Opposite Sides Bisectors Theory in Fundamental Science on General Problem Bisectors

by

© Ph. D. & Dr. Sc. Lev Gelimson

Academic Institute for Creating Fundamental Sciences (Munich, Germany)

Mathematical Journal

of the "Collegium" All World Academy of Sciences

Munich (Germany)

11 (2011), 46

The least square method (LSM) [1] by Legendre and Gauss only usually applies to contradictory (e.g. overdetermined) problems. Overmathematics [2-4] and fundamental science of solving general problems [5] have discovered many principal shortcomings [2-6] of this method, by methods of finite elements, points, etc. Additionally consider its simplest approach which is typical. Minimizing the sum of the squared differences of the alone preselected coordinates (e.g., ordinates in a two-dimensional problem) of the graph of the desired approximation function and of everyone among the given functions depends on this preselection, ignores the remaining coordinates, and provides no coordinate system rotation invariance and hence no objective sense of the result.

In fundamental science on general problem bisectors [5], opposite sides bisectors theory (OSBT) is valid by coordinate system rotation invariance. Show the essence of this theory in the simplest but most typical case providing linear solving with giving the unique best quasisolution [2-5] to a finite overdetermined quantiset [2-5] of n (n > 2; n ∈ N+ = {1, 2, ...}) linear equations

q(i)(ajx + bjy = cj) (j = 1, 2, ... , n) (Ei)

with their own quantities q(i) and 2 unknown variables x and y in a plane with the same coordinates x , y by any given real numbers q(i) > 0, aj , bj , and cj . Suppose that these n straight lines build a polygon P1P2...Pn with n corners (vertices) P1 , P2 , ... , Pn (and, naturally, n sides, or edges) either in the clockwise order or in the anticlockwise order which can be provided via preliminarily renumbering (reindexing) the equations. If necessary and possible, find and use such renumbering (reindexing) of these n corners that polygon P1P2...Pn is convex. If it is impossible, then provide the minimum number of nonconvex corners (vertices) (i.e., with convexity violation) and hence the maximum number of convex corners (vertices) (i.e., without convexity violation). Let Pi be the intersection of straight lines Ei and Ei+1 with their own quantities q(i) and q(i+1) which has its own quantity

p(i) = q(i)q(i+1).

Because of the clear possible periodicity of the indices i = 1, 2, ... , n with period n , consider

Pj+kn = Pj ,

q(i+kn) = q(i) ,

p(i+kn) = p(i)

for any integer k . Then the quantiset [2-5] of n polygon corners (vertices) is

p(1)P1 , p(2)P2 , ... , p(n)Pn .

Nota bene: A bisector of an angle (and of the corresponding angle sides) divides it into two equal parts and is the set of all the points equidistant from the both angle sides [1]. For two intersecting straight lines, there are two mutually perpendicular bisector straight lines. If for a polygon, there is an inscribed circumference, then the desired unique best quasisolution its simply the inscribed circumference center which is the intersection of the bisectors of all the internal corners of this polygon which in this case is always convex. This all holds for any triangle whose case is, therefore, trivial.

Hence further consider case n > 3 only when for a polygon, there is no inscribed circumference.

The main idea and essence of opposite sides bisectors theory (OSBT) is:

1) if number n is even, i.e. there is such a positive integer m > 1 that n = 2m , then for any j = 1, 2, ... , m , consider each disordered pair PjPj+1 and Pj+mPj+1+m of sides (edges) of this polygon as its opposite sides. There are m disordered pairs

(P1P2 , Pm+1Pm+2) , (P2P3 , Pm+2Pm+3) , ... , (PmPm+1 , P2mP1)

of the opposite sides of polygon P1P2...P2m . Follow substeps 1.1, 1.2, 1.3, and 1.4:

1.1) for each disordered pair PjPj+1 and Pj+mPj+1+m of the opposite sides of this polygon, determine the set of all the polygon area (including its interior and boundary) points equidistant from the two straight lines containing these sides, respectively. If these lines are parallel, then their equidistant line is the unique straight line which is parallel to those both straight lines and divides the distance between them into two equal parts. If the straight lines containing these sides are intersecting, then there are two mutually perpendicular bisector straight lines so that each of their points is equidistant from the two straight lines containing these sides, respectively. Now determine the intersection of the equidistant line(s) with the polygon area. We also obtain the quantiset [2-5] of all the m bisectors of the opposite sides of this polygon

q(1)q(m+1)B1 , q(2)q(m+2)B2 , ... , q(m)q(2m)Bm .

If polygon P1P2...P2m is convex, then all the polygon area points equidistant from the two straight lines containing these sides, respectively, build a segment whose ends belong to some sides of polygon P1P2...P2m ;

1.2) for each two disordered pairs

(PjPj+1 , Pj+mPj+1+m) , (PkPk+1 , Pk+mPk+1+m)

(where any k = 1, 2, ... , m does not coincide with j)

of the polygon opposite sides, determine the intersection of the both sets of all the polygon area points equidistant from the both pairs of the straight lines including the corresponding sides of polygon P1P2...P2m , respectively (see substep 1.1). If polygon P1P2...P2m is convex, then this intersection typically consists of the unique point which is the intersection of the both corresponding equidistant segments whose ends belong to some sides of polygon P1P2...P2m ;

1.3) build the quantiset which is the quantiunion [2-4] of the intersections (corresponding to all the different disordered pairs of disordered pairs of the polygon opposite sides, see substep 1.2) the both sets of all the polygon area points equidistant from the both pairs of the straight lines including the corresponding sides of polygon P1P2...P2m , respectively. The own quantity of each of these intersections is the product of the own quantities of the both bisectors (of the opposite sides of this polygon) building this intersection. Using namely a quantiset and a quantiunion with a quantity of each element is also necessary (because those different intersections can contain coinciding elements) to precisely consider the quantity (multiplicity in the simplest case) of such an element. Namely, by each coincidence, simply add the own quantities of the coinciding intersections;

1.4) pointwise consider the above quantiset (see substep 1.3). If it contains at all s (s ∈ N+ = {1, 2, ...}) intersections Ri which are points with their own quantities r(i) and coordinates

r(1)R1(x1 , y1), r(2)R2(x2 , y2), ... , r(s)Rs(xs , ys),

then determine their center (of gravity, mass, etc)

x = Σi=1s r(i)xi / Σi=1s r(i) ,

y = Σi=1s r(i)yi / Σi=1s r(i)

and consider this point (x , y) to be the unique best quasisolution [2-5] to a finite overdetermined set of n linear equations (Ei).

Nota bene: If all the intersections of the bisectors of the opposite sides of the polygon with n = 2m sides (and n = 2m vertices) are different, then there are precisely m pairs of the opposite sides (and their bisectors) and hence

s = m(m-1)/2

such intersections. Otherwise, we have

s < m(m-1)/2

and by each coincidence, simply add the own quantities of the coinciding intersections.

2) if number n is odd, i.e. there is such a positive integer m > 1 that n = 2m + 1, then for any i = 1, 2, ... , 2m + 1 , consider both each disordered pair (with adding m to the both indices in the expression for the 2nd element of such a pair)

(PiPi+1 , Pi+mPi+m+1)

and each disordered pair (with adding m + 1 to the both indices in the expression for the 2nd element of such a pair)

(PiPi+1 , Pi+m+1Pi+m+2)

of sides of this polygon as its quasiopposite sides. There are 2m + 1 disordered pairs

(P1P2 , Pm+1Pm+2) , (P2P3 , Pm+2Pm+3) , ... , (P2m+1P1 , PmPm+1)

of the opposite sides of polygon P1P2...P2m+1 .

Nota bene: We obtained the above 2m + 1 disordered pairs via using pair type (PiPi+1 , Pi+mPi+m+1) (with adding m to the both indices in the expression for the 2nd element of such a pair) only for every i = 1, 2, ... , 2m + 1 . Pair type (PiPi+1 , Pi+m+1Pi+m+2) (with adding m + 1 to the both indices in the expression for the 2nd element of such a pair) brings nothing new because for any i = 1, 2, ... , 2m + 1 , such a disordered pair can be represented due to the index periodicity as follows:

(PiPi+1 , Pi+m+1Pi+m+2) = (Pi+m+1Pi+m+2 , PiPi+1) = (Pi+m+1Pi+m+2 , Pi+2m+1Pi+2m+2)

(with adding m to the both indices in the expression for the 2nd element of such a pair) naturally already given by the previous pair type (PiPi+1 , Pi+m+1Pi+m+2) with i + m + 1 instead of i .

Now follow substeps 2.1, 2.2, 2.3, and 2.4 which are similar to substeps 1.1, 1.2, 1.3, and 1.4.

Nota bene: If all the intersections of the bisectors of the quasiopposite sides of the polygon with n = 2m + 1 sides (and n = 2m + 1 vertices) are different, then there are precisely n = 2m + 1 pairs of the quasiopposite sides (and their bisectors) and hence

s = n(n-1)/2 = m(2m + 1)

such intersections. Otherwise, we have

s < n(n-1)/2 = m(2m + 1)

and by each coincidence, simply add the own quantities of the coinciding intersections.

It is very useful to develop an explicit algorithm to obtain the equation of the unique straight line equidistant from the two straight lines including two segments which are nonintersecting themselves whereas these two straight lines can be either intersecting or parallel, this unique straight line intersecting with the tetragon with these two segments as its opposite sides (edges). In particular, these both segments can be either opposite (by even n) or quasiopposite (by odd n) sides of polygon P1P2...Pn . The algorithm is as follows:

1. Obtain the equations of the both straight lines including those two segments, respectively.

Let AB and CD be these both segments whose endpoints with coordinates x , y in a plane are

A(xA , yA), B(xB , yB), C(xC , yC), D(xD , yD).

The equation of straight line AB including segment AB is

(x - xA)/(xB - xA) = (y - yA)/(yB - yA),

(yB - yA)x + (xA - xB)y = xA(yB - yA) + yA(xA - xB).

The equation of straight line CD including segment CD is

(x - xC)/(xD - xC) = (y - yC)/(yD - yC),

(yD - yC)x + (xC - xD)y = xC(yD - yC) + yC(xC - xD).

2.1. If the main determinant of the obtained equations set

(yB - yA)x + (xA - xB)y = xA(yB - yA) + yA(xA - xB),

(yD - yC)x + (xC - xD)y = xC(yD - yC) + yC(xC - xD)

vanishes, then

(yB - yA)(xC - xD) = (xA - xB)(yD - yC)

and straight lines AB and CD are parallel.

2.1.1. By nonzero xA - xB and xC - xD , we transform the same equations to their equivalent forms

(yB - yA)(xC - xD)x + (xA - xB)(xC - xD)y = [xA(yB - yA) + yA(xA - xB)](xC - xD),

(xA - xB)(yD - yC)x + (xA - xB)(xC - xD)y = [xC(yD - yC) + yC(xC - xD)](xA - xB)

and obtain the equation

(yB - yA)(xC - xD)x + (xA - xB)(xC - xD)y =

{[xA(yB - yA) + yA(xA - xB)](xC - xD) + [xC(yD - yC) + yC(xC - xD)](xA - xB)}/2

of the unique straight line equidistant from the two straight lines including two segments.

2.1.2. By zero xA - xB and xC - xD , we use nonzero yB - yA and yD - yC , transform the same equations to their equivalent forms

(yB - yA)x = xA(yB - yA),

(yD - yC)x = xC(yD - yC);

x = xA ,

x = xC

and obtain the equation

x = (xA + xC)/2

of the unique straight line equidistant from the two straight lines including two segments.

2.2. If the main determinant of the obtained equations set

(yB - yA)x + (xA - xB)y = xA(yB - yA) + yA(xA - xB),

(yD - yC)x + (xC - xD)y = xC(yD - yC) + yC(xC - xD)

is nonzero, then:

2.2.1. Determine the unique intersection S of straight lines AB and CD.

2.2.2. Select one point of segment AB , e.g., A , as well as one point of segment CD , e.g., C .

2.2.3. Determine the lengths LSA and LSC of segments SA and SC .

2.2.4. Determine the intersection E of the desired unique straight line equidistant from the two straight lines including two segments with segment AC :

xE = xA + (xC - xA)LSA / (LSA + LSC).

yE = yA + (yC - yA)LSA / (LSA + LSC).

Nota bene: It is well-known [1] that in any triangle, a bisector divides the opposite side by their intersection in to two parts whose lengths are proportional to their adjacent sides. In this equation proportion itself, the ratio of one of the segment parts to the whole segment is considered.

2.2.5. Determine the equation of the desired unique straight line equidistant from the two straight lines including two segments AB and CD which includes points S and E already known:

(yE - yS)x + (xS - xE)y = xS(yE - yS) + yS(xS - xE).

Nota bene: These simple formulae at once give the desired result unlike the well-known formulae [1]

(Fx + Gy + H)/(F2 + G2)1/2 = ±(Px + Qy + R)/(P2 + Q2)1/2

for the both mutually perpendicular bisector straight lines of any pair of intersecting straight lines

Fx + Gy + H = 0

and

Px + Qy + R = 0

in a plane with coordinates x , y .

Compare applying opposite sides bisectors theory (OSBT), adjacent corners bisectors theory (ACBT), adjacent sides bisectors theory (ASBT) with one step only, distance quadrat theory (DQT), the least square method (LSM), least normed square method (LNSM), unierror equalizing method (EEM), and direct solution method (DSM) [2-4] to test equation set

29x + 21y = 50,

50x - 17y = 33,

x + 2y = 7,

2x - 3y = 0,

with the unit own quantities of the equations, see Figures 1, 2:

OpSdBis1.gif

Figure 1

DsQdPrb2.gif

Figure 2

The LSM gives x ≈ 1.0023, y ≈ 1.0075 practically ignoring the last two equations with smaller factors (unlike DQT, the EEM, DSM, and even LNSM).

Opposite sides bisectors theory (OSBT) providing simple explicit quasisolutions to even contradictory problems is very efficient by solving many urgent problems.

Acknowledgements to Anatolij Gelimson for our constructive discussions on coordinate system transformation invariances and his very useful remarks.

References

[1] Encyclopaedia of Mathematics. Ed. M. Hazewinkel. Volumes 1 to 10. Kluwer Academic Publ., Dordrecht, 1988-1994

[2] Lev Gelimson. Elastic Mathematics. General Strength Theory. The ”Collegium” International Academy of Sciences Publishers, Munich, 2004

[3] Lev Gelimson. Providing Helicopter Fatigue Strength: Flight Conditions. In: Structural Integrity of Advanced Aircraft and Life Extension for Current Fleets – Lessons Learned in 50 Years After the Comet Accidents, Proceedings of the 23rd ICAF Symposium, Dalle Donne, C. (Ed.), 2005, Hamburg, Vol. II, 405-416

[4] Lev Gelimson. Overmathematics: Fundamental Principles, Theories, Methods, and Laws of Science. The ”Collegium” All World Academy of Sciences Publishers, Munich, 2010

[5] Lev Gelimson. General Problem Fundamental Sciences System. The ”Collegium” All World Academy of Sciences Publishers, Munich, 2010

[6] Lev Gelimson. Corrections and Generalizations of the Least Square Method. In: Review of Aeronautical Fatigue Investigations in Germany during the Period May 2007 to April 2009, Ed. Dr. Claudio Dalle Donne, Pascal Vermeer, CTO/IW/MS-2009-076 Technical Report, International Committee on Aeronautical Fatigue, ICAF 2009, EADS Innovation Works Germany, 2009, 59-60