AREAS OF POLYGONS WITH COORDINATES OF VERTICES FROM HORADAM AND LUCAS NUMBERS. 1. Introduction

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1 SARAJEVO JOURNAL OF MATHEMATICS Vol.13 (26), No.2, (2017), DOI: /SJM AREAS OF POLYGONS WITH COORDINATES OF VERTICES FROM HORADAM AND LUCAS NUMBERS ZVONKO ƒerin In memory of my dear friends from Zagorje, tefa Fu kara i tefa Papeºa Abstract. We consider several classes of polygons in the plane with coordinates of points from the Horadam and (general) Lucas numbers. Our goal is to determine formulas for oriented areas of these polygons. Some of our formulas are for polygons with any number of vertices while others deal only with triangles, quadruples and pentagons. 1. Introduction A selection of a (rectangular) system of coordinates in the plane allows to build polygons by describing coordinates of its vertices. By restricting that these coordinates are all from a special kind of numbers (for example, from binary recursive sequences like Fibonacci and Lucas numbers,...) we obtain geometric objects that can have some special properties that depend on the choice of numbers. This idea was recently considered by Davala and Panda [6] (extending earlier author's results for triangles [2]). They discovered formulas for areas of nine classes of polygons with coordinates of vertices from Fibonacci and Lucas sequences. In this paper we improve their results by using Horadam and (general) Lucas numbers w n and x n (i. e., the most general form of a second order binary recurrence whose denitions we recall in the next section) and by using three parameters to dene our classes of polygons. One gets their nine classes by special selections of these parameters in four of our classes Mathematics Subject Classication. Primary 11H99. Key words and phrases. area, polygon, vertices,coordinates, Horadam, Lucas. Copyright c 2017 by ANUBIH.

2 180 ZVONKO ƒerin Our formulas are for oriented areas of some among sixteen classes of polygons that we dene here. In some cases we can not deal with arbitrary polygons and sometimes we are unable to nd any formulas (these are left as open problems). Analogous formulas (apparently also only for less general numbers) have been presented at the seventeenth Fibonacci Conference in Caen (France), June 27July 2, 2016, by Virginia Johnson [3] (see also [4]). 2. Horadam and (general) Lucas numbers The Horadam and (general) Lucas numbers w n and x n are dened by the recurrence relations and w 0 = s, w 1 = t, w n = p w n 1 + q w n 2 for n 2, x 0 = 2 t s p, x 1 = t p + 2 s q, x n = p x n 1 + q x n 2 for n 2, where s, t, p and q are real numbers such that p 0, q 0 and = p q 0. Let α = p+ 2, β = p 2, α 0 = t s β, β 0 = t s α. It is well-known that w n = α0 αn β 0 β n and x n = α 0 α n + β 0 β n, for every integer n 0. Since the right-hand sides of these equalities make sense for every integer n, it follows that both w n and x n are in this way dened for every integer n. The numbers w n and x n for the initial values s = 0 and t = 1 play and important role in this paper. Hence, we dene Let δ = q s 2 + p s t t 2. y n = w n (0, 1, p, q), and z n = x n (0, 1, p, q). 3. The first set of eight infinite series of polygons Recall (see [5, p. 51] and the above gure) that a polygon is dened as a geometric object consisting of a number of points (called vertices) and an equal number of line segments (called sides), namely a cyclically ordered set of points in a plane, with no three successive points collinear, together with the line segments joining consecutive pairs of points. In other words, a polygon is a closed broken line lying in a plane.

3 POLYGONS WITH HORADAM AND LUCAS COORDINATES 181 Figure 1: Some polygons: triangle (trigon), quadrangle (tetragon), pentagon and heptagon. Let n be a natural number and let = (a, b, c) be a triple of integers. In this paper we shall consider the following polygons A = A, B = B, C = C B j and D = D in the plane with n vertices A j = A, j = (w u, w v ), = B, j = (w u, x v ), C j = C, j = (x u, x v )and D j = D, j = (x u, w v ), where v = v(j) = a + b j + c, u = u(j) = a + b j and j = 0,..., n 1, respectively. In special cases our polygons reduce to families of polygons in the reference [6] as follows. When s = 0, t = 1, p = 1, q = 1 and a = k, then A (n+2) = n+2, k and C (n+2) = n+2, k for b = 1 and c = 1. A (n+2) = n+2, k and C(n+2) = n+2, k for b = 2 and c = 1. B (n+2) = n+2, k for b = 1 and c = 0 A (n+2) = Φ n+2, k, r, B (n+2) = Ψ n+2, k, r, C (n+2) = Φ n+2, k, r and D (n+2) = Ψ n+2, k, r for b = 1 and c = r. Besides these four polygons it is possible to dene another set of four polygons in the plane that we denote A = A While A has as vertices, B = B, C = C and D = D. (w u(0), w v(0) ), (w u(1), w v(1) ), (w u(2), w v(2) ), (w u(3), w v(3) ),... the polygon A has as vertices Similarly, B has (w u(0), w v(0) ), (x u(1), x v(1) ), (w u(2), w v(2) ), (x u(3), x v(3) ),.... (w u(0), x v(0) ), (x u(1), w v(1) ), (w u(2), x v(2) ), (x u(3), w v(3) ),...,

4 182 ZVONKO ƒerin C has (x u(0), x v(0) ), (w u(1), w v(1) ), (x u(2), x v(2) ), (w u(3), w v(3) ),... and D has (x u(0), w v(0) ), (w u(1), x v(1) ), (x u(2), w v(2) ), (w u(3), x v(3) ), Areas of polygons from the first set Our goal now is to nd (oriented) areas of the eight polygons from the rst set. Let q denote q. For an integer k, let T ± k Theorem 1. Let = (a, b, c) be a triple of integers. polyhedra A, B, C and D be q k ± 1. Let Q ± = δ q a. 2 T ± b The oriented areas of the for an integer n 3 and of the polyhedra (2 n+1) (2 n+2) (2 n+1) (2 n+2) A, A, C and C for a natural numbers n are A ( ) = y c Q y b T b(n 1) y b(n 1) T b B ( ) = z c Q z b T b(n 1) y b(n 1) T b C = 2 A, D = z c A y c (2 n+1) ( ) A = y c Q + z b T b(2 n) + y b(2 n) T + b (2 n+2) ( A = y c Q + z b T + b(2 n+1) z b(2 n+1) T + b (2 n+2) (2 n+2) C = A (2 n+1) ( C = Q + y b z c T b(2 n) + 2 y c y b(2 n) T + b, (1), (2), (3), (4) ), (5), (6) ). (7) Proof. We shall prove the formula (1) by induction on n. Other formulas in this theorem have similar proofs. Recall that the triangle ABC with the vertices whose coordinates are (a 1, a 2 ), (b 1, b 2 ), and (c 1, c 2 ) has the oriented area equal to a 1 a 2 1 ABC = 1 2 b 1 b 2 1 c 1 c 2 1. (8) When we substitute the coordinates of the vertices of the triangles A (3) and A (k+1) 0 A (k+1) k 1 A(k+1) k into (8), after some simplications, we obtain ( A (3) = δ 2 qa y c yb T + b y 2 b), (9)

5 POLYGONS WITH HORADAM AND LUCAS COORDINATES 183 A (k+1) 0 A (k+1) ( ) k 1 A(k+1) k = δ 2 qa y c yb q (k 1)b + y (k 1) b y k b. (10) Since T 2 b = T + T b, we see that (9) agrees with (1) for n = 3. Hence, (1) is true for b this initial value. On the other hand, the dierence of (1) for n = k + 1 and of (1) for n = k is precisely the right hand side of (10). This shows the inductive step that if (1) holds for n = k it will hold also for n = k + 1 because A (k+1) A (k+1) = 0 A (k+1) k 1 A(k+1) k + A (k). This concludes the proof of formula (1) by induction on n (the number of vertices of the polygon). Another approach to formulas (1)(7) is to apply the surveyor's formula for oriented area A of a polygon A = A 0 A 1... A n in the plane which says (see [1]) that if its vertices have coordinates (p j, q j ) (j = 0,..., n), then A = 1 2 p n 1 n p 0 q n q 0 + p j p j+1 q j q j+1. (11) j=0 For example, in order to get the formula (1), we let A = A becomes A Since it follows that = 1 2 w a+(n 1)b n 2 j=0 w a+(n 1)b+c w a+j b w a+j b+c w a+j b w a+j b+c On the other hand, w a+(n 1)b w a+(n 1)b+c w a w a+c n 2 + w a+j b j=0 w a+(j+1)b w a+(j+1)b+c w a+(j+1)b w a+(j+1)b+c w a w a+c From (12), (13) and (14) it follows that (1) is true. w a+j b+c = qa+j b δ y b y c, We do not have any formulas for the oriented areas w a+(j+1)b w a+(j+1)b+c in (11) so that it. (12) = qa δ y b y c T (n 1)b. (13) T b = qa δ y (n 1)b y c. (14) B and D for any integer n 3. However, their sums for n = 3, 4, 5, 6 are given as follows. Let

6 184 ZVONKO ƒerin Z 0 = δ 2 q a (1 2 )y b y c and Z j = B (j) + D (j) for j 3. [ (T Z 3 = Z 0 T + b, Z ) + 2 ] 4 = Z 0 b z 2 b, (15) Z 5 = Z 3 T + 2 b, Z 6 = Z 4 [ z2 b + q 2 b + T + b ]. (16) Proof of the rst formula in (15). The area Z 3 is 1 2 (D + ϕ(d)), where ϕ denotes the transformation that replaces w with x and x with w and w a x a+c 1 D = x a+b w a+b+c 1 w a+2 b x a+2 b+c 1. The expansion of D along the rst row is e 1 e 2 + e 3 e 4, where e 1 = w a (w a+b+c x a+2 b+c ), e 2 = x a+c (x a+b w a+2 b ), e 3 = x a+b x a+2 b+c and e 4 = w a+2 b w a+b+c. It is easy to see that and e 3 + ϕ(e 3 ) e 4 ϕ(e 4 ) = q a+b (1 2 )y b y c e 1 + ϕ(e 1 ) e 2 ϕ(e 2 ) = q a (1 2 )y b y c. Hence, Z 3 = q a+b (1 2 )y b y c + q a (1 2 )y b y c = Z 0 T + b. In order to get Z 4, Z 5 and Z 6 we subdivide a quadrangle into two triangles, a pentagon into a quadrangle and a triangle and a hexagon into two quadrangles and compute area of each part separately. 5. The second set of eight infinite series of polygons Let v = v(j) = a b j + c and u = u(j) = a b j. polyhedra in the plane is A = A with vertices Another innite family of (w u(0), w v(0) ), (w u(1), w v(1) ), (w u(2), w v(2) ), (w u(3), w v(3) ),..., for every natural number n 3 and any triple = (a, b, c) of integers. We can similarly dene polyhedra B, C, D, A, B, C and D.

7 POLYGONS WITH HORADAM AND LUCAS COORDINATES Areas of some polygons from the second set In this section, we consider oriented areas of the polygons A (j), B (j), C (j), D (j), A (j) and C (j) for j = 3, 4, 5. Unfortunately, we are unable to get formulas for these areas when j 6 and also the areas of B (j) and D (j) (for j 3) except the sums B (j) D (j) + (for j = 3, 4, 5). Let Q 3 = 1 2 q ( a b δ y b zb T + ) b (zb + 1), Q 4 = δ 2 q ( a 3 b y b z2 b + q b) [ zb 2 ( T + ) 2 ] b, Q 5 = δ 2 q ( a 3 b z 2 b zb T + ) ( b y4 b + T + b y 3 b + q b T + b y ) 2 b + q 2 b y b. The oriented areas of the polygons A (j) given in the following theorem., B (j), C (j) and D (j) for j = 3, 4, 5 are Theorem 2. Let = (a, b, c) be a triple of integers. Then A (j) = y c Q j, B (j) = z c Q j, C (j) = 2 A (j), D (j) = Proof of the rst formula A (3) = y c Q 3. The area A (3) w a w a+c 1 D = w a b w a b+c 1 w a+2 b w a+2 b+c 1. B (j). is 1 2 D, where The expansion of D along the second row is e 1 e 2 + e 3 e 4, where e 1 = w a b+c (w a w a+2 b ), e 2 = w a b (w a+c w a+2 b+c ), e 3 = w a+c w a+2 b and e 4 = w a w a+2 b+c. Now, and e 3 e 4 = q a δ y 2 b y c, zb 2 T + b q b z b = ( z b T + b Hence, from (17) and (18), we get A (3) = y c Q 3. e 1 e 2 = q a b δ y b y c ( z 2 b T + b ) (17) ) (zb + 1). (18) The other parts of Theorem 2 have similar proofs that are more demanding in case of quadrangles and pentagons.

8 186 ZVONKO ƒerin Let R 3 = δ 2 q a b y c, R 4 = [z R3 2 q 2 b b ( T + ) 2 ] b, R 5 = R3 z q 2 b 2 b, R A 3 = z 2 b + q b y b T + b, RC 3 = z 2 b + 2 q b y b T + b, RE 3 = z 2 b + T b, For the polygons A (j) R A 4 = z 2 b 3 q b, R C 4 = R A 4, R E 4 = z 2 b q b, R A 5 = z 4 b T + b z 2 b + q 3 b y b + q b T + b, R C 5 = z 4 b T + b z 2 b + 2 q 3 b y b + q b T + b, R E 5 = z 4 b + T b z 2 b + q b T b., B (j), C (j) and D (j) the following formulas hold (j = 3, 4, 5). Theorem 3. Let = (a, b, c) be a triple of integers. Then A (j) = z b R j Rj A, C (j) = z b R j Rj C, B (j) + D (j) = ( 2 1 ) y b R j Rj E. Proof of the rst formula A (3) = z b R 3 R3 A A. The area (3) w a w a+c 1 D = x a b x a b+c 1 w a+2 b w a+2 b+c 1. is 1 2 D, where The expansion of D along the second row is e 1 e 2 + e 3 e 4, where e 1 = x a b+c (w a w a+2 b ), e 2 = x a b (w a+c w a+2 b+c ), e 3 = w a+c w a+2 b and e 4 = w a w a+2 b+c. Now, and e 3 e 4 = q a δ y 2 b y c, e 1 e 2 = q a b δ z b y c ( z 2 b 3 q b 1 ), (19) zb 2 2 q b = z 2 b. (20) Hence, from (19) and (20), we get A (3) = z b R 3 R3 A. The other parts of Theorem 3 have similar proofs that are again more demanding in case of quadrangles and pentagons.

9 POLYGONS WITH HORADAM AND LUCAS COORDINATES Areas of some quadrangles Besides the above sixteen polygons in the plane whose coordinates are from the sequences w n and x n, it is possible to use these sequences in some other combinations. Here we compute the oriented areas of two quadrangles that are not from our two sets of polyhedra, yet the similar formulas hold. The oriented area Q of a quadrangle Q with the vertices (w u(0), w v(0) ), (w u(1), w v(1) ), (x u(2), x v(2) ), (x u(3), x v(3) ) is the following expression Q = 1 2 q [ a δ y b y ( c 1 2 y 2 b + q 2 b)], (21) while the one for the quadrangle R with the vertices (x u(0), x v(0) ), (x u(1), x v(1) ), (w u(2), w v(2) ), (w u(3), w v(3) ) has a similar form R = 1 2 q [ a δ y b y c 2 (y 2 b 1) + q 2 b]. (22) Proof of (21). The area Q is 1 2 (D + E), where w a w a+c 1 w a w a+c 1 D = w a+b w a+b+c 1 and E = x a+2 b x a+2 b+c 1 x a+2 b x a+2 b+c 1 x a+3 b x a+3 b+c 1. Let D 0 = y b + q b z b z 2 b and E 0 = z 2 b z 3 b 2 q 2 b y b. It is easy to see that D = q a δ y c D 0, E = q a δ y c E 0, (23) and D 0 + E 0 = y b [ 1 2 ( y 2 b + q 2 b)]. (24) From (23) and (24) we get (21). The formula (22) is proved similarly. References [1] B. Braden, The Surveyor's Area Formula, The College Mathematics Journal, 17 (4) (1986), [2] Z. ƒerin, On triangles with Fibonacci and Lucas numbers as coordinates, Sarajevo J. Math. 3 (15) (2007), 3-7 [3] C. Cook and V. Johnson, Areas of triangles, quadrilaterals and other polygons with vertices from various sequences, (conference talk, July 2016). [4] C. K. Cook and V. Johnson, Areas of triangles and other polygons with vertices from various sequences, (arxiv: v1, August 2016).

10 188 ZVONKO ƒerin [5] H. S. M. Coxeter and S. Greitzer, Geometry Revisited, Math. Assoc. Amer., Washington D. C., [6] R. K. Davala and G. K. Panda, On Polygons with Coordinates of Vertices from Fibonacci and Lucas Sequences, Journal for Geometry and Graphics 18 (2) (2014), (Received: September 08, 2016) (Revised: April 05, 2017) Zvonko ƒerin Department of Mathematics University of Zagreb Kopernikova 7, Zagreb CROATIA