-like geodesic triangle

π x axis (see Table 4.1) and its angle is ω1 = 2 in the SL^2R space (this angle is in the π Euclidean sense also 2 since A1 = E0). In order to determine the further interior angles of a fibre-like geodesic triangle

A1A2A3 we define translations TAi , (i ∈ {2, 3}) as elements of the isometry group of

SL2(R), (see 4.6) that maps the origin E0 onto Ai. For example the isometry TA2 and its inverse (up to a positive determinant factor) can be given by:

    1 0 y2 0 1 0 −y2 0      2  2  0 1 0 −y  −1  0 1 0 y  TA =   , T =   , (4.16) 2  2  A2  2  y 0 1 0  −y 0 1 0      0 −y2 0 1 0 y2 0 1 and the images TA2 (Ai) of the vertices Ai (i ∈ {1, 2, 3}) are the following (see Figure

59 4.3):

−1 2 2 −1 2 T (A1) = A = (1 : 0 : −y : 0), T = A (A2) = E0 = (1 : 0 : 0 : 0), A2 1 A2 2 (4.17) −1 2 3 2 3 2 TA2 (A3) = A3 = (1 : x : −y : x y ),

Similarly to the above cases we obtain:

−1 3 3 −1 3 3 2 3 2 T (A1) = A = (1 : −x : 0 : 0), T (A2) = A = (1 : −x : y : −x y ), A3 1 A3 2 (4.18) −1 3 TA3 (A3) = A3 = E0 = (1 : 0 : 0 : 0),

P3 Our aim is to determine the angle sum i=1(ωi) of the interior angles of the above

2 2 3 Figure 4.3: Translated image A1A2A3 of the fibre-like geodesic triangle A1A2A3 by translation TA2

π right angled fibre-like geodesic triangle A1A2A3. We have seen that ω1 = 2 and the angle of geodesic curves with common point at the origin E0 is the same as the Euclidean one therefore it can be determined by usual Euclidean sense. Moreover, the translations TA2 and TA3 are isometries in SL^2R geometry thus ω2 is equal to 2 2 2 2 2 2 2 2 the angle (g(A2,A1)g(A2,A3))∠ (see Figure 4.4) where g(A2,A1), g(A2,A3) are the 3 3 3 3 2 oriented geodesic curves and ω3 is equal to the angle (g(A3A1)g(A3A2))∠ (E0 = A2 = 3 A3 see 4.17, 4.18). The parametrization of a geodesic curve in the model is given by the geographic π π sphere coordinates (λ, α), as longitude and altitude, (−π < λ ≤ π, − 2 ≤ α ≤ 2 ), and the arc-length parameter s ≥ 0 (see Table 4.1 and 4.12). j We denote the oriented unit tangent vector of the oriented geodesic curves g(E0,Ai ) j j by ti where (i, j) = (2, 3), (3, 2), (1, 2), (1, 3). The Euclidean coordinates of ti are:

j j j j j j ti = (sin(αi ), cos(αi ) cos(λi ), cos(αi ) sin(λi )). (4.19)

60 3 2 Figure 4.4: The translated points A2 and A3 are antipodal related to the origin E0

Lemma 4.3.1 ([15]) The sum of the interior angles of a fibre-like right angled ge- odesic triangle is greater or equal to π.

2 3 Proof: It is clear, that t1 = (0, −1, 0) and t1 = (−1, 0, 0). Moreover, the points 3 2 3 2 A2 and A3 are antipodal relative to the origin E0 therefore the equation |α2| = |α3| 3 holds (i.e. the angle between the vector t2 and the [y, z] plane are equal to the angle 2 π 3 π 2 between the vector t3 and the [y, z] plane). That means, that ω3 = 2 −|α2| = 2 −|α3|. 2 The vector t1 lies in the [y, z] plane therefore the angle ω2 is greater or equal than 3 2 α2 = α3. Finally we obtain, that 3 X π π 3 (ωi) = + − |α2| + ω2 ≥ π.  i=1 2 2 In Table 4.3 we summarize some numerical data of geodesic triangles for given pa- rameters:

4.3.1.2 Hyperbolic-like right angled geodesic triangles

A geodesic triangle is hyperbolic-like if its vertices lie in the base plane (i.e. [y, z] coordinate plane) of the model. In this section we analyze the interior angle sum of the right angled hyperbolic-like triangles. We can assume without loss of generality that the vertices A1, A2, A3 of a hyperbolic-like right angled triangle (see Figure 4.5)

Tg have the following coordinates:

2 3 A1 = (1 : 0 : 0 : 0),A2 = (1 : 0 : y : 0),A3 = (1 : 0 : 0 : z ) (4.20)

61 Table 4.3: Fibre-like right angled geodesic triangles for x3 = 1/5

2 3 2 P3 y |α2| = |α3| d(A2A3) ω2 ω3 i=1(ωi) → → 0 → π/2 arctan(1/5) ≈ → π/2 → 0 → π ≈ 0.1974 1/1000 1.5657 0.1974 1.5658 0.0051 3.1417 1/3 0.4993 0.3970 0.3560 1.0715 3.1806 1/2 0.3170 0.5809 0.3560 1.2538 3.1806 3/4 0.1630 0.9891 0.2043 1.4078 3.1829 999/1000 0.0299 3.8032 0.0422 1.5409 3.1540 → 1 → 0 → ∞ → 0 → π/2 → π

The geodesic segment A1A2 lies on the y axis, the geodesic segment A1A3 lies on the

2 2 2 Figure 4.5: Hyperbolic-like geodesic triangle A1A2A3 and its translated copy A1A2A3.

π π z axis and its angle is ω1 = 2 in SL^2R (this angle is in Euclidean sense also 2 since

A1 = E0). In order to determine the further interior angles of fibre-like geodesic triangle

A1A2A3 similar to the fibre-like case we define translations TAi , (see 4.6) that map the origin E0 onto Ai. E.g. the isometry TA3 and its inverse (up to a positive

62 determinant factor) can be given by:

    1 0 0 z3 1 0 0 −z3      3   3   0 1 z 0  −1  0 1 −z 0  TA =   , T =   . (4.21) 3  3  A3  3   0 z 1 0   0 −z 1 0      z3 0 0 1 −z3 0 0 1

−1 Similarly to the above cases we find that the images TAj (Ai) of the vertices Ai (i ∈ {1, 2, 3}, j ∈ {2, 3}) are the following (see also Figure 4.5):

−1 2 2 −1 2 T (A1) = A = (1 : 0 : −y : 0), T = A (A2) = E0 = (1 : 0 : 0 : 0), A2 1 A2 2 (4.22) −1 2 2 3 2 3 TA2 (A3) = A3 = (1 : y z : −y : z ),

−1 3 3 −1 3 2 3 2 3 T (A1) = A = (1 : −z : 0 : 0), T (A2) = A = (1 : −y z : y : −z ), A3 1 A3 2 (4.23) −1 3 TA3 (A3) = A3 = E0 = (1 : 0 : 0 : 0), P3 We study similarly to the above fibre-like case the sum i=1(ωi) of the interior angles of the above right angled hyperbolic-like geodesic triangle A1A2A3.

It is clear, that the angle of geodesic curves with a common point at the origin E0 is the same as the Euclidean angle therefore it can be determined in the usual Euclidean way. The translations TA2 and TA3 preserve the measure of angles ωi (i ∈ {2, 3} 2 2 2 2 3 3 3 3 therefore ω2 = (g(A2,A1)g(A2,A3))∠ (see Figure 4.5) and ω3 = (g(A3A1)g(A3A2))∠. Similarly to the fibre-like case the Euclidean coordinates of the oriented unit j j tangent vectors ti of the oriented geodesic curves g(E0,Ai ) ((i, j) = (2, 3), (3, 2), (1, 2), (1, 3)) are given by (4.6). Finally, we get the following lemma:

Lemma 4.3.2 ([15]) The interior angle sums of hyperbolic-like geodesic triangles can be less than or equal to π.

Conjecture 4.3.3 ([15]) The sum of the interior angles of any hyperbolic-like right angled geodesic triangle is less than or equal to π.

In Table 4.4 we summarize some numerical data of geodesic triangles for given pa- rameters:

4.3.1.3 Geodesic triangles with interior angle sum π

In the above sections we discussed the fibre- and hyperbolic-like geodesic triangles P3 and proved that there are right angled geodesic triangles whose angle sum i=1(ωi) is P3 greater than or equal to π, less than or equal to π. The i=1(ωi) = π case is realized if one of the vertices of a geodesic triangle A1A2A3 tends to the infinity (see Table 3-4). Furthermore, we prove the following

63 Table 4.4: Hyperbolic-like geodesic triangles for y2 = 1/2

3 3 2 P3 z |α2| = |α3| d(A2A3) ω1 ω2 i=1(ωi) → → 0 → 0 arctanh(1/2) ≈ → 0 → π/2 → π ≈ 0.5493 1/10 0.0811 0.5638 0.1334 1.2830 2.9872 1/3 0.2103 0.6994 0.3613 0.7170 2.6491 999 1000 0.0649 4.0707 0.5817 0.0913 2.2438 (106−1) 106 0.0330 7.5174 0.0467 0.6112 2.2288

Lemma 4.3.4 ([15]) There are geodesic triangles A1A2A3 with interior angle sum

π where all the vertices are proper (i.e. Ai ∈ SL2R).

Proof: We consider a hyperbolic-like geodesic right angled triangle with vertices 2 h 3 A1 = E0 = (1 : 0 : 0 : 0), A2 = (1 : 0 : y : 0), A3 = (1 : 0 : 0 : z ) and a

fibre-like right angled geodesic triangle with vertices A1 = E0 = (1 : 0 : 0 : 0), 2 f 3 2 3 3 A2 = (1 : 0 : y : 0), A3 = (1 : x : 0 : 0) (0 < y , z < 1, 0 < x < ∞). We consider f h the straight line segment (in the Euclidean sense) A3 A3 ⊂ SL2R. f h We consider a geodesic right angled triangle A1A2A3(t) where A3 ∈ A3 A3 , (t ∈ h f h [0, 1]). A3 is moving on the segment A3 A3 and if t = 0 then A3(0) = A3 , if t = 1 f then A3(1) = A3 .

As in the above cases the interior angles of the geodesic triangle A1A2 A3(t) are P3 P3 denoted by ωi(t)(i ∈ {1, 2, 3}). The angle sum i=1(ωi(0)) < π and i=1(ωi(1)) > π.

Moreover the angles ωi(t) change continuously if the parameter t run in the interval P3 [0, 1]. Therefore there is a tE ∈ (0, 1) where i=1(ωi(tE)) = π.  We obtain by the lemmas of this section the following result.

Theorem 4.3.5 ([15]) The sum of the interior angles of a geodesic triangle of SL^2R space can be greater than, less than or equal to π.

4.3.2 Translation triangles

Considering a translation triangle we can assume that one of its vertices is the origin

(translate one if necessary) and the other two points are A(1 : x1 : y1 : z1) and

B(1 : x2 : y2 : z2). Despite the metric of SL^2R being non-homogeneous (therefore it is not so easy to determine both the distance and the angle), SL^2R is a homogeneous geometry, so that any point can be translated to the origin by (4.6). It turns out that the translation curve between A and B is also a straight line in the model since the

64 −1 isometry TA is a collineation so that applying it to the translation curve between −1 −1 TA (A) = O and TA (B) preserves linearity. It was observed, that the neighborhood of the origin behaves like Euclidean space, therefore the angle of two vectors (u and v) with the origin as their base point seems real size in the model. Now, we translate both edges A and B to the origin to determine the other two interior angles of the translation triangle:

Table 4.5: Translations of general points A and B

A= (1 : x1 : y1 : z1) B= (1 : x2 : y2 : z2) −1 −1 TA (A) = (1 : 0 : 0 : 0) TB (B) = (1 : 0 : 0 : 0) −1 −1 TA (O)= (1 : −x1 : −y1 : −z1) TB (O)= (1 : −x2 : −y2 : −z2) −1 −1 TA (B)= (1 : TB (A)= (1 : x − x − y z + y z x − x − y z + y z 1 2 1 2 2 1 : 2 1 2 1 1 2 : −x1x2 + y1y2 + z1z2 − 1 −x1x2 + y1y2 + z1z2 − 1 −x z + x z + y − y −x z + x z + y − y 1 2 2 1 1 2 : 2 1 1 2 2 1 : −x1x2 + y1y2 + z1z2 − 1 −x1x2 + y1y2 + z1z2 − 1 x y − x y + z − z ! x y − x y + z − z ! 1 2 2 1 1 2 2 1 1 2 2 1 . −x1x2 + y1y2 + z1z2 − 1 −x1x2 + y1y2 + z1z2 − 1

−1 −1 One can see from Table 4.5 that the point pairs (A, TA (O)), (B, TB (O)) and −1 −1 (TA (B), TB (A)) are antipodal as in the geodesic cases. Figure 4.6 shows the translated triangles in the hyperboloid model, as well as the plane containing the triangle twists during the translation.

Figure 4.6: Translations of the original (red) translation triangle.

65 Lemma 4.3.6 ([15]) A plane σ trough the origin with Euclidean normal vector v is invariant for translations perpendicular to v (in the Euclidean sense) if and only if v is light–like.

−1 Proof: It suffices to prove that TA (B) is contained in the plane.   y1z2 − y2z1 −→ −−→     v = OA × OB = x2z1 − x1z2   x1y2 − x2y1

Then we obtain the equation of σ:

σ :(y1z2 − y2z1)x + (x2z1 − x1z2)y + (x1y2 − x2y1)z = 0. (4.24)

−1 The point TA (B) satisfies the equation (4.24) if and only if the point is on σ:

x1 − x2 − y1z2 + y2z1 −x1z2 + x2z1 + y1 − y2 (y1z2 − y2z1) + (x2z1 − x1z2) + −x1x2 + y1y2 + z1z2 − 1 −x1x2 + y1y2 + z1z2 − 1 x1y2 − x2y1 + z1 − z2 1 + (x1y2 − x2y1) = · −x1x2 + y1y2 + z1z2 − 1 −x1x2 + y1y2 + z1z2 − 1  2 (x1 − x2)(y1z2 − y2z1) − (y1z2 − y2z1) + (y1 − y2)(x2z1 − x1z2)+

2 2 +(x2z1 − x1z2) + (z1 − z2)(x1y2 − x2y1) + (x1y2 − x2y1) =

2 2 2 − (y1z2 − y2z1) + (x2z1 − x1z2) + (x1y2 − x2y1) = 2 2 2 = −v1 + v2 + v3 = 0 ⇔ v is light − like. (4.25) We now have the following theorem:

Theorem 4.3.7 ([15]) The sum of the interior angles of a translation triangle in the SL^2R geometry is greater than or equal to π.

Proof: To get the angle it is sufficient to consider the projection of both the original and the translated vertices onto the unit sphere around the origin (see Figure 4.6) . The sum of the angles is equivalent to the distance on the sphere (see Figure 4.7). Due to antipodality these distances can be considered as three consecutive arcs from a point to its antipodal pair. Since the triangle inequality holds on the sphere, the sum of these arc lengths is greater than or equal to the half of the circumference of the main circle on the unit sphere in other words π.

Remark 4.3.8 The sum of the interior angle of the translation triangle is π if and only if the normal vector of the plane containing the triangle is light–like.

66 Figure 4.7: Antipodal point-pairs with main circle arcs.

4.4 Isoptic surface of a line segment with transla- tion curves

In this section we connect the theory of isoptic curves with the SL^2R geometry. Namely, we will determine the α and π−α isoptic surface of a line segment AB where A = (1 : 0 : a : 0) and B = (1 : 0 : −a : 0) a ∈ (0, 1). A point P = (1 : x : y : z) belongs to the α−isoptic surface if and only if the angle formed by the directed −→ −−→ straight lines PA and PB is a given angle α. Since geodesic and translation curves are different in this geometry, it is necessary to stipulate that we connect A and B with P using translation curves. −1 Since P is a general point, we recall that the TP translation which maps P to the origin can be found in (4.6). Applying the transformation to A and B:

T−1(A) = (1 − ay : −x − az : −y + a : −z − ax) P (4.26) −1 TP (B) = (1 + ay : −x + az : −y − a : −z + ax)

−1 −1 When P is at the origin, then the angle formed by TP (A)OTP (B)∠ seems in real size, because the geometry around the origin is locally Euclidean (see (4.10) with xi = 0). So that the angle can be computed using by (4.14), where u = (−x − az, −y + a, −z − ax) and v = (−x + az, −y − a, −z + ax). Of course, it −1 −1 can happen that either TP (A) or TP (B) ’slips through’ and ’comes back’ from the 1 opposite direction resulting the complementary angle. Moreover, if y = ± a , then one of the translated points becomes ideal. However, the angle still can be defined since the direction is given.

67 Simplifying the equation, we have the following theorem:

Theorem 4.4.1 (CsG) Given a translation line segment in the SL^2R geometry by its endpoints A = (1 : 0 : a : 0) and B = (1 : 0 : −a : 0) where a ∈ (0, 1). Then the α and π − α isoptic surfaces of a line segment AB have the equation:

(x2 + y2 + z2 − a2(1 + x2 + z2))2 cos2(α) = (4.27) (x2 + y2 + z2 + a2(1 + x2 + z2))2 − (4axz − 2ay)2

π Remark 4.4.2 If we set α = 2 , then we get the so called SL^2R Thales surface, which has the equation of a spheroid:

x2(1 − a2) y2 z2(1 − a2) + + = 1 a2 a2 a2 On the following figures, one can see the isoptic surface of the line segment:

π π (a) a = 0.75 and α = 3 (b) a = 0.8 and α = 2

Figure 4.8: Isoptic surface of line segment

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