SHAPE THEOREMS FOR QUADRANGLES * Georgi Hristov Georgiev, Radostina P. Encheva

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1 МАТЕМАТИКА И МАТЕМАТИЧЕСКО ОБРАЗОВАНИЕ, 2003 MATHEMATICS AND EDUCATION IN MATHEMATICS, 2003 Proceedings of the Thirty Second Sring Conference of the Union of Bulgarian Mathematicians Sunny Beach, Aril 5 8, 2003 SHAPE THEOREMS FOR QUADRANGLES * Georgi Hristov Georgiev, Radostina P Encheva A shae of a triangle is a comlex number corresonding to an orbit of triangles under the grou of the lane direct similarities A shae of a quadrangle is an ordered air of comlex numbers corresonding to an equivalence class of quadrangles under the same grou We aly the shaes of triangles for the examinations of the shaes of quadrangles In articular, we obtain some equalities for shaes by the use of some geometric constructions Comlex numbers are a ower tool to study the Euclidean lane There are several books which consider different alications of comlex number in the lane geometry (see [5] and [9]) June Lester created a new comlex analytic formalism based on a cross-ratio and introduced the notion of a shae of a triangle Many alications of this formalism are considered in triangle series (see [6], [7] and [8]) The notion of shae was extended by R Artzy in [1] He introduced a shae of olygons and examined some roerties of shaes of quadrangles In this aer, we aly shaes of triangles for examination of shaes of quadrangles First, we recall the basic definitions and assertions concerning shaes According to [6], if a, b and c are three distinct oints in the Gaussian lane the number abc = a c a b is called a shae of the oriented triangle The shae abc is real if and only if abc is degenerated, ie the oints a, b and c are collinear It is clear that abc C \ R for any non - degenerate triangle abc If = abc, then = bca = (1 ) 1 and = cab = 1 1 In [6], J Lester roves two main theorems which are a very useful tool for calculation of shaes We shall recall the arts of these theorems which will be used in our roofs First Shae Theorem Let a and b be distinct fixed oints and let, q and r be three arbitrary oints in the Euclidean lane If ab = λ, qab = µ, rab = ν, then the triangle qr has a shae (1 µ)(λ ν) qr = (1 ν)(λ µ) * The research is artially suorted by Shumen University under grant Math Subject Classification: 51M05, 51M15 162

2 Let ab = λ, abq = µ and abr = ν, then λ = 1 1 λ, µ = 1 1 µ, ν = 1 1 ν Using First Shae Theorem we obtain another more comfortable reresentation of the shae of the triangle qr 1 µ (1) qr = ( ν 1 λ 1 ) 1 ν ( µ 1 λ 1 ) = λ ν λ µ Second Shae Theorem Let abc be a non - degenerate triangle with shae abc, and let, q and r be three arbitrary oints If cb = λ, qac = µ and rba = ν, then qr = λ abc (1 λν)(1 ν) 1 (1 λµ)(1 µ) 1 abc 1 As above, if cb = λ, acq = µ and bar = ν, then λ = 1 1 λ, µ = 1 1 µ, ν = 1 1 ν and (2) qr = (λ 1) (λ + ν 1) (λ + µ 1) λ See [6] for more details Other alications of shaes are considered in [2] and [3] According to [1], the shae of an ordered quadrangle abcd in the Gaussian lane is the ordered air [, q], where = abc and q = acd We denote the shae of the quadrangle abcd by S(abcd) It is easy to see that similar triangles with the same orientation have the same shae, ie the shae of a triangle is invariant under the grou of direct lane similarities Now we shall rove the same assertion for quadrangles Proosition 1 Let abcd and a b c d be two convex quadrangles with shaes [, q] and [, q ], resectively Then i The quadrangles are similar if and only if they have the same orientation and =, q = q ii The quadrangles are antisimilar if and only if they have an oosite orientation and =, q = q Proof i If abcd and a b c d are similar, then there exists a similarity f Sim + (2) which reserves the orientation in the lane and f(a) = a, f(b) = b, f(c) = c, f(d) = d Thus, abc is similar to a b c and acd is similar to a c d, i e = and q = q Conversely, if = and q = q, then there is a similarity f Sim + (2) such that f(a) = a, f(b) = b and f(c) = c Suose that f(d) = d According to i, the quadrangles abcd and a b c d are similar and then q = acd = a c d = a c d Hence, d = d and the quadrangles abcd and a b c d are similar The roof of the second assertion is the same The roosal of this aer is to give generalizations of the first and second shae theorems For this we need formulas for calculating the shaes of quadrangles when the order of the vertices is changed Lemma 1 Let abcd be a[ convex quadrangle ] with shae [[, q], then i S(bcda) = 1, 1, S(cdab) = 1 1 q, 1 ], 163

3 ii S(dcba) = S(badc) = [ q 1 S(dabc) = q, [ ] (1 q), q q 1 [ 1, ] (1 q) [ ], S(cbad) = 1, 1 q, ] [, S(adcb) = 1q, 1 ] Proof Since = a a b c and q = a a d c, then abd = a a d b = q Thus, bda = () 1 and bad = On the other hand, bcd = b d ()(b a) = = = ()(1 ) 1 Hence, S(bcda) = [ bcd, bda ] = [ b c 1, 1 ] b c The roof of the rest equalities is analogous The first shae theorem can be extended to a formula for calculating the shae of a quadrangle Figure 1 Theorem 1 Let a and b be two distinct oints in the lane For arbitrary oints c i, i = 1, 2, 3, 4, we suose that abci = λ i Then, the shae of the quadrangle c 1 c 2 c 3 c 4 is [ λ3 λ 1 S(c 1, c 2, c 3, c 4 ) =, λ ] 4 λ 1 λ 2 λ 1 λ 3 λ 1 and Proof Since a c i a b = λ i, we have c i = a λ i (a b) (see Figure 1) Then c1c 2c 3 = c 1 c 3 c 1 c 2 = a λ 1(a b) a + λ 3 (a b) a λ 1 (a b) a + λ 2 (a b) = λ 3 λ 1 λ 2 λ 1 c1c 3c 4 = c 1 c 4 = a λ 1(a b) a + λ 4 (a b) c 1 c 3 a λ 1 (a b) a + λ 3 (a b) = λ 4 λ 1 λ 3 λ 1 This comletes the roof Theorem 2 Let abcd be a quadrangle with a shae [, q] On the sides of the quadrangle abcd construct triangles cda 1, dab 1, abc 1 and bcd 1 externally with shaes λ 1, λ 2, λ 3 and λ 4, resectively Then the shae of the quadrangle a 1 b 1 c 1 d 1 164

4 is [ λ1 (1 q) λ 3 (1 λ 1 )(1 q) + qλ 2, ] λ 1 (1 q) + (1 λ 4 )(1 ) λ 1 (1 q) + λ 3 Proof It follows from cda1 = λ 1, dab1 = λ 2, abc1 = λ 3 and bcd1 = λ 4 that a 1 = c λ 1 (c d), b 1 = d λ 2 (d a), c 1 = a λ 3 (a b) and d 1 = b λ 4 (b c) Figure 2 (3) Hence, we find that a1b 1c 1 = a 1 c 1 = c a λ 1(c d) + λ 3 (a b) = a 1 b 1 (1 λ 1 )(c d) + λ 2 (d a) = 1 (1 λ 1) dac λ 3 adb λ 2 (1 λ 1 ) dac Since dac = 1 adc = 1 1 q and adb = 1 dab = q 1 (see Figure 2), then relacing in (3) we obtain the first comonent of the shae of the quadrangle a 1 b 1 c 1 d 1 Similarly (4) a1c 1d 1 = a 1 d 1 = c b λ 1(c d) + λ 4 (b c) a 1 c 1 c a λ 1 (c d) + λ 3 (a b) = 1 λ 1 cbd λ 4 = 1 λ 1 cbd (1 λ 3 ) bca (q 1) From cbd = 1 and bca = (1 ) 1, relacing in (4) we get the second comonent of the shae of the quadrangle a 1 b 1 c 1 d 1 and this comletes the roof Some articular cases of Theorem 2 are used in [3] to solve of some roblems from [4] Theorem 3 Let abcd be a quadrangle with a shae [, q] Let ac bd = m, ab cd = k and bc ad = l Then Im q i abm = 2 Im q + Im, bcm = 1 1 q Im 2 Im q + Im, cdm = 1 1 ( Im q + Im Im q q 2 2, Im q + Im dam = 1 ) 2 Im q 2 Im q + Im, 165

5 abl = Im q 2 Im q Im q 2 Im q ; qim ( 2 Im q + Im ) Im q(im q 2 Im q) ii mkl = qim ( 2 Im q + Im ) Im q( 2 Im q + 2Im Im q) Im Im q Im q 2 Im q ; iii The shae of the quadrangle mkld is [ mkl, mld ], where mld = q( 2 Im q + Im ) Im q Im q( 2 Im q + Im ) Im q(im Im q Im q) ( 2 Im q Im q) Figure 3 Proof Alying the First Shae Theorem for the triangles abb, abd and abm with shaes 1, q and abm, resectively we find that bdm = abm 1 q 1 Since b, d, m are collinear (see Figure 3),then bdm = bdm Hence abm 1 (5) = abm 1 q 1 q 1 Similarly, alying the First Shae Theorem for the triangles aba, abc and abm with shaes 0, and abm, resectively and using that the oints a, c, m are collinear we get (6) abm = abm From (5) and (6), eliminating abm we find the first equality in i The shaes of the triangles bcm, cdm and dam are obtained from the shae of the triangle abm by cycling the vertices of the quadrangle abcd [ and Lemma 1 Alying the first equality in i for the quadrangle bacd with a shae 1, 1 1 q ] we get Im q bal = (1 ) Im q 2 Im q Using that abl = 1 bal we find the last equality in i Denote abm = λ 1, cbk = λ 2 and acl = λ 3 Hence m = a λ 1 (a b), 166

6 k = c λ 2 (c b) and l = a λ 3 (a c), and relacing in mkl = m m k l we find that (7) λ 3 λ 1 mkl = (1 λ 2 ) + λ 2 λ 1 [ It follows from[ the second equality in i for the quadrangles acbd and bacd with shaes 1, q] and 1, 1 1 q ] that (8) cbk (1 q) = λ 2 = 1 Im Im Im q (9) acl Im = λ 3 = q Im q 2 Im q Relacing (8), (9) and the first equality from i in (7) we get ii The last assertion in iii follows from the First Shae Theorem for the triangles abm, abl and abd with shaes abm, abl and abd = q, resectively REFERENCES [1]R Artzy Shaes of olygons J Geom, 50 (1994), [2]G Georgiev, R Encheva, M Sirova Shaes of inscribed and circumscribed quadrangles Mathematics and Education in Mathematics, 30 (2001), [3] G Georgiev, R Encheva Studying secial quadrangles by shaes Annuals of Konstantin Preslavski University, FMI, XVC (2002), 5 21 [4] V Gusev, V Litvinenko, A Mordkovich Solving Problems in Geometry Mir Publishers, Moscow, 1988 [5]L-S Hahn Comlex Numbers and Geometry MAA, Washington DC, 1994 [6]J A Lester Triangles I: Shaes Aequationes Math, 52 (1996), [7] J A Lester Triangles II: Comlex triangle coordinates Aequationes Math, 52 (1996), [8] J A Lester Triangles III: Comlex triangle functions Aequationes Math, 53 (1997), 4 35 [9]IM Yaglom Comlex Numbers in Geometry Academic Press, New York, 1968 Georgi H Georgiev Faculty of Mathematics and Informatics Shumen University Alen Mak Str, No Shumen, Bulgaria ggeorgiev@shu-bgnet Radostina P Encheva Faculty of Mathematics and Informatics Shumen University Alen Mak Str, No Shumen, Bulgaria rencheva@fmishu-bgnet 167

7 ШЕЙП ТЕОРЕМИ ЗА ЧЕТИРИЪГЪЛНИЦИ Георги Христов Георгиев, Радостина П Енчева Шейп на триъгълник е комплексно число, съответстващо на орбитата от триъгълници под действието на групата на подобностите в равнината, запазващи ориентацията Шейп на четириъгълник е наредена двойка от комплексни числа, съответствуваща на еквивалентен клас от четириъгълници по отношение на същата група Прилагаме шейпове на триъгълници за изучаване на шейпове на четириъгълници В частност получаваме тъждества за шейпове чрез използване на накои геометрични конструкции 168