The Cognitive Condition

Of late, regarding the Harmonious Pythagorean Tetrahedra, I thought it prudent to consider both apex singularity solutions inclusively. The resulting object is a concave hexahedron where the base right triangle exists in the medial plane. By manipulating the associated geometric construction, the requisite Pythagorean areas are distributed among the lateral faces of the upper and lower “tetrahedra”. In other words, I divided the area of each lateral face in the single-solution case (tetrahedron) and applied half to each in the two-solution case such that their sum achieves Pythagorean compliance for the developed hexahedra. This exploration led to a fun tangent...

As with most any problem, this too may be approached from a multitude of directions. The tack I have chosen distills to the determination of the discrete among a locus of points at the coping intersection of the two cylinders at right angles. For reference and context please see ALL prior posts associated with A Precursory Study of Pythagorean Theorem Generalization a Consequence of Three-Dimensions Example: Catheti a = 3 b = 4 Hypotenuse c  = √(a2 + b2) = 5 To a requisite number of decimal places, exhaustive iteration of D1z against the defined a,b will produce values for...

Presented is an equation for intercept Point “P” of a Line and Parabola where the Line passes through the Focus and is at angular displacement to the Latus Rectum. x = (tan θ ± √( (tan θ)2 + 4af) ) / 2a y = ax2

A Preface – As is often the case for myself, I can offer no conclusion… merely commentary with an invitation for scrutiny. A simple description of the Circle is all points on a plane equal in displacement to a discrete point. Extension then allows definition of radius, diameter, and circumference. This as basis, renders the Circle easy enough to construct. But what of the Circle’s origin. Unintentional as it may be, we tend to forsake the planar nature of the Circle. Ours is a three-dimensional world, and two-dimensional geometric shapes arise as planar sections of three-dimensional objects. Or, do they?...

At the suggestion of a colleague and in reference to the extension of the Pythagorean Theorem to a cuboid, I’ve begun to explore the analogous application of an+bn+cn=dn. Please reference prior post regarding the Harmonious Pythagorean Tetrahedra Conjecture: Pythagorean Theorem Generalization as Consequence of Three-Dimensions Solutions to an+bn=cn for non-right triangles with mutual exponents are requisite common a:b ratios and ∠C’s. Established are an a:b of 1:2 and the assignment of a=1, b=2 and c=4 (Ref. Fig. 1). This would seem to necessitate a solid of non-orthogonal construction with a parallelogram base that can divided from opposite corners into isosceles...

Preface Having written prior about much of the following, further contemplation on this specific topic remains unavoidably necessary. It is the exponent of 2 that clearly defines the square unit, that magic of 90° which affords the Pythagorean Theorem its achievement and just preeminence. The area of the parallelogram, brought to orthogonal compliance by Cavalieri, seems an excellent example of our insistence for conformity with the square. But this construct of square units, ingrained at a young age, makes difficult the notion of alternatives. The idea of area absent a unit “squared” seems quite alien. So much so, I have...

Representing triangle occurrence, the curves shown of ∠C on the graph cover a range from obtuse to the acute above the isosceles inversion. Regarding the scalene, the influence of both a:b ratio and ∠C is apparent when administered by an+bn=cn. Noted, the same will hold for the application of bn-an=cn to the acute remainder below the inversion. As a:b approaches 1:1, diminishing sensitivity to C becomes discernable. Contrastingly, the Pythagorean Theorem enjoys immunity to the a:b ratio and is held accountable only to the right angle. The degenerate locus curve delineates the regions of tetrahedral solutions (green) and that of...

This contribution in regards to the Harmonious Pythagorean Tetrahedra conjecture: PYTHAGOREAN THEOREM GENERALIZATION as CONSEQUENCE of THREE-DIMENSIONS My great appreciation to Walter Trump for his assistance on this topic as the trees often obscure the forest. Equally the same to William Walkington for facilitating our interaction. Mr. Walkington’s post “Magic Triangular Pyramids” found on his blog “Magic Squares, Spheres and Tori” provided motivation to seek a clearer understand of the degenerate pyramidal limits which Mr. Trump so graciously was able to provide. Please visit William’s blog for some exceedingly thought provoking content. The 3 construction cylinders intersect with the x-y-plane...

Please reference prior post: WHY DOES the PYTHAGOREAN THEOREM WORK? For any triangle ABC where c* is the long leg, an exponent n exists that will satisfy a n + b n = c n of non-Diophantine concern. *Alternatively, a n + b n = c n can be re-written b n – a n = c n at the isosceles inversion point to maintain constancy of convention. This is necessitated as ∠C changes between the degenerate triangle limits of zero and π radians, the transition from obtuse scalene to acute scalene. Constrained by this equation, as triangles of a : b < 1 approach isosceles, n becomes extreme. In the case of a : b = 1, the equation is invalid at the equilateral where c fails to...

Why does the Pythagorean Theorem work? Or perhaps a more pointed query… why only for the right triangle? Regrettably, and not surprisingly, I cannot advise with absolute conclusion. Nonetheless, I offer a vantage that may afford thought without mere recital of proofs. Regarding all triangles, the Law of Cosines notwithstanding, a more analogous with Pythagoras approach may be an exponent solution. For any triangle ABC where *c is the long leg, an exponent n exists that will satisfy a n + b n = c n of non-Diophantine concern. *Alternatively, a n + b n = c n can be re-written b n – a n = c n at the isosceles inversion point to maintain...

Concerning the Harmonious Pythagorean Tetrahedra conjecture, please reference the prior post: PYTHAGOREAN THEOREM GENERALIZATION as CONSEQUENCE of THREE-DIMENSIONS For any base right triangle of a:b ratio equal to or less than ¼:1, its area summed with that figured for each of its three sides compliant with the Pythagorean Theorem, is insufficient for three-dimensional existence. Any base right triangle exceeding an a:b ratio of 1:1 represents a reciprocal dilation in the form b:a. Thus, a ¼:1 ratio serves as the two-dimensional threshold or limit while a 1:1 ratio can be examined as a constructive constraint with the region between the two...

The ensuing description to serve as a possible path to a mathematical proof. For scaling to any size, a ratio of side a to b (a:b) is employed. Side b is maintained constant between the limits 1:4 and 4:1 for any base right triangle. Therefore, the radius of the cylinder with b as its axis is constant. Point D1 is defined as the apex of the evolved oblique triangular pyramid as per the prior conjecture: PYTHAGOREAN THEOREM GENERALIZATION as CONSEQUENCE of THREE-DIMENSIONS As point D1 exists in the locus of points of the two intersecting perpendicular cylinders whose axes are...

Given the Pythagorean Theorem a2 + b2 = c2, I interject that unsurprisingly for any right triangle, the square of the area corresponding to the hypotenuse less the same for that of the longer leg and again for that of the shortest leg, the resultant divided by eight, equals the square of the right triangle area. Reference prior post: Pythagorean Theorem Generalization as Consequence of Three-Dimensions (c2)2 – (b2)2 – (a2)2 / 8 = (.5ab)2 c4 – b4 – a4 = (.5ab)2 (8) √(c4 – b4 – a4) / 2√2 = .5ab √(c4 – b4 – a4) = ab√2 c=±(2b2...

Sectioning a conical surface with generator angle of 47.34508022773829+° yields a Latus Rectum bounded Parabola area equal to that of the Circle sectioned from the center of the Focal (Dandelin) Sphere. Pi is not requisite in determining this generator angle. Its value accomplished by creating a Parabola area equal that of its Focal Circle and placing it in correct geometric orientation forming the conic. Thus, Circle area may be found via its relationship with Parabola area. Subsequent simplification affirms with the value of pi. (Latus Rectum length)(Focal length)(2/3)        = Area of Parabola ((8)((tan47.345+°)2)(r2))/3           = Area of Circle          ((8)((1.0854+)2)(r2))/3         ...

As basis, the Pythagorean Theorem. The length of quadrilateral side opposite that coincident with the triangle equals: (√(a2+b2)) – (b/(2√(1+(b2/a2)))) (Ref. Fig. 1) A polygon, being a quadrilateral with right angle, containing an edge coincident with the side of a right triangle and having an area in accord with the square of the same is developed. Repeated for the adjacent side, each polygon is rotated about their respective vertex until coincident with the hypotenuse, their supplement forming the perimeter of an irregular pentagon with area compliant to the squared hypotenuse. A subsequent Pythagorean tiling or tessellation is also depicted. I...

Delineated is an oblique irregular tetrahedron of four right triangle faces where two vertices are formed by two right angle faces. The sum of the squares of the area ABO and ACO less the same of BCO equals that of ABC. A2ABC = A2ABO + A2ACO – A2BCO .25a2b2 = (.25a2(h2+b2)) + (.25b2 h2) – (.25c2h2) 0 = 25h2(a2+b2-c2) c2 = a2+b2

Effects of surface deviation on angle of reflection.

Φ = ( 1 + √5 ) / 2

First, allow me the following characterization of the Parabola and conical surfaces: There is only “the” Parabola, not “a” parabola in the sense that “a” implies variety of shape. “The” Parabola, like “the” Circle, has only a single shape representation. It can be depicted in any dilation, but its contour remains the same. In contrast, conical surfaces (if you like, right circular conical surface or cone) have infinite shape representation as consequence of their conic generator angle. Therefore, as an infinity of conical surfaces can arise from the Parabola (as detailed below), and whereas only the Parabola can come from...

Graphic depicting the region of 3D solutions for the prior post: Pythagorean Theorem Generalization as Consequence of Three-Dimensions Three-dimensional solutions, being tetrahedra in the shape of oblique triangular pyramids, for a base right triangle ABC with proportions given as ratio a:b occur within the limits of ¼:1 < ◿ABC < 4:1.

1. Strike line AB tangent to the Parabola2. Strike line CE ⊥ to line AB at tangent point P13. Strike line GH ⊥ to line AB at tangent point P24. Strike line IJ ⊥ to line GH at tangent point P25. Mark intersection points P3 and P4 (P4 lies on the Directrix) 6. Strike line through P3 and P4 (line is ∥ to Axis of Symmetry) 7. Strike line KM ⊥ to line P3P4 at point P4 (establishes Directrix) 8. Measure closest distance from line KM to the Parabola and mark points V and D (establishes Axis of Symmetry) 9....

α = Conic Generator AngleXv = X Coordinate Value, Conic VertexYv = Y Coordinate Value, Conic VertexXs = X Coordinate Value, Surface DataYs = Y Coordinate Value, Surface DataZs = Z Coordinate Value, Surface Data Zs = (1 / tan α) ((Xs – Xv)2 + (Ys – Yv)2) .5

From C = π D and A = π r2 to C = A D / r2 and A = C r2 / D. Constructed is a diagram depicting the sector relationship of both Circumference and Area of a Circle (Ref. Fig. 1) Circle data for example shown: radius                  = 1inDiameter             = 2in               = 2 x 1in                         = 2 rArea                    = 3.14159+in² = 3.14159+ x (1in x 1in) = π r²Circumference    = 6.28319+in   = 3.14159+ x 2in            = π D Three major sectors, each having both a circumferential perimeter equal to the Diameter, and an area...

The Harmonious Pythagorean Tetrahedra Delineated is base right triangle ABC with proportions given as ratio a : b. A point D1 exists above the plane where for any proportion ¼ : 1 < ◿ABC < 4 : 1, the surface area of ▽BCD1 = a 2, ▽ACD1 = b 2, and ▽ABD1 = c 2. Thus, occurs a four-sided polyhedron, being an oblique triangular pyramid, with right triangle base and corresponding lateral faces consonant with the Pythagorean Theorem (Fig. 1 of original manuscript). Areas, corresponding to the catheti and hypotenuse, need not necessarily hold similar shape to conform to a...