We also prove some interesting characterizations of the ellipse

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HYPERBOLIC IS THE ONLY HILBERT GEOMETRY HAVING CIRCUMCENTER OR ORTHOCENTER GENERALLY

J ÓZSEF KOZMA AND ÁRP ÁD KURUSA

Abstract. A Hilbert geometry is hyperbolic if and only if the perpendicular bisectors or the altitudes of any trigon form a pencil. We also prove some interesting characterizations of the ellipse.

1. Introduction

Hilbert geometries, introduced by David Hilbert in 1899 [6], are natural gen- eralizations of hyperbolic geometry, and hence the question immediately arises if some properties of a Hilbert geometry are specific to the hyperbolic geometry. For a recent survey on the results see [5].

To place our subject in a broader context we mention that it can also be considered as a so-calledellipsoid characterization problem in Euclidean space, which is often treated as characterization of Euclidean spaces (inner product spaces) among the normed spaces (See [1] and [15,16]). Further, the unitary imaginary unit sphere in generalized space-time model [7,8] can also be considered as a Hilbert geometry.

In this article we prove two results: the existence of a circumcenter in every trigon (Theorem 5.1) or the existence of an orthocenter in every trigon (Theo- rem5.2) renders Hilbert geometry hyperbolic. Moreover, we also prove two characterizations of ellipses in Section4.

2. Preliminaries

Points of Rⁿ are denoted as a,b, . . .; the line through different points a and bis denoted byab, the open segment with endpoints a andbis denoted by ab.

Non-degenerate triangles are calledtrigons.

For given different pointspandqinRⁿ, and pointsx,y∈pqone has the unique linear combinations x= λ₁p+µ₁q, y =λ₂p+µ₂q (λ₁+µ₁ = 1, λ₂+µ₂ = 1) which allows to define thecross ratio

(2.1) (p,q;x,y) = µ₁λ₂

λ1µ2

,

of the pointsp,q,xandy, provided thatλ₁µ₂6= 0 (see [2, page 243]).

1991Mathematics Subject Classification. 53A35; 51M09, 52A20.

Key words and phrases. Hilbert geometry, hyperbolic geometry, circumcenter, orthocenter, ellipsoid characterization.

1

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Definition 2.1 ([2, page 297]). LetH ⊂Rⁿ (n≥2) be an open and convex set with boundary∂H. The metricd_H:H × H →R_0≤ defined by

(2.2) d_H(x,y) = (₁

2

ln|(p,q;x,y)|

, ifx6=y, where {p,q}=xy∩∂H,

0, ifx=y,

is called theHilbert metriconH. The pair (H, d_H) is calledHilbert geometry.

Note that as all the defining conditions of a Hilbert geometry (H, dH) is projective invariant, two Hilbert geometries are isomorphic if there is a projective map between their sets of points. Further, the generalized Cayley–Klein model of the hyperbolic geometryHⁿ is, in fact, a special kind of Hilbert geometry (E, d_E) given by an ellipsoidE.

Let a,b be different points in H. For anyc ∈ H ∩(ab\ {b}) the hyperbolic ratio¹of the triplea,b,cis defined by

(2.3) ha,b;ciH =

(−^sinh_sinh_d^d^H^(c,a)

H(b,c), ifc∈ab,

sinhdH(c,a)

sinhdH(b,c), otherwise.

Perpendicularity of H-lines, non-empty intersections of Euclidean lines withH, in Hilbert geometry is defined in [2, pp. 119–121]². It is based on the notion of the foot of a point ofHon an H-line.

Let `be anH-line and let the point g∈ H be outside of`. The pointf ∈`is the`-foot of g, ifd_H(g,x)≥d_H(g,f) for everyx∈`.³

A line `⁰ intersecting the line`in a pointf is said to beH-perpendicular to` if f is an`-foot ofgfor everyg∈`⁰\ {f}. We denote this relation by`⁰ ⊥H`.⁴

It is proved in [2, (28.11)] that, ifHis strictly convex, then for any given point f ∈ HandH-line`there exists a uniqueH-line`⁰ such that it goes throughf and

`⁰⊥H`. Moreover, the Euclidean line containing`⁰ is the one that connectsf and the intersection of those tangents ofHthat touchHat the points∂`.

A set of lines is said to form a pencil if they have a common (maybe ideal) point. This point is called thecenter of the pencil. We say that a set ofH-lines forms a pencil with centerc, if the corresponding euclidean lines form a pencil with centerc. Thus the set of those lines that areH-perpendicular to an arbitrary fixed line`is a pencil.

Based on the foregoing, one can speak about the

• H-perpendicular bisector of a segment ab, as the unique line through the midpoint ofab, that isH-perpendicular to the lineab, and the

1The name ‘hyperbolic ratio’ comes from the hyperbolic sinus function in the definition.

2In fact, it is defined for projective metrics.

3Observe that a point may have more`-foots in general.

4Notice, that⊥_His not necessarily a symmetric relation. In fact it is symmetric if and only ifH is an ellipse [10].

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• H-altitude of a triangle4abc, as a line through one of the vertices of4abc, that isH-perpendicular to the corresponding opposite edge of4abc.

These definitions⁵extend the respective notion of the perpendicular bisector of a segment and the altitude of a triangle, as defined in hyperbolic geometry.

From now on, we assume thatHis strictly convex and hasC² boundary.

3. Utilities

The useful notations uτ = (cosτ,sinτ) and u^⊥_τ = (−sinτ,cosτ) are used all over this article. Also the following technical lemmas and the notations will be used in proving our main results.

Lemma 3.1 ([13, Lemma 2.3]). Let a,b and c be collinear points in a Hilbert geometry H, and let ab∩∂H = {p,q}, such that a separates p and b. Set a Euclidean coordinate system onab such that the coordinates ofpanda are0 and 1, respectively. Let q, b and c, with assumptions q > b >1 and0 < c < q, be the coordinates ofq,bandc, respectively, in this coordinate system. Then we have

(3.1) |ha,b;ciH|= |c−b|

|c−1|√ b

s

1 + b−1 q−b.

Lemma 3.2 ([2, Lemma 12.1, pp. 226]). A bounded open convex set H in Rⁿ (n≥2) is an ellipsoid if and only if every section of it by any 2-dimensional plane is an ellipse.

Lemma 3.3. LetH be a convex body in the plane. Then

(i) there exists an ellipseE circumscribed around Hwith at least three different contact pointse1,e2,e3 lying in ∂H ∩∂E such that the closed triangle 4e1e2e3 contains the centercofE, and

(ii) if H 6≡ E, then these contact points can be chosen so that in every neigh- borhood of one of them∂H \∂E 6=∅.

Lett1, t2, t3 be the common support lines at e1,e2,e3, respectively. Then

(iii) cis in the interior of 4e1e2e3 if and only if t1, t2, t3 form a trigon with verticesm1=t2∩t3,m2=t3∩t1 andm3=t1∩t2;

(iv) cis in one of the edges of 4e1e₂e₃, sayc∈e₂e₃, if and only if t₁, t₂, t₃ form a half strip with verticesm₂ =t₁∩t₃, m₃ = t₂∩t₁ and the ideal pointm₁=t₂∩t₃.

If b₁,b₂,b₃ are the midpoints of the segments e₂e₃, e₃e₁ and e₁e₂, respectively, then

(v) the straight linesm_ib_i (i= 1,2,3) meet inc.

5Notice that these notions could also be introduced by using⊥_Hin the reverse order.

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Proof. Take the unique minimal area ellipse E containingHand let the centerc ofE be the origino.

(i) By [3, Theorem 2 (ii)] there is an integer (5 ≥)m ≥3 such that there are contact pointse1, . . . ,emlying in∂H ∩∂Esuch that a positive linear combination of the contact points vanishes. This means that the origin is in the convex hull of these contact points, hence a (closed) trigon of three of them, say4e1e2e3, also contains the origin.

(ii) Transform the configuration given in (i) with a linear affinity µ so that D=µ(E) is the unit disc centered too. Lete⁰_i =µ(e_i) (i= 1,2,3) andH⁰=µ(H).

By (i)the centerois in the trigon4e⁰₁e⁰₂e⁰₃. Letε_i∈(−π, π] be such thate⁰_i= u_ε_i and let the support function ofH⁰ be denoted byh_H⁰. Defineα_i:= lim sup{α: h_H⁰([ε_i, ε_i+α]) ={1}}(i= 1,2,3). Ifα_i is infinite, thenH ≡ E, that is excluded.

Assume that α_k = min_i=1,2,3α_i for some k ∈ {1,2,3}. Set f_i = µ⁻¹(u_ε_i_+α_k) (i = 1,2,3). Then f1,f2,f3 are contact points of ∂H and ∂E, the trigon f1,f2,f3

contains the centero, and in every neighborhoodN offk (∂H \∂E)∩ N 6=∅.

(iii) and(iv)are easy consequences of the strict convexity of the ellipseE. (v) This readily follows if one transforms the ellipse into a circle by a linear

affinity.

Lemma 3.4. For a smallε >0letr,p: (−ε,0]→R²be twice differentiable convex curves such that p(τ) = p(τ)uτ and r(τ) = r(τ)uτ, where p, r: (−ε,0] → R+, λ(τ) := r(τ)/p(τ) takes its minimum value 1 at τ = 0, and max_(−δ,0]λ > 1 for everyδ∈(0, ε).

Let τn be a sequence in(−ε,0]tending to0such thatλ(τn)>1for every n∈N. Then the tangent lines of r andpat r(τ_n) andp(τ_n), respectively, intersect each other in a pointm(τ_n)that tends to p(0)asτ_n→0 so that it is on the same side of the line0p(τ_n)asp(0) is.

Proof. First we prove the statement with the assumption thatλtakes its minimum value 1 uniquely atτ = 0. This means that ˙λ(0) = 0, ¨λ(0) >0 and we have to prove that

(3.2)

the tangent lines ofrandpatr(τ) andp(τ), respectively, intersect each other in a pointm(τ) that tends top(0) asτ →0 so that it is on the same side of the line0p(τ) asp(0) is.

Since ˙r=λp˙+ ˙λp, ˙pkr˙ if and only if ˙λ= 0, thereforem(τ) exists uniquely for everyτ6= 0.

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o m0

p(τ)

r(τ)

m(τ)

u_τ u^⊥_τ

˙ p(τ)

˙ r(τ) τ <0, λ <˙ 0

Figure 3.1. The crossing of the tangent lines.

We clearly have

(3.3) ± |m−p| p˙

|p|˙ +p=m=±|m−r| r˙

|˙r|+r, that is±|m−p||r|˙p˙ +|r||˙ p|p˙ =±|m−r||p|˙ r˙+|p||˙ r|r.˙

Since ˙p= ˙pu_τ+pu^⊥_τ, ˙r= ˙ru_τ+ru^⊥_τ andu_τ⊥u^⊥_τ, we obtain

|m−p||˙r|p=|m−r||p|r˙ (3.4)

and

±|m−p||˙r|p˙+p|p||˙ r|˙ =±|m−r||p|˙ r˙+r|˙r||p|.˙ (3.5)

Multiplying (3.5) bypthen substituting (3.4) into the product results in

±|m−r||p|r˙ p˙+p²|p||˙˙ r|=±|m−r||p|p˙˙ r+pr|r||˙ p|,˙ hence

±|m−r|= p|˙r|(r−p)

rp˙−pr˙ = |r|p˙ ²(λ−1)

λpp˙−p( ˙λp+λp)˙ =|˙r|λ−1

−λ˙ . (3.6)

This implies lim_τ→0|m(τ)−r(τ)|= 0 via l’Hˆospital’s rule.

On the other hand, using (3.4) and putting (3.6) into (3.3) gives λ(λ−1)

−λ˙ p˙ +p=m= λ−1

−λ˙ r˙+r.

Asλ≥1, this implies that mis on the same side of0rand0p asm(0) =r(0) = p(0). This proves claim (3.2).

For the proof of the statement in the lemma we take the broken line ¯qwith ver- ticesp(τn) and edgesp(τn)p(τn+1). It is clearly convex and can easily be deformed into a twice differentiable convex curveq so that q(τn) = p(τn), ˙q(τn) = ˙p(τn) andr(τ)/|q(τ)| takes its minimum value 1 uniquely at τ = 0. Using claim (3.2) forqand rtherefore immediately implies the lemma.

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4. Characterizations of ellipses

The following configuration, construction, theorems, and the notations they introduce, are used in the next sections, but are interesting on their own too.

Definition 4.1. If a strictly convex bodyHis given in the plane, and the points e₁,e₂,e₃ are placed on its border∂H, then the following configuration is defined.

e1

e2

e3

∂H

`2

`3 `₁

t^H₁

t^H₂ t^H₃ f₂^H

f₃^H f₁^H

Figure 4.1. Construction for Definition4.1

For everyi= 1,2,3,`i denotes the lineejek, t^H_i denotes the tangent line of Hat the pointei, and f_i^H denotes the straight line through ei that forms a harmonic pencil with the lines`j, `k, t^H_i , where{i, j, k}={1,2,3}.

Theorem 4.2. ⁶Take a configuration given in Definition 4.1.

(i) For any ellipseE the linesf₁Ê, f₂Ê, f₃Ê form a pencil.

(ii) If the lines f₁^H, f₂^H, f₃^H form a pencil for any points e1,e2,e3∈∂H, then His an ellipse.

Proof. First note that not only keep projectivities the cross ratio, but takes any tangent line of a curve into a tangent line of the image curve.

6After this theorem was proved it turned out, that the dual of this statement is, via the theorems of Menelaus and Ceva equivalent to Segre’s result in [17,§3] which, as noted in [12, 6.15. T´etel], does not use the finiteness of the geometry but only the commutativity of the field; note that following [11, p. 133], the perspectivity of the circimscribed and inscribed triangle was named asπ-property in [12].

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e1

e₂

e₃

∂E

`2

`₃ `1

t^E₁

t^E₂

t^E₃ f₂^E

f₃^E f₁^E

π(e1)

π(e2)

π(e3)

∂D π(`₂)

π(`3) π(`₁) t^D₁

t^D₂

t^D₃ f₂^D

f₃^D f₁^D

π

Figure 4.2. Transforming the ellipse E into a disc D and the triangle4e1e2e3 into a regular one.

(i) Taking a suitable affinity we may assume that ellipse H is a disc D. The projective group is three-transitive⁷ on every conic, hence we may assume that e₁,e₂,e₃ forms a regular triangle on the circle ∂D. Then, obviously, the lines f₁, f₂, f₃ meet in the center ofC that proves statement(i).

(ii) The condition remains unchanged if the configuration is transformed by a projective map, therefore we may assume that the points e₁,e₂ and e₃ and f := f₁^H ∩f₂^H∩f₃^H are such that e₁ = (0,1), e₂ = (0,−1), e₃ = (1,0) and f = (√

2−1,0).

Then, the straight linesf₁^H, f₂^Handf₃^Hare determined and from the conditions

−1 = (`1, `2;t^H₃, f₃^H) = (`2, `3;t^H₁, f₁^H) = (`3, `1;t^H₂, f₂^H), we get the equations y= 1,y=−1 andx= 1 fort^H₁, t^H₂ andt^H₃, respectively.

Now choose a general point h ∈ ∂Hdifferent from e1,e2, and let ∂Eh be the unique ellipse through the pointse1,e2,hwith tangentst^E₁ :=t^H₁ andt^E₂ :=t^H₂ at e1 ande2, respectively.

Let us introduce some new notations (See Figure4.3.):

• t^H_h is the tangent ofHath;

• `i is the linehei fori= 1,2;`3 is the linee1e2;

• f_i^H is the line throughei fori= 1,2 such that−1 = (`j, `k;t^H_i , f_i^H), where {i, j, k}={1,2,3};

• f_h^H is the line throughhsuch that −1 = (`₁, `₂;t^H_h, f_h^H).

We denote the analogous objects for the ellipseEhin the same way except that the superscriptHis exchanged toE.

7This is easy to prove by using conic involutions.

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e1

e2

h

∂E_h

∂H

`2

`3

`1

t^H₁ =t^E₁

t^H₂ =t^E₂

t^E_h t^H_h

f_h^E f_h^H

f₁^E

f₂^E

Figure 4.3. Introducing harmonic pencils SincetÊ₁ =t^H₁ andtÊ₂ =t^H₂ we clearly havef_i^H=f_iÊ fori= 1,2.

As the lines f₁Ê, f₂Ê and f_hÊ form a pencil by (i), and f₁^H, f₂^H and f_h^H form a pencil by the condition in(ii), we deduce that the linesf_h^H and f_hÊ intersect each other not only inh, but also inf₁^H∩f₂^H=f₁Ê ∩f₂Ê, hence they coincide.

Thus, we have t^H_h =tÊ_h. So it makes sense to introduce the notationst_i:=tÊ_i = t^H_i fori= 1,2 andth:=tÊ_h=t^H_h.

Letr: (−π, π]→R+be such thath(ϕ) =r(ϕ)u_ϕis in∂H, for everyϕ∈(−π, π].

Then a tangent vector of∂Hath(ϕ) is ˙h(ϕ) = ˙r(ϕ)uϕ+r(ϕ)u^⊥_ϕ which is parallel to the tangent of the unique ellipse∂E_h(ϕ).

e1

e2

e3=h(0) h(ϕ)

∂Eh(ϕ)

∂H

x y

o t₁

t₂

t_h

r(ϕ)u^ϕ

ϕ a(h(ϕ))

Figure 4.4. Parametrization of ∂H

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The ellipse∂E_h(ϕ)goes through the pointse1,e2,h(ϕ) and it has tangentst1andt2

ate1ande2, respectively, therefore its equation is^x_a²2+y²= 1 for somea=a(h(ϕ)).

Putting the coordinate ofh(ϕ) into this equation we get (4.1) 1 =r²(ϕ)cos²ϕ

a² + sin²ϕ

, that is, a²= r²(ϕ) cos²ϕ 1−r²(ϕ) sin²ϕ.

On the other hand, the slope of the tangent of the ellipse at (x, y) is ^dy_dx = _ya^−x₂ which at the pointh(ϕ) is

˙

r(ϕ) sinϕ+r(ϕ) cosϕ

˙

r(ϕ) cosϕ−r(ϕ) sinϕ= dy dx = −x

ya² = −cosϕ a²sinϕ. This implies

˙ r(ϕ)

r(ϕ) =(1−a²) sinϕcosϕ a²sin²ϕ+ cos²ϕ =

1−_1−r^r²^{(ϕ) cos}2(ϕ) sin²²^ϕϕ

sinϕcosϕ

r²(ϕ) cos²ϕ

1−r²(ϕ) sin²ϕsin²ϕ+ cos²ϕ = (1−r²(ϕ)) tanϕ.

At everyϕ, where r(ϕ)6= 1, this gives

˙ r(ϕ)

r(ϕ)(1−r²(ϕ)) = tanϕ which, by integration, yields

−1

2 ln|1−r²(ϕ)|

r²(ϕ) =−ln|cosϕ|+c₀ for a constantc₀. An equivalent reformulation of this is

r(ϕ) = 1

p1±c₁cos²ϕ,

wherec1is a constant. Substituting this into (4.1),a²(1±c1) = 1 follows, hencea is the same constant for all ellipses∂E_h(ϕ), which are therefore a fixed ellipse∂E.

This means that∂His a subset of ∂E having equation (1±c1)·x²+y²= 1.

However, ∂Hcontains the point e3= (1,0) too, hencec1= 0 and therefore∂H is the unit circle centered at the origin. This proves statement(ii).

Definition 4.3. Take a configuration according to Definition4.1. We construct a set of geometric object in the following way: Chose a pointx_i close to e_i on the open segmentσ_i=e_je_k for every i= 1,2,3, where{j, k}={1,2,3} \ {i}.

Let the linese₂x₃, e₃x₁ ande₁x₂be denoted by`⁰₁, `⁰₂, `⁰₃, respectively.

Take the points v₁ =`⁰₂∩`⁰₃, v₂ =`⁰₃∩`⁰₁, v₃ =`⁰₁∩`⁰₂, and denote the open segmentsv2v3,v3v1andv1v2, byσ₁⁰,σ⁰₂ andσ₃⁰, respectively.

Further, we take the points x^t₁ = t1∩`⁰₂, x^t₂ = t2∩`⁰₃, x^t₃ = t3 ∩`⁰₁, and x^H₁ =∂H ∩(`⁰₂\ {e3}),x^H₂ =∂H ∩(`⁰₃\ {e1}),x^H₃ =∂H ∩(`⁰₁\ {e2}). These points of intersection do exist ifxi are chosen close enough toei (i= 1,2,3).

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H

e1

e₂

e3

x^H₂

x^H₃

x^H₁

`₁

`2

`3

`⁰₁

`⁰₂

`⁰₃

σ⁰₁ σ⁰₂ σ₃⁰

x₁

x₂

x3

v1

v3

v2

t₁ t2

t₃ x^t₂

x^t₃

x^t₁

ξ1

α₁ β1

ξ₃

α₃ β₃ ξ₂

α2

β₂

Figure 4.5. A construction for the triple asymptoticH-triangle.

Finally, let the magnitude of the angles∠x2e1e2,∠x3e2e3,∠x1e3e1be denoted by ξ1,ξ2, ξ3, respectively, that of the angles∠x^t₁e1e2,∠x^t₂e2e3,∠x^t₃e3e1 be denoted byα1,α2, α3, respectively, and that of the angles∠e3e1e2,∠e1e2e3,∠e2e3e1 be denoted byβ1,β2,β3, respectively.

Theorem 4.4. Take a construction according to definition 4.3. For every i = 1,2,3 denote the Euclidean midpoint of the segmentσi bybi and theH-midpoint of the segmentσ_i⁰ byb^H_i

The lines f₁^H, f₂^H, f₃^H form a pencil if and only if the pointsx1,x2 andx3 can be chosen for anyε, δ >0 so that

|b^H₁ −b1|+|b^H₂ −b2|+|b^H₃ −b3|< ε,

|x₁−e₁|+|x₂−e₂|+|x₃−e₃|< δ.

Proof. Since d_H(vj,b^H_i ) =d_H(b^H_i ,vk), where {i, j, k} ={1,2,3}, (2.2) implies (ej,x^H_k;vj,b^H_i ) = (ej,x^H_k;b^H_i ,vk), hence

(4.2)

1 = (ej,x^H_k;vj,b^H_i )

(ej,x^H_k;b^H_i ,vk) = (ej,x^H_k,vj)/(ej,x^H_k,b^H_i ) (ej,x^H_k;b^H_i )/(ej,x^H_k,vk)

=(ej,x^H_k,vk)(ej,x^H_k,vj)

(e_j,x^H_k;b^H_i )² = |ej−vk||ej−vj|

|x^H_k −v_k||x^H_k −v_j| 1 (e_j,x^H_k;b^H_i )². From now on, assume that ξ_i → 0 for every i = 1,2,3. Then x^H_k → e_k, hence the affine midpoint ofe_jx^H_k converges tob_i, and thereforeb^H_i →b_i if and only if (ej,x^H_k;b^H_i )→1.

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H

e1

e₂

e₃

x1

x2

x3

t1

t2

t3

f₁^H f₂^H

f₃^H

f₂

f1

f3

b^H₂ b^H₁

b^H₃

b2

b1

b3

φ₁ ψ1

φ₃ ψ3

φ₂ ψ₂

Figure 4.6. The construction with the midpoints.

As we have ^|e^j^−v^k^|

|x^H_k−vj| →1, (4.2) implies the asymptotic equation (e_j,x^H_k;b^H_i )²∼ |ej−v_j|

Using the law of sines we obtain

|x^H_k −v_k|

|ej−vj|

= |x^t_k−ej| − |vk−ej| − |x^H_k −x^t_k|

|ei−ej|sinξi/sin(βj−ξj+ξi)

=

|ej−ek|

sin(β_k+α_k+ξj)sin(βk+αk)−_sin(β^|e^j^−e^k^|

k−ξ_k+ξj)sin(βk−ξk)− |x^H_k −x^t_k|

|ei−ej|

sin(βj−ξj+ξi)sinξ_i

= sin(βj−ξj+ξi) sinξi

|ej−ek|

|ei−ej|×

× sin(βk+αk)

sin(βk+αk+ξj)− sin(βk−ξk)

sin(βk−ξk+ξj)−|x^H_k −x^t_k|

|ej−ek|

= sin(βj−ξj+ξi) sinξi

|σi|

|σk|×

× tan(β_k+α_k)

sinξj+ cosξjtan(βk+αk)− tan(β_k−ξ_k)

sinξj+ cosξjtan(βk−ξk)−|x^H_k −x^t_k|

|ej−ek|

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= sin(β_j−ξ_j+ξ_i) sinξi

|σ_i|

|σk|×

× sinξj(tan(βk+αk)−tan(βk−ξk))

(sinξ_j+ cosξ_jtan(β_k+α_k))(sinξ_j+ cosξ_jtan(β_k−ξ_k))−|x^H_k −x^t_k|

|e_j−e_k|

= sinξj

sinξi

|σi|

|σk|×

× sin(βj−ξj+ξi)(tan(βk+αk)−tan(βk−ξk)) (sinξ_j+ cosξ_jtan(β_k+α_k))(sinξ_j+ cosξ_jtan(β_k−ξ_k))−

−sin(β_j−ξ_j+ξ_i)|x^H_k −x^t_k|

|ej−ek|sinξj

∼ sinξj

sinξ_i

|σi|

|σk|

sinβj(tan(βk+αk)−tanβk)

tan(β_k+α_k) tanβ_k = sinξj

sinξ_i

|σi|

|σk|

sinβjsinαk

sin(β_k+α_k) sinβ_k. Putting this into (4.3) results in

b^H₁ →bi ⇐⇒ sinξ1

sinξ₂ ∼|σ1|

|σ₃|

sinβ2sinα3

sin(β₃+α₃) sinβ₃, b^H₂ →b₂ ⇐⇒ sinξ₂

sinξ3

∼|σ₂|

|σ1|

sinβ₃sinα₁ sin(β1+α1) sinβ1

,

b^H₃ →b3 ⇐⇒ sinξ₃ sinξ1

∼|σ3|

|σ2|

sinβ₁sinα₂ sin(β2+α2) sinβ2

.

Having arbitraryξ₂→0,ξ₁→0 andξ₃→0 can be so chosen, that the first two of these asymptotic equations are satisfied, and the fulfillment of the third one then depends exactly on their product. Such anglesξ₁, ξ₂, ξ₃ can be chosen therefore if and only if

(4.4) sinα1sinα2sinα3= sin(β1+α1) sin(β2+α2) sin(β3+α3).

Denote the magnitude of the angles ∠f₁^H`2, ∠f₂^H`3, ∠f₃^H`1 byψ1, ψ2, ψ3, respectively, and the angles∠f₁^H`3, ∠f₂^H`1, ∠f₃^H`2 by φ1, φ2, φ3, respectively (see Figure4.6). We clearly haveψ_i+φ_i=β_i for everyi= 1,2,3.

Observe that

−1 = (`3, `2;t1, f₁^H) =−sinα1/sin(β1+α1) sinφ₁/sinψ₁ ,

−1 = (`1, `₃;t₂, f₂^H) =−sinα₂/sin(β₂+α₂) sinφ2/sinψ2

,

−1 = (`2, `1;t3, f₃^H) =−sinα3/sin(β3+α3) sinφ3/sinψ3

, hence (4.4) is equivalent to

1 = sinφ1

sinψ1

sinφ2

sinψ2

sinφ3

sinψ3

.

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Letf_i^H∩σi=fifor every i= 1,2,3. Then the law of sines gives (e₁,e₂;f₃) = |σ₂|

|σ1| sinφ₃ sinψ3

, (e₂,e₃;f₁) = |σ₃|

|σ2| sinφ₁ sinψ1

, and (e₃,e₁;f₂) =|σ₁|

|σ3| sinφ₂ sinψ2

.

By Ceva’s theorem the product of these ratios equals to 1 if and only if the lines f₁^H, f₂^H andf₃^H form a pencil. This proves the theorem.

5. Circumcenter and orthocenter in Hilbert geometry

Existence of the circumcenter of a trigon, the common point of the three perpendicular bisectors, is a well known property in Euclidean plane. It can be formulated also for the hyperbolic plane [14, p. 350]: In hyperbolic geometry the perpendicular bisectors of any trigon form a pencil.

Theorem 5.1. If theH-perpendicular bisectors of any trigon in the Hilbert geom- etryHform a pencil, then (H, dH)is the hyperbolic geometry.

Proof. We need to show that H is an ellipsoid. By Lemma3.2 we only need to work in the plane, therefore from now on in this proofHis in a planeP.

Suppose thatHis not an ellipse. We shall have to arrive at a contradiction.

By (i)of Lemma 3.3there exists an ellipse E circumscribed aroundHwith at least three different contact pointse1,e2,e3lying in∂H ∩∂E such that the closed triangle4e1e₂e₃contains the origin.

Suppose we have the configuration described in(iv)of Lemma3.3.

Choose a planeP⁰such that one of its open halfspaceScontainst₁andE. Now choose a point P out of P⁰ ∪ P ∪ S. Let π be the perspective projection of P intoP⁰ through the point P. This projectionπclearly maps the configuration in P into a configuration in P⁰ that is described in (iii)of Lemma 3.3. Thus, since the statement of the theorem is of projective nature, it is enough to validate it for configurations described in(iii)of Lemma3.3.

Take a construction defined by Definition 4.3 (see Figure 5.1), and let ε =

|x1−e1|+|x2−e2|+|x3−e3|.

By (v)of Lemma3.3the straight linesmibi(i= 1,2,3) meet in the centeroof the ellipseE, which is in the interior of the trigon4e1e2e3, therefore the centero of the ellipseE is in the interior of the trigon4v1v2v3 too, ifεis small enough, which we assume from now on.

According to Lemma3.1the lines`⁰₁, `⁰₂, `⁰₃ contain the points

(5.1)

e2≺v2≺b^H₁ bÊ₁ ≺v3≺x3≺x^H₃ xÊ₃, e3≺v3≺b^H₂ bÊ₂ ≺v1≺x1≺x^H₁ xÊ₁, e₁≺v₁≺b^H₃ bÊ₃ ≺v₂≺x₂≺x^H₂ xÊ₂, respectively, in the given order (see Figure5.1).

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14 J. KOZMA AND ´A. KURUSA

H E

e1

e2

e3

x^E₂x^H₂

x^E₃ x^H₃

x^E₁ x^H₁x₁ x2

x3

v1

v3

v2

t1

t2

t3

m2

m1

m3

t^H₁ t^E₁ t^H₂

t^E₂

t^H₃ t^E₃

m^E₃

3

m^E₁ m^H₁

m^E₂

m^H₂ b^E₂

b^H₂ b^E₁ b^H₁

b^E₃ b^H₃

c

Figure 5.1. Constructions of circumcenters inHandE.

Take the tangent lines t^H_i andtÊ_i ofH andE at the pointsx^H_i andxÊ_i, respectively, for everyi∈ {1,2,3}. Letm^H₁ =t^H₂ ∩t₃,m^H₂ =t^H₃ ∩t₁,m^H₃ =t^H₁ ∩t₂, and mÊ₁ =tÊ₂ ∩t₃,mÊ₂ =tÊ₃∩t₁ andmÊ₃ =tÊ₁ ∩t₂ (see Figure5.1).

According to Lemma3.4, the tangentst₁, t₂, t₃ contain the points (5.2) m2≺e1≺m3≺m^E₃ m^H₃, m3≺e2≺m1≺m^E₁ m^H₁,

m₁≺e₃≺m₂≺m^E₂ m^H₂, in the given order, respectively (see Figure5.1).

Notice, that for well-chosen x1,x2,x3 we have x^H_i 6=xÊ_i for some i∈ {1,2,3}, sayx^H₁ 6=xÊ₁, and then we also havemÊ₃ 6=m^H₃ andbÊ₂ 6=b^H₂ by(ii)of Lemma3.3.

While the triangle 4v1v2v3 is in IntH ∩IntE, letting εtend to 0, it tends to the triangle4e1e2e3 in Euclidean meaning, hence Lemma3.4 implies

(5.3) m^H_i →m_i and m^E_i →m_i for every i∈ {1,2,3}.

On the other hand, for well-chosenx₁,x₂,x₃, Theorems 4.2and4.4imply that (5.4) b^E_i →bi and b^H_i →bi for everyi∈ {1,2,3}

asε→0.

By (5.3) and (5.4) the hyperbolic circumcentercof the triangle4v1v2v3tends to the centeroofE asε→0, and therefore the circumcentercis in the interior of the triangle4v1v2v3 ifεis small enough.

Suppose that the triangle4v1v₂v₃has also anH-circumcenter, sayc⁰. By (5.3) and (5.4) theH-circumcenter c⁰ tends also to the centeroof E asε→0, hence it is in the interior of the triangle4v1v₂v₃ ifεis small enough.

On account of (5.1) and (5.2), for everyi∈ {1,2,3}the closed segmentsm^H_i b^H_i andm^E_ib^E_i havek(i)≥1 points in common, which is on the same side of`⁰_i asmi

is. By the notice after the relations (5.1) and (5.2), we may assume thatk(2) = 1

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and k(3) = 1. This implies that c⁰ is in the left open half plane of the lines mÊ_ibÊ_i directed frommÊ_i tobÊ_i for everyi∈ {2,3}and it is in the left closed half plane of the lines mÊ₁bÊ₁ directed frommÊ₁ to bÊ₁. This contradicts the fact that the intersection of these half planes are empty, therefore the supposition of the existence ofc⁰ was wrong, and the theorem is proved.

Existence of the orthocenter of a trigon, the common point of the three altitudes, is well known in Euclidean plane. It is also known for the hyperbolic plane [9, Theorem 3]: In hyperbolic geometry the altitudes of any trigon form a pencil.

Theorem 5.2. If the altitudes of any trigon in a Hilbert geometryHform a pencil, then(H, d_H)is the hyperbolic geometry.

Proof. Following the proof of Theorem5.1we have the very same construction of the tangents and points, but without the midpoints for now.

H E

e1

e2

e3

x^E₁x^H₁

x^E₂ x^H₂

x^E₃ x^H₃

v1

v3

v₂

t1

t2

t3 m2

m1

m3

t^H₃ t^E₃ t^H₁

t^E₁

t^H₂ t^E₂ m^E₁

m^H₁

m^E₂

m^H₂ m^E₃

m^H₃

a

Figure 5.2. A triangle with altitudes intersecting in inner points

According to (5.3), the intersection a of the hyperbolic altitudes vim^E_i (i = 1,2,3) of the triangle4v1v2v3tends to the centeroofE asε→0, and therefore ais in the interior of the triangle4v1v₂v₃ ifεis small enough.

Suppose that the H-altitudesv_im^H_i (i = 1,2,3) of the triangle 4v1v₂v₃ also intersect in a pointa⁰.

By (5.3) the pointa⁰ also tends to the centeroofE asε→0, hence it is in the interior of the triangle4v₁v₂v₃ ifεis small enough.

On account of relations (5.2), vi = vim^H_i ∩vimÊ_i for every i ∈ {1,2,3}, and therefore a⁰ is in the left open half plane of the lines vimÊ_i directed from vi to mÊ_i for everyi∈ {1,2,3}. This contradicts the fact that the intersection of these halfplanes are empty, therefore the supposition of the existence ofa⁰ was wrong,

and the theorem is proved.

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Acknowledgement. This research was supported by the European Union and co-funded by the European Social Fund under the project “Telemedicine-focused research activi- ties on the field of Mathematics, Informatics and Medical sciences” of project number

‘T ´AMOP-4.2.2.A-11/1/KONV-2012-0073”.

The authors appreciate János Kincses, Gábor Nagy, György Kiss and Tamás Sz˝onyi for their helpful discussions.

Thanks goes also to the anonymous referee whose advises improved this paper and in particular figures5.1and5.2considerable.

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József Kozma, Bolyai Institute, University of Szeged, Aradi vértanúk tere 1, 6725 Szeged (Hungary); E–mail:kozma@math.u-szeged.hu .

Arp´´ ad Kurusa, Bolyai Institute, University of Szeged, Aradi v´ertan´uk tere 1, 6725 Szeged (Hungary); E–mail:kurusa@math.u-szeged.hu .