On the Geometry of a Triangle in the Elliptic and in the Extended Hyperbolic Plane 3

Home , Concurrent lines, Lemoine point, Orthocentric system, Spieker center, Trilinear coordinates

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A suitable l is, for example, l =∗(p×q)=(p×q)∗, where × stands for the canonical cross product in V = ℝ3 and ∗ for the isomorphism V → V ∗, ∗(r) = r⋅(.). The linear form l is uniquely determined up to a nonzero real factor, so there is a 1∶1-correspondence between the lines in the projective plane and the elements of ∗ = (V ∗−{0∗})∕ℝ×. We identify the line l = P ∨ Q with the element ∗ ∗ (p2q3 − p3q2 ∶ p3q1 − p1q3 ∶ p1q3 − p2q1) ∈  . ∗ ∗ In the projective plane, two different lines k = (k1∶k2∶k3) , l = (l1∶l2∶l3) always meet at one point k ∧ l = Π((k1, k2, k3)×(l1, l2, l3)), the so-called meet of these lines. 1.2. Collineations, correlations and polarities The automorphism group Aut() of  can be identified with the projective linear group PGL(V ) = GL(V )∕ℝ×. These automorphisms on  are called collineations because the image of a line under an automorphismis again a line. A correlation is a bijective mapping ∗ ∗ c∶  ∪  →  ∪  with the following property: The restriction c  of c to  is a point-to-line transformation that maps collinear points to concurrent lines, while c ∗ is a line-to-point transformation that maps concurrent lines to collinear points.ð Thus there are → ∗ ∗ → for each correlation c uniquely determined linear mappings c1∶   , c2∶  ð  with c1 = c  and c2 = c ∗ . If c2 is the inverse of c1, this correlation is an involution. ∗ In this case, c1, c2 are symmetric mappings since  and  have even dimensions, and the correlation cðis a polarity. ð 1.3. Elliptic and hyperbolic metric structures on . We now fix a nonzero real number . By we denote the polarity which maps a point ∗ ∗ P = (p1∶p2∶p3) to its dual line P = (p1∶p2∶p3) and a line l = (l1∶l2∶l3) to its dual point l =(l1∶l2∶l3). ∗ 1.3.1. We make use of to introduce orthogonality. A line k = (k1∶k2∶k3) is orthog- ∗ onal (or perpendicular) to a line l = (l1∶l2∶l3) when the dual k of k is a point on l; this is precisely when k1l1 + k2l2 + k3l3 = 0. Obviously, if k is orthogonal to l, then l is orthogonal to k. Two points P = (p1∶p2∶p3) and Q = (q1∶q2∶q3) are orthogonal to each other if p1q1 + p2q2 + p3q3 = 0. This is the case exactly when their dual lines P and Q are orthogonal. Self-orthogonal points and lines are called isotropic. The isotropic points form the so-called absolute conic abs. This is either the empty set, in which case the geometry is called elliptic, or it is a properconic and the geometryis called hyperbolic. Each symmetric real 3x3-matrix M with det M ≠ 0 determines a scalar product (symmetric bilinear form) [M]∶ V ×V → ℝ by w1 M M (v1, v2, v3) [ ] (w1, w2, w3) = (v1, v2, v3) w2 . ⎛w3⎞ The orthogonality of points and lines can be expressed with the help⎜ ⎟ of the scalar products S S−1 S ⎜ ⎟ [ ] and [ ] where = diag(, 1, 1). Two points P = (p1∶p2∶⎝p3)⎠and Q = (q1∶q2∶q3) ∗ are orthogonal precisely when (p1,p2,p3) [S](q1,q2,q3)=0, two lines k =(k1∶k2∶k3) and ∗ −1 l =(l1∶l2∶l3) are orthogonal precisely when (k1,k2,k3) [S ] (l1,l2,l3)=0.

∗ 1.3.2. Consider some point P =(p1∶p2∶p3) and some line l =(l1∶l2∶l3) with l ≠ P . The perpendicular from P to l is the line ∗ perp(l, P ) ∶= P ∨ l =(p2l3−p3l2 ∶ (p3l1−p1l3) ∶ (p1l2−p2l1)) , and the point on l orthogonal to P is perp(P , l) ∶= l ∧ P =((p2l3−p3l2) ∶ p3l1−p1l3 ∶(p1l2−p2l1)). ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 3

The line through P parallel to l is par(l, P ) ∶= perp(perp(l, P ), P ). The point ped(P , l) ∶= perp(perp(P , l), l), is the orthogonal projection of P on l, also called the pedal of P on l.

1.3.3. Line segments and angles. Deﬁne the function ∶ V → {−1, 0, 1} by

0, if (p1,p2,p3)=(0, 0, 0) , (p0,p1,p2)= 1, if (p1,p2,p3) > (0, 0, 0) with respect to the lexicographic order, ⎧−1, if (p ,p ,p ) < (0, 0, 0) with respect to the lexicographic order. ⎪ 1 2 3 For an anisotropic⎨ point P =(p ∶p ∶p ) let P ◦ ∈ V be the vector ⎪ 1 2 3 ⎩ ◦ (p1,p2,p3) (p1,p2,p3) P ∶= (p1,p2,p3)= (p1,p2,p3). (p1,p2,p3) [S] (p1,p2,p3) p2 p2 p2 1 + 2 + 3 ◦ P is obviously√ uniquely determined by P . t ð ð For any two different points P and Q we introduce two lineð segments ð[P,Q]+, [P,Q]−, thesearetheclosuresof thetwo connectedcomponentsof the set P ∨Q−{P,Q}. If R, S are ◦ ◦ ℝ two different anisotropic points in [P,Q]+, then [R, S]+ = {Π(sR + tS ) s,t ∈ ,st ≥ 0} is a subset of [P,Q]+, and if R, S are two different anisotropic points in [P,Q]−, then ◦ ◦ ℝ [R, S]− = {Π(sR + tS ) s,t ∈ ,st ≤ 0} is a subset of [P,Q]−. ð We define angles as subsets of the pencil of lines through a point, which is the vertex of this angle: Given three noncollinearð points Q,R,S, put

∠+QSR ∶= {S ∨ P P ∈[Q, R]+} , ∠−QSR ∶= {S ∨ P P ∈[Q, R]−} . The union of these two angles is the complete pencil of lines through S. ð ð 1.3.4. The length of a segment and the measure of an angle. We introduce the measure of segments and angles using concepts that were developed by C. Vörös in Hungary at the beginning of the 20th century and more recently (2014) described by Horváth in two papers [13, 14], where he applied these to various conﬁgurations in hyperbolic geometry. To each line segment s and to each angle is assigned its measure (s) resp. () which is a complex number with a real part in ℝ̄ = ℝ ∪ {−∞, ∞} and an imaginary part in the interval [0, ].

Deﬁne for each anisotropic point P =(p1∶p2∶p3) a number "P ∈ {1, −i} by ◦ ◦ 1) "P ∶= 1∕ P [S]P . First we describe the measure ([P,Q]±) of√ line segments with anisotropic endpoints P,Q, P ≠ Q. (The remaining cases will be treated afterwards.) Here, the function satisﬁes the following rules:

(1) ([P,Q]+) and ([P,Q]−) are ﬁnite complex numbers with imaginary parts in the interval [0, ] satisfying ([P,Q]+)+ ([P,Q]−)= (P ∨ Q)= i. ◦ ◦ (2) cosh([P,Q]+)= "P "QP [S] Q . (3) If R is an anisotropic inner point of [P,Q]+,

1) We follow the convention r exp(i) = r exp( 1 i) for r ∈ ℝ+and 0 ≤ < 2. 2 √ √ 4 MANFRED EVERS

then ([P , R]+)+ ([R, Q]+)= ([P,Q]+). (4) Q = perp(P , P ∨Q) precisely when ([P,Q] )= ([P,Q] )= i. + − 2 Let us look at special cases.

In the elliptic case, all points R are anisotropic with "R =1. For two diﬀerent points P and ◦ ◦ Q we have −1 < P [S] Q < 1 and ([P,Q]±)= si with a real number s, 0 1. We have the choice of ([P,Q]+) being positive or negative and decide for the positive value. Subcase 2: One of the points P,Q is inside, the other one is outside the absolute conic. Let us assume P is outside; then the point R ∶= perp(P , P ∨Q) lies inside the absolute conic. In this case, ([P,Q] )= ([R, Q] )+ i, ([P,Q] )= −([R, Q] )+ i, if R ∈[P,Q] , and + + 2 − + 2 + ([P,Q] )=−([R, Q] )+ i, ([P,Q] )= ([R, Q] )+ i, if R ∈[P,Q] . + + 2 − + 2 − Subcase 3: Both, P and Q, are points outside the absoluteconic. With R ∶= perp(P , P ∨Q) and S ∶= perp(Q, P ∨Q) we get ◦ ◦ ◦ ◦ cosh(([P,Q]+)= "P "QP [S] Q = "R"SR [S] S = cosh(([R, S]+). The rules (1) and (4) require ([P,Q]+)=−([R, S]+) < 0. ◦ ◦ Ananalysisofthediﬀerentcasesshowsthatbyknowingthetwo numbers "P "Q and P [S] Q we can determine ([P,Q]±).

We now set the lengths of the segments [P,Q]± if at least one endpoint is isotropic. In this ℂ case, ([P,Q]±) is not an element in , the real part of ([P,Q]±) is either +∞ or −∞. Again we "decline" various cases. If P ∨ Q is an isotropic line and P is an isotropic point, then Q is anisotropic and P = perp(Q, P ∨ Q), ([P,Q] )= ([P,Q] )= i. + − 2 If P ∨ Q is an anisotropic line and just one of the points P,Q, let us say P , is isotropic, then: ([P,Q] )=∞+ 1 i, ([P,Q] )=−∞+ 3 i in case of " =−i. + 4 − 4 Q ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 5

([P,Q] )=−∞+ 1 i, ([P,Q] )=∞+ 3 i if " =1 and P is the only isotropic + 4 − 4 Q point in [P,Q]+, ([P,Q] )=∞+ 3 i, ([P,Q] )=−∞+ 1 i if there are two isotropic points + 4 − 4 in [P,Q]+. If P and Q are isotropic, then ([P,Q] )=±∞+ i. ± 2 The measure of angles: We use the same symbol for the measure of angles as for line segments. The angle ∠+QSR is either identical with {P P ∈ [(S ∨ Q) , (S ∨ R) ]+} or with {P P ∈ [(S ∨ Q) , (S ∨ R) ]−}. In the ﬁrst case we set (∠±QSR) = ([(S ∨ Q) , (S ∨ R) ]±), in the second (∠±QSR) = ([(S ∨ Qð) , (S ∨ R) ]∓). If the points R,S,Q andð the lines S∨Q, S∨R are anisotropic, then cosh((∠+QSR)) = −cosh((∠−QSR)) ◦ ◦ ◦ ◦ (S × Q ) [S−1] (S × R ) = (S◦ × Q◦) [S−1] (S◦ × Q◦) (S◦ × R◦) [S−1] (S◦ × R◦) ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ (S [S] S )(Q [S] R )−(S [S] Q )(S [S] R ) = √ √ . (S◦ [S] S◦)(Q◦ [S] Q◦)−(S◦ [S] Q◦)2 (S◦ [S] S◦)(R◦ [S] R◦)−(S◦ [S] R◦)2 √ √ 1.3.5. The distance between points and the (angle) distance between lines. We introduce an order ≺ on ℝ̄ + ℝi by b1 < b2 a1 + b1i≺a2 + b2i iﬀ or T b1 = b2 and a1

1.3.6. Barycentriccoordinatesofapointona line. Let P and Q be two diﬀerent anisotropic points on an anisotropic line and R± ∈[P,Q]± be two other anisotropic points on this line, ◦ ◦ then R± = Π(xP ± yQ ) with any two real numbers x, y satisfying (∗) x ∶ y = sinh(d(R±,Q)) "P ∶ sinh(d(R±, P )) "Q . Real numbers x, y satisfying (∗) are called barycentric coordinates of P± with respect to the points P,Q. Proof of (∗) ∶ Let R be an anisotropic pointon P ∨Q. Assume R◦ = xP ◦+ yQ◦, x, y ∈ ℝ. Then: 6 MANFRED EVERS

sinh2(d(P , R)) = cosh2(d(P , R))−1 ◦ ◦ 2 2 (R [S] P ) " = P −1 R◦ [S] R◦ ◦ ◦ ◦ 2 2 ((xP + yQ ) [S]P ) " = P −1 (xP ◦+ yQ◦) [S](xP ◦+ yQ◦) 2 2 ◦ ◦ 2 ◦ ◦ 2 2 x " +2xyP [S] Q + y (P [S] Q ) " = P P −1 2 2 ◦ S ◦ 2 2 x "P +2xyP [ ] Q + y "Q 2 ◦ S ◦ 2 2 2 y ((P [ ] Q ) "P − "Q ) = . R◦ [S] R◦

2 ◦ S ◦ 2 2 2 x ((P [ ] Q ) "Q − "P ) In the same way we get sinh2(d(Q, R)) = . R◦ [S] R◦ 2 2 2 ◦ ◦ 2 2 2 [S] 2 2 2 "Q sinh (d(P , R)) y ((P Q ) "P "Q − 1) y (cosh (d(P,Q))−1) y Thus: = = = . 2 2 2 2 2 2 x2((P ◦ [S] Q◦)2" " − 1) 2 x "P sinh (d(Q, R)) P Q x (cosh (d(P,Q))−1) Now we look at diﬀerent cases. If all points of P ∨ Q are anisotropic, then "P = "Q = "R = 1 and d(P , R), d(Q, R) ∈ [0, 1 i]. Therefore, (x, y)∈ ℝ(−i" sinh(d(Q, R)), −i" sinh(d(P , R))). 2 P Q If P ∨ Q is a line passing through two isotropic points, we consider four subcases: ℝ (1) ∶ "P = "Q = "R. Then d(P , R), d(Q, R)∈ and ℝ (x, y)∈ ("P sinh(d(Q, R)), "Q sinh(d(P , R))) if "P =1, ℝ (x, y)∈ (−i"P sinh(d(Q, R)), −i"Q sinh(d(P , R))) if "P =−i. (2) ∶ " = " ≠ " , hence d(P , R), d(Q, R)∈ 1 i + ℝ. P Q R 2 ℝ If "P = "Q =1, then (x, y)∈ (−i"P sinh(d(Q, R)), −i"Q sinh(d(P , R))). ℝ If "P = "Q =−i, then (x, y)∈ ("P sinh(d(Q, R)), "Q sinh(d(P , R))). (3) ∶ If " = " ≠ " , then d(P , R)∈ ℝ and d(Q, R)∈ 1 i + ℝ. P R Q 2 ℝ In this case, (x, y)∈ (−i"P sinh(d(Q, R)), −i"Q sinh(d(P , R))) if "P =1, ℝ and (x, y)∈ ("P sinh(d(Q, R)), "Q sinh(d(P , R))) if "P =−i. (4) ∶ If " ≠ " = " , then d(P , R)∈ 1 i + ℝ and d(Q, R)∈ ℝ. P R Q 2 ℝ In this case, (x, y)∈ ("P sinh(d(Q, R)), "Q sinh(d(P , R))) if "P =1, ℝ and (x, y)∈ (−i"P sinh(d(Q, R)), −i"Q sinh(d(P , R))) if "P =−i. □

◦ ◦ 1.3.7. Semi-midpoints. The points R± = P ± Q ∶= Π(P ± Q ) we call semi-midpoints of P and Q. These points were introduced and investigated by Wildberger and Alkhaldi [29] under the name smydpoints. Properties of semi-midpoints: If "P = "Q, then P + Q and P − Q are (proper) midpoints: d(P ± Q, P )= d(P ± Q,Q). If "P ≠ "Q, then d(P ±Q, P ) ≠ d(P ±Q,Q). So, P +Q and P −Q are not proper midpoints of P and Q. We will call them pseudo-midpoints; in [29] they are called sydpoints. If P ≠ perp(Q, P ∨ Q), then P + Q and P − Q are anisotropic and d(P ± Q, P )"P ±Q"P = d(P ∓ Q,Q)"P ∓Q"Q. If P = perp(Q, P ∨ Q), then P + Q and P − Q are isotropic and d(P ± Q, P )=−"2 ∞+ i , d(P ± Q,Q)=−"2 ∞+ i. P 4 Q 4 P,Q,P + Q, P − Q always form a harmonic range. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 7

1.4. The use of barycentric coordinates. 1.4.1. Barycentric coordinates with respect to a reference triple Δ of points. We ﬁx a reference triple Δ= ABC of non-collinear, anisotropic points A,B,C ∈ . For every point P we can ﬁnd a triple (p1,p2,p3) of real numbers such that P = p1A + p2B + p3C ∶= ◦ ◦ ◦ Π(p1A + p2B + p3C ). Such a triple of real numberswill be called (triple of) barycentric coordinates of P with respect to Δ. The triple (p1,p2,p3) is not uniquely determined by P , buteveryothertripleofbarycentriccoordinatesis a realmultiple of (p1,p2,p3). Thepoint P is determined by the homogenoustriple (p1∶p2∶p3) and Δ, and we write P =[p1∶p2∶p3]Δ. ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ The triple can be calculated by p1∶p2∶p3 = P ⋅ (B × C ) ∶ P ⋅ (C × A ) ∶P ⋅ (A × B ). Remark: As barycentric coordinates we also accept a triple of complex numbers as long as it is a complex multiple of a real triple.

1.4.2. Lines. Let P =[p1 ∶ p2 ∶ p3]Δ and Q =[p1 ∶ p2 ∶ p3]Δ be two diﬀerent points, then the line P ∨ Q consists of all points X =[x1 ∶ x2 ∶ x3]Δ satisfying the equation ((p1,p2,p3)×(p1,p2,p3)) ⋅ (x1,x2,x3)=0.

If R ∨ S is a line through R =[r1 ∶r2 ∶r3]Δ and S =[s1 ∶s2 ∶s3]Δ, diﬀerent from P ∨ Q, then both lines meet at a point T =[t1 ∶t2 ∶t3]Δ with (t1,t2,t3)=((p1,p2,p3)×((q1, q2, q3))×((r1,r2,r3)×(s1,s2,s3)). 1.4.3. Triangles. The reference triple Δ = ABC determines four triangles that share the same vertices A,B,C and the same sidelines a ∶= B ∨ C, b ∶= C ∨ A, c ∶= A ∨ B. These triangles are the closures in  of the four connected components of  −{a, b, c}:

Δ0 ∶= {[p1 ∶ p2 ∶ p3]Δ p1,p2,p3 > 0}∪[B,C]+ ∪[C,A]+ ∪[A,B]+, Δ1 ∶= {[p1 ∶ p2 ∶ p3]Δ −p1,p2,p3 > 0}∪[B,C]+ ∪[C,A]− ∪[A,B]−, Δ2 ∶= {[p1 ∶ p2 ∶ p3]Δð p1, −p2,p3 > 0}∪[B,C]− ∪[C,A]+ ∪[A,B]−, Δ3 ∶= {[p1 ∶ p2 ∶ p3]Δð p1,p2, −p3 > 0}∪[B,C]− ∪[C,A]− ∪[A,B]+.

Remark: We do not considerð the closures of the sets  −Δk, k =0, 1, 2, 3, as triangles. ð 1.4.4. Note: In the following, we always assume that not only the vertices but also the sidelines of Δ are anisotropic. 1.4.5. The characteristic matrix of Δ. We introduce the symmetric matrix ◦ ◦ ◦ ◦ ◦ ◦ A [S] A A [S] B A [S] C ℭ c ◦ S ◦ ◦ S ◦ ◦ S ◦ =( ij)i,j=1,2,3 ∶= B [ ] A B [ ] B B [ ] C . ◦ ◦ ◦ ◦ ◦ ◦ ⎛C [S] A C [S] B C [S] C ⎞ ℭ is a regular matrix with det ℭ = (A⎜ ◦⋅ (B◦ × C◦))2 ≠ 0 and inverse⎟ ⎜ ⎟ c c⎝ − c2 c c − c c c ⎠c − c c 1 22 33 23 23 31 12 33 12 23 31 22 D =(d ) = c c − c c c c − c2 c c − c c . ij i,j=1,2,3 ℭ 23 31 12 33 33 11 31 31 12 23 11 det c c c c c c c c c c c2 ⎛ 12 23 − 31 22 31 12 − 23 11 11 22 − 12 ⎞ ℭ is the Gram matrix of the basis⎜ A◦,B◦,C◦ of the vectorspace V with the scalar⎟ product ⎜ ⎟ [S] . We call ℭ the characteristic⎝ matrix of Δ, since from this matrix, as we will see,⎠ one can read oﬀ the lengths of the sides and the measures of the angles of the triangles Δk, k = 0, 1, 2, 3.

Let us denote the lengths of the sides of Δ0 by a∶=([B,C]+), b∶=([C,A]+), c∶=([A,B]+), then 8 MANFRED EVERS

◦ ◦ 1 1 cosh(a)= " " B [S] C = c , B C c c 23 22 33 ◦ ◦ ◦ ◦ cosh(b) = " " C [S] A , cosh(c)= " " A [S] B . C A √ √ A B ℭ By knowing , we also know "A, "B, "C, cosh(a), cosh(b), cosh(c). With the help of these numbers we can calculate the lengths of the sides of Δ0, but also the lengths of the sides of the triangles Δ1, Δ2, Δ3. For example, the sidelengths of Δ1 are a,i − b,i − c.

Denote the measures of the (inner) angles of Δ0 by ∶= (∠+BAC), ∶= (∠+CBA), ∶= (∠+ACB).

We can calculate cosh( ), cosh( ), cosh( ) from ℭ and get for example, d c c − c c cosh( )=− 23 =− 31 12 23 11 . d d 22 33 c c c2 c c c2 11 22 − 12 33 11 − 31 √ √ t t Of course, when we know , and , we also know the measures of the angles of the other triangles Δk, k =1, 2, 3.

1.4.6. Some useful trigonometric formulae.

2 c c c c 2 2 (" " " ( 31 12 − 23 11)) cosh(b) cosh(c) − cosh(a) cosh2( ) = A B C = (" 2" 2(c 2 − c c ))(" 2" 2(c 2 − c c )) sinh(b) sinh(c) A B 12 11 22 A C 13 11 33 0 1 cosh(i − ) cosh(i − ) − cosh(i − ) 2 cosh2(a) = cosh2(i − a)= sinh(i − ) sinh(i − ) 0 1 cosh( ) cosh( ) + cosh( ) 2 = sinh( ) sinh( ) 0 1 cosh( ) cosh( ) + cosh( ) − sinh( ) sinh( ) 2 cosh2(a) = 1+ sinh( ) sinh( ) 0 1 cosh( ) + cosh( + ) 2 = 1+ sinh( ) sinh( ) 0 1 2 2 cosh() cosh(− ) 1 = 1+ with = ( + + ) sinh( ) sinh( ) 2 0 1 2 sinh2( ) cosh2( )−1 1 cosh(b) cosh(c) − cosh(a) = = −1 2 2 2 sinh(b) sinh(c) sinh (a) sinh (a) sinh (a) H0 1 I −2 cosh(a) cosh(b) cosh(c)+ cosh2(a) + cosh2(b) + cosh2(c)−1 = sinh2(a) sinh2(b) sinh2(c) sinh2( ) sinh2( ) = = sinh2(b) sinh2(c) cosh(b) cosh(c) − cosh(a) cosh(b) cosh(c) − cosh(a) sinh2( ) = −1 +1 sinh(b) sinh(c) sinh(b) sinh(c) 0 1 0 1 (cosh(b + c) − cosh(a))(cosh(b − c) − cosh(a)) = sinh2(b) sinh2(c) ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 9

4 sinh(s) sinh(s − a) sinh(s − b) sinh(s − c) 1 =− with s = (a + b + c) sinh2(b) sinh2(c) 2

1.4.7. The distance between two points which are given by barycentric coordinates. A point S =[s1∶s2∶s3]Δ is isotropic precisely when (s1,s2,s3) [ℭ](s1,s2,s3)=0. Given two diﬀerent anisotropic points P =[p1 ∶ p2 ∶ p3]Δ and Q =[p1 ∶ p2 ∶ p3]Δ, then 2 ((p ,p ,p ) [ℭ](q , q , q )) cosh2(d(P,Q)) = 1 2 3 1 2 3 ((p1,p2,p3) [ℭ] (p1,p2,p3))((q1, q2, q3) [ℭ](q1, q2, q3)) ℭ ℭ 2 (p1,p2,p3) [ ](p1,p2,p3) 2 (q1, q2, q3) [ ](q1, q2, q3) "P = , "Q = . (p1,p2,p3) [ℭ](p1,p2,p3) (q1, q2, q3) [ℭ](q1, q2, q3) The three numbers cosh2(d(P,Q)), " 2, " 2 determine d(P,Q). ð P Q ð ð ð

1.4.8. The dual of Δ. Put A¨ ∶= a,B¨ ∶= b,C¨ ∶= c. The triple Δ¨ = A¨B¨C¨ is called the dual of Δ. The representation of A¨ by barycentric coordinates with respect to Δ is ◦ ◦ −1 ◦ ◦ ◦ ◦ −1 ◦ ◦ ◦ ◦ −1 ◦ ◦ [(B × C ) [S ] (B × C ) ∶ (B × C ) [S ] (C × A ) ∶ (B × C ) [S ] (A × B )]Δ, with S = diag(, 1, 1). It can be easily checked (see also 1.4.16) that A¨ d d d c c c2 c c c c c c c c . =[ 11∶ 12∶ 13]Δ =[ 22 33 − 23 ∶ 23 31 − 12 33 ∶ 12 23 − 31 22]Δ In the same way the barycentric coordinates of B¨ and C¨ can be calculated, getting ¨ d d d ¨ d d d B =[ 21∶ 22∶ 23]Δ,C =[ 31∶ 32∶ 33]Δ, and we conclude: c c c c c c c c c A =[ 11∶ 12∶ 13]Δ¨ , B =[ 21∶ 22∶ 23]Δ¨ , C =[ 31∶ 32∶ 33]Δ¨ .

¨ ¨ 1.4.9. The dual triangles. Just like Δ, the triple Δ determines four triangles Δ k, k = ¨ 0, 1, 2, 3. But, in general, triangle Δ k is not the dual of Δk. In fact, the dual of Δk is Δk ∶=  − P . P inner point of Δk ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨ We put a ∶= B ∨C , b ∶= C ∨A , c ∶=⋃ A ∨B and denote by a , b , c the lengths of the sides and by ¨, ¨, ¨ the measures of the angles of triangle Δ0. Then (see 1.3.4), a¨ = i − , b¨ = i − , c¨ = i − , ¨ = i − a, ¨ = i − b, ¨ = i − c .

1.4.10. Reﬂections. For each anisotropic point M = [m1∶m2∶m3]Δ we deﬁne a mapping → M ∶  − abs  as follows: The image of an anisotropic point P =[p1∶p2∶p3]Δ under M is the point Q =[q1∶q2∶q3]Δ with

(m1,m2,m3) [ℭ](p1,p2,p3) (q1, q2, q3)=(p1,p2,p3)−2 (m1,m2,m3). (m1,m2,m3) [ℭ] (m1,m2,m3) The points P,M,Q are obviously collinear, and it can be easily checked that P and Q have the same distance from M. In particular, Q is also an anisotropic point, lying together with P in the same connected component of  − abs. It can also be verified that M (Q) = P , and that P = Q only if either P = M or d(P , M)= i. 2 We extend M to a mapping which is continuous on : If P is the only isotropic point on the (isotropic) line P ∨ M, then M (P ) = P ; if the line P ∨ M contains still another isotropic point Q besides P , then M (P )= Q. → This mapping M ∶   is an involution. We call it the reflection in the point M and call Q the mirror image of P under M . It is not difficult to show that this reflection preserves 10 MANFRED EVERS the distance between points and the (angle) distance between lines.

Remark: A reflection M in an anisotropic point M can also be interpreted as a reflection l in the line l = M . If a nonempty subset  of  is invariant under the reflection in an anisotropic point M, then M is called a symmetry point of ; theline M is a symmetry axis of . Aspecialcase: If P and Q are two different anisotropic points, the points P + Q and P − Q are symmetry points of {P,Q} precisely when they are (proper) midpoints.

1.4.11. The pedals and antipedals of a point. We calculate the coordinates of the pedals of a point P =[p1 ∶ p2 ∶ p3]Δ on the sidelines of Δ: ¨ d d d d A[P ] ∶= ped(P,a)=(P ∨ C ) ∧ a =[0∶ 11p2 − 12p1 ∶ 11p3 − 31p1]Δ, d d d d B[P ] =[ 22p1 − 12p2 ∶ 0 ∶ 22p3 − 23p2]Δ, d d d d C[P ] =[ 33p1 − 31p3 ∶ 33p2 − 23p3 ∶ 0]Δ. The antipedal points of P are d p − d p d p − d p A[P ] ∶= perp(B ∨ P,B) ∧ perp(C ∨ P,C)=[−1∶ 22 1 12 2 ∶ 33 1 31 3 ] , d d d d Δ 11p2 − 12p1 11p3 − 31p1 d p − d p d p − d p B[P ] =[ 11 2 12 1 ∶−1∶ 33 2 23 3 ] , d d d d Δ 22p1 − 12p2 22p3 − 23p2 d p − d p d p − d p C[P ] =[ 11 3 31 1 ∶ 22 3 23 2 ∶ −1] , d d d d Δ 33p1 − 31p3 33p2 − 23p3

1.4.12. The cevian and the anticevian triple of a point. If P = [p1∶p2∶p3]Δ is a point diﬀerent from A,B,C, then the lines P ∨ A, P ∨ B, P ∨ C are called the cevian lines of P with respect to Δ. The cevian lines meet the sidelines a, b, c at the points AP = [0∶p2∶p3]Δ,BP = [p1∶0∶p3]Δ,CP = [p1∶p2∶0]Δ, respectively. These points are called the cevian points or the traces, and the triple AP BP CP is called the cevian triple of P with respect to Δ. The anticevian triple AP BP CP consists of the harmonic associates P P P A = [−p1∶p2∶p3]Δ, B =[p1∶− p2∶p3]Δ, C =[p1∶p2∶− p3]Δ of P .

1.4.13. Tripolar and tripole. Given a point P =[p1∶p2∶p3]Δ, thenthe point [0∶−p2∶p3]Δ is the harmonic conjugate of AP with respect to {B,C}. Correspondingly, the harmonic conjugates of the traces of P on the other sidelines are [−p1∶0∶p3]Δ and [p1∶− p2∶0]Δ. Thesethreeharmonicconjugatesarecollinear; theequationoftheline l is p2p3x1+p3p1x2+ p1p2x3 =0. This line is called the tripolar line or the tripolar of P and we denote it by P . P is the tripole of l and we write P = l .

1.4.14. Isoconjugation. Let P =[p1∶p2∶p3]Δ be a point not on a sideline of Δ and let the point Q =[q1∶q2∶q3]Δ be not a vertex of Δ, then the point R =[p1q2q3∶p2q3q1∶p3q1q2]Δ is called the P -isoconjugate of Q with respect to Δ. Special cases: If P = G = [1∶1∶1]Δ (the centroid of Δ0), P -isoconjugation is called d d d isotomic conjugation, and for P = K =[ 11∶ 22∶ 33]Δ (the symmedian of Δ0) it is called isogonal conjugation. The isotomic (resp. isogonal) conjugate of a point P with respect to Δ agrees with the isogonal (resp. isotomic) conjugate of P with respect to Δ¨. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 11

1.4.15. The area of a triangle. For a triangle  2) we deﬁne its area (also called its excess) by area( )= + + − i, where , , are the measures of the inner angles of  . This function is additive: If we dissect  in ﬁnitely many triangles, then the sum of the areas of these parts equals the area of  . Adding up all the areas of the triangles Δk, k =1, 2, 3, 4, we get 2i for the area of the whole plane .

M m M 1.4.16. Conics. Let = ( ij)i,j=1,2,3 be a symmetric matrix, then we denote by ( ) the quadratic form q(x ,x ,x )= m x2 + m x2 + m x2 +2m x x +2m x x +2m x x . 1 2 3 11 1 22 2 33 3 23 2 3 31 3 1 12 1 2 M If is indeﬁnite, the set of points [x1∶x2∶x3]Δ which satisfy the quadratic equation M q(x1,x2,x3)=0 is a real conic which we denote by ( ). This conic is called nondegenerate if det(M) ≠ 0. M The polarof a point P =[p1∶p2∶p3]Δ with respect to ( ) is the line with the equation (p1,p2,p3) [M] (x1,x2,x3)=0, and the pole of the line l∶ l1x1 + l2x2 + l3x3 = 0 with M M# respect to ( ) is the point P =[p1∶p2∶p3]Δ with (p1,p2,p3)=(l1, l2, l3) , where m m m2 m m m m m m m m 22 33 − 23 23 31 − 12 33 12 23 − 31 22 M# m m m m m m m2 m m m m = 23 31 − 12 33 33 11 − 31 31 12 − 23 11 m m − m m m m − m m m m − m2 ⎛ 12 23 31 22 31 12 23 11 11 22 12 ⎞ is the adjoint of⎜M. ⎟ ⎜ ⎟ A special case:⎝ In the hyperbolic plane (< 0), the conic (ℭ) is the absolute conic⎠ which consists of all the isotropic points. In the elliptic plane, ℭ is deﬁnite and no real isotropic points exist.

A point P is a symmetry point of a nondegenerate real conic (M), M ≠ ℭ, precisely when the polar of P equals P . Proof : Take any line through P that meets (M) in two points Q and R. This line meets the polar of P at the harmonic conjugate S of P with respect to Q and R. Precisely when P is a proper midpoint of {Q, R}, the point S is also a proper midpoint of {Q, R} and lies on P . □

If M is a diagonal matrix, then the polar lines of A,B,C with respect to (M) are the lines B ∨ C, C ∨ A, A ∨ B, respectively. If M is not diagonal, then the poles of B ∨ C, C ∨ A, A ∨ B with respect to (M) form a triple, perspective to Δ at perspector 1 1 1 [ ∶ ∶ ] . m m m m m m m m m m m m Δ 11 23 − 31 12 22 31 − 12 23 33 12 − 23 31 This perspector is called the perspector of (M) with respect to Δ and, if (M) is a real conic, the perspector of (M) with respect to Δ. Let (M) be a nondegenerate real conic. The set {p p is a tangent of (M)} is the conic (ℭM♯ℭ); we call it the dual of (M). Both conics, (M) and its dual, share the same symmetry points and symmetry axes. ð There are other conics named "dual of (M)". One, for example, is (M♯). To avoid a name collision, we call this one the adjoint conic.

2) Implicitly it is always assumed that the vertices and the sidelines of a triangle are anisotropic. 12 MANFRED EVERS

1.4.17. Circles. We assume that (M), M ≠ ℭ, is a nondegeneratereal conic. Then (M) is a circle if there exists a line l consisting of symmetry points. In this case, the point l is also a symmetry point of (M) and is called the center of the circle. Remarks: ∙ A circle may have (up to two) isotropic points. ∙ In the elliptic plane, the center of a circle always lies inside the circle. In the hyperbolic plane, the center of a circle lies inside the circle precisely when this center also lies inside the absolute conic. If P and Q are two distinct anisotropic points on a circle with center M, then d(P , M)= d(Q, M), because: Let R be the intersection of the lines P ∨ Q and M, then the triple PMQ is mapped onto the triple QMP by the (distance preserving) reflection R. So we can define the radius of a circle as the common distance between its center and any of its anisotropic points. Given two anisotropic points M and P , then there exists a unique nondegenerate circle C (M, P ) with center M through the point P precisely when 0 ≺ d(M, P ) ≺ i. 2 Besides the nondegenerate circles, one can regard the following degenerate conics as circles: It is common to look at a double line l (a line with multiplicity 2) as a circle with center l and radius i. A double point can be considered as a circle with radius 0 around 2 this point. And also a degenerate conic consisting of two different isotropic lines can be seen as a circle; its center is the meet of these lines and its radius is 0.

2. SPECIAL TRIANGLE CENTERS, CENTRAL LINES, CONICS AND CUBICS

2.1. Triangle centers and central triangles based on orthogonality.

2.1.1. The common orthocenter of Δ and Δ¨. The perspector of (ℭ) is the common orthocenter H of Δ and Δ¨, 1 1 1 H =[ ∶ ∶ ] =[d d ∶ d d ∶ d d ] . c c c c c c c c c c c c Δ 31 12 12 23 23 31 Δ 31 12 − 23 11 12 23 − 31 22 23 31 − 12 33 Since A∨H is orthogonal to B∨C, B∨H orthogonal to C∨A and C∨H orthogonal to A∨B, the points A,B,C,H together with the lines a,b,c,A∨H, B∨H,C∨H form an orthocentric system; each of the points is the orthocenter of the other three. The same applies to A¨,B¨,C¨, H, together with the lines through any two of them. If we combine these two systems and add the points a∧a¨, b∧b¨, c∧c¨ and the line H, we get a system which - if Δ0 has neither a right angle nor a right side - consists of 10 points and 10 lines such that the dual of each of its points is one of its lines and the dual of each of its lines is one of its points, and each pointis incidentwith 3 lines and each line incident with 3 points.

¨ ¨ 2.1.2. Another triple perspective to Δ and Δ at perspector H. Deﬁne Ab ∶= b ∧ a , ¨ Ba ∶= a ∧ b , and the points Ac,Bc,Ca,Cb accordingly. We calculate the coordinates of c c c c ̃ ̃ ̃ these points: Ab =[ 31 ∶0∶− 11]Δ,Ba = [0∶ 23 ∶− 22]Δ, etc. The point-triple ABC, d c c Ã ∶= (A ∨ B )∧(A ∨ C )=[ 11 31 12 ∶ d ∶ d ] , b a c a c c 12 13 Δ 11 23 d c c B̃ ∶= (B ∨ C )∧(B ∨ A )=[d ∶ 22 12 23 ∶ d ] , c b a b 21 c c 23 Δ 22 31 d c c C̃ ∶= (C ∨ A )∧(C ∨ B )=[d ∶ d ∶ 33 23 31 ] , a c b c 31 32 c c Δ 33 12 ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 13 is perspective to Δ and Δ¨ at perspector H, cf. Vigara [27] section 8.2.

FIGURE 1. The triples Δ, Δ¨ and ÃB̃C̃. The light blue circle is the absolute conic in the hyperbolic plane. All ﬁgures were created with the software program GeoGebra [31].

The points Ba,Ca,Cb,Ab,Ac,Bc lie on the conic 3 c c j+1,j+1 j+2,j+2 {[x ∶x ∶x ] c 2 x 2 + (c + )x x =0} (indices mod 3) 1 2 3 Δ j,j j j+1,j+2 c j+1 j+2 j=1 j+1,j+2 with perspectorÉ[c (1+ c 2 )(1 + c 2 )−2c c (1+ c 2 ) ∶ ⋯ ∶ ⋯] . ð 23 12 31 12 31 23 Δ ̃ ̃ ̃ ̃ ̃ ̃ The three points Aa ∶= (B ∨ C) ∧ a,Bb ∶= (C ∨ A) ∧ b, Cc ∶= (A ∨ B) ∧ c lie on the line c c c c c c {[x1∶x2∶x3]Δ 11 23x1 + 22 31x2 + 33 12x3 =0.} By a routine calculation we can determine the coordinates of the orthocenter of the triple Δ̃ = ÃB̃C̃. The expression forð it is rather complicated, so we do not present it here. But it might be worth mentioning that the euclidean limit ( → ±∞) of this point is the 3) de Longchamps point X20 of Δ. ̃ ¨ ̃ + d d d d d d The dual Δ of Δ is perspective to Δ. The perspector is O =[ 11 23∶ 22 31∶ 33 12]Δ. ̃ ¨ The euclidean limit of Δ is the mirror image of Δ in the circumcenter of Δ0. The six points (B∨A¨)∧(A∨C¨), (C∨A¨)∧(A∨B¨), (C∨B¨)∧(B∨A¨), (A∨B¨)∧(B∨A¨), (A∨C¨)∧(C∨B¨), (B∨C¨)∧(C∨A¨) lie on a conic. The perspector of this conic with d d d respect to Δ is the point [ 11 ∶ 22 ∶ 33 ] , it is d d d d d d d d d d d d Δ 11 23 − 31 12 22 31 − 12 23 33 12 − 23 31

3) The designation of the triangle centers corresponds to ETC [16]. 14 MANFRED EVERS the isogonal conjugate of the deLongchamps point L, which will be introduced in the next subsection. The center of the euclidean limit of the conic is the circumcenter of Δ0. 2.1.3. The double triangle. We assume that the orthocenter H of Δ is not a vertex. Wild- berger [28] showed that the antipedals A[H],B[H],C[H] of H are the harmonic associates of the point + c c c d d d d d d d d d d d d G =[ 23∶ 31∶ 12]Δ =[ 11 23 − 12 31 ∶ 22 31 − 12 23 ∶ 33 12 − 23 31]Δ. It can be easily checked that G+ is the tripole of the dual line of H. The points A[H],B[H], C[H] are the vertices of a triangle which contains G+. In [28], this triangle is called the double triangle of Δ, and G+ is called the double point. Furthermore,Wildbergerprovedthat the points A,B,C arepropermidpointsof {B[H],C[H]}, {C[H],A[H]}, {A[H],B[H]}, respectively. Thus, the point H is the center of a circle through the points A[H],B[H],C[H]. The perspector of this circle with respect to Δ is H, the perspector with respect to the triple A[H]B[H]C[H] (the Lemoine point of the double d c 2 d c 2 d c 2 triangle) is the point [ 23 23∶ 31 31∶ 12 12]Δ. The point O+ is identical with a pseudo-circumcenter, introduced in [27] as the meet of ¨ ¨ ¨ perpendicular pseudo-bisectors AG+ ∨ A ,BG+ ∨ B ,CG+ ∨ C . And it is also identical with the basecenter introduced in [28]. The triple Δ[H] = A[H]B[H]C[H] is perspective to Δ¨ at perspector d d d d d d d d d d d d L =[ 11 23 + 31 12 ∶ 22 31 + 12 23 ∶ 33 12 + 23 31]Δ. This point L we will call the deLongchamps point of Δ.

2.1.4. The orthic triangle. We assume again that the orthocenter H of Δ is not a vertex. The orthic triangle has the vertices A[H],B[H],C[H], it contains H, anditisthedualofthe double triangle of Δ¨. The point H is the incenter and the points A,B,C are the excenters of the orthic triangle. The pedals of H on the sidelines of Δ[H] are the traces with respect to Δ[H] of the point d d d 2 d d d d d 2 d d d 2 d 2 d d d d d 2 d [ 31 12( 31 22 −2 31 12 23 + 12 33)(2 11 23 + 31 22 −4 31 12 23 + 12 33) ∶ ⋯ ∶ ⋯]Δ.

This point has euclidean limit X52.

+ [O+] d c d c d c [O+] 2.1.5. The antipedal triple of O . The points A = [− 11∕ 23, 22∕ 31, 33∕ 12]Δ, B , + + C[O ] form a triple Δ[O ], which is perspective to Δ at the isogonal conjugate of G+ and perspective to the orthic triple Δ[H] at d (d (d d − d 2 )+ d d 2 + d d 2 −2d d d d [ 23 11 22 33 23 22 31 33 12 22 33 31 12 ∶ ⋯ ∶ ⋯] . d c Δ 23 23 2.1.6. The star of a point with respect to the triple Δ. Deﬁne the mapping → ⋆ = ⋆Δ ∶  −{A,B,C}  by ⋆ D P = ⋆([p1∶p2∶p3]Δ)=[q1∶q2∶q3]Δ , (q1, q2, q3)=(p2p3,p3p1,p1p2) . P ⋆ is the dual of the tripolar of P , so the preimage of a point Q ∈  −{A¨,B¨,C¨} under ⋆ is the point Q . If Δ0 is not right-angled, then we get for the special case P = H: ⋆ d d d d d d d d d d d d H =[ 11 23 +2 31 12 ∶ 22 31 +2 12 23 ∶ 33 12 +2 23 31]Δ.

The orthic axis and its dual are deﬁned even if Δ0 has one or two right angles, so we extend ⋆ in this case. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 15

The point H⋆ is introduced in [28] under the name orthostar; we adopt the name.

FIGURE 2. Trianglecentersontheorthoaxis. Thethreeconicsaretheab- solute conic (light blue), the circumcircle of Δ[H] with center H (brown) and the bicevian conic of the points H and G+ (black).

2.1.7. Centers on the orthoaxis. Wildberger [28] showed that the line H ∨ H⋆, d d d 2 d d 2 d d d 2 d d 2 d d d 2 d d 2 23( 22 31 − 33 12)x1 + 31( 33 12 − 11 23)x2 + 12( 11 23 − 22 31)x3 =0, which he named orthoaxis, is incident with the points L,O+,G+. Furthermore, he proved that L,O+,H,H⋆ as well as L,O+,G+, H form harmonic ranges. The euclidean limits of + + ⋆ the points L,O ,G ,H,H are L = X20,O = X3,G = X2, H = X4, the Euler inﬁnity point X30, respectively. The orthoaxis is a symmetry axis of the bicevian conic throughthe traces of H and G+. The points H +O+ and H −O+ are, in additionto the dual point of the orthoaxis,symmetry points of this conic precisely when these two points are proper midpoints of H and O+.

Proof of the last two statements: For two points P = [p1∶p2∶p3]Δ and Q = [q1∶q2∶q3]Δ, not on any of the lines of Δ0, the conic which passes through the traces of P and Q has the matrix (cf. [30]) M m m 2 m 1 1 =( ij)i,j=1,2,3 with ii =− , ij = + for i, j =1, 2, 3,i ≠ j. piqi piqi pjqj (Remark: For P = Q, this is the matrix of an inconic with perspector P .) 16 MANFRED EVERS

It can now be checked that for the special case P = H and Q = G+, the pole of the orthoaxis with respect to the bicevian conic is the point c d d 2 d 2 d c d 2 d d d 2 c d d 2 d d 2 [ 23( 22 31 − 12 33), 31( 12 33 − 11 23), 12( 11 23 − 22 31)]Δ. But this is also the dual point of the orthoaxis. We show that H ± O+ is a point on the polar of H ∓ O+ with respect to the bicevian conic (M) of H and G+ by proving the correctness of the equation ◦ +◦ ◦ +◦ (H + O ) [M] (H − O )=0. We transform this equation equivalently: ◦ +◦ ◦ +◦ (H + O ) [M](H − O )=0 ◦ ◦ +◦ +◦ ⇔ H [M] H − O [M] O =0 + + + + ⇔ (h [M] h)(o [ℭ] o )−(h [ℭ] h)(o [M] o )=0 d d d d d d + d d d d d d with h =( 31 12 , 12 23 , 23 31) and o =( 11 23 , 22 31 , 33 12). The last equation holds, as can be checked with the help of a CAS. Now it follows: Neces- sary and suﬃcient for H + O+ and H − O+ to be symmetry points of (M) is that their distance is i. But this holds precisely when H + O+ and H − O+ are proper midpoints 2 of {H,O+}. □

The orthoaxes of the triples ABC,AHB,BHC,CHA meet at the point + d d d d d d d d d d d d N ∶= [ 11 23 −2 12 31, 22 31 −2 12 23, 33 12 −2 23 31]Δ. The coordinates of H,N+,O+, H⋆ indicate that these four points form (in this order) a + harmonic range. The euclidean limit of N is N = X5.

The orthic triple of the orthic triple Δ[H] is perspective to Δ at perspector d d d d 2 d d 2 d d 2 d d d P =[ 31 12( 11 23 − 22 31 − 33 12 +2 12 23 31) ∶ ⋯ ∶ ⋯]Δ.

This is a point on the orthoaxis with euclidean limit X24. The deLongchamps point, the double point, the pseudo-circumcenter and the orthostar of Δ¨ are O+, H⋆,L and G+, respectively.

In case of "A = "B = "C and "A¨ = "B¨ = "C¨ , the coordinates of the above given points on the orthoaxis can be written

+ G = [sinh( ) cosh( ) + cosh( ) cosh( ) ∶ ⋯ ∶ ⋯]Δ + O = [sinh(2 ) ∶ ⋯ ∶ ⋯]Δ, H = [tanh( ) ∶ ⋯ ∶ ⋯]Δ, + N = [sinh( ) cosh( ) +2cosh( ) cosh( ) ∶ ⋯ ∶ ⋯]Δ, ⋆ H = [sinh( ) cosh( ) −2cosh( ) cosh( ) ∶ ⋯ ∶ ⋯]Δ, L = [sinh( ) cosh( ) − cosh( ) cosh( ) ∶ ⋯ ∶ ⋯]Δ, 2 2 2 P = [tanh( ) cosh ( ) − cosh ( ) − cosh ( )− 2cosh( ) cosh( ) cosh( ) ∶ ⋯ ∶ ⋯]Δ. 2.1.8. The Taylor conic. By projecting the pedals A[H],B[H],C[H], lying on a resp. b resp. c, onto the other sidelines of Δ, we get altogether six points ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 17

B = [d 2 ∶0∶d d − d d ] , C = [d 2 ∶d d − d d ∶0] , C , A , [A[H]] 12 12 23 22 31 Δ [A[H]] 13 31 23 33 12 Δ [B[H]] [B[H]] A ,B which all lie on a conic with the equation [C[H]] [C[H]]

3 d d d d 2 d d ( i,i i+1,i+2 − i+2,i i,i+1) + i+2,i i,i+1 d x 2 + x x =0 (indices mod 3). i+1,i+2 i d d − d d i+1 i+2 i=1 i,i i+1,i+2 i+2,i i,i+1 É We like to call this conic Taylor conic, since its euclidean limit is the Taylor circle.

Let HA, HB, HC be the orthocenters of AB[H]C[H],BC[H]A[H],CA[H]B[H], respectively. The triple HAHBHC is perspective to A[H]B[H]C[H]; the perspector is d d d d 2 d d 2 d d d d [ 11( 23( 22 31 + 33 12)− 22 33 31 12) ∶ ⋯ ∶ ⋯]Δ.

The euclidean limit of this point is X389, the center of the Taylor circle. But, in general, this point is not a symmetry point of the Taylor conic.

2.1.9. Orthotransversaland orthocorrespondentof a point. For any point P =[p1∶p2∶p3]Δ which is not a vertex of Δ, the three points perp(A ∨ P , P ) ∧ a, perp(B ∨ P , P ) ∧ b and perp(C ∨ P , P ) ∧ c are collinear on the line 1 x =0 (indices mod 3). p (d p − d p − d p )+ d p p i i=1,2,3 i i+1,i+2 i i+2,i i+1 i,i+1 i+2 i,i i+1 i+2 É The line is called the orthotransversal (line) of P , and its tripole is called the orthocorrespondent of P . As a special case, the orthocorrespondentof H is G+. The points perp(A ∨ P , P ) ∧ a¨, perp(B ∨ P , P ) ∧ b¨ and perp(C ∨ P , P ) ∧ c¨ are also collinear; they lie on the dual of P . We dualize these statements: The three lines perp(a ∧ P , P ) ∨ A, perp(b ∧ P , P ) ∨ B and perp(c ∧ P , P ) ∨ C p meet at the point Q =[ 1 ∶ ⋯ ∶ ⋯] . c c c c Δ p1( 11p1 − 12p2 − 31p3)+ 23p2p3 A special case: If P = G+, then Q = H.

If P is not a vertex of Δ¨, then the points par(a, P ) ∧ a, par(b, P ) ∧ b, par(c, P ) ∧ c lie on the line P . 2.2. Center functions / triangle centers and their mates. ℝ6 → ℝ Given a triangle center T of Δ0, then there exists a homogenous function f∶ such that ∙ f(x1,x2,x3,x4,x5,x6)= f(x1,x3,x2,x4,x6,x5) and d d d d d d d d d d d d d d d d d d ∙ T =[f( 11, 22, 33, 23, 31, 12)∶f( 22, 33, 11, 31, 12, 23)∶f( 33, 11, 22, 12, 23, 31)]Δ. A function satisfying these conditions is called a center function of T with respect to the D triangle Δ0 and the matrix . We give two examples: Center functions of H and L are fH (x1,x2,x3,x4,x5,x6)= x5x6 and fL(x1,x2,x3,x4,x5,x6)= x1x4 + x5x6. Obviously, by knowing a center function of a triangle center of Δ0, one also knows the barycentric coordinates of that center. The triangle center T = T0 of Δ0 is accompanied by its three mates T1,T2,T3; these are the correspondingtrianglecenters of Δ1, Δ2, Δ3, respectively. Their representationby barycentric coordinates are 18 MANFRED EVERS

d d d d d d T1 = [−f( 11, 22, 33, 23,− 31, − 12) d d d d d d d d d d d d ∶f( 22, 33, 11, − 31, − 12, 23)∶f( 33, 11, 22, − 12, 23, − 31)]Δ, d d d d d d T2 =[f( 11, 22, 33, − 23, 31, − 12) d d d d d d d d d d d d ∶− f( 22, 33, 11, 31, − 12, − 23)∶f( 33, 11, 22, − 12, 23, − 31)]Δ, d d d d d d T3 =[f( 11, 22, 33, − 23, − 31, 12) d d d d d d d d d d d d ∶f( 22, 33, 11, − 31, 12, − 23)∶− f( 33, 11, 22, 12, − 23, − 31)]Δ. All triangle centers in the last subsection are absolute triangle centers, their mates do not diﬀer from the main center. Remark: Without giving a formal introduction, we also use (and used already) center func- ℭ tions which depend on the matrix , or on the sidelengths of Δ0, or on its inner angles.

2.3. Circumcircles, incircles and related triangle centers.

FIGURE 3. The four pairs of twin circumcircles (green, purple, brown, orange) of ABC, their centers and the dual lines of these centers. The blue circle is the absolute conic.

2.3.1. Twin circles, circumcircles and incircles. The points G = G0 = [1, 1, 1]Δ, G1 = [−1, 1, 1]Δ,G2 =[1, −1, 1]Δ, G3 =[1, 1, −1]Δ are the centroids of triangle Δ0, Δ1, Δ2, Δ3, ⋆ respectively. The point Ok = Gk is called circumcenter of Δk (k =0, ⋯, 3),cf. [29]. Inthe elliptic plane, there always exists a circle around Ok, passing through the vertices A,B,C. The situation is different in the hyperbolic plane. Unless "A = "B = "C , there does not exist any circle which passes through all three vertices A,B,C. If for example "A = "B ≠ "C , then a circle with center Ok through A will also pass through B, but it differs from a circle centered at Ok passing through C. Wildberger and Alkhaldi discovered a special relation ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 19 between these two circles: They form a couple of twin circles. Wildberger and Alkhaldi [29] proved that, given any circle  with center P and radius r ∈ ℂ in the hyperbolic plane, there exists a uniquely determined twin circle of . This is a circle with center P and radius r¨, satisfying the equation cosh2(r¨) = −cosh2(r). Thus, in the hyperbolic plane, one can always find exactly one pair of twin circles with center Ok such that their union contains the vertices A,B,C. The coordinates of the circumcenters of Δ are d d d d d d d d d O = O0 =[ 11+ 12+ 31, 12+ 22+ 23, 31+ 23+ 33]Δ, d d d d d d d d d O1 =[− 11+ 12+ 31, − 12+ 22+ 23, − 31+ 23+ 33]Δ, etc.

For the distances d(O,A) and d(O1,A) we get: 2 c d d d d d d cosh (d(O,A)) = 1∕( 11( 11+ 22+ 33 + 2( 23+ 31+ 12))), 2 c d d d d d d cosh (d(O1,A)) = 1∕( 11( 11+ 22+ 33 + 2( 23− 31− 12))). The circumcenters of Δ¨ are the incenters of Δ. To be more precise, the circumcenter of k Δ is the incenter Ik of Δk, k =0, ⋯ , 3. We calculate the coordinates: d d d d d d I = I0 =[ 11 , 22 , 33 ]Δ, I1 = [− 11 , 22 , 33 ]Δ, ⋯ . In the elliptic plane,√ the√ incenters√ are the centers of√ incircles√ of Δ, and√ this is also true in the hyperbolic planeð ð if "ðA¨ =ð"B¨ ð= "Cð¨ . But in general,ð twinð ð circlesð withð ð center Ik are needed to touch all sidelines a, b, c. We calculate the distances d(I,a)= d(I, ped(I,a)) = ¨ i − d(I,A ) and d(I1,a)∶ 2 d c d c d d cosh (d(I,a)) = sgn( 11)∕ ( ii ii +2 i,i+1 ii i+1,i+1 ) , i=1,2,3 É t 2 d c d c d d c d d c d d cosh (d(I1,a)) = sgn( 11)∕ ( iið iið )+2( 23 ð 22 33 − ð31 33 11 − 12 11 22 ) . i=1,2,3 É √ √ √ ¨ The triple O1O2O3 is perspective toð Δð at perspectorð O0, theð tripleð I1I2Ið3 is perspectiveð ð to Δ at perspector I0. c c c c We now assume "A = "B = "C and put ∶= 11 = 22 = 33 ∈ {−1, 1} , then : cc = cosh(a), cc +1 = 2cosh2( a ), cc −1 = 2sinh2( a ), etc. 23 23 2 23 2 and c c c c c c c c c c c c c c c c c c O0 = [( 23− )( 23− 31− 12+ )∶( 31− )(− 23+ 31− 12+ )∶( 12− )(− 23− 31+ 12+ )]Δ = [(sinh2( a ))(− sinh2( a ) + sinh2( b ) + sinh2( c )) ∶ ⋯ ∶ ⋯] , 2 2 2 2 Δ c c c c c c c c c c c c c c c c c c O1 = [( 23− )( 23+ 31+ 12+ )∶( 31+ )(− 23− 31+ 12+ )∶( 12+ )(− 23+ 31− 12+ )]Δ, = [(sinh2( a ))(sinh2( a )+ cosh2( b )+cosh2( c )) ∶ (cosh2( b ))(− sinh2( a )−cosh2( b ) 2 2 2 2 2 2 2 +cosh2( c )) ∶ (cosh2( c ))(− sinh2( a )+ cosh2( b )−cosh2( c ))] , ⋯ . 2 2 2 2 2 Δ

The four points O0,O1,O2,O3 form an orthocentric system.

The triple O1O2O3 is perspective to Δ at the isogonal conjugate of O = O0, and as a consequence of this, the triples O0O3O2, O3O0O1, O2O1O0 are perspective to Δ at the isogonal conjugates of O1,O2,O3, respectively.

Let Ck denote the circumcircle with center Ok, k = 0, ⋯ , 3. All inner points of Δ0 lie inside C0, which is not true for any other circumcircle Ck, k ≠ 0. Therefore, we call C0 the main circumcircle of Δ0, while the others will be regarded as its mates. 20 MANFRED EVERS

̃ The perspector of Ck is called the Lemoine point Kk of Δk ∶ K̃ =[c − c ∶ c − c ∶ c − c] = [sinh2( a ) ∶ sinh2( b ) ∶ sinh2( c )] , 0 23 31 12 Δ 2 2 2 Δ K̃ =[c − c ∶ c + c ∶ c + c] = [−sinh2( a ) ∶ cosh2( b ) ∶ cosh2( c )] , .... 1 23 31 12 Δ 2 2 2 Δ ̃ ̃ The isogonal conjugate of Kk is Gk : G̃ =[c + c ∶ c + c ∶ c + c] = [cosh2( a ) ∶ cosh2( b ) ∶ cosh2( c )] , 0 23 31 12 Δ 2 2 2 Δ G̃ =[c + c ∶ c − c ∶ c − c] = [−cosh2( a ) ∶ sinh2( b ) ∶ sinh2( c )] , .... 1 23 31 12 Δ 2 2 2 Δ ̃ ̃ ̃ ̃ ̃ ̃ ̃ The triple K1K2K3 is perspective to Δ at G0 and, as a consequence, the triple G1G2G3 is ̃ ̃ ̃ ̃ ̃ perspective to Δ at K0. The triple G2G3G0 is perspective to Δ at K1, etc. The four lines ̃ ̃ Kj ∨ Gj, 0 ≤ j ≤ 3, meet at the point G = [1, 1, 1]Δ. ̃ ̃ ̃ ̃ The euclidean limits of K = K0 and G = G0 are the symmedian K = X6 and the centroid G = X2 of Δ0, respectively. The euclidean limit of the circle C1 is the union of two lines, one is the sideline a = B ∨ C, the other its parallel through A.

Remark: In euclidean geometry, the perspector of the circumcircle C0 coincides with isogonal conjugate of G0. This is not the case in elliptic and in hyperbolic geometry; so we have to find different names for the two points. As proposed by Horváth, the perspector of the circumcircle will be called Lemoine point, the isogonal conjugate K of G symmedian. In euclideangeometry, the centroidof the pedal triangle Δ[K],0 of K agrees with K. Neither of the two points K, K̃ has this property in elliptic and in hyperbolic geometry. The Lemoine point K̃ we can get by the following construction, cf. [10] for the euclidean ¨ case: Let lA be the line that connects the point AG with the harmonic conjugate of A with ̃ respect to A,AH , and define the lines lB, lC accordingly. The lines lA, lB, lC meet at K.

If  is any conic and P a point on , then let TP  denote the tangent of  at P . The points A,TBC2 ∧ TCC3,TBC1 ∧ TC C1 are collinear.

Let Q0j bethenontrivialintersectionpointof C0 and Cj, j=1, 2, 3. The triple Q01Q02Q03 is perspective to Δ at G0.

If we assume "A¨ = "B¨ = "C¨ , we get: d d d sgn( 11)= sgn( 22)= sgn( 33) and d d d I = I0 =[ 11 ∶ 22 ∶ 33 ]Δ = [sinh(a) ∶ sinh(b) ∶ sinh(c)]Δ, I = [− d ∶ d ∶ d ] = [− sinh(a) ∶ sinh(b) ∶ sinh(c)] ,... 1 11√ 22√ 33√ Δ Δ I0,I1,I2,I3 form√ an orthocentric√ √ system.

Let ℐk denote the incircle with center Ik, k = 0, ⋯ , 3. The center Ik of ℐk is always a point inside of Δk. We call ℐk the proper incircle of Δk, while the others will be called the excircles of Δk. Caution: The inner points of ℐk can completely lie outside of Δk, cf. Figure 4. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 21

The perspector of ℐk is called the Gergonne point Gek of Δk: d d d d d d d d d Ge0 =[ 22 33 + 23 ∶ 33 11 + 31 ∶ 11 22 + 12]Δ d d d d d d d d d d d d = [(√ 33√ 11 − 31)(√ 11 √ 22 − 12)∶(√ 11√ 22 − 12)( 22 33 − 23) d d d d d d ∶(√ 33√ 11 − 31))(√ √22 33 − 23)]Δ√ √ √ √ cosh( )−1 cosh( )−1 cosh( )−1 =[ √c c√ √∶ c√ c ∶ c c ] , 22 33 sinh(a) 33 11 sinh(b) 11 22 sinh(c) Δ √ d√ d d √d √d d d √d √d Ge1 = [−( 22 33 + 23) ∶ 33 11 − 31 ∶ 11 22 − 12]Δ

We introduce√ the√ Nagel point Na√k of√Δk, k =0, 1, 2√, 3 , by√ d d d d d d Na0 =[ 22 33 − d23 ∶ 33 11 − d31 ∶ 11 22 − d12]Δ d d d d d d d d = [( √33 √11 + d31)( √11 √22 + d12)∶( √ 11 √ 22 + d12)( 22 33 + d23) d d d d ∶(√ 33√ 11 + d31))(√ 22√ 33 + d23)]Δ√, √ √ √ d d d d d d d d d Na1 = [−(√ √22 33 − 23√) ∶ √33 11 + 31 ∶ 11 22 + 12]Δ.

Nak and Gek are√ isotomic√ conjugates√ for 0√≤k≤ 3. The√ triples√Ge1Ge2Ge3 and Na1Na2Na3 are perspective to Δ at Na0 and Ge0, respectively.

The lines I0 ∨ Ge0,I1 ∨ Ge1,I2 ∨ Ge2,I3 ∨ Ge3 concur at the deLongchamps point L, + the lines I0 ∨ Na0,I1 ∨ Na1,I2 ∨ Na2,I3 ∨ Na3 at the point G , and the lines Ge0 ∨ Na0, Ge1 ∨ Na1,Ge2 ∨ Na2,Ge3 ∨ Na3 meet at the isotomic conjugate of the orthocenter H.

2.3.2. Note: From now on, we always assume "A = "B = "C and "A¨ = "B¨ = "C¨ . This is the "classical case". The points A,B,C lie either in the elliptic plane or in a special part of the extended hyperbolic plane.

The classical Cayley-Klein cases in the extended hyperbolic plane:

"A = "B = "C =−i and "A¨ = "B¨ = "C¨ =1 proper hyperbolic / Lobachevsky "A = "B = "C =1 and "A¨ = "B¨ = "C¨ =1 double-hyperbolic / deSitter "A = "B = "C =1 and "A¨ = "B¨ = "C¨ =−i dual-hyperbolic / anti-deSitter

In this case, the following trigonometric formulae apply: cosh(b) cosh(c) − cosh(a) cosh( ) = sinh(b) sinh(c) cosh( ) cosh( ) + cosh( ) 2 sinh() cosh(− ) 1 cosh(a)= =1+ , = area(Δ ). sinh( ) sinh( ) sinh( ) sinh( ) 2 0 sinh( ) sinh( ) sinh( ) = = . sinh(a) sinh(b) sinh(c) The coordinates of I,Ge,Na,O, G,̃ K̃ can now be written as functions of the angles:

I = [sinh( ) ∶ ⋯ ∶ ⋯]Δ,

Ge = [tanh( ) ∶ ⋯ ∶ ⋯] , Na = [coth( ) ∶ ⋯ ∶ ⋯] , 2 Δ 2 Δ O = [sinh( ) cosh( -) ∶ ⋯ ∶ ⋯]Δ, ̃ ̃ K = [sinh( ) sinh( -) ∶ ⋯ ∶ ⋯]Δ, G = [sinh( )∕sinh( -) ∶ ⋯ ∶ ⋯]Δ. 22 MANFRED EVERS

The isotomic conjugate of I is a point on the line Ge ∨ Na. The isogonal conjugate of O is − − the point H = [sinh( )∕cosh( -) ∶ ⋯ ∶ ⋯]Δ. The cevian line A ∨ H is perpendicular to the sideline BG ∨ CG of the medial triangle.

FIGURE 4. An anti- de Sitter triangle together with its circumcircle C0 (red), its incircle ℐ0 (green) and excircles (light-green).

2.3.3. Other triangle centers related to the circumcenters and the incenters. ¨ ∙ The triple I1I2I3 is perspective to Δ at the Bevan point d d d d d d d d d d Be = Be0 =[ 11( 11 22 33 − 11 23 + 22 31 + 33 12) ∶ ⋯ ∶ ⋯]Δ = [sinh(√ )(1√ +√ cosh( √) − cosh(√ ) − cosh(√ )) ∶ ⋯ ∶ √⋯]Δ = [sinh( )(sinh2( 1 ) − sinh2( 1 ) − sinh2( 1 )) ∶ ⋯ ∶ ⋯] . 2 2 2 Δ The euclidean limit of Be is the point X40. Be is the incenter of the extangents triangle of Δ0 (see 2.4.9). But, diﬀerent from its euclidean limit, it is in general not the circumcenter of the excentral triangle (I1I2I3)0, neither a point on I ∨ O. + The four lines Ik ∨ Bek, 0 ≤ k ≤ 3, meet at the point O .

∙ The four lines Ok ∨ Ik, k =0, ⋯ , 3, meet at the point d d d d d d d d d d d d d [ 11( 22( 11 33 + 31( 23 − 33)) + 33( 11 22 + 12( 23 − 22))) d d d d d d d d d d 2 +√ 22√ 33 11( 11 − 12 − 31)+ 11(√22 33 − 23) ∶ ⋯ ∶ ⋯]Δ.

This is a second√ point√ with euclidean limit X40 and, in general, it also diﬀers from all the circumcenters of excentral triangles of Δ.

∙ The triples Δ[O] =ΔG and I1I2I3 are perspective at the Mittenpunkt of Δ0, ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 23

d d d d Mi = Mi0 =[ 11(− 11 + 22 + 33) ∶ ⋯ ∶ ⋯]Δ = [sinh( )(−sinh( ) + sinh( ) + sinh( )) ∶ ⋯ ∶ ⋯] , √ √ √ √ Δ a point on the line I ∨ K. It has euclidean limit X9.

∙ Deﬁne the points Pa and Qa by Pa ∶= (I0 ∨ Mi0)∧(I1 ∨ Mi1) and Qa ∶= (I2 ∨ Mi2) ∧ (I3 ∨ Mi3) and the points Pb,Qb, Pc,Qc accordingly. The points B, Pa,C,Qa form a har- sinh( ) monic range. The triple P P P is perspective to Δ at the perspector [ ∶ a b c sinh( )−sinh( ) ⋯ ∶ ⋯]Δ, a point with euclidean limit X100.

¨ ∙ Vigara [27] proved that the triples Δ[O] and Δ [I] are perspective; the perspector he named pseudo- Spieker center. But this point is in fact a good candidate for the Spieker center in elliptic and in hyperbolic geometry, as it is one of four radical centers of the three 0 incircles ℐk, k =1, 2, 3; and it is the one that lies inside Δ . Its coordinates are Sp = Sp0 = [(s1+s2+s3)(s1c2c3+s2c3c1+s3c1c2+s1s1s3)+(s2c3+s3c2)(s2+s3−2s1) + s1(2s2+2s3−s1)∶ ⋯ ∶ ⋯]Δ, with s1 ∶= sinh(a), c1 ∶= cosh(a), s2 ∶= sinh(b),.... ¨ More general, each triple Δ[O ] is perspective to each triple Δ , j, k ∈ {0, 1, 2, 3}. j [Ik] By this, we get altogether 16 perspective centers. Let P be the perspector for Δ and kj [0k] Δ¨ . GeoGebra-constructions indicate: [Ij ] - The six points P12, P21, P23, P32, P31, P13 lie on a singular conic (union of two lines). - Put Q1 =(P12∨P21)∧(P31∨P13),Q2 =(P23∨P32)∧(P12∨P21),Q3 =(P31∨P13)∧(P23∨P32); ¨ the triple Q1Q2Q3 is perspective to Δ and to Δ at points which lie on the line O ∨ I. ∙ The points H−,Mi,Be,Sp are collinear.

+ ∙ Define the point S1 by S1 ∶= (I1 ∨ O ) ∧ a and the points S2,S3 likewise. The triple S1S2S3 is perspective to Δ at the point sinh( ) sinh( ) sinh( ) [ ∶ ∶ ] . cosh( ) + cosh( ) cosh( ) + cosh( ) cosh( ) + cosh( ) Δ We call this point pseudo- Schiffler point; the euclidean limit of this point is the Schiffler point X21, cf. [6].

[O ] ¨ ∙ The antipedal triple Δ 0 of O0 is perspective to O1O2O3 and to Δ at O0.

Let P be the point sinh( ) 1 [ ∶ ⋯ ∶ ⋯] =[ 2 ∶ ⋯ ∶ ⋯] , d c c d c c Δ − Δ ( 22( − 12)− 33( − 31) sinh( ) 2 [O ] [O ] [O ] [O ] then: (B√0 ∨ I2)∧(C √0 ∨ I3)= AP , (C 0 ∨ I3)∧(A 0 ∨ I1)= BP , [O ] [O ] and (A 0 ∨ I1) ∧ B 0 ∨ I3)= CP . P is a point on the circumcircle C0, its euclidean limit is X100.

∙ The pedal triple Δ of I is perspective to I I I at [I0] 0 1 2 3 24 MANFRED EVERS

d d d d d d d d d d [ 11(− 23 11 + 31 22 + 12 33 − 11 22 33) ∶ ⋯ ∶ ⋯]Δ

= [sinh(√ ) − cosh(√ ) + cosh(√ ) + cosh(√ ) ∶√⋯ ∶√⋯] √. 2 2 2 Δ

This point is also the orthocorrespondent of I. Its euclidean limit is X57.

c c c c c c ̃ + ∙ The tripole [1∕( 31− 12) ∶ 1∕( 12− 23) ∶ 1∕( 23− 31)] of the line K ∨ G is a point on the circumcircle and has euclidean limit X99.

∙ The circumcenters of the triangles (O0BC)0, (O0CA)0, (O0AB)0 form (in this order) a triple which is perspective to Δ at the Kosnita point d d d d d d d d d [1∕ ( 12 + 22)( 31 + 33)− 23( 11 + 23 + 31 + 12 − c) ∶ ⋯ ∶ ⋯]Δ

D d d d d d d with = (1, 1, 1)[ ](1, 1, 1) = 11 + 22 + 33 + 2( 12+ 23 + 31) .

The euclidean limit√ of this point is X54. √ ð ð ð ð ∙ Put P1 ∶= (O2 ∨ C)∧(O3 ∨ B), P2 ∶= (O3 ∨ A)∧(O1 ∨ C), P3 ∶= (O1 ∨ B)∧(O2 ∨ A). ¨ The triple P1P2P3 is perspective to Δ at O and to Δ at c c c 2 c c c c c c c c 2 c c c c 2 c c [( 23 + )( 23( 23 + 31 + 12 + )− 23( 31 + 12 + ) −( 31 + 12)(( 31 − 12) + 31 + 12 + c c ) − )∶ ⋯ ∶ ⋯]Δ. The euclidean limit of this point is the deLongchamps point X20.

∙ The incenters of the triangles (I0BC)0, (I0CA)0, (I0AB)0 form (in this order) a triple perspective to Δ. The perspector is the 1st deVilliers point 1 ∶ ⋯ ∶ ⋯]Δ d d d d d c d d 2 22 33( 22 33 − 23)+ 22 33 u tsinh(t ) t tsinh( ) t sinh(t ) = ∶ ∶ ]Δ. 2 cosh( 1 )+1 2 cosh( 1 )+1 2 cosh( 1 )+1 2 2 2 It has euclidean limit X1127. Experimental constructions using GeoGebra indicate that there also exist elliptic and hy- nd perbolic analogues of the 2 de Villiers point X1128 and of the three Stevanovic points X1130, X1488 and X1489.

̃ 2.3.4. Kimberlings "Four-Triangle Problem". Let T, T be triangle centers of Δ0, and let ̃ ̃ ̃ ̃ T1, T2, T3 be the T -centers of the triangulation triangles (ATB)0, (BTC)0, (CTA)0, respec- ̃ ̃ ̃ ̃ tively. If the triple T1T2T3 is perspective to Δ0, we will say that T #T is well-defined, and T #T̃ will stand, in this case, for the perspector. There is a problem, which are the centers T and T̃ such that T #T̃ is well-defined. Kimberling posed this problem (the "Four-Triangle Problem") in [15] for the special case T = T̃ , and as far as I know, this problem is still open. It was shown above that O#O is the Kosnita point and that I#I is the de Villiers point. Experiments with GeoGebra indicate that T #T is well-defined for the absolute centers T = H,G+,O+,N+, L,H⋆ and that L#L = H, but for the absolute center P on the orthoaxis (see 2.1.7) it is not well-defined. If T ∈{K,̃ G̃}, then T #T is well-defined, whereas for T ∈{Ge, Na} it is not. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 25

2.4. Circles, radical centers and centers of similitude.

The following two theorems together with their proofs were presented by Ungar [26] for the proper hyperbolic case.

2.4.1. The Inscribed Angle Theorem. Let be the measure of the angle ∠+(BOC), then

(1, 1, 1) [D](1, 1, 1) sinh( )= sinh(∕2) and b c √ cosh( ) cosh( ) ð 2 2 ð 1 1 sinh( − area(Δ )) = sinh( ( − area((BOC) ))). 2 0 2 0

A special case: If B,C,O are collinear, then − 1 area(Δ )= 1 and = + . 2 0 2

2.4.2. Tangent-Secant Theorem. The tangent at A of the circumcircle C0 of Δ0 meets the line a = B ∨ C at the point P =[0 ∶ c − c ∶ c − c] = [0 ∶ sinh2( b ) ∶ −sinh2( c )] , 13 12 Δ 2 2 Δ the harmonic conjugate of AK̃ with respect to {B,C}, and a sinh2(([P,A] )) cosh2( ) = sinh(([P,B] )) sinh(([P,C] )). + 2 + +

FIGURE 5. 26 MANFRED EVERS

2.4.3. It is easy to dualize these two theorems. We present euclidean limit versions of these dualizations, see Figure 5: 2 measure(∠ (BI C)) = measure(∠ (B I C )), i =0, 1. + i + [Ii] i [Ii]

Between the angles 2, 3, in Figure 5, the following relationship applies: 2 sin( ) sin( ) sin( ) sin( ) sin2()= 2 3 = 2 3 . 2 1 1 − cos( 2+ 3) sin ( ) 2 A spherical version of the following theorem together with a proof can be found in [24] ch. IX. 2.4.4. A second Tangent-Secant Theorem. Let  be a nondegenerate circle with center M and radius r, and let P be an anisotropic point. We introduce two sets of lines: SP ∶= {l l is line through P , intersecting the circle  at two distinct anisotropic points } TP ∶= {l l is line through P touching  at an anisotropic point }. Since thereð are at most two isotropic points on , there are inﬁnitely many lines in SP , and if P is a pointð on or outside the circle, then there is at least one line in TP . For each line l in SP we deﬁne a number p(P , l) as follows: If Q and R are points of intersection of l and , then p(P , l) ∶= tanh 1 ([P,Q] ) tanh 1 ([P , R] ) 2 + 2 + = tanh 1 ([P,Q] ) tanh 1 ([P , R] ) . 2 − 2 − For a tangent l ∈ TP with touchpoint T we put 2 p(P , l) ∶= tanh( 1 d(P,T )) . 2 The second Tangent-Secant Theorem states that for lines l , l ∈ S ∪ T : 1 2 P P p(P , l1)= p(P , l2). Thus, there exists a number p(P, ) such that p(P, )= p(P , l) for all l ∈ SP ∪ TP . This number is called the power of the point P with respect to the circle . This power can be calculated by p(P, ) = tanh 1 (d(P , M)+ r) tanh 1 (d(P , M)− r) 2 2 cosh(d(P , M)) − cosh(r) = . cosh(d(P , M)) + cosh(r)

2.4.5. Radical lines of two circles. Let 1, 2 be two nondegenerate circles with centers M1, M2, d(M1, M2) ≠ 0, and radii r1,r2. We want to find out which anisotropic points have the same power with respect to both circles. First, we define a "modified power" of an anisotropic point P with respect to a circle  having center M and radius r by p̃(P, ) ∶= cosh(d(P , M))∕cosh(r). In comparison to p, this modified power is easier to handle, on the other hand, an anisotropic point P has the same power with respect to 1 and 2 precisely when p̃(P, 1)= p̃(P, 2). We put ℛ = ℛ(1, 2) ∶= {P P is anisotropic and p̃(P, 1) = p̃(P, 2)}. One recog- ¨ nizes immediately that the point M ∶= (M1∨M2) belongs ℛ, as well as all anisotropic points of intersection of the twoð circles. Moreover, whenever P is a point in ℛ, different from M¨, then every other anisotropic point Q on M¨∨P is also a point in ℛ, because: cosh( 1 i−d(Q, M¨)) cosh( 1 i−d(Q, M¨)) 2 2 p̃(Q, 1)= p̃(P, 1) = p̃(P, 2) = p̃(Q, 2). cosh( 1 i−d(P , M¨)) cosh( 1 i−d(P , M¨)) 2 2 ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 27

There are exactly two points in ℛ ∩(M1 ∨ M2), P1 = cosh(d) cosh(r2)− cosh(r1) M1 + cosh(d) cosh(r1)− cosh(r2) M2 and P = cosh(d) cosh(r )+ cosh(r ) M − cosh(d) cosh(r )+ cosh(r ) M , 2 2 1 1 1 2 2 with d = d(M , M ). 1 2 ¨ ¨ Thus, ℛ is the union of the lines l1 = P1 ∨ M and l2 = P2 ∨ M . These lines are called radical lines   l , M , l , M of the circles 1 and 2. The lines 1 1 2 2 form a harmonic pencil.

2.4.6. Radical centers of three circles. We draw a circle around each vertex of triangle Δ0, around A a circle 1 with radius r1, etc. Then there are exactly four radical centers, points of equal powers with respect to the three circles. One of these points, R0, is a point inside the triangle Δ0, with coordinates d d d R0 = R0(r1,r2,r3) = [cosh(r1) 11 + cosh(r2) 12 + cosh(r3) 13 ∶ ⋯ ∶ ⋯]Δ. ¨ The other three points R1, R2, R3 form the anticevian triple of R0 with respect to Δ . R1 has coordinates: d d d R1 = [cosh(r1) 11− cosh(r2) 12−cosh(r3) 13 ∶ d d d d d d −cosh(r1) 21+ cosh(r2) 22− cosh(r3) 23 ∶−cosh(r1) 31− cosh(r2) 32+ cosh(r3) 33]Δ. ̃ A radical circle k can be drawn around each point Rk; this circle meets the circles ̃ 1, 2, 3 orthogonally. The radius of k is ̃rk = p̃(Rk, 1)= p̃(Rk, 2)= p̃(Rk, 3).

By taking special values for r1,r2,r3, we can ﬁnd triangle centers of Δ0. First, independently of the choice of the radii, we have lim R0(tr1,tr2,tr3)= O0. We also t→0 get the circumcenter O0 as a result for R0 when we take r1 = r2 = r3. If r = 1 (−a + b + c),r = 1 (a − b + c),r = 1 (a + b − c), then R = I . 1 2 2 2 3 2 0 0 When we choose r1 = a,r2 = b,r3 = c, then R0 = L (deLongchamps point). ℝ≥0 Remark: In the euclidean plane, when taking radii r1 = ta,r2 = tb,r3 = tc,t ∈ , one 2 2 2 2 2 2 2 2 2 2 2 2 gets points R0(t)=[t (a (−2a +b +c )+(b − c ) )+ a (a −b −c ) ∶ ⋯ ∶ ⋯]Δ, all lying on the Euler-line.

2.4.7. Centers of similitude of two circles. Given two nondegenerate circles 1, 2 with centers M1, M2, d(M1, M2) ≠ 0, and radii r1,r2, then there exist two points L1,L2 which are the duals of the two radical lines of the duals of 1 and 2. These points are called centers of similitude of 1, 2, cf. [24]. These two centers lie on the line M1 ∨ M2, and M1,L1, M2,L2 form a harmonic range. If the two circles have common tangents, then each of these tangents passes either through L1 or through L2.

2.4.8. Dualizing 2.4.6. Given circles 1, 2, 3 with centers A,B,C and radii r1,r2,r3 (respectively), then there exist six centers of similitude of these circles, taken in pairs. Three of these are the verticesof the cevian triangle of the point T = T (r1,r2,r3) = [1∕sinh(r1) ∶ 1∕sinh(r2) ∶ 1∕sinh(r3)]Δ, while the other three centers lie on the tripolar of T . For r1 = r2 = r3, we have T = G0. When we choose for the radii r1 = a,r2 = b,r3 = c, then the point T is the isotomic conjugate of the incenter I0, and for r1 = 1 (−a+b+c),r = 1 (a−b+c),r = 1 (a+b−c) we get T = Ge . 2 2 2 3 2 0

2.4.9. The excentral triangle and the extangents triangle. The triangle (I1I2I3)0 is called the excentral triangle of Δ0. The radical centers of the threeexcircles are the Spieker center I¨I¨I¨ (see 2.3.3) together with its harmonic associates with respect to the dual 1 2 3 of the triple I1I2I3. 28 MANFRED EVERS

FIGURE 6. Illustration of 2.4.8. This Figure and the following with the exception of Figure 12 show the situation in the elliptic plane. An indication is the grey dotted cir- ̃ cle. Since the absolute conic abs has no real points, this circle  = {(x ∶x ∶x ) x 2 = x 2 + x 2} serves as a substitute for constructions. 0 1 2 0 1 2

The triple I1I2I3 is perspectiveð to the orthic triple Δ[H] at the perspector d d d d [ 11( 23 − 31 − 12) ∶ ⋯ ∶ ⋯]Δ = [sinh( )(cosh( ) − cosh( ) − cosh( )) ∶ ⋯ ∶ ⋯]Δ, a point with euclidean limit X . √ 46 We introduce three points d d d d d d d d d d d d d d E1∶= [− 11( 22 33+ 23)∶ 22( 11 23+ 33 12)∶ 33( 11 23+ 22 31)]Δ, = [− sinh( )(1+cosh( ))∶sinh( )(cosh( )+ cosh( )) ∶ sinh( )(cosh( )+ cosh( ))] , √ √ √ √ √ √ √ √ Δ d d d d d d d d d d d d d d E2 ∶= [ 11( 33 12+ 22 31)∶− 22( 33 11+ 31)∶ 33( 11 23+ 22 31)]Δ, d d d d d d d d d d d d d d E3 ∶= [√ 11(√ 33 12+√ 22 31)∶ 22(√ 11√23+ 33 12√)∶− √33( 11 √22+ 12)]Δ. Thesepointsaretheverticesof√ √ √ the extangents√ √ triangle√of Δ0, a trianglewith√ followingprop-√ erties: ∙ The sideline Ej ∨Ek of this triangleis a tangentof the excircles ℐj and ℐk, 1 ≤ j < k ≤ 3. ∙ It has the Bevan point Be (s.2.3.3 ) as its incenter.

The triple E1E2E3 is perspective to Δ at perspector d d d d d d d d d d d d d d d [ 11( 22 31 + 33 12) ∶ 22( 33 12 + 11 23) ∶ 33( 11 23 + 22 31)]Δ = [sinh( )(cosh( )+cosh( )) ∶ sinh( )(cosh( )+cosh( )) ∶ sinh( )(cosh( )+cosh( ))] . √ √ √ √ √ √ √ √ √ Δ ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 29

This point is the isogonal conjugate of the pseudo- Schiﬄer point (see 2.3.3) and has euclidean limit X65, but in contrast to the euclidean case this perspector diﬀers from the orthocenter of the intouch triangle. The triple E1E2E3 is also perspective to the orthic triple Δ[H], theperspectoristhe Clawson point d d d d d d d d d d d d d d d [ 11 31 12( 22 33 + 23)( 11 22 33 − 11 23 + 22 31 + 33 12) ∶ ⋯ ∶ ⋯] = [tanh( ) cosh2( )(cosh2( ) − cosh2( ) − cosh2( ) ∶ ⋯ ∶ ⋯] √Δ √ √ 2 √2 √ √2 √ 2 √ Δ √ with euclidean limit X19.

2.5. Orthology, pedal-cevian points and cevian-pedal points.

2.5.1. Orthologic triples. A point-triple P QR, P ≠ A¨,Q ≠ B¨, R ≠ C¨, is orthologic to the triple Δ = ABC if the lines P ∨ A¨,Q ∨ B¨, R ∨ C¨ concur at some point S, which is then called the center of this orthology. If P QR is orthologic to Δ, then Δ is orthologic to P QR; the linesperp(Q∨R, A), perp(R∨ P,B), perp(P ∨ Q,C) meet at some point T . Outline of a proof : If P QR is orthologic to Δ with center S = [s1∶s2∶s2]Δ, then there exist real numbers x, y, z such that

P =[s1 + x d11 ∶ s2 + x d12 ∶ s3 + x d31]Δ,

Q =[s1 + y d12 ∶ s2 + y d22 ∶ s3 + y d23]Δ, R =[s1 + z d31 ∶ s2 + z d23 ∶ s3 + z d33]Δ.

Deﬁne vectors p, q, r by p = (s1+x d11,s2+x d12,s3+x d31), q = (s1+x d12,s2+x d22, s3+x d23), r = (s1+z d31,s2+z d23,s3+z d33), then ((q × r)D) × A, ((r × p)D) × B, ((p×q)D) × C formalineardependentsystem (useCAStocheck). Thelinesperp(Q∨R, A), perp(R∨P,B), perp(P ∨Q,C) meet at the point T = [1∕x, 1∕y, 1∕z]Δ. □ Remark: In euclidean geometry, the coordinates of S with respect to the triple P QR are the same as thecoordinatesof T with respect to Δ (cf.[2]). This is not true in elliptic/hyperbolic geometry. On the other hand, still applies the

Addition: If P = A[S], Q = B[S], R = C[S] form the pedal triple of S, then T is the isogonal conjugate of S.

2.5.2. Pedal-cevian points and the Darboux cubic. A point P is a pedal-cevian point of Δ if its pedal triple Δ[P ] is perspective to Δ; the perspector we call cevian companion of P . A point P =[p1∶p2∶p3]Δ is a pedal-cevian point precisely when its coordinates satisfy the cubic equation (d d + d d )p (d p 2 − d p 2 )=0. jj j+1,i+2 j,j+1 j+2,j j j+1,j+1 j+2 j+2,j+2 i+1 j=1,2,3 As in the euclidean∑ case, this cubic is a self-isogonalcubic with pivot point L. Onthiscubic + - we call it Darboux cubic as its euclidean limit - lie the points Ok,Ik(k =0, 1, 2, 3),O , H, + L, Be and their isogonal conjugates. The cevian companions of Ok,Ik, O ,H,Be,L are + ¨ ¨ ¨ Gk,Gek, G , H, Na and the isotomic conjugate of H, respectively. The points A ,B ,C c c c are also lying on the Darboux cubic. Their cevian companions are [− ∶ 1∕ 12 ∶ 1∕ 31]Δ, c c c c c c [ 1∕ 12 ∶ − ∶ 1∕ 32]Δ, [ 1∕ 31 ∶ 1∕ 23 ∶ − ]Δ, respectively. These three points form a triple which is perspective to Δ at G+. 30 MANFRED EVERS

2.5.3. Cevian-pedal points and the Lucas cubic. A point P is a cevian-pedal point of Δ if ¨ its cevian triple ΔQ is perspective to Δ ; the perspector we call the pedal companion of Q. The cevian-pedal points P =[p1∶p2∶p3]Δ form a cubic with the equation d p p 2 p 2 . j+1,j+2 j( j+2 − j+1)=0 j=1,2,3 This cubic, we call it Lucas cubic∑ , is a self-isotomic pivotal cubic; the pivot is the isotomic + conjugate of H. On this cubic lie the points Gk,Gek, Nak(k = 0, 1, 2, 3),G ,H,L and their isotomic conjugates. The pedal companion of L is d d d d d d d d 2 d d 2 d d 2 d d d ⋯ ⋯ [ 11(2 22 33 31 12 + 23(− 11 23 + 22 31 + 33 12 + 11 22 33)) ∶ ∶ ]Δ This is another point on the Darboux conic; it has euclidean limit X1498. 2.5.4. The Darboux cubic and the Lucas conic of Δ¨. The pedal-cevian points of Δ are the cevian-pedal points of Δ¨ and vice versa. Therefore,the Darbouxcubic and the Lucas cubic of Δ are the Lucas cubic and the Darboux cubic of Δ¨, respectively. 2.5.5. The Thomson cubic. In euclidean geometry, the Thomson cubic is the locus of perspectors of circumconics such that the normals at the vertices A,B,C meet at one point. An equation of the elliptic/hyperbolic analog is d d d d d 2 d 2 ( j,j j+1,j+2 − j,j+1 j+2,j)pj( j+2,j+2pj+1 − j+1,j+1pj+2)=0. (⋆) j=1,2,3 É This cubic is an isogonal cubic with pivot G+. Besides the vertices of Δ and the point G+, + ̃ ̃ it passes through the points H,O and Ik, Kk, Gk, k =0, 1, 2, 3. In the above deﬁnition of the Thomson cubic, the word "perspectors" may be replaced by "centers" without changing the euclidean curve, see [8]. But this is not the case in elliptic and hyperbolic geometry; here we get a diﬀerent curve of higher degree. A center [z1∶z2∶z3]Δ of a circumconic belongs to this curve precisely when the coordinates of the corresponding perspector [p1∶p2∶p3]Δ, p z z 2 d d d 2 z 2 d d d 2 j = j j ( j+1,j+1 j+2,j+2 − j+1,j+2)− j+1( j+2,j+2 j,j − j,j+1) −z 2 (d d − d 2 )+2z z (d d − d d ) , j+2 j,j j+1,j+1 j+2,j j+1 j+2 j+1,j+2 j,j j+2,j j,j+1 satisfy the equation (⋆). Points on this curve are: H and Ik,Ok, k =0, ⋯ , 3. 2.6. Conway’s circle, Kiepert perspectors, Hofstadter points, and related objects.

2.6.1. Forallrealnumbers x, y, z deﬁnethepoints Pa(x),Qa(x), Pb(y),Qb(y), Pc(z), Qc(z) by Pa(x) = [0∶x∶1]Δ, Pb(y) = [1∶0∶y]Δ, Pc(z)=[z∶1∶0]Δ,

Qa(x) = [0∶1∶x]Δ,Qb(y)=[y∶0∶1]Δ, Qc(z) = [1∶z∶0]Δ. These six points lie on the conic 1 1 1 {[p ∶p ∶p ] p2 + p2 + p2 −(x + )p p −(y + )p p −(z + )p p = 0}. 1 2 3 Δ 1 2 3 x 2 3 y 3 1 z 1 2 The points Pa, Pb, Pc are collinear precisely when xyz = −1. ð Put

X = X(x, y, z)=(Qc(z) ∨ Pa(x))∧(Qa(x) ∨ Pb(y)), Y =(Qa(x) ∨ Pb(y))∧(Qb(y) ∨ Pc(z)), Z =(Qb(y) ∨ Pc(z))∧(Qc(z) ∨ Pa(x)). ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 31

FIGURE 7. Illustration of 2.6.1

1 1 1 The triple XY Z is perspective to Δ at R(x, y, z)=[ , , ] . x + yz y + zx z + xy Δ If xyz ≠ 0, then R(x, y, z)= R(1∕x, 1∕y, 1∕z).

Put ¨ X =(Qb(y) ∨ Pc(z)) ∧ a, ¨ ¨ Y =(Qc(z) ∨ Pa(x)) ∧ b, Z =(Qa(x) ∨ Pb(y)) ∧ c. ¨ ¨ ¨ The points X , Y , Z are collinear on the tripolar line of [x, y, z]Δ.

Put ¨¨ X =(Qc(z) ∨ Pb(y)) ∧ a, ¨¨ ¨¨ Y =(Qa(x) ∨ Pc(z)) ∧ b, Z =(Qb(y) ∨ Pa(x)) ∧ c. ¨¨ ¨¨ ¨¨ The points X , Y , Z are collinear on the tripolar line of [yz, zx, xy]Δ. Special cases: ∙ If x = y = z, then R = G and X¨, Y ¨, Z¨,X¨¨, Y ¨¨, Z¨¨ lie on the tripolar of G. ∙ Assume d d d − d d d d d − d d y = 22 x + 23 11 31 22 and z = 33 x + 23 11 12 33 . d d d d d d d d d d √ 11 11√( 12 − 11√ 22) √ 11 11√( 31 − 11√ 33) √ √ √ √ √ √ √ √ 32 MANFRED EVERS

In this case, d(Pa(x),B)= d(Pb(y),C)= d(Pc(z),A) and: - The points R(x, y, z),x ∈ ℝ, lie on a circumconic of Δ through I, Ge and Na. - For all x, the tripolar of [yz, zx, xy]Δ passes through the point (I ∨ Ge) . ∙ Assume sinh((1 − k)a) sinh((1 − k)b) sinh((1 − k)c) 0

2.6.2. The Conway circle. Reflect B in I1 ∨C and reflect the mirror image in BG to get the d d d d d d d d point [ 23 11 + 31 22 ∶ 0 ∶ 11( 12 + 11 22)]Δ, denoted by R23. Reflect C in I ∨ B and this point in C to get the point R = [ d (d + d d ) ∶ 1 √ √ G √ √ √32 11 31 33 11 d d +d d ∶ 0] . Both, R and R , have distance a from A. Construct likewise 23 11 12 33 Δ 23 32 √ √ √ the points R , R with distance b from B, and R , R with distance c from C. The six √ 31√ 13 12 21 points R12, R21, R23, R32, R31, R13 lie on a circle with center I. The radius rConway of this circle can be calculated by 1 cosh(r ) = cosh(s) cosh(r) with s = (a + b + c)= semiperimeter of Δ Conway 2 0 and r = d(I,A[I])= radius of the incircle ℐ0.

Proof : By reflecting R23 in the line I ∨ A we get R32, and therefore d(I, R23)= d(I, R32). Accordingly, we have d(I, R31)= d(I, R13) and d(I, R12)= d(I, R21). The distance between A and R31 is b + c and agrees with the distance between A and R21. By reflecting R31 in the line I ∨ A we get R21, thus d(I, R31) = d(I, R21). In the same way, we get d(I, R12)= d(I, R32) and d(I, R13)= d(I, R23). By reflecting R13 in the line I∨A[I] we get R12. Therefore, A[I] is amidpointof {R13, R12}. The radius rConway can be calculated by applying the elliptic resp. hyperbolic version of Pythagoras’ theorem. □ The three points (R23 ∨ R32) ∧ a, (R31 ∨ R13) ∧ b, (R12 ∨ R21) ∧ c are collinear on a line with the equation d d d d d d d ( j,j+1 j+2,j+2 + j+2,j j+1,j+1)( j+1,j+2 − j+1,j+1 j+2,j+2)xj =0 . j=1,2,3 É t t t t The euclidean limit of this line is the tripolar of X86.

2.6.3. A dual of Conway’s circle theorem. Let Q1,Q2,Q3 be the apices of isosceles triangles, erected outwardly on the sides a, b, c of Δ0 with base angles , , , respectively. Then there exists a circle with center O which touches the lines A ∨ Q2,A ∨ Q3, B ∨ Q3,B ∨ Q1, C ∨ Q1,C ∨ Q2, see Figure 8. The euclidean limit of this circle is the circumcircle of Δ0. Instead of base angles , , ,wecan also takebase angles i− − ,i− − ,i− − to get the same touching circle.

The triple Q1Q2Q3 is perspective to Δ at a point with euclidean limit X6.

2.6.4. Dualizing 2.6.1. Let x, y, z be real numbers, and deﬁne the points P1, P2, P3 by d d d P1 =[yz 11∶y 22∶z 33]Δ, d d d d d d P2 =[x √11∶zx√ 22∶z√ 33]Δ, P3 =[x 11∶y 22∶xy 33]Δ. √ √ √ √ √ √ ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 33

FIGURE 8. A description of the red circle is given in 2.6.3.

Then: (∠+BAP3)= (∠+CAP2), (∠+CBP1)= (∠+ABP3), (∠+BCP1)= (∠+ACP2). d d d The triple P1P2P3 is perspective to Δ at the point R(x, y, z)=[x 11, y 22, z 33]Δ. √ √ √ Put Q1 =(P3 ∨ B)∧(P2 ∨ C),Q2 =(P1 ∨ c)∧(P3 ∨ A),Q3 =(P2 ∨ A)∧(P1 ∨ B). The triple Q1Q2Q3 is perspective to Δ at the isogonal conjugate of R(x, y, z).

Special cases: ∙ If x = y = z =1, then R = I. ∙ Assume d d d d y = 11 33 and z = 11 22 . d d d d d d d d 22 √33 +√x( 31 − 23) 22 √33 +√x( 12 − 23) In this case, the√ points√P1, P2, P3 are the apices of isosceles√ √ triangles, erected on the sides a, b, c of Δ0 (either all inwardly or all outwardly) with base angles which have all the same measure. The points R(x, y, z),x ∈ ℝ, are called Kiepert perspectors; they lie on the conic d d [p1,p2,p3]Δ ( j+2,j − j,j+1)pj+1pj+2 =0 , j=1,2,3 É which is a circumconic of Δ throughð G and H. The euclidean limit of this conic is the Kiepert hyperbola. The isogonal conjugates of the Kiepert perspectors lie on the line through K (= isogonal conjugate of G) and O+ (= isogonal conjugate of H). This line also passes through ̃ ̃ ⋆ the triangle centers O, K, K . The tripole of O ∨ K is a point on the circumcircle C0 with 34 MANFRED EVERS euclidean limit X110. ∙ Let k be a real number, diﬀerent from 0 and 1, and assume sinh(k ) sinh(k ) sinh(k ) x = , y = , z = . sinh((1 − k) ) sinh((1 − k) ) sinh((1 − k) ) In this case, we call the point R(x, y, z) Hofstadter k-point, according to the euclidean case. The isogonal conjugate of the Hofstadter k-point is the Hofstadter (1-k)-point. Here are some examples of Hofstadter points: The 1 -point is I, the (-1)-point H and the 2-point is 2 + O . The limes of the k-point, as k approaches zero, is [ ∶ ∶ ]Δ. 2.7. The Lemoine conic, the Lemoine axis and Tucker circles.

2.7.1. Lemma. Let P = [p1 ∶ p2 ∶ p3]Δ be a point not on a sideline of Δ, and let l = {[x1 ∶ x2 ∶ x3]Δ q1x1 + q2x2 + q3x3 = 0} be a line that does not contain any vertex of Δ. Put P11 ∶= l ∧ a, P12 ∶= (P11 ∨ P ) ∧ b, P13 ∶= (P11 ∨ P ) ∧ c, and deﬁne the points P22, P23, P21, P33ð, P31, P32 accordingly. The points P12, P13, P21, P23, P31, P32 lie on the conic qj(pj+1qj+1 + pj+2qj+2) [x ∶x ∶x ] x2 1 2 3 Δ p j j=1,2,3 j É ð pjqj (pjqj + pj+1qj+1 + pj+2qj+2)+2pj+1qj+1pj+2qj+2 − xj+1xj+2 =0 . pj+1pj+2 We call this conic the conic associated with the pair (P, l). Its perspector is p1 [ ∶ ⋯ ∶ ⋯]Δ. (2p1q1(p2q2 + p3q3)+ p2p3q2q3) Examples: ∙ The perspector of the conic associated with (P , P ) is the point P . ∙ The conic associated with (K,̃ K̃ ) is the Lemoine conic which is described in [5]; its euclidean limit is the First Lemoine circle. The line O ∨ K (= O ∨ K̃ ) is a symmetry axis of this conic. ∙ The conic associated with G and H is a conic with perspector c c c c c c c c c [1∕( 23 +2 31 +2 12) ∶ 1∕(2 23 + 31 +2 12) ∶ 1∕(2 23 +2 31 + 12)]Δ. ∙ The conic associated with incenter I and the orthotransversal of I has I as a symmetry point and a perspector with euclidean limit X10390. ∙ In euclidean geometry, the conic associated with the Lemoine point K and the line at inﬁnity is the First Lemoine circle; the Second Lemoine circle is associated with K and the tripolar of X25. 2.7.2. The Lemoine axis and the apollonian circles. The tripolar of the Lemoine point K̃ is called Lemoine axis. This axis is perpendicular to the line O ∨ K (= O+ ∨ K̃ ). Let L1,L2,L3 be the intersection points of the Lemoine axis with the sidelines a, b, c, respectively. The circle C (L1,A) with center L1 through vertex A meets the circumcircle C0 perpendicularly. Mutatis mutandis, this is also true for the circles C (L2,B), C (L3,C). We will call these circles apollonian circles, as they are called in the euclidean case. There are two points, J+ and J+, J = [(c − c)( c + c − c − c ± 1 det(ℭ) ) ∶ ⋯ ∶ ⋯] , ± 23 23 31 12 3 Δ on the line O ∨ K at which all three apollonian circlest meet; their euclidean limits are the ̃ ⋆ ̃ isodynamic points. The points K and K ∧(K ∨ O) areð the midpointsð of J− and J+, and ̃ J−,O,J+, K form a harmonic range. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 35

Remark: An elliptic/hyperbolic version of Apollonius’ Theorem (see [24] ch. VIII for the elliptic/spherical version together with its proof) states that 1 1 sinh( ([P,B]+)) sinh( c) C (L ,A)= P 2 = 2 . 1 1 1 sinh( ([P,C]+) sinh( b) 2 2 ð

2.7.3. Tucker hexagons and Tucker circles. Let P1Q2P3Q1P2 Q3P1 be a closed polygonal chain with vertices on the sidelines of Δ0, P1,Q1 on a, P2,Q2 on b, P3,Q3 on c, and assume that none of these points is a vertex of Δ. In euclidean geometry, the polygon P1Q2P3Q1P2Q3 is called a Tucker hexagon of Δ0, if its line segments are alternately parallel/antiparallel to the sidelines of Δ, P1Q2 parallel (or antiparallel) to c, Q2P3 antiparallel (parallel) to a, etc. The vertices of a Tucker hexagon are always concyclic and the corresponding circle is called a Tucker circle. We will show that Tucker hexagons and Tucker circles also exist in the elliptic and in the hyperbolic plane. The concept of antiparallelism of lines with respect to an angle can be transferred from euclidean geometry to elliptic and to hyperbolic geometry. Let us explain this for the angle ∠+BAC of the triangle Δ0. (See Akopyan [1] for a more detailed description). Given two lines g,ℎ such that #{b,c,g,ℎ}=4, then g is antiparallel to ℎ with respect to ∠+BAC iff one of the following two conditions holds: condition 1: g ∧ ℎ = A and ℎ is the mirror-image of g in the angle-bisector A ∨ I0 of ∠+BAC. condition 2: Define P1∶= b ∧ g, P2∶= g ∧ c, P3∶= c ∧ ℎ, P4∶= ℎ ∧ b. #{P1, P2, P3, P4} ≥ 3, and the segments [P1, P2]+, [P2, P3]+, [P3, P4]+, [P4, P1]+ are cords of a circle. (We recall thata cord of a circle is a closed segmentwhose boundarypoints lie on the circleand whose inner points lie inside the circle.) It can be easily verified that, if A is neither a point on g nor on ℎ, then g is antiparallel to ℎ with respectto ∠+BAC precisely when (∠+(P1P2A))−(∠+(P2P1A)) = (∠+(P3P4A))− (∠+(P4P3A)). Now we define parallelism between lines with respect to the angle ∠+BAC: Two lines g,ℎ are parallel with respect to ∠+BAC precisely when the mirror image of ℎ in the angle bisector A ∨ I0 is antiparallel to g. Choose a point P1 on a, different from B and C, and construct successively points Q2 on b, P3 on c, Q1 on a, P2 on b and Q3 on c such that P1 ∨ Q2 is parallel to c, Q2 ∨ P3 antiparallel to a, P3 ∨ Q1 parallel to b, Q1 ∨ P2 antiparallel to c, P2 ∨ Q3 parallel to a. Then, Q3 ∨ P1 is antiparallel to b and the points P1, ⋯ ,Q3 are concyclic. Proof : Since P1 ∨Q2 is parallel to c, Q2 ∨P3 antiparallel to a, and P3 ∨Q1 parallel to b, we have the equations − = "−, − = −, − = −, see Figure9. We canprovethat the four points P1,Q2, P3,Q1 are concyclic on a circle  by showing − = − (⋆). Proof of (⋆) ∶ − =(i − " − )−(i − − ) =( − )+ − " =( − )+( + − )−( + − ) = −

Q1 ∨ P2 is antiparallel to P1 ∨ Q2 and to c precisely when P2 is a point on , concyclic with P1,Q2,Q1. P2 ∨ Q3 is antiparallel to Q2 ∨ P3 and therefore parallel to a precisely when Q3 is a point on , concyclic with P1,Q1, P2. The point Q3 is a point on , together with P1,Q1, P3. Therefore, Q3 ∨ P1 is antiparallel to P3 ∨ Q1 and to b. The polygon 36 MANFRED EVERS

FIGURE 9. A Tucker hexagon and its circle.

P1Q2P3Q1P2Q3 is a Tucker hexagon of Δ0; its vertices are concyclic. □

Starting from Q3 =[q31∶1∶0]Δ, we calculate the coordinatesof the other vertices of the Tucker hexagon. We get, for example, q (s − c ) P = [0∶1∶p ] , p = 31 12 , 1 13 Δ 13 c s − 23

c ℭ c s = (q31, 1, 0) [ ](q31, 1, 0) − q31 √c √ c q 2 c q c q , = ( 31 +1)+2 12 31 − 31 c c c c c c (s −t)(s ( 31 − )+ (2 12 − 31)−1) P =[p ∶0∶1] , p = √ . 2 21 Δ 21 c c c c 2 ( − 12)(s − 12) The center T of the Tucker circle can be calculated by T = ((P1 − Q3)∨(P2 − Q3)) . By using CAS, it can be veriﬁed that T lies on the line O ∨ K (= O ∨ K̃ ). The coordinates of T can then be expressed by c c c c c c c c c c c c c c c c c c T = [( − 23)( + 23− 31− 12+ t)∶( − 31)( + 31− 12− 23+ t)∶( − 12)( + 12− 23− 31+ t)]Δ with c c c c c 2 c 2 c 2 (2 12 23 31 − ( + + −1))((p13+1) r31 −(q31+1)r13) t = 12 23 31 , r31 s13 + r13 s31 r c q 2 c q , r c p 2 c p , 31 = ( 31+1)+2 12 31 13 = ( 13 +1)+2 23 13 t t ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 37

c c c c c c c c c c c c s31 = q31( ( 12− 23+ 31)−2 12 31+1) + ( 12+ 23− 31)−2 12 23+1 , c c c c c c c c c c c c s13 = p13( ( 12− 23− 31)+2 23 31−1) − ( 12+ 23− 31)+2 12 23−1 . Constructions with GeoGebra show that the perspector of the Tucker circle is, in general, not a point on the line O ∨ K̃ .

Deﬁne the points R1, R2, R3,T1,T2,T3 by R1 =(P1 ∨ Q2)∧(Q1 ∨ P3), R2 =(P2 ∨ Q3)∧(Q2 ∨ P1), R3 =(P3 ∨ Q1)∧(Q3 ∨ P2), T1 =(P1 ∨ Q3)∧(Q1 ∨ P2),T2 =(P2 ∨ Q1)∧(Q2 ∨ P3),T3 =(P3 ∨ Q2)∧(Q3 ∨ P1).

The triples R1R2R3 and T1T2T3 are both perspectiveto Δ, theperspectorbeingtheLemoine point K̃ . This can be shown by calculations with the help of CAS.

Constructions indicate: The points a∧(P2 ∨Q3), b∧(P3 ∨Q1), c ∧(P1 ∨Q2) are collinear on a line which we name g, and the points a ∧(Q2 ∨ P3), b ∧(Q3 ∨ P1), c ∧(Q1 ∨ P2) are collinear on a line r. The lines g and r meet at a point on the dual of T .

̃ Let l1, l2, l3 bethreelinespassing through K and parallelto a, b, c with respectto ∠+BAC, ∠+CBA, ∠+ACB, respectively. The six points l1 ∧ b, l1 ∧ c, l2 ∧ c, l2 ∧ a, l3 ∧ a, l3 ∧ b lie st on a circle, the 1 Lemoine circle. The point Q3 = l1 ∧ c has coordinates (c − c )(c (c + c − c )+ c −1)+2(1− c c )(c − c ) [q ∶1∶0] =[ 12 23 12 23 31 12 23 23 ∶1∶0] , 31 Δ c c c c Δ 4(1 − 12 23√)( − 31) c c 2 c 2 c 2 c c c c c c c c c c c c = 12+ 23+ 31 − 2( 12 23+ 23 31+ 31 12)+5 +4 12 23 31 − 2( 12+ 23+ 31). ̃ Constructionst indicate: The ﬁrst Lemoine circle is the conic associated with (K,g), where

FIGURE 10. The 3rd Lemoine circle (blue) 38 MANFRED EVERS

FIGURE 11. The Apolloniuscircle (lightblue). The smaller blue circle is the Hart-Feuerbachcircle which touches the (brown)excircles externally. The Spieker center is the radical center of the excircles. g is the line described above. ̃ Let l1, l2, l3 be three lines passing through K and antiparallel to a, b, c, respectively. The nd six points l1 ∧ b, l1 ∧ c, l2 ∧ c, l2 ∧ a, l3 ∧ a, l3 ∧ b lie on a circle, the 2 Lemoine circle.The point Q3 = l2 ∧ c has coordinates (c − c )(c (c + c − c )+ c −1)+2(1− c c )(c − c ) [q ∶1∶0] =[ 12 23 12 23 31 12 23 31 ∶1∶0] , 31 Δ c 2 c c cc c c c c c Δ 31( + 12)−2( 23 31√)− 23 + 31)+ − 12 as above. Constructions indicate: The second Lemoine circle is the conic associated with (K,r̃ ). ̃ Let Q2 and P3 be the second intersections of the circumcircle of the triangle (BKC)0 with the sidelines b and c, respectively. Define the points Q3, P1 and Q1, P2 likewise. The rd hexagon P1Q2P3Q1P2Q3 is a Tucker hexagon. The associated Tucker circle is the 3 Lemoine circle. See Figure 10. We do not present the coordinates of Q3, they are very complicated. In euclidean geometry, a circle which internally touches the three excircles of a triangle is called the Apollonius circle of this triangle. Grinberg and Yiu [9] showed that this Apollonius circle is a Tucker circle. As constructions indicate, this seems to be true also in elliptic and in hyperbolic geometry, see Figure 11. 2.7.4. Dualizing Tucker hexagons and Tucker circles. We can easily formulate a definition of parallelism and antiparallelism of points P,Q with respect to a segment which is dual to the definition of parallelism and antiparallelism of lines with respect to an angle. (We use ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 39

FIGURE 12. A triangle in the euclidean plane with duals of a Tucker hexagon and a Tucker circle. the names "parallelism/antiparallelism", even though they do not ﬁt well.) A dual version of a Tucker hexagon and a Tucker circle is shown in Figure 12.

2.8. Pseudoparallels of the sidelines and their duals.

2.8.1. Lemma / part 1. Let Q = [q1∶q2∶q3]Δ be a point not on a sideline of Δ, and let Q1,Q2,Q3 be the intersections of the tripolar Q with the sidelines a, b, c, respectively. Let R1, R2, R3 be any points in , with coordinates Rk = [rk,1∶rk,2, ∶rk,3]Δ, k = 1, 2, 3. Deﬁne three lines l1, l2, l3 by lk = Rk ∨ Qk. Then: The points l2 ∧ l3, l3 ∧ l1, l1 ∧ l2 form a triple perspective to Δ. The perspector is the point r11 r22 r33 rk1 rk2 rk3 P =[ ∶ ∶ ]Δ, ̃rk = + + , k =1, 2, 3. ̃r1 ̃r2 ̃r3 q1 q2 q3 We look at special cases: ∙ If R1R2R3 is the anticevian triple of Q, then P = Q. ∙ Let R =[r1∶r2∶r3]Δ be a point not on a sideline of Δ and R1R2R3 its anticevian triple. If Q is a point on the circumconic of Δ with perspector R, then P = Q. + ∙ If R1R2R3 is the anticevian triple of G and Q = H , then P is a triangle center with euclidean limit X2996.

2.8.2. A special case: pseudoparallels of the sidelines. If in the previous lemma Q = G, we call the lines l1, l2, l3 pseudoparallels of the sidelines a, b, c. We look at different cases 40 MANFRED EVERS for the triple R1R2R3. ¨ ¨ ¨ ∙ We put P1 ∶= sinh(a) A + cosh(a) B, P2 ∶= sinh(b) B + cosh(b) C, P3 ∶= sinh(c) C + cosh(c) A. In this case, d(l1,B) = d(l1,C) = a, d(l2,C) = d(l2,A) = b, d(l3,A) = d(l3,B)= c and sinh2(a) P =[ ) ∶ ⋯ ∶ ⋯] . D d d d Δ cosh(a)det( )+ 11 + 12 + 31 The euclidean limit of this point is X6. cc cc cc ∙ We use the abbreviations t1 ∶=1− 23,t2 ∶=1− 31,t3 ∶=1− 12,t123 ∶= t1−t2−t3, t231 ∶= t2−t3−t1,t312 ∶= t3−t1−t2 and define the points P1, P2, P3 by P1 =[t1t123 ∶ t2t231 − t2t3 − k t2 t3 t123 ∶ t3t312 − t2t3 − k t2 t3 t123]Δ, P =[t t − t t − k t t t ∶ t t ∶ t t − t t − k t t t ] , 2 1 123 3 1 3 1√231√ 2 231 3 312 3 1 √ 3√ 1 231 Δ P =[t t − t t − k t t t ∶ t t − t t − k t t t ∶ t t ] . 3 1 123 1 2 √ 1√ 2 312 2 231 1 2 1 2√312√ 3 312 Δ P , P , P are the vertices of the first circumcircle-midarc-triangle of Δ for k =1 and the 1 2 3 √ √ √ √ 0 vertices of the second circumcircle-midarc-triangle for k =−1. The perspector P has euclidean limit X56 resp. X55 if k =1 resp. k =−1, but P need not lie on the line I ∨ O.

2.8.3. Lemma / part 2. Let R = [r1∶r2∶r3]Δ be a point not on a sideline of Δ and let R1 = [r1+ t1∶ r2 ∶ r3]Δ, R2 = [r1∶ r2+ t2 ∶ r3]Δ, R3 = [r1 ∶ r2 ∶ r3+ t3]Δ be points on the lines R ∨ A, R ∨ B, R ∨ C, respectively. Then the three points (R2 ∨ R3) ∧ a, (R3 ∨ R1) ∧ b, (R1 ∨ R2) ∧ c are collinear on the tripolar of the point P =[t1 ∶ t2 ∶ t3]Δ. Examples: ∙ If Q is the perspector of a circumconic of Δ and R1, R2, R3 are the second intersections of the lines R ∨ A, R ∨ B, R ∨ C with this circumconic, then P is the Ceva point [1∕(r2p3+r3p2) ∶ 1∕(r3p1+r1p3) ∶ 1∕(r1p2+r2p1)]Δ of Q and R. ∙ A line l is a pseudoparallel of the sideline a¨ of the triangle Δ0 precisely when its dual l is a point on the bisector of the inner angle ∠+CAB of Δ0. If we choosefor R1, R2, R3 the vertices of the ﬁrst resp. second tangent-midarc-triangle,thus points on the angle bisectors, then the coordinates of the point P are 1 1 1 P =[ ∶ ∶ ] , sinh(d(I,A)− kr) sinh(d(I,B)− kr) sinh(d(I,C)− kr) Δ with r = radiusof theincenter and k =1 incase oftheﬁrst and k = −1 in case of thesecond tangent-midarc-triangle. The euclidean limits of these two centers are X8091 and X8092, but R1R2R3 and Δ are, in general, not orthologic triples.

2.9. Isoptics and isogonic points.

2.9.1. Isoptics and thaloids. Given three noncollinear points P,Q,R, the isoptic (curve) of the segment [P ; Q]+ through R is the point set isoptic(P,Q; R)={X d(X ∨ P,X ∨ Q)= d(R ∨ P , R ∨ Q)}. In the euclidean plane, such an isoptic is, in general, the union of two circles. But these two circles merge into one single circleð when ∠P RQ is a right angle (Thales’ theorem). The situation is similar in the elliptic and in the extended hyperbolic plane. An isoptic of a segment is an algebraic curve of degree 4: Let p = (p1,p2,p3), q, r, x be vectors with P =[p1∶p2∶p3]Δ, etc, then the equation of the isoptic is 2 (x × p) [D] (x × q) (r × p) [D] (r × p) (r × q) [D] (r × q) 2 = (r × p) [D] (r × q) (x × p) [D] (x × p)(x × q [D] (x × q). ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 41

If the angle ∠P RQ is right, this equation reduces to (x × p) [D] (x × q)=0, which is the equationofa conic but, in general,nottheequationof a circle. Thisconicis called orthoptic or Thales conic or thaloid.

FIGURE 13. The three apollonian thaloids of Δ0.

2.9.2. The apollonian thaloids of a triangle. Let P1, P2, P3 be the points where the tripolar oftheincenter I meets the sidelines a, b, c, respectively. Thethreecurvesisoptic(AI , P1; A), isoptic(BI , P2; B), isoptic(CI , P3; C) are thaloids. If two of these thaloids meet at a point, then this point is also a point of the third, see [28]. The euclidean limits of these thaloids are the apollonian circles of Δ0, so these thaloids are called apollonian thaloids of Δ0 in [28].

2.9.3. Isogonic points of a triangle. A point P is called isogonic point of Δ0 iﬀ d(P ∨ A, P ∨ B) = d(P ∨ B, P ∨ C) = d(P ∨ C, P ∨ A) = 1 i. Not every triangle has an isogo- 3 nic point. But in the elliptic plane every triangle has two isogonic points, and the same is true for a proper hyperbolic triangle, i.e. a triangle lying inside the absolute conic. In the euclidean case, there is a well known procedure how to ﬁnd the isogonic points of a triangle by a geometric construction: The 1st(2nd) isogonic point is the intersection of the circumcircles of the equilateral triangles erected outwardly (inwardly) on the sides of Δ0. A similar constructioncan be used in the elliptic and in the hyperbolic plane, but now using isoptics (instead of circles) and isosceles triangles with angles of measure 1 i at the apex, 3 see Figure 14. Suppose that all the sides of Δ are smaller than 1 i and all the angles smaller than 0 2 2 i (with respect to ≺) and that isogonic points exist for Δ . Then one isogonic point 3 0 lies inside the triangle and is a Fermat point of Δ0, a point which minimizes the function X → d(X,A)+ d(X,B)+ d(X,C). See [7] for a proof in the elliptic case. 42 MANFRED EVERS

FIGURE 14. Upper part: The isogonic points F1 and F2 are constructed with the help of isoptics. Lower part: The dual lines of F1 (green) and F2 (blue).

The lower part of Figure 14 shows a dual version of the upper. The green and the blue line are the duals of the isogonic points F1 and F2. Eachofthese linesis cut by the sidelines ¨ ¨ ¨ ¨ a , b , c of Δ into equidistant parts. Since F1 is a Fermat point of Δ0, the green line, the ¨ dual of F1, minimizes the sum of the angle distances to the sidelines of Δ . Problems: 1. Determine the coordinates of the isogonic points. 2. Experiments with GeoGebra sug- gest that, as in the euclidean case, a point P is isogonic precisely when its orthocorrespondent is identical with P . A proof is missing.

2.10. Reﬂection triangles. Euclidean analogues of the theorems presented in this section can be found in [12].

2.10.1. The reflection-triple of a point with respect to Δ. Given a point P and a line l, the reflection of P in l will be denoted by P (l). The reflections of P in the sidelines of Δ, the points P (a), P (b), P (c), are collinear precisely when P is a point on the cubic x x x d d d d d d d d 2 1 2 3 11 22 33 −4 23 31 12 + j,j j+1,j+2 j=1,2,3 2 d d d d d + xj xj+1 j+2,j+2( ∑j,j+1 j+1,j+2 − j+1,j+1 j+2,j) j=1,2,3 ∑ ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 43

d d d d d − xj+2 j+1,j+1( j+1,j+2 j+2,j − j+2,j+2 j,j+1) = 0 The euclidean limit of this cubic is the union of the circumcircle and the line at infinity. The triple P (a)P (b)P (c) is called the reflection-triple of P with respect to Δ. (a) (b) (c) The circumcenter of the reflection-triangle (P P P )0 of P is the isogonal conjugate of P .

The anticevian triple ΔP and the reﬂection-triple of P are perspective at the cevian quo- tient of H and P , d d d d d d d d d [p1(− 23p1 + 31p2 + 12p3) ∶ p2( 23p1 − 31p2 + 12p3) ∶ p3( 23p1 + 31p2 − 12p3)]Δ.

If P is not a vertex of Δ, the reﬂection triple of P with respect to the cevian triple ΔP is perspective to Δ at the point p p 2 d p 2 d p p d p 2 d p 2p 2 ⋯ ⋯ . Q= [ 1 1 ( 33 2 −2 23 2 3 + 22 3 )− 11 2 3 ∶ ∶ ]Δ For P = H or P = L , the perspector Q is a point on the orthoaxis.

2.11. The Neuberg cubic and two related cubics. The reﬂection triple of a point Q is perspective to Δ precisely when Q lies on the Neuberg cubic p(K, H⋆), d d d d x d x 2 d x 2 ( j,j j+1,j+2)+2 j+2,j j,j+1) j ( j+2,j+2 j+1 − j+1,j+1 j+2)=0. j=1,2,3 This is a self-isogonal∑ cubic with pivot H⋆. On this cubic lie the incenter, the excenters, + ⋆ the triangle centers H,O , H ,J+,J−, the points W1 = [sinh( )(1 + 2(cosh( ) − cosh( ) − cosh( ))) ∶ ⋯ ∶ ⋯]Δ. 2 2 2 W2 = [sinh( )(4 cosh( )(cosh ( )−cosh ( )−cosh ( ))−cosh( )+4cosh( ) cosh( )) ∶ ⋯ ∶ ⋯]Δ, 2 2 2 W3 = [sinh( )(4(cosh ( )− cosh ( ) − cosh ( )) + 1)∕(cosh( ) +2cosh( ) cosh( )) ∶ ⋯ ∶ ⋯]Δ and their isogonal conjugates. W1 is the Evans perspector, the perspector of the two triples (a) (b) (c) I1I2I3 and A B C ; its euclidean limit is X484. W2 has euclidean limit X399, and W3 is a point on the line through O+ and the isogonal conjugate of N+ and has euclidean limit X1157.

We introduce another cubic which consists of all points R that satisfy one of the following three equivalent conditions: (1) The reﬂection triple of R and the cevian triple of R are perspective. (2) The reﬂection triple of R and the triple A(a)B(b)C(c) are perspective. (3) The cevian triple of R and the triple A(a)B(b)C(c) are perspective.

This cubic is p(W4, W5),a W4-isocubic with pivot W5, 2 2 W4 = [sinh ( )∕(1 − 4cosh ( )) ∶ ⋯ ∶ ⋯]Δ, 2 W5 = [sinh( ) cosh( )∕(1 − 4cosh ( )) ∶ ⋯ ∶ ⋯]Δ. The euclidean limits of W4, W5 are X1989,X265, respectively. ⋆ → ↦ There is a bijective mapping p(K, H ) p(W4, W5), given by Q R, where R is the perspector of the two triples Q(a)Q(b)Q(c), Δ and Q is the perspector of the two triples (a) (b) (c) R R R , ΔR. + ⋆ Q H O H I W1 W2 W3 + ⋆ R H N W5 W6 W7 H W8 44 MANFRED EVERS

W6 = [sinh( )∕(1 + 2cosh( )) ∶ ⋯ ∶ ⋯]Δ with euclidean limit X79, W7 = [sinh( )∕(1 − 2cosh( )) ∶ ⋯ ∶ ⋯]Δ with euclidean limit X80, 2 W8 = [sinh( )∕((1 − 4cosh ( ))(cosh( ) + 2cosh( ) cosh( )) ∶ ⋯ ∶ ⋯]Δ with euclidean limit X1141, The points P whose anticevian triple ΔP is perspective to the triple A(a)B(b)C(c) form the cubic p(K,N+), the isogonal cubic with pivot N+. There is a bijective mapping p(K,N+) → p(K, H⋆), P ↦ Q = perspector of ΔP and A(a)B(b)C(c).

+ + P H O W9 I W10 N W11 + ⋆ Q H W2 O W1 I H W3 2 2 2 W9 = [sinh( )(4 cosh( )(cosh ( )−cosh ( )−cosh ( ))−cosh( )−4 cosh( ) cosh( )) ∶ ⋯ ∶ ⋯]Δ with euclidean limit X195, W10 = [sinh( )(1 + 2(−cosh( ) + cosh( ) + cosh( ))) ∶ ⋯ ∶ ⋯]Δ with euclidean limit X3336, + W11 = isogonal conjugate of N .

2.12. Substitutes for the Euler line. 2.12.1. The Euler line in euclidean geometry. In euclidean geometry, the centroid G and the orthocenter H of a triangle have a common cevian circle, the nine-point-circle(or Euler circle). G, H andthecenter N ofthiscirclearecollinear. TheylieontheEulerline, together with the circumcenter of the triangle and several other interesting triangle centers. In elliptic and in hyperbolic geometry, the points G and H do not have a common cevian circle, so there is no direct analog of an Euler line in these geometries, but there are several central lines that can serve as a substitute. We list four of these. 2.12.2. The orthoaxis. One of these lines, the orthoaxis, we treated already in subsection 2.1.7. As was shown by N. Wildberger [28], there are triangle centers with euclidean limits Xi,i =2, 3, 4, 20, 30 lying on this line. So Vigara proposed to call it Euler-Wildberger line.

2.12.3. The orthoaxis of the medial triangle. The orthoaxis of the medial triangle ΔG,0 is the line G ∨O. Besides G and O, it passes throughthe isogonal conjugate of O, throughthe Lemoine-isoconjugateof O, throughL and several other points of interest. A more detailed description of this line is given in [5]. G O G The line ∨ can also be a substitute for an Euler line of the anticevian triangle Δ 0 G of G. G can serve as a pseudo-centroid, C0 as a pseudo-Euler-circle and the lines A ∨ A¨,BG B¨,CG C¨ G ∨ ∨ as pseudo-altitudes of Δ 0. These pseudo-altitudes meet at a point G O G G on ∨ which is the circumcevian conjugate of with respect to Δ 0. Problem: Taking ΔG as a reference triple, what are the coordinates of the point G? 2.12.4. The line through G and H. The line G ∨ H has the equation c c cc cc c c ( j+2,j j,j+1 − j+1,j+2)(1 + j+1,j+2)( j,j+1 − j+2,j)xj =0, j=1,2,3 É d d d which is equivalent to j+1,j+2( j+2,j − j,j+1)xj =0. j=1,2,3 É ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 45

It passes through the center of a circle which touches internally the incircle and externally the excircles of Δ0. This was shown for a triangle on a sphere in 1864 by G. Salmon [22], and Salmon also calculated the trilinear coordinates of this center, which we denote by N̂ . It has barycentric coordinates: ̂ d d D d d D d d D N =[ 31 12 − det( ), 12 23 − det( ), 23 31 − det( )]Δ cc c c 2 c c c 2 c c c ⋯ ⋯ = [(1+ 23)( ( 31 − 31 12 + 12 +1)+ 23( 31 12 −1)) ∶ ∶ ]Δ = [sinh( ) cosh( − ) ∶ sinh( ) cosh( − ) ∶ sinh( ) cosh( − )]Δ.

An equation of the circle in trilinear coordinates was given for the elliptic (spherical) case in 1861 by A.S. Hart [11]; Salmon therefore used the name Hart’s circle. c(c −2c + c )+ c c −1 The Hart circle is a conic (M) with m = 12 23 31 12 31 , m = 11 cc 23 23 +1 cc 1− 23, ⋯ . Salmon has shown the following relation between the radius H of the Hart circle and the radius R of the circumcircle: tanh( )= 1 tanh(R). H 2

The touchpoint of the incircle ℐk with the Hart circle we denote by F0,k; these Feuer- bach points of triangle Δ0 have coordinates d d D d d d D d d d D d F0,0 =[ 31 12 −det( )−f 11 ∶ 12 23 −det( )−f 22 ∶ 23 31 −det( )−f 33]Δ, − − − = [sinh( ) sinh2( √) ∶ sinh( ) sinh2( ) ∶ sinh(√ ) sinh2( )] , √ 2 2 2 d d d 2 D f = 12 23 31 det ( ), d d D d d d D d d d D d F0,1 =[ 31 √12 −det(√ √)− f 11 ∶ 12 23 −det( )+ f 22 ∶ 23 31 −det( )+ f 33]Δ, etc. √ √ √

The triple F0,1F0,2F0,3 is perspective to Δ; the perspector is the point d d D d d d D d d d D d [ 31 12 −det( )+f 11 ∶ 12 23 −det( )+f 22 ∶ 23 31 −det( )+f 33]Δ, = [sinh( ) cosh2( − ) ∶ sinh( ) cosh2( − ) ∶ sinh( ) cosh2( − )] . 2 √ 2 √ 2 Δ √ ̂ It is the harmonic conjugate of F0,0 with respect to {I0, N} and has euclidean limit X12. The dual of the Hart circle is a circle with center N̂ which internally touches the circumcircle C0 and externally the circumcircles Ck, k = 1, 2, 3 , with touchpoints lying on ̂ the lines N ∨ Ok, k =0, ⋯ , 3, respectively. 2.12.5. The Akopyan line. The line O ∨ H⋆ has the equation c c cc cc c j+2,j − j,j+1 (1− j+2,j)(1 − j,j+1)−2(1− c j+1,j+2) xj =0. j=1,2,3 There∑ are several important triangle centers on this line. First of all, a point Ĝ whose cevian lines divide the triangle area in equal parts. The existence of such a point was already shown (for a spherical triangle) 1827 by J. Steiner [23]. The coordinates of this point are 1+ cc 1+ cc [ 23 ∶ 31 cc cc cc cc cc cc 2 1+ 23√+ 1+ 31 1+ 12 2 1+ 31√+ 1+ 12 1+ 23 cc √ √ 1+√ 12 √ √ √ √ √ ∶ ]Δ cc cc cc 2 1+ 12√+ 1+ 23 1+ 31 √ √ √ √ 46 MANFRED EVERS

FIGURE 15. The Akopyan line, the Hart circle (red) and incircle (brown)

cosh( 1 a) = [ 2 cosh( 1 a) + cosh( 1 b) cosh( 1 c) 2 2 2 cosh( 1 b) cosh( 1 c) 2 2 ∶ ∶ ]Δ. cosh( 1 b) + cosh( 1 c) cosh( 1 a) cosh( 1 c) + cosh( 1 a) cosh( 1 b) 2 2 2 2 2 2 Another point on O ∨ H⋆ is the center N̂ of the Hart circle. Akopyan [1] proved in 2011 that this Hart circle is a circumcircle of the traces of Ĝ and also a circumcircle of the traces of another point Ĥ , which he called pseudo-orthocenter. Moreover, he showed that the three points G,̂ N,̂ Ĥ are collinear on a line which passes also throughthe circumcenter O. It is easy to check by calculation that this line is O ∨ H⋆. Vigara proposed to call this line Akopyan-Euler-line. Akopyan used the name Euler line. We introduce the Akopyan-measure ofaninnerangleofa triangleasthesum oftheother two inner angles diminished by a half of the triangle area 4). For example, the Akopyan- measure of the angle ∠ ACB of triangle Δ is ̂(∠ ACB)= 1 ( + − + i). By adding + 0 + 2 up the Akopyan-measuresof the three inner angles of triangle Δ we get 1 area(Δ )+2i. 0 2 0 With the help of the Akopyan-measure we can formulate the following version of the In- scribed Angle Theorem, cf. [1]: ̂(∠ ACB) = 1 i ± (∠ OAB), depending on whether the circumcenter O lies outside + 2 +

4) See [20]. ON THE GEOMETRY OF A TRIANGLE IN THE ELLIPTIC AND IN THE EXTENDED HYPERBOLIC PLANE 47 the triangle Δ0 or not. We like to shortly present Akopyan’s explanation for Ĥ having properties similar to those of the orthocenter. For this we also make use of the Akopyan measure: While the altitude C∨H meets the line c at the point CH such that the angles ∠+(ACHC) and ∠ (BC C) have both measures equal to 1 i, the pseudo-altitude C∨Ĥ meets the line + H 2 c in C such that ̂(∠ AC C)= ̂(∠ BC C)= 1 (i+ + + ). Ĥ + Ĥ + Ĥ 4 1+ cc ̂ 23 The coordinates of H are [ ∶ ⋯ ∶ ⋯]Δ cc cc cc 2 1+ 23√− 1+ 31 1+ 12 cosh( 1 a) √ √ √ 2 √ = [ ∶ ⋯ ∶ ⋯]Δ. cosh( 1 a) − cosh( 1 b) cosh( 1 c) 2 2 2

The points H,̂ N,̂ G,Ô form a harmonic range. Proof : We use the following abbreviations: c1 ∶= cosh(a∕2), c2 ∶= cosh(b∕2), c3 ∶= cosh(c∕2), s1 ∶= sinh(a∕2),s2 ∶= sinh(b∕2),s3 ∶= sinh(c∕2). Deﬁne vectors g, h, n, o by cj cj g =(g1,g2,g3), gj = , h =(ℎ1,ℎ2,ℎ3), ℎj = , cj + cj+1cj+2 cj − cj+1cj+2 2 2 2 2 2 2 2 2 2 2 n =(n1, n2, n3), nj = cj cj (2cj+1cj+2 − cj+1 − cj+2)− cj+1sj+1 − cj+2sj+2 2 2 2 2 o =(o1, o2, o3), oj =−sj(1 + cj − cj+1 − cj+2), and deﬁne real numbers r,s,t by r = c1c2c3, 2 2 2 s =(c1 + c2c3)(c2 + c3c1)(c3 + c1c2)(1+2r − c1 − c2 − c3 ), 2 2 2 t =(c1 − c2c3)(c2 − c3c1)(c3 − c1c2)(1−2r − c1 − c2 − c3 ). ̂ ̂ ̂ Then G =[g1∶g2∶g3]Δ, H =[ℎ1∶ℎ2∶ℎ3]Δ, N =[n1∶n2∶n3]Δ,O =[o1∶o2∶o3]Δ and ro + n = sg , ro − n = th. □

̂ ¨ Thecevian line C∨G meets the perpendicularbisector CG∨C at a pointwithcoordinates

1+ cc 1+ cc [ 23 ∶ 31 cc cc cc cc cc cc 2 1+ 23√+ 1+ 31 1+ 12 2 1+ 31√+ 1+ 31 1+ 23 cc cc √ √ √ √(1− 12) √1+√ 12 √ √ ∶ ]Δ. cc cc cc cc cc cc ( 2 1+ 23 + 1+ 31 1+ 12)(√ 2 1+ 31 + 1+ 31 1+ 23) √ √ √ √ √ √ √ √ This point is the center of a circle throughthe points A,B,AĤ ,BĤ . Thus,the line AĤ ∨BĤ is antiparallel to A ∨ B with respect to the angle ∠+ACB, and since AĤ ,BĤ ,AĜ ,BĜ are points on the Hart circle, the line AĜ ∨ BĜ is parallel to A ∨ B . We give a short hint of how to construct the cevians Ĝ ∨ C and Ĥ ∨ C; see [5] for a more detailed explanation. Let m be the sideline AG∨BG of the medial triangle. Deﬁne points R1 = ped(A,m), R2 = ped(B,m), S1 = m ∧ perp(c,A), S2 = m ∧ perp(c,B). The ̂ bisectors of the angles ∠−R1AS1 and ∠−R2BS2 meet at a point on G ∨ C, the bisectors of 48 MANFRED EVERS

̂ the angles ∠+R1AS1 and ∠+R2BS2 meet at a point on H ∨ C.

The Hart circle meets each cevian line of Ĝ and Ĥ twice. One intersection point is the correspondingcevian point, the other one is also a signiﬁcant point. The description of this second point generalizes as follows: If P and Q are two points not on the sidelines of Δ, then for each vertex R ∈{A,B,C},the cevian line R∨RP meets the bicevian conic of P and Q in a point which is the harmonic conjugate with respect to {R, P } of the intersection point of R ∨ RP with the tripolar of Q. Proof : Without loss of generality we take R = C. For the matrix of the bicevian conic of P = [p1∶p2∶p3]Δ and Q = [q1∶q2∶q3]Δ, see 2.1.7. The second intersection of the cevian line C ∨ CP with this conic is the point S =[p1q1q2 ∶ p2q1q2 ∶ p1q2q3+q1p2q3+2q1q2p3], the intersection of C ∨ CP with Q is the point T =[p1q1q2 ∶ p2q1q2 ∶−q3(p1q2 + p2q1)]. It can be easily checked that P,S,C,T form a harmonic range. □ Addition: The pole of Q with respect to the bicevian conic of P and Q, [q1(2p1q2q3 + q1(p2q3 + p3q2)) ∶ ⋯ ∶ ⋯]Δ, is a pointontheline P ∨ Q. If P is a perspector of an inconic of Δ, then its polar with respect to this conic agrees with its tripolar.

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Current address, Manfred Evers: Bendenkamp 21, 40880 Ratingen, Germany E-mail address, Manfred Evers: [email protected]