AN INNOVATIVE ANALYSIS TO DEVELOP NEW THEOREMS ON IRREGULAR POLYGON

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1 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran AN INNOVATIVE ANALYSIS TO DEVELOP NEW THEOREMS ON IRREGULAR POLYGON *Kalaimaran Ara Construction & Civil Maintenance Unit, Central Food Technological Research Institute, Mysore-0, Karnataka, India *Author for Correspondence ABSTRACT The irregular Polygon is a four sided polygon of two dimensional geometrical figures. The triangle, square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon, parallelogram, rhombus, rhomboid, trapezium or trapezoidal, kite and dart are the members of the irregular polygon family. A polygon is a two dimensional example of the more general prototype in any number of dimensions. However the properties are varied from one to another. The author has attempted to develop two new theorems for the property of irregular polygon for a point anywhere inside of the polygon with necessary illustrations, appropriate examples and derivation of equations for better understanding. Key Words: Irregular Polygon, Triangle, Right-angled triangle, Perpendicular and Vertex INTRODUCTION Polygon (Weisstein, 003) is a closed two dimensional figure formed by connecting three or more straight line segments, where each line segment end connects to only one end of two other line segments. Polygon is one of the most all-encompassing shapes in two- dimensional geometry. The sum of the interior angles is equal to 180 degree multiplied by number of sides minus two. The sum of the exterior angles is equal to 360 degree. From the simple triangle up through square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon (Weisstein, 003,) and beyond is called n-gon. Depending on its interior vertex (Weisstein, 003) angle, the polygon is divided into various categories viz., Convex polygon (Weisstein, 003) a polygon with all diagonals in the interior of the polygon. All vertices point outward concave, at least one vertex point inward towards the center of the polygon. A convex polygon is defined as a polygon with all its interior angles less than 180. Concave polygon (Weisstein, 003) a non-convex. The concave polygon is a simple polygon having at least one interior angle greater than 180. Simple polygon (Weisstein, 003) it does not cross itself. Complex polygon (Weisstein, 003) its path may self-intersect. It looks as combined polygons. Equiangular polygon: all its corner angles are equal. Cyclic polygon: all corners lie on a single circle. Isogonal or vertex-transitive polygon: all corners lie within the same symmetry orbit. The polygon is also cyclic and equiangular. Equilateral polygon or Regular Polygon (Weisstein, 003) all edges are of the same length. (A polygon with five or more sides can be equilateral without being convex). Isotoxal or edge-transitive polygon: all sides lie within the same symmetry orbit. The polygon is also equilateral. Tangential polygon: all sides are tangent to an inscribed circle. The particular cases of the polygons are: Square (Weisstein, 003) and Rhombus (Weisstein, 003). NEW THEOREM ON IRREGULAR POLYGON Theorem 1: Suppose P is a point anywhere inside of the irregular polygon, points A 1, A, A 3, A n are the vertices of the polygon and points M 1, M, M 3, M n are the perpendicular projection of the point P on s 73

2 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran A 1 A, A A 3, A 3 A 4 A n A 1 of the polygon. If M 1 A 1, M A, M 3 A 3,, M n A n are left part line segment and M 1 A, M A 3, M 3 A 4,, M n A 1 are right part of the line segment of s respectively then the sum of the squares of all the left side line segments is equal to the sum of the squares of all the right side line segments (ref. fig.1). The mathematical form of the theorem is M 1 A 1 + M A + M 3 A M n A n = M 1 A + M A 3 + M 3 A M n A 1 Figure 1: A diagram of irregular cyclic Polygon A 1 A A n 1 A n Derivation of equations for theorem-1 Point P is anywhere in the ellipse, points A 1, A, A 3, A n are the vertices of the polygon and points M 1, M, M 3, M n are the projection of the point P on s of the polygon A 1 A, A A 3, A 3 A 4 A n A 1 respectively. Therefore, PM 1 A 1 A, PM A A 3 PM n A n A 1. Point P is joined with vertices of the polygon. PA 1, PA, PA 3, PA n are the distance of the vertices from point P. Let, PA 1 = a 1, PA = a, PA 3 = a 3, PA n = a n Let, PM 1 = h 1, PM = h, PM 3 = h 3, PM n = h n Let, A 1 A = b 1, A A 3 = b, A 3 A 4 = b 3, A n A 1 = b n Let, M 1 A 1 = c 1, M 1 A = d 1, M A = c, M A 3 = d, M n A n = c n, M n A 1 = d n Referring fig., In right triangle PM 1 A 1, PA 1 = PM1 + M1 A 1 Substituting, PA 1 = a 1, PM 1 = h 1, M 1 A 1 = c 1, we get a 1 = h 1 + c 1 [1] In right triangle PM 1 A, PA = PM1 + M1 A Substituting, PA = a, PM 1 = h 1, M 1 A = d 1, we get a = h 1 + d 1 [] Subtracting [] from [1], 74

3 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran a 1 a = c 1 d 1 [3] Rreferring fig. 3, Figure : A diagram of irregular polygon with PA 1 A Figure 3: A diagram of irregular polygon with PA A 3 In right triangle PM A, PA = PM + M A Substituting, PA = a, PM = h, M A = c, we get a = h + c [4] 75

4 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran In right triangle PM A 3, PA 3 = PM + M A 3 Substituting, PA 3 = a 3, PM = h, M A 3 = d, we get a 3 = h + d [5] Subtracting [5] from [4], a a 3 = c d [6] Similarly referring fig. 4, In right triangle PM n A 1, PA n = PMn + Mn A n Substituting, PA n = a n, PM n = h n and M n A n = c n we get a n = h n + c n [7] In right triangle PM n A 1, PA 1 = PMn + Mn A 1 Substituting, PA 1 = a 1, PM n = h n and M n A 1 = d n we get a 1 = h n + d n [8] Figure 4: A diagram of irregular polygon with PA n A 1 Subtracting [8] from [7], a n a 1 = c n d n [9] Adding equations [3], [6] [9], we get a 1 a + a a 3 + a n a 1 = c 1 d 1 + c d + + c n d n 0 = c 1 d 1 + c d + + c n d n Therefore, c 1 d 1 + c d + c 3 d c n d n = 0 Therefore, c 1 + c + c c n = d 1 + d + d d n Rewritting, c 1 = M 1 A 1, c, = M A, c 3 = M 3 A 3, c n = M n A n, d 1 = M 1 A, d = M A 3, d 3 = M 3 A 4, d n = M n A 1 M 1 A 1 + M A + + M n A n = M 1 A + M A M n A 1 [10] 76

5 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran The equation 10 is the mathematical form of the theorem-1. Theorem : Suppose P is a point anywhere inside of the irregular polygon, points A 1, A, A 3, A n are the vertices of the polygon and points M 1, M, M 3, M n are the perpendicular projection of the point P on s A 1 A, A A 3, A 3 A 4 A n A 1 of the polygon. If M 1 A 1, M A, M 3 A 3,, M n A n are left part line segment and M 1 A, M A 3, M 3 A 4,, M n A 1 are right part line segment of s respectively then the sum of the squares of all s of the polygon and twice the sum of the squares of all the perpendiculars is equal to twice the sum of the squares of the distance of point P from the vertices and twice the sum of the product of right side line segments and right side line segments (ref. fig.1). The mathematical form of the theorem is A 1 A + A A 3 + A3 A An A 1 + PM1 + PM + + PMn = PA 1 + PA + M1 A 1 M 1 A + M A M A M n A n M n A 1 Derivation of equations for theorem- Adding [1] and [], a 1 + a = h 1 + c 1 + h 1 + d 1 [11] Adding [4] and [5], a + a 3 = h + c + h + d [1] Adding [7] and [8], a n + a 1 = h n + c n + h n + d n [13] Adding [11] to [13], we get a 1 + a + a + a a n + a 1 = h 1 + c 1 + h 1 + d 1 + h + c + h + d + +h n + c n + h n + d n a 1 + a + a a n = h 1 + h + h 3 + +h n + c 1 + d 1 + c + d + + c n + d n a 1 + a + a a n = h 1 + h + h 3 + +h n + c 1 + d 1 c 1 d 1 + c + d c d + c 3 + d 3 c 3 d c n + d n c n d n Let, c 1 + d 1 = A 1 A, c + d = A A 3, c n + d n = A n A 1 and substituting in above eqn. a 1 + a + a a n a 1 + a + a a n = h 1 + h + h 3 + +h n + A 1 A c1 d 1 + A A 3 c d +A 3 A 4 c3 d A n A 1 cn d n = h 1 + h + h 3 + +h n + A 1 A + A A 3 + A3 A An A 1 c 1 d 1 + c d + c 3 d c n d n A 1 A + A A 3 + A3 A An A 1 = a 1 + a + a a n + c 1 d 1 + c d + c 3 d c n d n h 1 + h + h 3 + +h n A 1 A + A A 3 + A3 A An A 1 = PA 1 + PA + M1 A 1 M 1 A + M A M A M n A n M n A 1 77

6 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran A 1 A + A A 3 + A3 A An A 1 PM 1 + PM + + PMn = PA 1 + PA + M1 A 1 M 1 A + M A M A M n A n M n A 1 PM 1 + PM + + PMn [14] Eqn.[14] is the mathematical form of theorem- RESULTS AND DISCUSSION Example-1 An irregular hexagon is considered for an example (ref. fig. 5). Figure 5: An AutoCAD drawing of irregular hexagon with dimensions Table 1: Dimesions of an irregular hexagon M 1 A M 1 A A 1 A PA PM M A.3800 M A A A PA PM M 3 A M 3 A A 3 A PA PM M 4 A M 4 A M 4 A PA PM M 5 A M 5 A M 5 A PA PM M 6 A M 6 A M 6 A PA PM

7 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran Theorem-1 (eqn.10) is: M 1 A 1 + M A + M 3 A M n A n = M 1 A + M A 3 + M 3 A M n A 1 LHS = M 1 A 1 + M A + M 3 A M n A n Substituting the values given in table-1 in the above equation, LHS = LHS = sq. units [15] RHS = M 1 A 1 + M A + M 3 A M n A n Substituting the values given in table-1 in the above equation, RHS = RHS = sq. units [16] Comparing eqns. [15] and [16], LHS = RHS Hence, the theorem-1 is proved. Theorem- (eqn.14) is: A 1 A + A A 3 + A3 A An A 1 + PM1 + PM + + PMn = PA 1 + PA + M1 A 1 M 1 A + M A M A M n A n M n A 1 LHS = A 1 A + A A 3 + A3 A A6 A 1 + PM1 + PM + + PM6 Substituting the values given in table-1 in the above equation, LHS = LHS = = sq. units [17] RHS = PA 1 + PA + + PA6 + M1 A 1 M 1 A + M A M A M 6 A 6 M 6 A 1 Substituting the values given in table-1 in the above equation, RHS = RHS = = sq. units [18] Comparing eqns. [17] and [18], LHS = RHS Hence, the theorem- is also proved. Example- A scalene triangle is considered for an example (ref. fig. 6). Table : Dimesions of scalene triangle M 1 A M 1 A A 1 A PA PM M A M A A A PA PM M 3 A M 3 A A 3 A PA PM

8 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran (i) Theorem-1 (eqn.10) is: Figure 6: An AutoCAD drawing of scalene triangle with dimensions M 1 A 1 + M A + M 3 A M n A n = M 1 A + M A 3 + M 3 A M n A 1 LHS = M 1 A 1 + M A + M 3 A M n A n Substituting the values given in table- in the above equation, LHS = LHS = sq. units [19] Substituting the values given in the table-1 are substituted in the above equation, RHS = M 1 A + M A 3 + M 3 A M n A 1 Substituting the values given in table- in the above equation, RHS = RHS = sq. units [0] Comparing eqns. [19] and [0], LHS = RHS Hence, the theorem-1 is proved. (ii) Theorem- (eqn.14) is: A 1 A + A A 3 + A3 A An A 1 + PM1 + PM + + PMn = PA 1 + PA + M1 A 1 M 1 A + M A M A M n A n M n A 1 LHS = A 1 A + A A 3 + A3 A 1 + PM1 + PM + + PMn Substituting the values given in table- in the above equation, LHS = LHS = LHS = sq. units [1] RHS = PA 1 + PA + PA3 + M1 A 1 M 1 A + M A M A 3 + M 3 A 3 M 3 A 1 Substituting the values given in table- in the above equation, RHS = RHS = RHS = sq. units [] 80

9 International Journal of Physics and Mathematical Sciences ISSN: (Online) 013 Vol. 3 (1) January-March, pp.73-81/kalaimaran Comparing eqns. [1] and [], LHS = RHS Hence, the theorem- is also proved. Conclusion In this article, a new theorem on irregular cyclic polygon with appropriate illustration where it is necessary, including necessary equations derived for new properties including step by step derivations of equations and illustrations where ever necessary. The property discussed in this article has been proved with two appropriate examples. The theorems, which have been defined in this article, may be useful for whose work is related to geometry, research or further study in the irregular Polygon. REFERENCES Weisstein Eric W (003). CRC Concise Encyclopedia of Mathematics. nd edition CRC Press of Wolfram Research Inc. New York. 81