@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 The Construction of (k,3)-Arcs in PG(2,9) by Using Geometrical Method Adil Mahmod Amal .SH.Al-Mukhtar Fatema.F. Kareem Dept. of Mathematics/College of Education For Pure Science(Ibn Al-Haitham)/ University of Baghdad Received in : 27 March 2001 , Accepted in :28 May 2001 Abstract In this work, we construct projectively distinct (k,3)-arcs in the projective plane PG(2,9) by applying a geometrical method. The cubic curves have been been constructed by using the general equation of the cubic. We found that there are complete (13,3)-arcs, complete (15,3)-arcs and we found that the only (16,3)-arcs lead to maximum completeness. Keywords: arcs, secant, Projective , Plane 239 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 Introduction A (k,n)-arc K in PG(2,q) is a set of K points, s.t. some n, but no n + 1 of them are collinear. A (k,n)-arc K is complete if there is no (k+1,n)-arc containing it, [1]. A line m of the plane containing exactly 3 points of K is called a trisecant of K. Let K be a (k,3)-arc. The points of PG(2,9)\K which are not on any trisecant of K will be called the points of index zero for K. The (k,3)-arc is complete iff there is no point of index zero for it, so every point of the plane lies on a trisecant of K, [2]. The projective plane PG(2,9) contains 91 points, 91 lines, every line contain 10 points, and every point is on 10 lines, [3]. Let Pi and Li, i = 1, …, 91, be the points and lines of PG(2,9), respectively. Any point can be represented by a triple (x0,x1,x2) ≠ (0,0,0), x0,x1,x2 are in GF(9) and two triples (x0,x1,x2) and (y0,y1,y2) represent the same point if there exists λ in GF(9)\{0} s.t. xi = λyi, i = 0, 1, 2. Also any line in PG(2,9) can be represented by a triple [x0,x1,x2] ≠ [0,0,0], x0,x1,x2 are in GF(9). Two triples [x0,x1,x2] and [y0,y1,y2] represent the same line if there exists λ in GF(9)\{0} s.t. xi = λyi, i = 0, 1, 2. The addition and multiplication’s operation of GF(9), [3] are given in the tables. The point (x0,x1,x2) is on the line [y0,y1,y2] iff x0y0 + x1y1 + x2y2 = 0, [2]. Let i stands for the point Pi. All the points and lines of PG(2,9) are given in the table. The Construction of (k,3)-Arcs for k ≥ 7: [3] We construct projectively distinct (K,3)-arcs in projective plane PG(2,9) by using a geometrical method. This method having the following steps: The general equation of the cubic is: 3 2 2 2 2 3 2 2 3 1 0 2 0 1 3 0 2 4 0 1 5 0 1 2 6 0 2 7 1 8 1 2 9 1 2 10 2F x x x x x x x x x x x x x x x x x x 0= α + α + α + α + α + α +α + α + α + α = …(1) By substituting the points 1,2,11,12,20 of the (7,3)-arc which is a quadrangle whose vertices the points 1, 2, 11, 21 and the diagonal points 3, 12,20, we get: 1 = (1,0,0) ⇒ α1 = 0 2 = (0,1,0) ⇒ α7 = 0 11 = (0,0,1) ⇒ α10 = 0 3 = (1,1,0) ⇒ α2 + α4 = 0 12 = (1,0,1) ⇒ α3 + α6 = 0 20 = (0,1,1) ⇒ α8 + α9 = 0 21 = (1,1,1) ⇒ α5 = 0 So (1) becomes: 2 2 2 2 2 2 4 0 1 0 1 6 0 2 0 2 9 1 2 1 2(x x x x ) (x x x x ) (x x x x ) 0α − + α − + α − = …(2) If α4 = 0, then 2 2 2 26 0 2 0 2 9 1 2 1 2(x x x x ) (x x x x ) 0α − + α − = 2 6 0 2 0 9 1 2 1x { x (x x ) x (x x )} 0α − + α − = is reducible, therefore α4 ≠ 0. So α6 ≠ 0 and α9 ≠ 0. Dividing equation (2) by α4, we get: 2 2 2 2 2 2 0 1 0 1 0 2 0 2 1 2 1 2x x x x (x x x x ) (x x x x ) 0− + α − +β − = where: 6 9 4 4 0, 0 α α α = ≠ β = ≠ α α . Let x2 = 1, then: 240 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 2 2 2 2 0 1 0 1 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 0 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 0 0 1 1 1 1 x x x x (x x ) (x x ) 0 x x (x x ) x (1 x ) x (1 x ) 0 x x {(x 1) (1 x )} x x (1 x ) x (1 x ) 0 x x (x 1) x x (1 x ) x (1 x ) x (1 x ) 0 x x (x 1) x x (1 x ) x (1 x ) x (x 1) 0 x (x − + α − +β − = − + α − +β − = − + − + α − +β − = − + − + α − −β − = − + − + α − +β − = 0 0 0 11)(x ) x (1 x )(x ) 0− −β + − + α = 1 1 0 0 0 1x (x 1)(x ) x (x 1)(x ) 0− −β − − + α = …(3) 0 01 1 1 0 x (x 1)x (x 1) x x −− = + α −β , where α ≠ – 1 = 2 (mod(9), β ≠ 1. The cubic curves have been constructed by implementing equation (3) with deferent values of α and β where α, β = 1,2,…,8 and α, β ≠ 0, α ≠ 2, β ≠ 1, α ≠ – β, because this will form a degenerated cubic. The Set of all the Cubic Curves: [3] 1. If α = 1, β = 3, then the cubic curve C1 is: x1(x1 – 1)(x0 – 3) – x0(x0 – 1)(x1 + 1) = 0 C1 = {1,2,11,21,3,12,20,32,40,45,80,82} which is incomplete cubic, and by adding to C1 is the point 36, which is the point of index zero for C1, we obtain a complete (13,3)-arc. 2. If α = 1, β = 4, then the cubic curve C2 is: x1(x1 – 1)(x0 – 4) – x0(x0 – 1)(x1 + 1) = 0 C2 = {1,2,11,21,3,12,20,33,50,53,86,89}, which is incomplete cubic, and by adding to it the point 37, which is the point of index zero for C2, we obtain a complete (13,3)-arc. 3. If α = 1, β = 5, then the cubic curve C3 is: x1(x1 – 1)(x0 – 5) – x0(x0 – 1)(x1 + 1) = 0 C3 = {1,2,11,21,3,12,20,34,68,73}, which is incomplete cubic, and by adding to it the points of index zero for C3, which are: 32, 40, 42, 58, 79, we obtain a complete (15,3)-arc. 4. If α = 1, β = 6, then the cubic curve C4 is: x1(x1 – 1)(x0 – 6) – x0(x0 – 1)(x1 + 1) = 0 C4 = {1,2,11,21,3,12,20,35,69,72}, which is incomplete cubic, and by adding to it the points 32, 43, 54, we obtain a complete (13,3)-arc. 5. If α = 1, β = 7, then the cubic curve C5 is: x1(x1 – 1)(x0 – 7) – x0(x0 – 1)(x1 + 1) = 0 C5 = {1,2,11,21,3,12,20,36,42,43,76,77}, which is incomplete cubic, and by adding to it the point 32, we obtain a complete (13,3)-arc. 6. If α = 1, β = 8, then the cubic curve C6 is: x1(x1 – 1)(x0 – 8) – x0(x0 – 1)(x1 + 1) = 0 C6 = {1,2,11,21,3,12,20,37,52,54}, which is incomplete cubic, and by adding to it the points 32, 40, 43, 77, 78, we obtain a complete (15,3)-arc. 7. If α = 3, β = 2, then the cubic curve C7 is: x1(x1 – 1)(x0 – 2) – x0(x0 – 1)(x1 + 3) = 0 C7 = {1,2,11,21,3,12,20,52,53,67,79,80}, which is incomplete cubic, and by adding to it the point 58, we obtain a complete (13,3)-arc. 8. If α = 3, β = 3, then the cubic curve C8 is: x1(x1 – 1)(x0 – 3) – x0(x0 – 1)(x1 + 3) = 0 241 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 C8 = {1,2,11,21,3,12,20,34,42,60,68,88}, which is incomplete cubic, and by adding to it the point 63, we obtain a complete (13,3)-arc. 9. If α = 3, β = 4, then the cubic curve C9 is: x1(x1 – 1)(x0 – 4) – x0(x0 – 1)(x1 + 3) = 0 C9 = {1,2,11,21,3,12,20,69}, which is incomplete cubic, and by adding to it the points 32, 33, 40, 43, 49, 54, 68, we obtain a complete (15,3)-arc. 10. If α = 3, β = 5, then the cubic curve C10 is: x1(x1 – 1)(x0 – 5) – x0(x0 – 1)(x1 + 3) = 0 C10 = {1,2,11,21,3,12,20,50,55,70,77,82}, which is incomplete cubic, and by adding to it the point 62, we obtain a complete (13,3)-arc. 11. If α = 3, β = 7, then the cubic curve C11 is: x1(x1 – 1)(x0 – 4) – x0(x0 – 1)(x1 + 3) = 0 C11 = {1,2,11,21,3,12,20,72}, which is incomplete cubic, and by adding to it the points 32, 33, 40, 45, 50, 67, 79, we obtain a complete (15,3)-arc. 12. If α = 3, β = 8, then the cubic curve C12 is: x1(x1 – 1)(x0 – 8) – x0(x0 – 1)(x1 + 3) = 0 C12 = {1,2,11,21,3,12,20,32,33,40,44,58,62,73,86,87}, which is a complete (16,3)-arc. 13. If α = 4, β = 2, then the cubic curve C13 is: x1(x1 – 1)(x0 – 2) – x0(x0 – 1)(x1 + 4) = 0 C13 = {1,2,11,21,3,12,20,60,64,78,82,85}, which is incomplete cubic, and by adding to it the point 49, we obtain a complete (13,3)-arc. 14. If α = 4, β = 3, then the cubic curve C14 is: x1(x1 – 1)(x0 – 3) – x0(x0 – 1)(x1 + 4) = 0 C14 = {1,2,11,21,3,12,20,33,43,52,69,86}, which is incomplete cubic, and by adding to it the point 54, we obtain a complete (13,3)-arc. 15. If α = 4, β = 4, then the cubic curve C15 is: x1(x1 – 1)(x0 – 4) – x0(x0 – 1)(x1 + 4) = 0 C15 = {1,2,11,21,3,12,20,59,62,77,80,87}, which is incomplete cubic, and by adding to it the point 55, we obtain a complete (13,3)-arc. 16. If α = 4, β = 5, then the cubic curve C16 is: x1(x1 – 1)(x0 – 5) – x0(x0 – 1)(x1 + 4) = 0 C16 = {1,2,11,21,3,12,20,88}, which is incomplete cubic, and by adding to it the points 32,33,40,42,50,55,58, we obtain a complete (15,3)-arc. 17. If α = 4, β = 6, then the cubic curve: C17 = {1,2,11,21,3,12,20,32,34,40,46,49,55,68,70,80}, which is a complete (14,3)-arc. 18. If α = 4, β = 7, then the cubic curve: C18 = {1,2,11,21,3,12,20,90}, which is incomplete cubic, and by adding to it the points 32, 33, 40, 42, 49, 53, 70, we obtain a complete (15,3)-arc. 19. If α = 5, β = 2, then the cubic curve: C19 = {1,2,11,21,3,12,20,36,45,50,59,76}, which is incomplete cubic, and by adding to it the point 40, we obtain a complete (13,3)-arc. 20. If α = 5, β = 3, then the cubic curve: C20 = {1,2,11,21,3,12,20,35,37,44,46,49,54,58,63,77}, which is a complete (16,3)-arc. 21. If α = 5, β = 4, then the cubic curve: C21 = {1,2,11,21,3,12,20,72,73,78,85,88}, which is incomplete cubic, and by adding to it the point 46, we obtain a complete (13,3)-arc. 22. If α = 5, β = 5, then the cubic curve: C22 = {1,2,11,21,3,12,20,67,69,79,89,90}, which is incomplete cubic, and by adding to it the point 44, we obtain a complete (13,3)-arc. 23. If α = 5, β = 6, then the cubic curve: 242 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 C23 = {1,2,11,21,3,12,20,80}, which is incomplete cubic, and by adding to it the points 32, 33,42,49,58,76,86, we obtain a complete (15,3)-arc. 24. If α = 5, β = 8, then the cubic curve: C24 = {1,2,11,21,3,12,20,82}, which is incomplete cubic, and by adding to it the points 32, 33,40,43,49,54,78, we obtain a complete (15,3)-arc. 25. If α = 6, β = 2, then the cubic curve: C25 = {1,2,11,21,3,12,20,32,40,72,77,90}, which is incomplete cubic, and by adding to it the point 76, we obtain a complete (13,3)-arc. 26. If α = 6, β = 4, then the cubic curve: C26 = {1,2,11,21,3,12,20,42}, which is incomplete cubic, and by adding to it the points 32, 33,49,53,58,68, we obtain a complete (15,3)-arc. 27. If α = 6, β = 5, then the cubic curve: C27 = {1,2,11,21,3,12,20,43}, which is incomplete cubic, and by adding to it the points 32, 33,42,49,53,58,68, we obtain a complete (15,3)-arc. 28. If α = 6, β = 6, then the cubic curve: C28 = {1,2,11,21,3,12,20,44,50,52,58,64}, which is incomplete cubic, and by adding to it the point 79, we obtain a complete (13,3)-arc. 29. If α = 6, β = 7, then the cubic curve: C29 = {1,2,11,21,3,12,20,33,34,45,67,68,78,79,85,86}, which is a complete (16,3)-arc. 30. If α = 6, β = 8, then the cubic curve: C30 = {1,2,11,21,3,12,20,46,49,53,59,60}, which is incomplete cubic, and by adding to it the point 78, we obtain a complete (13,3)-arc. 31. If α = 7, β = 2, then the cubic curve: C31 = {1,2,11,21,3,12,20,43,44,58,88,89}, which is incomplete cubic, and by adding to it the point 67, we obtain a complete (13,3)-arc. 32. If α = 7, β = 3, then the cubic curve: C32 = {1,2,11,21,3,12,20,59}, which is incomplete cubic, and by adding to it the points 32, 33,40,43,49,53,87, we obtain a complete (15,3)-arc. 33. If α = 7, β = 4, then the cubic curve: C33 = {1,2,11,21,3,12,20,36,37,54,55,60,67,70,76,79}, which is a complete (16,3)-arc. 34. If α = 7, β = 6, then the cubic curve: C34 = {1,2,11,21,3,12,20,42,62,87,90}, which is incomplete cubic, and by adding to it the point 70, we obtain a complete (13,3)-arc. 35. If α = 7, β = 7, then the cubic curve: C35 = {1,2,11,21,3,12,20,35,53,63,73,82}, which is incomplete cubic, and by adding to it the point 68, we obtain a complete (13,3)-arc. 36. If α = 7, β = 8, then the cubic curve: C36 = {1,2,11,21,3,12,20,64}, which is incomplete cubic, and by adding to it the points 32, 33, 42, 49, 50, 58, we obtain a complete (15,3)-arc. 37. If α = 8, β = 2, then the cubic curve: C37 = {1,2,11,21,3,12,20,42,46,49,69,73}, which is incomplete cubic, and by adding to it the point 85, we obtain a complete (13,3)-arc. 243 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 38. If α = 8, β = 3, then the cubic curve: C38 = {1,2,11,21,3,12,20,50}, which is incomplete cubic, and by adding to it the points 32, 33, 40, 42, 49, 63, 79, we obtain a complete (15,3)-arc. 39. If α = 8, β = 5, then the cubic curve: C39 = {1,2,11,21,3,12,20,35,36,52,62,63,76,78,85,87}, which is a complete (16,3)-arc. 40. If α = 8, β = 6, then the cubic curve: C40 = {1,2,11,21,3,12,20,53}, which is incomplete cubic, and by adding to it the points 32, 33, 40, 42, 49, 70, 73, we obtain a complete (15,3)-arc. 41. If α = 8, β = 7, then the cubic curve: C41 = {1,2,11,21,3,12,20,37,54,64, 80,89}, which is incomplete cubic, and by adding to it the point 86, we obtain a complete (13,3)-arc. 42. If α = 8, β = 7, then the cubic curve: C42 = {1,2,11,21,3,12,20,43,45,55,70,72}, which is incomplete cubic, and by adding to it the point 87, we obtain a complete (13,3)-arc. when we checked each set of the cubic curves to find, if the points of each curve lead to maximum completeness, we found that the only (16,3)-arcs are complete, and we found that there exist complete (13,3)-arcs and complete (15,3)-arcs. References 1. Hirschfeld, J. W. P. (1979) Projective Geometries Over Finite Fields, Second Edition, Oxford University Press. 2. Kwaam, A.A. (1999) Classification and Construction of (k,3)-arcs in PG(2,11), M.Sc. Thesis, University of Baghdad, Iraq. 3. Kareem, F.F. (2000) Classification and Construction of (k,3)-arcs in PG(2,9), M.Sc. Thesis, University of Baghdad. (GF(9),∗) ∗ 1 2 3 4 5 6 7 8 1 1 2 3 4 5 6 7 8 2 2 1 6 8 7 3 5 4 3 3 6 4 7 1 8 2 5 4 4 8 7 2 3 5 6 1 5 5 7 1 3 8 2 4 6 6 6 3 8 5 2 4 1 7 7 7 5 2 6 4 1 8 3 8 8 4 5 1 6 7 3 2 244 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 (GF(9),+) + 0 1 2 3 4 5 6 7 8 0 0 1 2 3 4 5 6 7 8 1 1 2 0 4 5 3 7 8 6 2 2 0 1 5 3 4 8 6 7 3 3 4 5 6 7 8 0 1 2 4 4 5 3 7 8 6 1 2 0 5 5 3 4 8 6 7 2 0 1 6 6 7 8 0 1 2 3 4 5 7 7 8 6 1 2 0 4 5 3 8 8 6 7 2 0 1 5 3 4 Points and Lines of PG(2,9) i Pi ℓi 1 (1,0,0) 2 11 20 29 38 47 56 65 74 83 2 (0,1,0) 1 11 12 13 14 15 16 17 18 19 3 (1,1,0) 4 11 22 30 44 55 63 68 9 87 4 (2,1,0) 3 11 21 31 41 51 61 71 81 91 5 (3,1,0) 9 11 27 34 40 53 60 66 82 86 6 (4,1,0) 6 11 24 37 45 49 59 70 80 84 7 (5,1,0) 8 11 26 32 46 52 58 69 75 90 8 (6,1,0) 7 11 25 36 39 50 64 67 78 89 9 (7,1,0) 5 11 23 35 42 54 57 73 76 88 10 (8,1,0) 10 11 28 33 43 48 62 72 77 85 11 (0,0,1) 1 2 3 4 5 6 7 8 9 10 12 (1,0,1) 2 13 22 31 40 49 58 67 76 85 13 (2,0,1) 2 12 21 30 39 48 57 66 75 84 14 (3,0,1) 2 18 27 36 45 54 63 72 81 90 15 (4,0,1) 2 15 24 33 42 51 60 69 78 87 16 (5,0,1) 2 17 26 35 44 53 62 71 80 89 17 (6,0,1) 2 16 25 34 43 52 61 70 79 88 18 (7,0,1) 2 14 23 32 41 50 59 68 77 86 19 (8,0,1) 2 19 28 37 46 55 64 73 82 91 20 (0,1,1) 1 29 30 31 32 33 34 35 36 37 21 (1,1,1) 4 13 21 29 46 54 62 70 78 86 22 (2,1,1) 3 12 22 29 42 52 59 72 82 89 23 (3,1,1) 9 18 25 29 44 51 58 73 77 84 24 (4,1,1) 6 15 28 29 40 50 63 71 75 88 25 (5,1,1) 8 17 23 29 43 49 64 66 81 87 26 (6,1,1) 7 16 27 29 41 55 57 69 80 85 27 (7,1,1) 5 14 26 29 45 48 60 67 79 91 28 (8,1,1) 10 19 24 29 39 53 61 68 76 90 29 (0,2,1) 1 20 21 22 23 24 25 26 27 28 30 (1,2,1) 3 13 20 30 43 50 60 73 80 90 31 (2,2,1) 4 12 20 31 45 53 64 69 77 88 32 (3,2,1) 7 18 20 34 46 48 59 71 76 87 33 (4,2,1) 10 15 20 37 44 52 57 67 81 86 34 (5,2,1) 5 17 20 32 39 51 63 70 82 85 245 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 35 (6,2,1) 9 16 20 36 42 49 62 68 75 91 36 (7,2,1) 8 14 20 35 40 55 61 72 78 84 37 (8,2,1) 6 19 20 33 41 54 58 66 79 89 38 (0,3,1) 1 74 75 76 77 78 79 80 81 82 39 (1,3,1) 8 13 28 34 45 51 57 68 74 89 40 (2,3,1) 5 12 24 36 43 55 58 71 74 86 41 (3,3,1) 4 18 26 37 42 50 61 66 74 85 42 (4,3,1) 9 15 22 35 41 48 64 70 74 90 43 (5,3,1) 10 17 25 30 40 54 59 69 74 91 44 (6,3,1) 3 16 23 33 46 53 63 67 74 84 45 (7,3,1) 6 14 27 31 39 52 62 73 74 87 46 (8,3,1) 7 19 21 32 44 49 60 72 74 88 47 (0,4,1) 1 47 48 49 50 51 52 53 54 55 48 (1,4,1) 10 13 27 32 42 47 64 71 79 84 49 (2,4,1) 6 12 25 35 46 47 60 68 81 85 50 (3,4,1) 8 18 24 30 41 47 62 67 82 88 51 (4,4,1) 4 15 23 34 39 47 58 72 80 91 52 (5,4,1) 7 17 22 33 45 47 61 73 75 86 53 (6,4,1) 5 16 28 31 44 47 59 66 78 90 54 (7,4,1) 9 14 21 37 43 47 63 69 76 89 55 (8,4,1) 3 19 26 36 40 47 57 70 77 87 56 (0,5,1) 1 65 66 67 68 69 70 71 72 73 57 (1,5,1) 9 13 26 33 39 55 59 65 81 88 58 (2,5,1) 7 12 23 37 40 51 62 65 79 90 59 (3,5,1) 6 18 22 32 43 53 57 65 78 91 60 (4,5,1) 5 15 27 30 46 49 61 65 77 89 61 (5,5,1) 4 17 28 36 41 52 60 65 76 84 62 (6,5,1) 10 16 21 35 45 50 58 65 82 87 63 (7,5,1) 3 14 24 34 44 54 64 65 75 85 64 (8,5,1) 8 19 25 31 42 48 63 65 80 86 65 (0,6,1) 1 56 57 58 59 60 61 62 63 64 66 (1,6,1) 5 13 25 37 41 53 56 72 75 87 67 (2,6,1) 8 12 27 33 44 50 56 70 76 91 68 (3,6,1) 3 18 28 35 39 49 56 69 79 86 69 (4,6,1) 7 15 26 31 43 54 56 68 82 84 70 (5,6,1) 6 17 21 34 42 55 56 67 77 90 71 (6,6,1) 4 16 24 32 40 48 56 73 81 89 72 (7,6,1) 10 14 22 36 46 51 56 66 80 88 73 (8,6,1) 9 19 23 30 45 52 56 71 78 85 74 (0,7,1) 1 38 39 40 41 42 43 44 45 46 75 (1,7,1) 7 13 24 35 38 52 63 66 77 91 76 (2,7,1) 9 12 28 32 38 54 61 67 80 87 77 (3,7,1) 10 18 23 31 38 55 60 70 75 89 78 (4,7,1) 8 15 21 36 38 53 59 73 79 85 79 (5,7,1) 3 17 27 37 38 48 58 68 78 88 80 (6,7,1) 6 16 26 30 38 51 64 72 76 86 81 (7,7,1) 4 14 25 33 38 49 57 71 82 90 82 (8,7,1) 5 19 22 34 38 50 62 69 81 84 83 (0,8,1) 1 83 84 85 86 87 88 89 90 91 246 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 84 (1,8,1) 6 13 23 36 44 48 61 69 82 83 85 (2,8,1) 10 12 26 34 41 49 63 73 78 83 86 (3,8,1) 5 18 21 33 40 52 64 68 80 83 87 (4,8,1) 3 15 25 32 45 55 62 66 76 83 88 (5,8,1) 9 17 24 31 46 50 57 72 79 83 89 (6,8,1) 8 16 22 37 39 54 60 71 77 83 90 (7,8,1) 7 14 28 30 42 53 58 70 81 83 91 (8,8,1) 4 19 27 35 43 51 59 67 75 83 247 | Mathematics @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ÚÓ‘Ój�n€a@Î@Úœäñ€a@‚Ï‹»‹€@·rÓ:a@Âig@Ú‹©@Ü‹1a26@@ÖÜ»€a@I2@‚b«@H2013 Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 26 (2) 2013 باستخدام طریقة PG(2,9)في المستوي إالسقاطي (k,3)بناء االقواس ھندسیة عادل محمود أحمد آمال شھاب المختار فاطمة فیصل كریم ( ابن الھیثم ) / جامعة بغدادللعلوم الصرفة قسم الریاضیات / كلیة التربیة 2001آیار 28، قبل البحث في : 2001آذار 27استلم البحث في : الخالصة بتطبیق طریقة ھندسیة. PG(2,9)اسقاطیة مختلفة في المستوي االسقاطي (k,3)یتضمن ھذا البحث، بناء االقواس كاملة، وكذلك (13,3) –ویتم بناء المنحنیات التكعیبیة باستخدام المعادلة العامة التكعیبیة. لقد وجدنا أن االقواس .(16,3) –كاملة، و وجدنا ان اكبر االقواس الكاملة ھي االقواس (15,3) –االقواس اطي اقواس ،قاطع ،مستوي اسق : الكلمات المفتاحیة 248 | Mathematics قسم الرياضيات / كلية التربية للعلوم الصرفة ( ابن الهيثم ) / جامعة بغداد استلم البحث في :27 آذار 2001 ، قبل البحث في :28 آيار 2001 يتضمن هذا البحث، بناء الاقواس (k,3) اسقاطية مختلفة في المستوي الاسقاطي PG(2,9) بتطبيق طريقة هندسية. ويتم بناء المنحنيات التكعيبية باستخدام المعادلة العامة التكعيبية. لقد وجدنا أن الاقواس – (13,3) كاملة، وكذلك الاقواس – (15,3) كاملة، و وجد...