Harmonic quadrangle in isotropic plane

Transcription

1 Turkish Journal of Mathematics journals. tubitak. gov. tr/ math/ Research Article Turk J Math (018) 4: c TÜBİTAK doi: /mat Harmonic quadrangle in isotropic plane Ema JURKIN 1 Marija ŠIMIĆ HORVATH Vladimir VOLENEC 3 Jelena BEBAN-BRKIĆ 4 1 Faculty of Mining Geology and Petroleum Engineering University of Zagreb Zagreb Croatia Faculty of Architecture University of Zagreb Zagreb Croatia 3 Department of Mathematics University of Zagreb Zagreb Croatia 4 Faculty of Geodesy University of Zagreb Zagreb Croatia Received: Accepted/Published Online: Final Version: Abstract: The concept of the harmonic quadrangle and the associated Brocard points are introduced and investigated in the isotropic plane by employing suitable analytic methods. Key words: Isotropic plane cyclic quadrangle harmonic quadrangle 1. Introduction The isotropic plane is a real projective metric plane whose absolute figure is a pair consisting of an absolute point Ω and an absolute line ω incident to it. If T = (x 0 : x 1 : x ) denotes any point in the plane presented in homogeneous coordinates then usually a projective coordinate system where Ω = (0 : 1 : 0) and the line ω with the equation x = 0 is chosen. Isotropic points are the points incident with the absolute line ω and the isotropic lines are the lines passing through the absolute point Ω. Metric quantities and all the notions related to the geometry of the isotropic plane can be found in [8] and [7]. Now we recall a few facts that will be used further on wherein we assume that x = x 0 x and y = x 1 x. Two lines are parallel if they have the same isotropic point and two points are parallel if they are incident with the same isotropic line. For T 1 = (x 1 y 1 ) and T = (x y ) two nonparallel points a distance between them is defined as d(t 1 T ) := x x 1. In the case of parallel points T 1 = (x y 1 ) and T = (x y ) a span is defined by s(t 1 T ) := y y 1. Both quantities are directed. Two nonisotropic lines p 1 and p in the isotropic plane can be given by y = k i x+l i k i l i R i = 1 labeled by p i = (k i l i ) i = 1 in line coordinates. Therefore the angle formed by p 1 and p is defined by φ = (p 1 p ) := k k 1 being directed as well. Any two points T 1 = (x 1 y 1 ) and T = (x y ) have the ( 1 midpoint M = (x 1 + x ) 1 ) (y 1 + y ) and any two lines with the equations y = k i x + l i (i = 1 ) have the bisector with the equation y = 1 (k 1 + k )x + 1 (l 1 + l ). Correspondence: marija.simic@arhitekt.hr 010 AMS Mathematics Subject Classification: 51N5 666

2 A triangle in the isotropic plane is called allowable if none of its sides are isotropic (see [4]). The classification of conics in the isotropic plane can be found in [1] and [7]. Recall that the circle in the isotropic plane is the conic touching the absolute line ω at the absolute point Ω. The equation of such a circle is given by y = ux + vx + w u 0 u v w R. As the principle of duality is valid in the projective plane it is preserved in the isotropic plane as well.. Cyclic quadrangle in isotropic plane The geometry of the cyclic quadrangle in the isotropic plane appeared first in [9]. The diagonal triangle and diagonal points were introduced and several properties concerning them were discussed. Let ABCD be the cyclic quadrangle with y = x (1) as its circumscribed circle ([9] p. 67). Choosing A = (a a ) B = (b b ) C = (c c ) D = (d d ) () with a b c d being mutually different real numbers where a < b < c < d the next lemma is obtained. Lemma 1 ([9] p. 67) For any cyclic quadrangle ABCD there exist four distinct real numbers a b c d such that in the defined canonical affine coordinate system the vertices have the form () the circumscribed circle has the equation (1) and the sides are given by AB... y = (a + b)x ab BC... y = (b + c)x bc CD... y = (c + d)x cd DA... y = (a + d)x ad AC... y = (a + c)x ac BD... y = (b + d)x bd. (3) Tangents A B C D of the circle (1) at the points () are of the form A... y = ax a B... y = bx b C... y = cx c D... y = dx d. (4) The points of intersection of the tangents in (4) are T AB = A B = ( a+b ab) T BC = B C = ( b+c bc) T AC = A C = ( a+c ac) T BD = B D = ( b+d bd) T AD = A D = ( a+d ad) T CD = C D = ( c+d cd). (5) The diagonal triangle of the cyclic quadrangle is formed by the intersection points of the opposite sides of the quadrangle: U = AC BD V = AB CD and W = AD BC. An allowable triangle introduced in [4] concerns each triangle whose sides are nonisotropic lines. According to [9] the allowable cyclic quadrangle is the cyclic quadrangle having the allowable diagonal triangle. Hence: 667

3 Lemma The diagonal points U V W of the allowable cyclic quadrangle ABCD are of the form U = V = W = ( ) ac bd ac(b + d) bd(a + c) a + c b d a + c b d ( ) ab cd ab(c + d) cd(a + b) a + b c d a + b c d ( ) ad bc ad(b + c) bc(a + d) a + d b c a + d b c (6) and the sides of the diagonal triangle are given with UV... y = (ad bc) ad(b + c) bc(a + d) x a + d b c a + d b c UW... y = (ab cd) ab(c + d) cd(a + b) x a + b c d a + b c d (7) V W... y = (ac bd) ac(b + d) bd(a + c) x a + c b d a + c b d where a + c b d 0 a + b c d 0 a + d b c 0. Note 1 Conditions a + c b d 0 a + b c d 0 and a + d b c 0 are the conditions for the cyclic quadrangle ABCD to be allowable. 3. On the harmonic quadrangle in the isotropic plane In this section we investigate the cyclic quadrangle with a special property. Theorem 1 Let ABCD be an allowable cyclic quadrangle with vertices given by () sides by (3) and tangents of its circumscribed circle (1) at its vertices given by (4). These are the equivalent statements: 1. the point T AC = A C is incident with the diagonal BD ;. the point T BD = B D is incident with the diagonal AC ; 3. the equality holds; d(a B) d(c D) = d(b C) d(d A) (8) 4. the equality holds. (ac + bd) = (a + c)(b + d) (9) 668

4 Proof Let us first prove the equivalence of statements 1 and 4. The point T AC = ( a+c ac) is obviously incident with the tangents A and C from (4) and therefore T AC = A C. On the other hand T AC is incident with the line BD from (3) providing being statement 4. ac = (b + d) a + c bd The equivalence of statements and 4 can be proved in an analogous way. The following equality d(a B) d(c D) d(b C) d(a D) = (b a)(d c) (c b)(d a) = (ac + bd) (a + c)(b + d) proves the equivalency between 3 and 4. A cyclic quadrangle will be referred to as a harmonic quadrangle if it satisfies one and hence all of the equivalent conditions presented in Theorem 1. Since properties 1 3 have completely geometrical sense property 4 does not depend on the choice of the affine coordinate system. Choosing the y -axis to be incident with the diagonal point U because of (6) ac = bd follows. Since ac < 0 and bd < 0 we can use the notation ac = bd = k. (10) Thus the diagonal point U turns into Statement 4 yields Next we will show that U = (0 k ). (11) (a + c)(b + d) = 4k. (1) (a c) (b d) (a + c b d) = 4k (13) i.e. (a c) (b d) 4k (a + c b d) = 0. Indeed the left side of the equality given above owing to (10) and (1) is equal to [(a + c) 4ac][(b + d) 4bd] 4k [(a + c) + (b + d) (a + c)(b + d)] = [(a + c) + 4k ][(b + d) + 4k ] 4k [(a + c) + (b + d) + 8k ] = (a + c) (b + d) 16k 4 = 0. Hence (a c)(b d) a + c b d = ±k. As the fraction of the left side is a negative real number we can choose k being k > 0 i.e. let (a c)(b d) a + c b d = k. (14) 669

5 Because of (10) a numerator of the fraction included in a constant term of the equation (7) of the line V W amounts to k (b + d a c) and hence this constant term equals k. Therefore the line V W is given by y = k. From (5) we get T AC = There are two more valid identities: ( ) ( ) a + c b + d k T BD = k. (a b)(c d) = (a c)(b d) (a d)(b c) = (a c)(b d). (15) Let us consider for example the first identity in (15). By using (10) and (1) we get (a b)(c d) = (ac ad bc + bd) = 4k (ad + bc) = (a + c)(b + d) (ad + bc) = (a c)(b d). Furthermore by adding the identities in (15) and dividing the result by the equality (a c)(b d) = (a b)(c d) + (a d)(b c) is obtained being Ptolemy s theorem. Due to the above discussion such a harmonic quadrangle is said to be in a standard position or it is a standard harmonic quadrangle (see Figure 1). Every harmonic quadrangle can be transformed into one in standard position by means of an isotropic transformation. In order to prove geometric facts for each harmonic quadrangle it is sufficient to give a proof for the standard harmonic quadrangle. The diagonal points from (6) turn into U = (0 k ) V = ( ) ( ) ab cd ad bc a + b c d k W = a + d b c k. (16) 4. Properties of the harmonic quadrangle in the isotropic plane In this section we will prove several theorems dealing with the properties of the harmonic quadrangle. For the Euclidean version of the next theorem see [6]. Theorem Let ABCD be a harmonic quadrangle and lines A B C and D the tangents to the circle (1) at the points A B C and D respectively. Assume that M AC and M BD are the midpoints of the diagonals AC and BD respectively. Then the points M AC B D and T BD = B D are incident with a circle. The same is valid for the points M BD A C and T AC = A C. Proof Let us prove the theorem for the points M AC B D and T BD = B D. Points B D and T BD are incident with a circle of the form y = x (b + d)x + bd. (17) 670

6 Figure 1. The harmonic quadrangle. Applying the coordinates of M AC = ( a+c a +c ) in the equation given above we obtain a + c = (a + c) 1 (a + c)(b + d) + bd i.e. (ac + bd) = (a + c)(b + d) being the condition (9) for the cyclic quadrangle to be harmonic. Hence M AC lies on the circle (17). Because of symmetry on the real numbers a b c and d the same is valid for M BD A C and T AC = A C whose circumscribed circle is of the form y = x (a + c)x + ac. (18) Note Circles (17) and (18) from Theorem intersect in (0 k ) the point parallel to the diagonal point U and incident to the line V W. Theorem 3 Let ABCD be a harmonic quadrangle and U AB U BC U CD U DA be the intersections of the isotropic line through U with the sides AB BC CD DA respectively. Then the following equalities hold: s(u U AB ) d(a B) = s(u U BC) d(b C) = s(u U CD) d(c D) = s(u U DA) d(d A). (19) Proof The point U AB has the coordinates (0 ab) and therefore s(u U AB ) = k ab. Let us prove s(u U AB ) d(a B) = s(u U BC) d(b C). 671

7 It holds precisely when is valid i.e. k + ab a b = k + bc b c k (b a c) + (a + c)b abc = 0. The latter equality holds if and only if 4bk k (a + c) + b (a + c) = 0 being after inserting 4k = (a + c)(b + d) equivalent to b(b + d) k + b = 0 i.e. This completes the proof. k + bd = 0. Theorem 4 Let ABCD be a harmonic quadrangle and let the lines ã b c d be incident with the vertices A B C D respectively and form equal angles with the sides AB BC CD DA respectively. The lines ã b c d form a harmonic quadrangle as well. Proof The lines ã... y = (a + b h)x + a(h b) b... y = (b + c h)x + b(h c) c... y = (c + d h)x + c(h d) d... y = (d + a h)x + d(h a) fulfill the condition of the theorem since (ã AB) = ( b BC) = ( c CD) = ( d DA) = h while (a + b h)a + a(h b) = a (analogously for b c d). Denoting by à = d ã B = ã b C = b c and D = c d the accuracy of the following equalities is obvious: 67 à = B = C = D = ( a + a d d b h a + ( b + b a ( c + c b a c h b + b d h c + ( d + d c c a h d + (a d)(a + b) h a d d b d b h (b a)(b + c) h b a a c a c h (c b)(c + d) h c b b d b d h (d c)(d + a) h d c c a c a h ) ) ) ).

8 Next some computing shows that the points à B C D are incident with the circle Indeed for the point Ã: (h + k)y khx + h x + hk = (h + k) = a h = a h + = a h + = a h + = a h + Therefore [ a (a + b)(a d) h + a d ] [ b d b d h k (a + b)(a d) h (a + b)(a d) hk + a d b d b d (h + k)y = khx h x hk. (0) a a a d (a d) h + b d (b d) h b d h k + 4a a d b d bd a b d h + hk ( a ab ad + bd ) (a b)(a d) b d (b d) h k + hk bd a b d h + hk ( a ab ad + ac ) a ab ad + ac b d (b d) h k + hk bd a b d h + a(a c)hk a b + c d (a c)(b d) ah k a c b d a b + c d (a c)(b d) + hk bd a b d h a(a c)h + ah a c b d + hk = ach + hk bd ac + b d h = 0. Due to (a b)(c d) = ab + cd + k within the following calculation we get On the other hand ] [ + h a a d ] b d h + hk hk (a d) (b d) h k + ah + hk d(ã B) = b + b a a c h a a d d b h = b a a + b ab + ab + cd ac bd h (a c)(d b) = b a (a b) + (a b)(c d) h = b a + (a c)(d b) = b a + (a b) k h = (b a)(1 + h k ). d(ã B) d( C D) = (b a)(d c)(1 + h k ). d( B C) d( D Ã) = (c b)(a d)(1 + h k ). (a b)(a b + c d) h (a c)(b d) Since (a b)(c d) = (a d)(b c) the equality d(ã B) d( C D) = d( B C) d( D Ã) is fullfiled by the points à B C D and the claim of the theorem is proved. The Euclidean case of the theorem given above can be found in []. Corollary 1 For the harmonic quadrangles ABCD and à B C D the following equalities are applicable: d(ã B) d( B C) d( C D) d( D Ã) = = = d(a B) d(b C) d(c D) d(d A) = 1 + h k. 673

9 Note that for h = k all four points à B C D coincide with one point say P 1 having the coordinates P 1 = ( k 3k ). (1) The point P 1 called the first Brocard point of the quadrangle ABCD is the point whose connection lines with the vertices A B C D form the equal angles with the sides AB BC CD and DA respectively. Similarly the second Brocard point P is defined as the point such that its connection lines with the vertices A B C D form the equal angles with sides AD DC CB and BA respectively. Observations similar to those in the proof of Theorem 4 and Corollary 1 result in P = ( k 3k ). () Some nice geometric properties of the Brocard points are described in the following two theorems and depicted in Figure. Theorem 5 Let ABCD be a harmonic quadrangle and M AC M BD be midpoints of the line segments AC BD respectively. Two Brocard points P 1 and P diagonal point U and two midpoints M AC and M BD lie on a circle. Furthermore d(p 1 U) = d(u P ) holds and the line P 1 P is parallel to the line V W. Proof The points P 1 P and U with the coordinates given by (1) () and (16) apparently lie on a circle with the equation y = x + k. It remains to prove that M AC and M BD are incident with the same circle. For example that is true for the point M AC = ( a + c a + c ) because of ( a + c ) + k = a k + c 4 + k = a + c. The second statement from Theorem 5 holds since d(p 1 U) = k = d(u P ). Lines P 1 P and V W have the equations y = 3k y = k respectively and therefore are parallel. Let us now prove that the line M AC M BD has the equation y = (a + b + c + d)x + 3k. Indeed y (a + b + c + d)x 6k = a + c (a + c)(a + b + c + d) 6k = ac (a + c)(b + d) 6k = k + 4k 6k = 0. The line M AC M BD passes through the point U = (0 3k ) which also lies on the line P 1 P. Theorem 6 Let ABCD be a harmonic quadrangle M AC and M BD midpoints of the line segments AC and BD respectively and P 1 and P two Brocard points of ABCD. Then P 1 = W M AC V M BD and P = V M AC W M BD. 674

10 Figure. The visualization of Theorem 5. Proof We prove the collinearity of the points P 1 W and M AC. Referring to (16) and (1) the slopes of lines W M AC and P 1 M AC are obtained to be (a + c + k )(a + d b c) (a + c)(a + d b c) (ad bc) = (a + c ac)(a + d b c) a c = (a c)(b + d) a + c 6k a + c k = (a + c) 4k a + c k and they are equal precisely when = a + c + k (a c)(a b c + d) = (a + c + k)(a b + c d) (a c)(a b c + d) a + c b d i.e. ad + bc ac bd = k(a b + c d). Because of (14) this is equivalent to ad + bc ac bd = (a c)(b d) and this is the first equality (15). The other three collinearities can be proved in a similar manner. For the Euclidean version of the following theorems see [5] (for Theorems 7 and 9) and [6] (for Theorem 8). Theorem 7 Let ABCD be a harmonic quadrangle and M AC be the midpoint of the line segment AC. Then the equality d(m AC A) = d(m AC B) d(m AC D) (3) holds. The line AC is the bisector of the lines M AC B and M AC D. 675

11 Proof The point M AC is of the form ( a + c a + c ). According to (9) d(m AC B) d(m AC D) d(m AC A) = ( a+c b)( a+c d) ( a+c a) = bd a + a+c (a b d) = bd + ac 1 (a + c)(b + d) = 0 is valid. Further on the equations of the lines M AC B M AC D are given by M AC B... y = (a + b + c d)x + b(d a c) M AC D... y = (a b + c + d)x + d(b a c). (4) In that case the equalities (M AC B M AC C) = (a + c) (a + b + c d) = d b (M AC C M AC D) = (a b + c + d) (a + c) = d b prove the second part of the theorem. Theorem 8 Let ABCD be a harmonic quadrangle and M AC be the midpoint of the line segment AC. The triangles M AC DA M AC AB and CDB have equal corresponding angles. Proof By using (3) and (4) it is easy to prove that (DA AM AC ) = (AB BM AC ) = (DB BC) = c d (M AC D DA) = (M AC A AB) = (CD DB) = b c and (AM AC M AC D) = (BM AC M AC A) = (BC CD) = d b. Corollary The diagonal line BD is a symmedian for the triangles ABC and CDB while the diagonal line AC is a symmedian for the triangles ABD and CDB. Theorem 9 Let ABCD be a harmonic quadrangle. The lines l AB l BC l CD l DA incident with the diagonal point U and parallel to AB BC CD DA respectively intersect the sides of the quadrangle ABCD in eight points l CD AD l CD BC l AB BC l AB AD l BC AB l AD AB l AD CD and l BC CD which lie on a circle. Proof It is easy to prove that the line l CD has the equation of the form y = (c + d)x + k. (5) Then the coordinates of l CD AD are given with ( ad + k a c a(d + k ) ). a c 676

12 Because of ( ) ad + k = a d + 4adk + k 4 a c (a c) = ad(ad + 4k + bc) (a c) the point given above is incident with the circle We can prove the same for the other seven points. = = ad(ad ab ad bc cd + bc) ad(ad ab + bc cd) (a c) = (a c) ad(a c)(d b) ad(d b) (a c) = a c y = x. = a(d + k ) a c The structure of the harmonic quadrangle allows us to obtain a rich structure of collinear points and concurrent lines. The Euclidean version of the following theorem can be found in [3]. Theorem 10 Let ABCD be a harmonic quadrangle. If C AB = AB C D AB = AB D A BC = BC A D BC = BC D A CD = CD A B CD = CD B B AD = AD B C AD = AD C are considered the four triples of points {C AB A BC T BD } {A CD C AD T BD } {D AB B AD T AC } {D BC B CD T AC } are collinear. Proof The points ( cd a A CD = acd (c + d)a c + d a c + d a ) ( ) ad c acd (a + d)c C AD = a + d c a + d c are lying on the line given by y = a (c + d) + c (a + d) + d (a + c) 6acd (a c) + (a d)(c d) x + acd(a + c + d) a c a d c d (a c). + (a d)(c d) The coordinates of the point T BD = ( b + d bd) satisfy the equation above if and only if (a d)(c d)(ab ac + bc + ad bd + cd) = 0 which happens exactly in the case of ab + bc + ad + cd = ac + bd. References [1] Beban-Brkić J Šimić M Volenec V. On foci and asymptotes of conics in isotropic plane. Sarajevo J Math 007; 3: [] Bernhart A. Polygons of pursuit. Scripta Math 1959; 4: [3] Bradley CJ. A Singular Cyclic Quadrangle. Available online at: [4] Kolar-Šuper R Kolar-Begović Z Volenec V Beban-Brkić J. Metrical relationships in standard triangle in an isotropic plane. Math Commun 005; 10: [5] Langley EM. A note on Tucker s harmonic quadrilateral. Math Gaz 193; 11: [6] Pamfilos P. The Associated Harmonic Quadrilateral. Forum Geom 014; 14:

13 [7] Sachs H. Ebene isotrope Geometrie. Braunschweig Germany: Vieweg-Verlag 1987 (in German). [8] Strubecker K. Geometrie in einer isotropen Ebene. Math Naturwiss Unterr ; 15: (in German). [9] Volenec V Beban-Brkić J Šimić M. The cyclic quadrangle in the isotropic plane. Sarajevo J Math 011; 7: