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Extend the Center Line
as required
Draw a Vector Line EI that is equal
to Line GH from Point I
to the Center Line.
Draw a Vector Line DF that is equal
to Line FI from Point F to the Center Line. |
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Transpose Pythagorean Triangles FGJ and
HIK as indicated.
Extend the Center Line
as required. |
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Inscribe Circles (in-circles) FGJ and HIK
with centers at
and
respective diameters LM and NO.
Bisect Lines FG and HI with Center Lines. |
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Construct a paired Pythagorean Triangle FGJ in accordance with base, side, and hypotenuse as indicated in the below Table of Formulas.
A pair of Pythagorean Triangles have short sides, radius (r) and base (b), which vary by the Elliptical Constant (EC) =
. |
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Construct a Pythagorean Triangle
HIK in accordance with base, side, and hypotenuse as indicated in the below Table of Formulas.
Pythagorean Triangles are triangles with
a right angle; and, all sides that are integers. |
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Pulsoid Theorem (PTm) |
V |
= |
|
P2 |
|
|
|
|
|
= |
Elliptical
Constant (EC) |
= |
|
One |
x |
|
= |
Integer
(base 10)=b/2 |
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All
Vector
Lines are
equal. |
. |
DF & EI |
|
=
|
Vector
(V)=(b/2)2 |
= |
|
x2 |
FGJ |
|
= |
Pythagorean
radius
Triangle |
FG |
|
=
|
wave
(w)=b2/2–b |
= |
|
2(x2–x) |
GJ |
|
=
|
radius
(r)=b– |
= |
|
2x– |
FJ |
|
=
|
hypotenuse
(h)=b2/2–b+ |
= |
|
2x2–2x+ |
HIK |
|
= |
Pythagorean
Vector
Triangle |
HI |
|
=
|
base
(b), V-wave,
(Vw) |
= |
|
2x |
HK |
|
=
|
V-radius
(Vr)=b2/4
– |
= |
|
x2– |
IK |
|
=
|
V-hypotenuse
(Vh)=b2/4+ |
= |
|
x2+ |
FGJ |
|
=
|
radius-Resoloid
(in-circle) |
= |
|
rR |
LM |
|
= |
r-diameter
(rd)=b–2 |
= |
|
2(x–) |
L |
|
=
|
r-radius
(rr)=b/2– |
= |
|
x– |
HIK |
|
=
|
Vector-Resoloid
(in-circle) |
= |
|
VR |
NO |
|
=
|
VR-diameter
(VRd)=b–2 |
= |
|
2(x–) |
N |
|
=
|
VR-radius
(VRr)=b/2– |
= |
|
x– |
The Pythagorean Theorem, a2 + b2
= c2,
is an an incredible coincidence
of
number theory, which arises from
the
Emergent Ellipsoid and
the Natural source of
Mathematics. |
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A Pythagorean Triangle Pair (PTP) can be mapped to
every integer greater
than One.
Four angles (two of the six angles are right angles) and all six sides of a Pythagorean Triangle Pair are of unlike values. |
The simplest equation
for the in-diameter of a circle inscribed within any right triangle is:
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An in-diameter of a right
triangle equals side "a" plus side "b" minus the hypotenuse "c". |
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Amazingly, the diameters of both circles inscribed within each
different
Pythagorean Triangle, of a given pair, are even integers, that are . . .
identical to one another.
Thus, the radii of the inscribed circles are also equal integers.
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