arkain101 Posted December 16, 2006 Report Posted December 16, 2006 pythagoras rule [math]a^2 + b^2 = c^2 [/math] A well known concept in mathamatics. I was mapping out some shapes to understand why this was. In the process I drew out some shapes and saw something. I assume this not neccesaraly unknown. The pythagoran theorem can be expressed as; [math](2)a + (2)b = (4)c [/math] Quote
arkain101 Posted December 16, 2006 Author Report Posted December 16, 2006 What got me interested in this was the physics aspects of nature. The differential of distance(ds) in cartesian 3D space is defined as: Which translates into. [math](6)ds = (2)dx_1 + (2)dx_2 + (2)dx_3[/math] Quote
arkain101 Posted December 16, 2006 Author Report Posted December 16, 2006 So I would predict that; In the geometry of special relativity, a fourth dimension, time, is added, with units of c, so that the equation for the differential of distance becomes: would translate into. [math](2)ds = (2)dx_1 + (2)dx_2 + (2)dx_3 - (2)c(2)dt[/math] Not positive.. Quote
ughaibu Posted December 16, 2006 Report Posted December 16, 2006 Arkain101: Your formula only applies to isosceles triangles. Here's a page with "72 approaches to proving the theorem": Pythagorean Theorem and its many proofs Quote
ughaibu Posted December 16, 2006 Report Posted December 16, 2006 In fact, your formula doesn't work even for isosceles triangles. Quote
arkain101 Posted December 16, 2006 Author Report Posted December 16, 2006 Okay what the heck.. I feel retarded now. Quote
ughaibu Posted December 16, 2006 Report Posted December 16, 2006 Dont worry about it. I think things like the Pythagorean theorem are great fun, simple but with incomprehensible ramifications. Quote
arkain101 Posted December 16, 2006 Author Report Posted December 16, 2006 You know what I think I meant to be doing is this. AB/BC = BD/AB and AC/BC = DC/AC. If you see my drawing, that is what I was comprehending, of course thought I confused the maths by accident. Quote
arkain101 Posted December 17, 2006 Author Report Posted December 17, 2006 What about this here? Did I find a constant for squares? The hypotenuse of a right triangle with equal lengths for a and b, or the hypotinuse of a square is has a constant, 1.4142136.... [math]a^2 + b^2 = c^2[/math][math]5^2 + 5^2 = c^2[/math][math]5^2 + 5^2 = 7.0710678^2[/math] Now, [math] 1.4142136 \times length(a,b) = 7.0710678[/math] Note Pie = 3.141And the constant I found is 1.41 In other words the distance of a square corner to corner is; [math] 1.4142136^2 = 2[/math] Constant = [math]\sqrt 2[/math] Sure hope I didnt make another blind mistake. :doh: Quote
arkain101 Posted December 17, 2006 Author Report Posted December 17, 2006 :( The length of the polygon diagonal of the unit square is [math]\sqrt 2[/math] , sometimes known as Pythagoras's constant. Quote
Turtle Posted December 17, 2006 Report Posted December 17, 2006 :( The length of the polygon diagonal of the unit square is [math]\sqrt 2[/math] , sometimes known as Pythagoras's constant. One can also derive this constant from the vesica piscis. http://hypography.com/forums/physics-mathematics/1902-vesica-piscis-real-sacred-geometry.html Quote
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