Each allowable triangle of an isotropic plane can be set in a
standard position, in which it is possible to prove geometric properties analytically in a simplified and easier way by means of
the algebraic theory developed in this paper

J. Beban-Brkić

R. Kolar-Šuper

V. Volenec

Z. Kolar-Begović

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Mathematical Communications 10(2005), 149-157 149Metrical relationships in a standard triangle in anisotropic planeR. Kolar–Sˇuper∗, Z. Kolar–Begovic´†, V. Volenec‡ and J. Beban–Brkic´§Abstract. Each allowable triangle of an isotropic plane can beset in a standard position, in which it is possible to prove geometricproperties analytically in a simplified and easier way by means of thealgebraic theory developed in this paper.Key words: isotropic plane, triangle, standard triangleAMS subject classifications: 51N25Received November 11, 2005 Accepted December 20, 20051. Terms of elementary geometry in I2Let P2(R) be a real projective plane, f a real line in P2, and A2 = P2 \ f theassociated affine plane. The isotropic plane I2(R) is a real affine plane A2 wherethe metric is introduced with a real line f ⊂ P2 and a real point F incidental withit. The main facts about the isotropic plane can be found in [1], [2], [3].We will first define some terms and point out some properties of triangles and circlesin I2 that are going to be used further on.All straight lines through the point F are called isotropic straight lines (isotropiclines). All other straight lines are simply called straight lines. Two points A,B (A =B) are called parallel if they lie on the same isotropic line. For two non-parallelpoints A = (a1, a2), B = (b1, b2), the isotropic distance is defined by d(A,B) =b1 − a1. Note that the isotropic distance is directed. For two parallel points A =(a1, a2), B = (b1, b2), a1 = b1, the quantity known as an isotropic span is defined bys(A,B) = b2−a2. A straight line P through two points A and B will be denoted byP = AB. Furthermore, it will be recognized from the context whether, for exampleBC, refers to the straight line passing through points B and C or to the length ofthe line segment with endpoints B,C.∗Teacher Training College, University of Osijek, Lorenza Ja¨gera 9, HR-31 000 Osijek, Croatia,e-mail: rkolar@vusos.hr†Department of Mathematics, University of Osijek, Gajev trg 6, HR-31 000 Osijek, Croatia,e-mail: zkolar@mathos.hr‡Department of Mathematics, University of Zagreb, Bijenicˇka c. 30, HR-10 000 Zagreb, Croatia,e-mail: volenec@math.hr§Department of Geomatics, Faculty of Geodesy, Kacˇic´eva 26, University of Zagreb, HR-10 000Zagreb, Croatia, e-mail: jbeban@geof.hr150 R.Kolar–Sˇuper, Z.Kolar–Begovic´, V.Volenec, J. Beban–Brkic´Each non-isotropic straight line G ⊂ I2 can be written in the normal formy = ux + v, that is, in line coordinates, G = (u, v). For two straight linesG1 = (u1, v1), G2 = (u2, v2) the isotropic angle is defined by ϕ = ∠(G1,G2) = u2−u1.Note that the isotropic angle is directed as well. The Euclidean meaning of theisotropic angle can be understood from the affine model given in Figure 1.For two parallel straight lines G1 = (u1, v1), G2 = (u2, v2) there exists anisotropic invariant defined by ϕ∗(G1,G2) = v2 − v1 (see Figure 2).Figure 1. Figure 2. Figure 3.An isotropic normal to the straight line G = (u, v) from the point P = (p1, p2),P /∈ G is an isotropic line through P . The inverse holds as well, i.e. each straightline G ⊂ I2 is a normal for each isotropic straight line.Denoting by S = (s1, s2) the point of intersection of the isotropic normal fromthe point P with the straight line G, the isotropic span of the point P from the lineG is given by s(S, P ) = p2 − s2 = p2 − up1 − v (see Figure 3).2. Triangle in a standard positionUnder a triangle in I2 an ordered set of three non-collinear points (A,B,C) isunderstood. A,B,C are called vertices, and BC, CA, AB sides of a triangle. Atriangle is called allowable if none of its sides is isotropic. In an allowable trianglethe lengths of the sides are defined by BC = d(B,C), CA = d(C,A), AB = d(A,B),with BC = 0, CA = 0, AB = 0, BC + CA + AB = 0. For the directed angles wehave ∠A = ∠(AB,AC), ∠B = ∠(BC,BA), ∠C = ∠(CA,CB), ∠A+∠B+∠C = 0(see Figure 4).Isotropic altitudes HBC , HCA, HAB associated with sides BC, CA and AB areisotropic straight lines passing through vertices A, B, C, i.e. normals to sides BC,CA and AB. Their lengths are defined by ha = s(Ah, A) where Ah = BC ∩ HBC ,etc. The Euclidean meaning is given in Figure 5.Metrical relationships in a standard triangle ... 151Figure 4. Figure 5.According to [1], to any allowable triangle in I2 exactly one circle can be cir-cumscribed. The equation of the circumcircle is of the formy = ux2 + vx+ w, u = 0.The coordinate system can be moved to the position that reduces above equationto the form y = ux2. Without loss of generality, by using the substitution yu → y,we will assume that the equation of this circle is given byKc . . . y = x2. (1)Excepting as arranged, vertices of a given allowable triangle ABC areA = (a, a2), B = (b, b2), C = (c, c2), (2)a, b, c being mutually different numbers. In writing relations, equations, etc., thefollowing abbreviations will be useful:a+ b+ c = s, abc = p, ab+ bc+ ca = q. (3)Besides, by choosing without loss of generality, thats = a+ b+ c = 0 (4)is fulfilled, it follows that the y–axis coincides with the diameter of the circle passingthrough the centroidG =(a+ b+ c3,a2 + b2 + c23)=(0,−23q)(5)of ABC, while the x–axis is a tangent line of the circumscribed circle at theendpoint of the diameter through G.For each allowable triangle ABC it can be achieved in the described way, thatits circumscribed circle has equation (1) and its vertices are of the form (2) whileequalities (3) and (4) are satisfied. We shall say that this triangle is in the standardposition or shorter triangle ABC is a standard triangle. To prove geometric factsfor each allowable triangle it is sufficient to give a proof for a standard triangle.152 R.Kolar–Sˇuper, Z.Kolar–Begovic´, V.Volenec, J. Beban–Brkic´3. Algebraic relationships in the standard triangleFor the systematic studying of the triangle in an isotropic plane various conse-quences of equations (3) and (4) will be useful.First of all, the following equalitiesa2 = bc− q, b2 = ca− q, c2 = ab− q (6)are valid, because of e.g.a2 = −a(b+ c) = bc− q,so it followsa2 + b2 + c2 = −2q, (7)and it implies that q < 0. Equality (7) is used in (5). Then we haveq = −(b2 + bc+ c2) = −(c2 + ca+ a2) = −(a2 + ab+ b2) (8)because of e.g.b2 + bc+ c2 = (b+ c)2 − bc = a2 − bc = −qis valid according to (6).The formulaBC2 + CA2 +AB2 = (b− c)2 + (c− a)2 + (a− b)2 = −6q (9)is also valid because of2(a2 + b2 + c2)− 2(bc+ ca+ ab) = −4q − 2q = −6q.The following equalities are also interesting(b− c)2 = −(q + 3bc), (c− a)2 = −(q + 3ca), (a− b)2 = −(q + 3ab), (10)(c−a)(a−b) = 2q−3bc, (a−b)(b−c) = 2q−3ca, (b−c)(c−a) = 2q−3ab, (11)and they are obtained for example in this way(b− c)2 = (b+ c)2 − 4bc = a2 − 4bc = bc− q − 4bc = −(q + 3bc),(c− a)(a− b) = −a2 − bc+ ca+ ab = −(bc− q) + q − 2bc = 2q − 3bc.By means of (10), the equality(b− c)2(c− a)2(a− b)2 = −(27p2 + 4q3) (12)follows, and therefore the inequality 27p2 + 4q3 < 0 is valid. Indeed, the left-handside of (12) is equal to−(q + 3bc)(q + 3ca)(q + 3ab)=−q3−3q2(bc+ ca+ ab)− 9qabc(a+ b+ c)−27a2b2c2= −27p2 − 4q3.Metrical relationships in a standard triangle ... 1534. Elements of the standard triangleThe standard triangle ABC has vertices given by equalities (2). The sides of thattriangle have the equationsBC . . . y = −ax− bc,CA . . . y = −bx− ca, (13)AB . . . y = −cx− ab,since, for example, for the point B = (b, b2) and the first line (13) owing to (4) theequality b2 = −ab− bc is valid.With x = a from the equation of the line BC owing to (6) it followsy = −a2 − bc = q − 2bc,and we obtain the first of the three feet of altitudesAh = (a, q − 2bc), Bh = (b, q − 2ca), Ch = (c, q − 2ab) (14)of the triangle ABC, and the remaining two are obtained analogously. Now, becauseof (6) and (11), we obtain, for example, for the altitude haha = s(Ah, A) = a2 − (q − 2bc) = 3bc− 2q = −(c− a)(a− b),then equalitiesha = −(c− a)(a− b), hb = −(a− b)(b− c), hc = −(b− c)(c− a) (15)are valid. Triangle ABC with vertices (2) has the (oriented) area , for which weobtain2 =∣∣∣∣∣∣a a2 1b b2 1c c2 1∣∣∣∣∣∣ = (b − c)(c− a)(a− b) = −BC · CA ·AB, (16)because of BC = c− b, CA = a− c, AB = b− a. From (15) and (16) the followingequalities, analogous to those in Euclidean geometry, follow immediately2 = BC · ha = CA · hb = AB · hc. (17)If we want the equality 4R = −BC ·CA ·AB to be as in Euclidean geometry, wehave to arrange that the radius of the circumscribed circle of the standard triangleis equal to R = 12 , i.e. this is the radius of the circle with equation (1). (In [1]the arrangement is slightly different; it is arranged that circle (1) has radius 1.)In various metric formulae, which are not seemingly homogenous, i.e. all membersare not of the same dimension, it is necessary to replace factor 1 with 2R at someplaces so that the formula becomes homogenous. So, for example the first equation(15) should be written in the form 2Rha = −CA ·AB.As the absolute point F has the role of the orthocenter and the center of thecircumscribed circle too, then the isotropic line through the centroid G of that154 R.Kolar–Sˇuper, Z.Kolar–Begovic´, V.Volenec, J. Beban–Brkic´triangle will be called an Euler line of that triangle (Figure 6). Because of (5), thestandard triangle ABC has Euler line E with the equation x = 0.The line with the equationy = 2ax+ 2bc− qpasses through points Bh and Ch from (14) since e.g. for the point Bh the equality2ab+2bc− q = q− 2ca is valid. For that reason it is the equation of the line BhCh.Triangle AhBhCh is called an orthic triangle of triangle ABC. Its sides have theequationsBhCh . . . y = 2ax+ 2bc− q,ChAh . . . y = 2bx+ 2ca− q, (18)AhBh . . . y = 2cx+ 2ab− q.Theorem 1. The corresponding sides of the triangle and its orthic triangleintersect at three points which lie on the same line (Figure 6).Proof. If we add to the first equation in (18) the first equation from (13)multiplied by 2, we obtain the equation of the line 3y = −q on which the pointBC ∩BhCh lies, and the same is valid for the points CA ∩ChAh and AB ∩AhBh.✷The line from Theorem 1 is called, by the analogy with the Euclidean case,orthic axis of the observed triangle.Corollary 1. The orthic axis H of the standard triangle ABC has the equationH . . . y = − q3. (19)The orthic axis of the triangle has one more interesting property.Theorem 2. The centroid of three points at which an arbitrary isotropic straightline intersects sides of an allowable triangle lies on the orthic axis of this triangle.Proof. Really, the arithmetic mean of the right-hand sides of equalities (13) isequal to − q3 for each x, because of (4). ✷In some way, the next theorem is related to Theorem 2.Theorem 3. The sum of the spans from vertices of the triangle to the line Gis equal to zero if and only if that line passes through the centroid of that triangle.Proof. The line G with equation y = ux+v passes through the centroid G from(5) of the standard triangle ABC under the assumption v = − 23q. Spans of pointsA,B,C from (2) to the line G are, respectively, equal toa2 − au− v, b2 − bu− v, c2 − cu− vand because of (4) and (7) we have the sum −2q− 3v, which is equal to zero underthe same assumption v = − 23q. ✷Out of all the lines through the centroid of the triangle the one which is parallelto its orthic axis is specially interesting (Figure 6). That line will be called inertialaxis of triangle.Corollary 2. The inertial axis Gof the standard triangle has the equationG . . . y = −23q. (20)Metrical relationships in a standard triangle ... 1555. Complementarity and anticomplementarityIf G is the centroid of the allowable triangle ABC, then the homothecy (G,− 12 )and its inverse homothecy (G,−2), by analogy with the Euclidean case, will becalled complementarity resp. anticomplementarity with regard to triangle ABC.The image of the point T (or the curve K) by these homothecies will be called acomplementary resp. anticomplementary point (curve) of the point T (of the curveK, respectively).Theorem 4. The point (−x2 ,− y2 − q) is a complementary point and the point(−2x,−2y− 2q) is an anticomplementary point to the point T = (x, y), with regardto triangle ABC.Proof. If the point T ′ = (x′, y′) is complementary to the point T = (x, y)then the equality T + 2T ′ = 3G is valid, i.e. because of (5) we have equalitiesx + 2x′ = 0 and y + 2y′ = −2q, wherefrom x′ = −x2 , y′ = − y2 − q follows. If thepoint T ′′ = (x′′, y′′) is anticomplementary to the point T = (x, y), then the equality2T +T ′′ = 3G is valid, i.e. we have equalities 2x+x′′ = 0 and 2y+ y′′ = −2q, fromwhich x′′ = −2x and y′′ = −2y − 2q follow. ✷Theorem 5. If the curve K has the equation K(x, y) = 0, then its complemen-tary and anticomplementary curve, with regard to the standard triangle ABC, havesuccessively the equationsK(−2x,−2y − 2q) = 0 and K(−x2,−y2− q)= 0. (21)Proof. If the point T ′ = (x′, y′) is complementary to the point T = (x, y) onthe curve K, then x = −2x′ and y = −2y′ − 2q is valid according to the proof ofTheorem 4, K(−2x′,−2y′−2q) = 0 is the equation of the curve which is described bythe point T ′. If we write x and y instead of x′ and y′, then we get the first equation(21). If the point T ′′ is anticomplementary to the point T = (x, y) then we obtainx = −x′′2 , y = − y′′2 −q by the proof of Theorem 4 hence K(−x′′2 ,− y′′2 −q) = 0 is theequation of the curve which is described by the point T ′′. With the substitutionsx′′ → x, y′′ → y we obtain the second equation (21). ✷The complementary point to the point A = (a, a2), because of−a22− q = −12(bc− q)− q = − q2− bc2and according to Theorem 4, is the first of the three analogous pointsAm = (−a2 ,−q2− bc2), Bm = (− b2 ,−q2− ca2), Cm = (− c2 ,−q2− ab2). (22)Anticomplementary point to the same point is the first one of the three pointsAn = (−2a,−2bc), Bn = (−2b,−2ca), Cn = (−2c,−2ab) (23)because of −2a2 − 2q = −2bc. Triangles AmBmCm and AnBnCn are complemen-tary and anticomplementary triangles of the standard triangle ABC. The points156 R.Kolar–Sˇuper, Z.Kolar–Begovic´, V.Volenec, J. Beban–Brkic´Am, Bm, Cm are in fact the midpoints of the sides BC, CA, AB of triangle ABC.Really, points B and C from (2) have the midpoint Am because of12(b+ c) = −a2,12(b2 + c2) =12(−q − bc)which is valid according to (4) and (8).A complementary resp. anticomplementary line to the line BC with the firstequation (13) has the equation −2y−2q = −a(−2x)−bc and − y2−q = −a(−x2 )−bcaccording to Theorem 4, and this is the first of the three analogous equations of thesides of the triangle AmBmCm and AnBnCn in the formBmCm . . . y = −ax− q + bc2 ,CmAm . . . y = −bx− q + ca2 , (24)AmBm . . . y = −cx− q + ab2 ,respectivelyBnCn . . . y = −ax+ 2a2,CnAn . . . y = −bx+ 2b2, (25)AnBn . . . y = −cx+ 2c2,because of bc − q = a2. The lines (25) are the middle lines of the triangle ABC,and the lines (26) are parallels with the lines BC, CA, AB, which obviously passsuccessively through the points A, B, C from (2).A complementary resp. anticomplementary circle to the circumscribed circle oftriangle ABC with the equation (1) is the circle with the equation −2y − 2q =(−2x)2 resp. − y2 − q = (−x2 )2, i.e. y = −2x2− q respectively y = − 12x2− 2q. Thuswe have proved these two theorems.Theorem 6. The complementary triangle AmBmCm to the standard triangleABC has vertices (22), sides with equations (24) and the circumscribed circle withthe equationy = −2x2 − q. (26)Theorem 7. The anticomplementary triangle AnBnCn to the standard triangleABC has vertices (23), sides with equations (25) and the circumscribed circle withthe equationy = −12x2 − 2q. (27)As midpoints Am, Bm, Cm of the sides of triangle ABC have the abscissas −a2 ,− b2 , − c2 , then the perpendicular bisectors of sides BC, CA, AB have successivelythe equationsx = −a2, x = − b2, x = − c2. (28)Metrical relationships in a standard triangle ... 157Figure 6.References[1] H. Sachs, Ebene isotrope Geometrie, Vieweg–Verlag, Braun-schweig/Wiesbaden, 1987.[2] I.M.Yaglom, Galilean principle of relativity and Non–Euclidean geometry (inRussian), Nauka, Moscow, 1969.[3] I.M.Yaglom, A Simple Non–Euclidean Geometry and Its Physical Basis,Springer, 1979.

Metrical relationships in a standard triangle in an isotropic plane

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