On Brocard Points of Harmonic Quadrangle in Isotropic Plane

Original scientific paper Accepted 31. 10. 017. EMA JURKIN MARIJA ŠIMIĆ HORVATH VLADIMIR VOLENEC On Brocard Points of Harmonic Quadrangle in Isotropic Plane On Brocard Points of Harmonic Quadrangle in Isotropic Plane ABSTRACT In this paper we present some new results on Brocard points of a harmonic quadrangle in isotropic plane. We construct new harmonic quadrangles associated to the given one study their properties dealing with Brocard points. Key words: isotropic plane harmonic quadrangle Brocard points MSC010: 51N5 O Brocardovim točkama harmoničnog četverokuta u izotropnoj ravnini SAŽETAK U radu se prikazuju neki novi rezultati o Brocardovim točkama harmoničnog četverokuta u izotropnoj ravnini. Konstruiraju se novi harmonični četverokuti pridruženi danom četverokutu te se proučavaju njihova svojstva vezana uz Brocardove točke. Ključne riječi: izotropna ravnina harmonični četverokut Brocardove točke 1 Introduction The geometry of harmonic quadrangle has been discussed in [3]. The harmonic quadrangle is a cyclic quadrangle ABCD see [5]) with the following property: the point of intersection of the tangents at the vertices A C lies on the line BD the intersection point of the tangents at the vertices B D lies on the line AC. If one of the requests is fulfilled the other one automatically follows. The quadrangle ABCD is chosen to have y x as a circumscribed circle the vertices A aa )B bb )C cc ) D dd ) with abcd being mutually different real numbers where a < b < c < d. In that case sides of ABCD are AB...y a + b)x ab BC...y b + c)x bc CD...y c + d)x cd tangents to y x at ABCD are DA...y a + d)x ad AC...y a + c)x ac BD...y b + d)x bd. A...y ax a B...y bx b C...y cx c D...y dx d. 1) ) Equations 1) ) can be found within Lemma 1 in [5]. In that paper the diagonal triangle of a cyclic quadrangle was introduced where diagonal points U AC BD V AB CD W AD BC are given by ac bd acb + d) bda + c) U a b + c d a b + c d ab cd abc + d) cda + b) V a + b c d a + b c d ad bc adb + c) bca + d) W. a b c + d a b c + d Taking ac bd k k > 0 we deal with harmonic quadrangle in a stard position. As every harmonic quadrangle can be represented in the stard position in order to prove geometric facts for each harmonic quadrangle it is sufficient to give a proof for the stard harmonic quadrangle. The diagonal points given by 3) in the case of 3) 11

stard position turn into ab cd U 0k ) V a + b c d k ad bc W a b c + d k. Due to Theorem 1 in [3] more characterizations of harmonic quadrangles among the cyclic ones are given. Theorem 1 Let ABCD be an allowable cyclic quadrangle with vertices A aa ) B bb ) C cc ) D dd ) sides 1) tangents of its circumscribed circle y x at its vertices are given by ). These are the equivalent statements: 4) 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 dab) dcd) dbc) dda) 5) holds; 4. the equality ac + bd) a + c)b + d) 6) holds. In the sequel we will deal with Brocard points of these quadrangles defined in [3] show several properties of them connected to the harmonic quadrangles associated to ABCD. For that purpose the following lemma will be very useful. Lemma 1 Let ABCD be an allowable cyclic quadrangle in the stard position with vertices A aa ) B bb ) C cc ) D dd ). The following equalities are valid ab + k kb a) cd + k kd c) bc + k kc b) da + k ka d). Proof. Let us prove ab + k kb a). Taking see [3]) )) a + c in ka b) ab k k 7) we get a b+ab a d+b c 4abc+4abd+abc abd+acd bcd a b + c d k i.e. a b + ab a d + b c 4k a b) a b + c d + k k. After employing a+c)b+d) 4k from [3] the upper equality turns into a b + ab a d + b c + a+c)b+d)a b) a b + c d Immediately k a b + c d) a b + c d + k k + k k. follows the claim of Lemma is proved. Further on we will always deal with the harmonic quadrangle in the stard position. Brocard points There is an interesting result discussed in [3]: whole family of harmonic quadrangles can be obtained out of the given harmonic quadrangle. Namely let ABCD be a harmonic quadrangle. Lines a b c d are taken in a way to be incident to vertices ABCD respectively make equal angles to the sides AB BCCD DA respectively. The quadrangle formed by lines a b c d is a harmonic quadrangle as well. Furthermore denoting obtained quadrangle by A B C D the ratios of the corresponding sides of given quadrangle ABCD obtained quadrangle A B C D are equal. Only in one case points A B C D coincide with one point P 1 the first Brocard point. In similar manner the second Brocard point P is obtained as well. In the latter case lines P A P B P C P D form the equal angles with the sides AD DCCB BA respectively. Brocard points are of the form P 1 k3k ) P k3k ). 8) The Brocard points can easily be constructed by using the fact that P 1 WM AC V M BD P V M AC WM BD 9) where M AC M BD are midpoints of the line segments AC BD respectively. The Euclidean case of Theorem can be found in [1] the Euclidean analogue of Theorem 3 is stated in [] [4]. 1

Theorem Let ABCD be a harmonic quadrangle P 1 P its Brocard points. The following equality holds: D dap 1 ) dbp 1 ) dcp 1 ) ddp 1 ) dap ) dbp ) dcp ) ddp ). Proof. Since M DA dap 1 ) dbp 1 ) dcp 1 ) ddp 1 ) k a)k b)k c)k d) k ka + c) + ac ) k kb + d) + bd ) k a + c)b + d) 4k 4 A P M BD P 1 M CD M AB M AC dap ) dbp ) dcp ) ddp ) k a) k b) k c) k d) k + ka + c) + ac ) k + kb + d) + bd ) k a + c)b + d) 4k 4 the theorem is proved. Theorem 3 Let ABCD be a harmonic quadrangle P 1 P its Brocard points. The four points AP 1 BP BP 1 CP CP 1 DP DP 1 AP lie on a circle that is incident with the points U P 1 P M AC M BD as well. Furthermore they are parallel with the midpoints of the line segments AB BC CD DA respectively. Proof. Due to Theorem 5 in [3] the circle incident to the points UP 1 P M AC M BD is given by y x + k. 10) Let us take for example the point AP 1 BP. Lines AP 1 BP have the equations see Theorem 4 in [3]) y a + b + k)x ak + b) y b + a k)x b k + a). Thus the point AP 1 BP is of the form a + b kb a) + a + b ). Due to Lemma 1 the equality kb a) ab + k is valid therefore the point AP 1 BP is incident to 10). B U M BC Figure 1: Visualization of Theorem 3 In the sequel we will study two quadrangles associated to the given harmonic quadrangle ABCD. Let A i B i C i D i be the intersection points of its circumscribed circle y x with the lines AP i BP i CP i DP i respectively i 1. According to [3] Theorem 4 for h k) the line AP 1 has the equation y a + b + k)x ak + b). Hence A 1 has coordinates b + kb + k) ). Similarly we get the other intersections: A 1 C 1 b + kb + k) ) B 1 d + kd + k) ) D 1 C c + kc + k) ) a + ka + k) ) 11) A d kd k) ) B a ka k) ) C b kb k) ) D c kc k) ). 1) Some interesting properties of the obtained quadrangles are stated in Theorems 4-9 that follow. The authors haven t found their Euclidean counterparts in the literature available to them but they are convinced in their validity in the Euclidean plane as well. Theorem 4 The quadrangles A 1 B 1 C 1 D 1 A B C D associated to the harmonic quadrangle ABCD are harmonic quadrangles as well. Proof. From 11) we get equations see Theorem 4 in [3]) da 1 B 1 ) dc 1 D 1 ) c b)a d) dbc) dda) db 1 C 1 ) dd 1 A 1 ) d c)b a) dcd) dab). x 13

Obviously da 1 B 1 ) dc 1 D 1 ) db 1 C 1 ) dd 1 A 1 ) precisely when dab) dcd) dbc) dda). Therefore A 1 B 1 C 1 D 1 is a harmonic quadrangle. Similar procedure gives the proof for A B C D. B C 1 therefore D A A B ). The equality ABBC) c a obtained from 1) completes the first part of the proof. The second part of the theorem follows directly from 11) 1). Theorem 6 For the quadrangles A 1 B 1 C 1 D 1 A B C D associated to the harmonic quadrangle ABCD the following statements on their diagonal points U i A i C i B i D i i 1 are valid: 1. du 1 U) duu ).. The connection line U 1 U is parallel to the connection lines P 1 P VW. 3. The points U 1 U are incident with the lines UP 1 UP respectively. C A D P 1 P A 1 B D 1 Figure : Harmonic quadrangles A 1 B 1 C 1 D 1 A B C D. Theorem 5 For the quadrangles A 1 B 1 C 1 D 1 A B C D associated to the harmonic quadrangle ABCD the equalities D A A B ) ABBC) B 1 C 1 C 1 D 1 ) A B B C ) BCCD) C 1 D 1 D 1 A 1 ) B C C D ) CDDA) D 1 A 1 A 1 B 1 ) C D D A ) DAAB) A 1 B 1 B 1 C 1 ) da 1 B 1 ) dbc) dc D ) db 1 C 1 ) dcd) dd A ) dc 1 D 1 ) dda) da B ) dd 1 A 1 ) dab) db C ) hold. Proof. Let us prove D A A B ) ABBC). From 1) we get A B... y d + a 4k)x d k)a k) A D... y d + c 4k)x d k)c k) C A B 1 D x Proof. Directly from 11) 1) the coordinates of U 1 U are obtained to be U 1 k5k ) U k5k ). Thus du 1 U) k duu ). The connection line U 1 U has the equation y 5k it is parallel to the lines P 1 P VW having the equations y 3k y k. The last part of the theorem holds since the coordinates of the points UU 1 P 1 satisfy the equation y kx+k while the coordinates of the points UU P satisfy the equation y kx + k. Theorem 7 The diagonal points V i A i B i C i D i W i A i D i B i D i of the quadrangles A i B i C i D i are incident with the lines V P i WP i respectively i 1. Proof. Let us prove that V 1 A 1 B 1 C 1 D 1 is incident with V P 1 i.e. V 1 VP 1 are collinear points. From 3) 11) we get V 1 ad bc a b c + d + k4k ad bc a b c + d + 3k ). Now the slopes of the lines V P 1 V 1 P 1 are obtained to be 4k a + b c d) ab cd ka + b c d) respectively are equal precisely when 4kad bc) ad bc + ka b c + d) ka + b c d)[ad bc + ka b c + d)] i. e. ad bc)[ab cd ka + b c d)] k a + b c d)a b c + d) ad bc)ab cd). 14

This is true since k a ac + c b + bd d ) a k + d k + b k c k. The other three collinearities can be proved in a similar way. Theorem 8 Let M A i C i M B i Di be the midpoints of the line segments A i C i B i D i respectively i 1. Then the following quadruples of points are collinear: {M A 1 C 1VM BDP 1 } {M B 1 D 1WM ACP 1 } {M A C WM BDP } {M B D VM ACP }. Proof. Let us prove the claim of Theorem for the quadruple of points {M A 1 C 1VM BDP 1 }. According to 9) the points VM BD P 1 are collinear. Therefore it is sufficient to show that the points M b+d BD b +d ) M b+d+4k A 1 C 1 b+k) +d+k) ) P 1 k3k ) are collinear. Their coordinates satisfy the equation y b + d + k)x b + d)k + k. Indeed b + d + k) b + d b + d)k + k b + d) + b + d)k b + d)k + k b + d. Similarly b + d + k) b + d + 4k b + d)k + k b + d) + kb + d) + kb + d) + 4k b + d)k + k b + k) + d + k) By using two Brocard points of the given quadrangle ABCD we constructed two new quadrangles. We can continue that procedure by using the Brocard points of the obtained quadrangles construct further four quadrangles. Actually we will get only two new quadrangles since two of them coincide with the referent quadrangle ABCD. Indeed if we use A 1 B 1 C 1 D 1 its first Brocard point we will get a new quadrangle but if we use its second Brocard point which is P 1 then we will get ABCD. So in each step of this procedure we have to use the first Brocard points or we always have to use the second Brocard points. Some computations verify that if we start a pattern with the quadrangle ABCD in the step 0 the quadrangle A 1 B 1 C 1 D 1 in the step 1 then the first Brocard point in the step n has the coordinates n + 1)kn + 1) k + k ). Therefore we conclude that all Brocard points of the quadrangles associated to the quadrangle ABCD lie on the circle y x + k. We will now focus on the other two quadrangles associated to the given harmonic quadrangle ABCD. Let the quadrangles A 3 B 3 C 3 D 3 A 4 B 4 C 4 D 4 be defined in the following way: The lines l AB l BC l CD l DA incident with the diagonal point U parallel to AB BC CD DA respectively intersect the sides of the quadrangle ABCD in eight points A 3 l AB AD B 3 l BC AB C 3 l CD BC D 3 l AD CD A 4 l AD AB B 4 l AB BC C 4 l BC CD D 4 l CD AD. D 4 D b + d + k)k b + d)k + k 3k. The other three collinearities can be proved in a similar way. A A 3 Theorem 9 Let P1 i Pi be Brocard points of the quadrangle A i B i C i D i i 1. Then the second Brocard point P 1 of the quadrangle A1 B 1 C 1 D 1 coincides with P 1 while the first Brocard point P1 of the quadrangle A B C D coincides with P. Proof. According to 9) P 1 V 1 M A 1 C 1 W 1 M B 1 D1. Now from Theorems 7-8 we get P 1 V M BD WM AC P 1. Similarly P1 W M A C V M B D WM BD V M AC P. B 3 B U A 4 C 3 Figure 3: Harmonic quadrangles A 3 B 3 C 3 D 3 A 4 B 4 C 4 D 4 B 4 C C 4 D 3 15

It was shown in [3] that the constructed eight points lie on a circle y x. The properties of the quadrangles A 3 B 3 C 3 D 3 A 4 B 4 C 4 D 4 in the Euclidean plane have been discussed in [4]. Here we present the similar study for the isotropic plane. Since the lines l AB AD are given by l AB... y a + b)x + k AD... y a + d)x ad the first coordinate of their intersection point A 3 is ad+k d b. According to Lemma 1 it equals ka d) d b. Similarly all intersection points are obtained as ) A 3 ka d) ka d) d b 13) d b ) B 3 kb a) kb a) ) C 3 kc b) kc b) ) D 3 kd c) kd c) c a c a ) A 4 kb a) kb a) ) B 4 kc b) kc b) c a c a ) C 4 kd c) kd c) d b d b ) D 4 ka d) ka d). Theorem 10 For the quadrangles A i B i C i D i i 34 the equalities Proof. In order to show 14) is valid we give a proof for da 3 B 3 ) dab) 1. Indeed da 3 B 3 kb a) ka d) ) + b a)) + )a d) k )) a b)a b + c d) a b)a b + c d) k k )) ka b + c d) b a dab) Further on let us prove A 3 B 3 B 3 C 3 ) ABBC). Out of 13) the slopes of the lines A 3 B 3 B 3 C 3 are obtained to be k a d d b + b a respectively. Thus ) b a k + c b ) A 3 B 3 B 3 C 3 ka b + c d) )) ) c a ABBC). Theorem 11 The quadrangles A 3 B 3 C 3 D 3 A 4 B 4 C 4 D 4 are harmonic quadrangles as well. Proof. It follows directly from 14) Theorem 1. Theorem 1 Let U i A i C i B i D i be the diagonal points of the quadrangles A i B i C i D i i 34. The following statements are valid: 1. The diagonal point U is the midpoint of the line segment U 3 U 4.. The connection line U 3 U 4 is parallel to the connection lines P 1 P VW. Proof. From 14) we get the equation of the line A 3 C 3 da i B i ) dab) dbi C i ) dbc) dci D i ) dcd) ddi A i ) dda) 1 D i A i A i B i ) DAAB) A i B i B i C i ) ABBC) B i C i C i D i ) BCCD) C i D i D i A i ) CDDA) hold. 16 14) y ka + b c d) x + k b c)a d) ). By using a d)b c) )) it turns into y ka + b c d) x k ). Similarly we get the equation of the line B 3 D 3 as y ka b c + d) x k ).

Therefore their intersection has coordinates U 3 k ) k. Analogously the diagonal point U 4 A 4 C 4 k B 4 D 4 k is obtained. The point U0k ) is obviously the midpoint of the points U 3 U 4. The line U 3 U 4 is given by the equation y k therefore parallel to the connection lines P 1 P VW. Theorem 13 Let V i A i B i C i D i W i A i D i B i D i be the diagonal points of the quadrangles A i B i C i D i i 34. Then the points V 3 W 4 are incident with BD the points V 4 W 3 are incident with AC. Proof. To prove the theorem we will show that V 3 lies on BD. The other statements can be shown in the similar manner. Out of 13) elementary but long calculation results with coordinates of V 3 A 3 B 3 C 3 D 3 in the form V 3 c a a + b c d k a b c + d a + b c d k ) 15) The equality ab bc cd + da k) that can be obtained from Lemma 1 will be used to prove that coordinates given by 15) satisfy the equation y b + d)x bd of the line BD. Indeed c a)k b + d) a + b c d bd bc ab + cd ad)k a + b c d k ) a + b c d + k a b c + d a + b c d k. + k Theorem 14 Let M A i C i M B i Di be the midpoints of the line segments A i C i B i D i respectively i 34. Then M A i C i are incident with AC Bi D i are incident with BD i 34. Proof. Let us for example prove that M A 3 C 3 is incident with AC. It is sufficient to prove that the coordinates of the midpoint a + b c d M A 3 C 3 k d b a d) + c b) ) d b) k 16) satisfy the equation y a + c)x ac of the line AC. Indeed a + c)a + b c d)k d b) precisely when + k k a d) + c b) d b) a + c)a + b c d) k k a d) + c b) d b) d b) d b). This is true if only if a + c)a + b c d) d b k a d) + c b) d b) d b) which is by using 7) equivalent to a + c)a + b c d)a b + c d) )a + c ad bc + bd). Taking ac bd k we get a + c)d b ) 4k d b). This is valid due to a + c)b + d) 4k. Theorem 15 Let P1 i Pi be Brocard points of the quadrangle A i B i C i D i i 34. Then P1 3 U P3 P P1 4 P 1 P 4 U. Proof. The facts P1 3 U P4 U follow directly from Theorems 13 14. Indeed P1 3 W 3 M A 3 C 3 V 3 M B 3 D 3 AC BD U P 4 V 3 M A 3 C 3 W 3 M B 3 D 3 BD AC U. It is left to prove P 3 P P1 4 P 1. For the illustration we will prove P 3 P. It is sufficient to show that P lies both on V 3 M A 3 C 3 W 3 M B 3 D3. Let us check that it lies on V 3 M A 3 C 3 i. e. that P V 3 M A 3 C3 are collinear points. From 8) 15) we get the following equation of the line P V 3 : y From ka + b c d) x + k) + 3k. 17) ka + b c d) a + b c d d b ) k + k + 3k k b d) a + b c d)a b c + d) + 3k k b d) a +b +c +d ac bd+ab ad bc+cd ) k b d) a ad + d + b bc + c ) k b d) a d) + b c) ) we conclude that the midpoint M A 3 C3 with coordinates 16) is incident with the line P V 3. We can start a construction of a sequence of quadrangles with the quadrangle ABCD in step 0 quadrangle A 4 B 4 C 4 D 4 in step 1. Some computations verify that a quadrangle constructed ) in step n) has Brocard points k3k ) 1 1 k 3 n k ). Obviously all n 1 n quadrangles in the sequence have the same first Brocard point P 1 k3k ) while their second Brocard points approach the point P 1. 17

References [1] A. BERNHART Polygons of Pursuit Scripta Math. 4 1959) 3 50. [] F.G.W. BROWN The Brocard Tucker Circles of a Cyclic Quadrangles Proc. Edinb. Math. Soc. 36 1917) 61 83. [3] E. JURKIN M. ŠIMIĆ HORVATH V. VOLENEC J. BEBAN-BRKIĆ Harmonic Quadrangle in Isotropic Plane Turk. J. Math. to appear. [4] E.M. LANGLEY A Note on Tucker s Harmonic Quadrilateral Math. Gaz. 11 193) 306 309. [5] V. VOLENEC J. BEBAN-BRKIĆ M. ŠIMIĆ Cyclic Quadrangle in the Isotropic Plane Sarajevo J. Math. 70) 011) 65 75. Ema Jurkin orcid.org/0000-000-8658-5446 e-mail: ema.jurkin@rgn.hr Faculty of Mining Geology Petroleum Engineering University of Zagreb Pierottijeva 6 HR-10000 Zagreb Croatia Marija Šimić Horvath orcid.org/0000-0001-9190-5371 e-mail: marija.simic@arhitekt.hr Faculty of Architecture University of Zagreb Kačićeva 6 HR-10000 Zagreb Croatia Vladimir Volenec e-mail: volenec@math.hr Faculty of Science University of Zagreb Bijenička cesta 30 HR-10000 Zagreb Croatia 18