July 8, 2010
Application of Advanced Analytic Geometry to, “Real World” Projects
History shows that advancements in analytic geometry have a myriad of applications to, “real world” projects and problems. The illustrations contained in this document are of collaborative work in the field of analytical geometry as it pertains to two historical problems: Al Hazan’s Billiard Problem; and the Circle Squaring Problem of ancient Greece.
The solutions to both problems have been independently verified by mathematical analysis performed by a professor of mathematics and computer science. What the solutions to these problems do is push the limits of traditional geometry to the point of transition from Euclidean Geometry to Non-Euclidean Geometry. It is at this point that the effects of transcendental numbers such as Pi and quantum behaviors begin to manifest themselves.
Because of the nature of Euclidean Geometry, and the effect of transcendental numbers, a finite definitive solution, to such problems as circle squaring, are impossible to achieve. The work depicted in this document recognizes this limit, however, pushes the solution to circle squaring to the ultimate degree of precision possible within the confines of Euclidean Geometry. By doing so, this solution has illustrated underlying related mathematical principles dealing with inverse processes and symmetry which may have potential application to the field of quantum mechanics and the design of quantum computing logic gates.
With regards to Al Hasan’s Billiard Problem, the illustrated solution may have potential application to improving the design and efficiency of spherical mirror devices. These devices have a variety of applications within the field of optical technology. https://www.bhphotovideo.com/c/product/1324606-REG/insta360_in36p_pro_vr_reimagined_360.html
Al Hazan’s Billiard Problem Solution (Team Geometric Solution)
Given two points on a circle draw an isosceles triangle with the lines joining these two points as base and a point on the circle as apex.
Given the radius of a circle C with radius r.
Points A and B are given.
With A as center draw circle C1 with radius r. This would cut the given circle at points A1 and A2.
Select the point, which makes B lie in the arc.
That is select A1 so that A1 B A is an arc of the given circle.
Draw circle C2 with A1 as center and radius r. This circle passes through the center of C.
[Proof: A1 is on C and thus the distance between the center of C and A1 is r. Hence the circle with A1 as center and radius r passes through the center of C.]
Draw another circle C3 with B as center and radius r. This circle cuts circle C at B1 and B2.
Select B1 such that B1 A B is an arc of the circle.
Draw a circle C4 with B1 as center and r as radius. This circle also passes through the center of C.
Call the center of C as point X.
Call the intersection of circles C2 and C4 other than X as Y.
The quadrilateral AXBY has sides AX, XB, BY, YA of length r.
AB is a diagonal.
XY is the other diagonal.
Let the intersection of AB and XY be Z.
Consider the triangles AXY and BXY. These are mirror image of each other along XY as AX = BX = r
and AY = BY = r. XY is common. Hence angle AXY = angle BXY.
The triangles AXZ and BXZ are congruent. AX = BX = r; Angle AXZ = Angle BXZ; side XZ is common.
Therefore AZ = BZ. Z is a mid point of AB. Angle AZX = Angle BZX = 90 degrees.
Thus the line XZ is a perpendicular bisector of the line AB.
Extend the line XZ to intersect C at T1 and T2.
The triangle A T1 B is an isosceles triangle. So also the triangle A T2 B.
(Team Geometrical Solution to Determination of Pi)
The 'Golden Ratio' is between the longer side of a right-angled triangle and the sum of the hypotenuse and the smaller side with sides of 1 and 2 adjoining the right angle, and hypotenus of sqrt(5), the ratio being 2/(1 + sqrt(5)). The same could be written as 2/(1 + sqrt(1 + sq(2))). The above ratio, which is easy to derive geometrically, could be generalised by replacing 2 with a variable x, giving x/(1 + sqrt(1 + sq(x))).
The difference between the subscribed and subscribing squares of a circle with unit radius is (2 - sqrt(2)). Let f(x) be (2 - sqrt(2)) (x/(1+ sqrt(1 + sq(x)))). Adding sqrt(2) to f(x) gives the length of a square in between the subscribing and subscribed squares.
With x = 1.9538, the side is 1.77245014939367929102264937734 giving an area of square equal to 3.1415795320856760372176022994069 which is 0.99999582329551789353099862852891 of 'pi.'
With x = 1.9539, the side is 1.7724585029533131400713535602691 with area of
3.1416091446914999652843809521262 which is 1.0000052492807073135307175048082 times 'pi.'
Hence the correct size of the square lies between these two lengths. The solution to the following equation
a l(1.9539) + (1 - a) l(1.9538) = sqrt(pi) gives a = 0.44310593315663435328511447202328 giving an exact match for the available precision. Such a fraction would be difficult to achieve with compass and scale, but not impossible. We consider the accuracy with a as 0.4431.
0.4431 l(1.9539) + 0.5569 l(1.9538) = 1.772453850855953049530495361302007958 giving an area of 3.1415926534140970568717650000139. Comparing the above with the area of the circle, 'pi' given by 3.1415926535897932384626433832795, we see that the first 9 digits after the decimal point are the same. The area of the square is 0.99999999994407416843488088398166 of the area of the circle.
(Team Geometrical Solution for Pi Derivative Algebraic Equation for Pi needed to design Quantum Logic Gate Qubit Processors)
From this equation we get an algebraic equation for Π.
Π = 4 / r2 = 4 / ((xS – xB)2 + (yS – yB)2)
(Literature review now indicates that the geometric accuracy achieved may allow for the construction of better quantum circuits)