AP's 240th book that ushered in my 253rd book--- New True Geometry starting with cycloid correction and Geometry-of-Motion // math research by Archimedes Plutonium This is AP's 240th published book of science published on Internet, Plutonium-Atom
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Archimedes Plutonium
Jun 1, 2026, 4:33:02 PM (6 days ago) Jun 1
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*New True Geometry starting with cycloid correction and Geometry-of-Motion
// math research*
by Archimedes Plutonium
This is AP's 240th published book of science published on Internet,
Plutonium-Atom-Universe, PAU newsgroup is this. Please read this textbook
for free on Internet.
https://groups.google.com/forum/?hl=en#!forum/plutonium-atom-universe
Preface: Motion Geometry such as producing the Cycloid curve in Old Math
was in error. Looking at the Limacon we see this error in Motion Geometry
at its worst. And the fix of the error has to come from physics of
Electromagnetic theory where magnetic field is always perpendicular to
electric field. The mistake made in Old Math Geometry is that they had two
motions in a cycloid and limacon construction. They had the motion of a
circle, but had arbitrary motion for the Pointer-Marker. This book
addresses the Old Math Geometry mistakes and opens up the entire field of
new math of Motion Geometry. In the midst of that correction several major
conjectures were discovered and proven in this book which has created a
complete overhaul of Old Math's conic sections. For the parabola and
hyperbola are not open curves but closed loop circuits of ellipse, oval,
circle. The nasty mistake of Old Math to think cones are apex to apex is
ridiculous to the nth degree, for true conics are base to base <>, not apex
to apex ><. And I finish this book with a stunning proof that 3 arbitrary
non-collinear points in the plane not only produces a unique circle, but
produces a unique ellipse, yet Old Math claimed it requires 5 points to
produce a unique ellipse. I suspect Old Math's 5 point requirement is based
on the square or rectangle, whereas my 3 point requirement is based on a
geometry of 2 identical right-triangles that forms a HourGlass figure which
can be seen in any given specific ellipse.
Cover Picture: My photograph of holding two cones base to base. Two cones
base to base forms the foundation of Conic Sections, and not the silly apex
to apex.
-------------------------------
Table of Contents
-------------------------------
1) My history behind this subject matter.
2) Repeating the history of how I corrected the cycloid and limacon and
geometry-of-motion.
3) Geometry-in-Motion speaks to Light Waves of physics.
4) My history of discovery of the error in cycloid geometry of motion.
5) Old Math's Limacon, Cycloid and similar curves in error.
6) Proof the Cycloid is a ellipse and smooth curve.
7) Defining what a Smooth Curve is in New Geometry.
8) Researching the Cycloid and Geometry of Motion, rolling one figure over
another.
9) Conic sections with base to base <>, rather than apex to apex ><.
10) Proof that all Smooth Curves are obtained from conic sections base to
base <>.
11) The oval in conic sections, base to base <>.
12) Proof the Parabola is a conic ellipse or oval.
13) Pi is actually square root of 10 = 3.16... when factoring in
Geometry-of-Motion.
14) Conic Sections in 2nd dimension.
15) Three non-collinear points determine a unique plane in space.
16) Takes only 3 arbitrary points to determine a unique ellipse, not Old
Math's 5.
17) AP's proof that 3 arbitrary points determine a unique circle, and the 3
same arbitrary points determine a unique ellipse.
18) Synthetic proof that 3 arbitrary non-collinear points forms unique
ellipse since those points form a unique circle.
19) Computers in geometry often just mess things up and can never be
trusted for geometry proofs, for they are "artwork" in geometry, not
mathematics.
------------
Text
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*1) My history behind this subject matter.*
I suppose my history of this subject matter intensely starts in 2016 with
the discovery that the slant cut of right-circular cone was not a ellipse,
but rather instead a oval with only 1 axis of symmetry. Which makes all the
common sense in the world for a ellipse has 2 axes of symmetry, but a
single cone has 1 axis of symmetry. You need a cylinder to provide 2 axes
of symmetry for a slant cut to be an ellipse.
And in late 2016, I was working on exposing the fakery of the sinusoid
curve, that sine and cosine functions yielded not a sinusoid curve, but
rather yielded semicircle waves. It was this research into the sinusoid
curve that lead to my discovery of the mistake of conic sections. This
often happens in science research. We dive into one subject and come out
with several different new researches.
And as a example of this new discoveries while doing something else, as a
case in point, I was doing my 238th book of science on Principle of Maximum
Electricity Production that I stumbled upon this big error of Geometry of
Motion of the cycloid curve. In science, one research begets another
research.
--- quoting my 238th book of science ---
Principle of Maximum Electricity Production by muon-as-electron inside
proton toruses// physics research
by Archimedes Plutonium
Preface: This is my 238th published book of science for AP // Proving the
Principle of Maximum Electricity Production is done by Atoms. That is an
alternative title, for this book is about proving this principle with
supporting evidence. A proof in physics is far different from a proof in
mathematics. For in physics we use experimental evidence to make a proof.
In mathematics there is no experimental proof, no deciding experiment, for
in physics we use experiments to tell us if one line of reasoning is
correct and the other is wrong. In physics it is all about deciding
experiments and data and facts in evidence. And in the course of doing this
book of a proof of the Principle of Maximum Electricity Production, happily
and by happenstance I was able to correct the cycloid curve of geometry
mathematics and the rolling motion geometry. For the cycloid curve is
actually a half ellipse, not a new independent curve different than a
ellipse. The cycloid construction failed to make the Pointer-Marker be
always perpendicular to the rolled over surface. So that when you roll a
ellipse over a identical ellipse you end up with a circle.
Cover Picture: Is my iphone photograph of page 76 of Seaborg & Loveland
"The Elements beyond Uranium" 1990, showing the torus geometry structure of
protons of F orbital shapes. Many of those shapes are torus shapes, but one
must keep in mind that those are hydrogen model constructs. Even so, the
majority of shapes are toruses. Also, one can see from these Dirac
orbitals, that the Dirac equation is flawed and in error. For the Dirac
Equation was an attempt to make the Schrodinger equation be relativistic.
But EM theory is already relativistic, so the world never needed a Dirac
equation. So what Dirac ended up doing, was, doubling the Schrodinger
equation. And we see this in the idea that the Dirac orbitals as shown in
this page 76 are two toruses joined as a figure 8, when all we needed was
one torus as shown as O with donut hole.
--- end quoting my 238th book of science ---
*2) Repeating the history of how I corrected the cycloid and limacon and
geometry-of-motion.*
It is very much worth repeating here, the history of how I found a huge
mistake in geometry of motion, the cycloid and limacon curves. And to make
a long story short, I needed to roll a cycloid over the back of another
cycloid and see what figure that produces. I could not find that in past
history literature, but I did find the rolling over the back of a ellipse
on a identical ellipse. And the literature says it is close to a circle but
not a circle, missing it by a tiny amount.
I was rather dazzled that it was _not a circle_ in Old Math, and felt
something was wrong here. For my intuition strongly believed a ellipse
rolled on a identical ellipse just had to be a circle traced out. And
exploring further, I came across the limacon, of the rolling of a circle
upon another circle. I saw a animation of the limacon in Wikipedia, and I
somewhat immediately saw where the mistake of the ellipse rolling over an
identical ellipse, saw the error. For in the Wikipedia limacon animation,
one can see two motions going on simultaneously. One can see the circle
rolling motion, but the Pointer-Marker was seen having arbitrary motion
itself.
I thought to myself-- ah ha!! I know where the mistake of the rolling of
ellipse over identical ellipse and not being a circle, where that mistake
comes from.
For I am a physicist and knows that magnetism is always perpendicular to
electric field.
I said to myself while seeing the animation of limacon. What if the Pointer
Marker had to be always perpendicular to the tangent of the surface the
circle was rolling over. If this requirement of perpendicular was imposed
upon ellipse rolling over identical ellipse, then the trace would indeed be
exactly a circle.
If the perpendicular requirement was imposed upon the limacon, the circle
rolling over another circle, then the trace is not a limacon but a larger
circle. And here I asked the logical question of Old Math geometry in
motion, how do they in Old Math get another circle by rolling a circle over
a identical circle? That question exposes their mistake of the Pointer
Marker having arbitrary motion. Because if they cannot get a circle from
rolling one circle over an identical circle, then they have a logical abyss.
Then my attention turned to the cycloid curve of Old Math, asking whether
they made a mistake here with having the Pointer Marker not always
perpendicular to the tangent of surface, a straight-line rolled over, and
thus the cycloid is a Half Ellipse, and not something different from an
ellipse.
But here is far more details of correcting the cycloid curve. For I wrote
much of this in the 238th book wanting answers of physics. I need a second
book, this one, wanting answers of math.
*3) Geometry-in-Motion speaks to Light Waves of physics.*
I knew of the particle wave duality of quantum mechanics physics in High
School and College in the late 1960s of the particle and wave nature of
light. But while doing my 238th book I realized a connection in producing
the cycloid wave and the circle rolling over a straight line. Suppose I
took the circle as a particle of Light, what is normally called a photon,
and took the curve it was tracing out as the wave nature of Light. So here
I was drawing both particle and wave of Light.
And taking that even further into physics for physics waves are divided
into Transverse waves and Longitudinal waves.
So, I have the world divided between either a Transverse Cycloid wave or a
Longitudinal Wave. And producing the cycloid wave, consider the circle as
the particle and consider the curve as the wave produced by particle.
With Cycloid versus Longitudinal, a ring of the longitudinal wave acts as a
circle rolling on a straightline forms the cycloid wave. And with AP's
cycloid versus longitudinal, I build all particles from cycloid waves if we
allow cycloid to be semicircle.
Interesting, I have to see how to form a semicircle transverse wave from a
cycloid structure. Or even perhaps a semiellipse transverse wave from a
cycloid structure.
A proton torus is merely its windings. A muon as electron is merely its
windings, both proton and muon perpendicular to one another. If we consider
all these windings as 2 times semicircles we see that windings can be
cycloid waves forming windings.
So if the world is cycloid waves, how do I form semicircle waves for the
cycloid is a semiellipse?
I am looking on how to recover a Semicircle wave in the manner that a
cycloid wave is constructed.
The natural first exploration would be to roll a ellipse. Perhaps on a
level surface or perhaps on another ellipse. In this book of research, I
found that rolling a ellipse upon another identical ellipse produces a
circle or semicircle.
*4) My history of discovery of the error in cycloid geometry of motion.*
This appeared in MathStackExchange showing that the ellipse rolled on
identical ellipse does not form a circle (semicircle), according to them.
But I found a mistake in their animation and ideas of rolling. And with the
correction of that mistake, the rolling of identical ellipses one on the
other does produce a semicircle.
--- quoting MathStackExchange ---
I'm struggling to prove the following.
Set one ellipse in contact with a congruent one so that the minor axis of
one is aligned with the major axis of the other. Now roll one round the
other. The locus of the centre of the rolling ellipse is a circle centre
the centre of the other, radius a + b.
Is there an obvious line of attack?
(Counterexamples in replies.)
--- end quoting MathStackExchange ---
Sometimes a Google search hinders one in finding what they want. For
example I am looking for how to get a circle from rolling, but Google hits
are all about using the circle and getting a cycloid.
So AP wants a hit where we use a cycloid given and roll upon that cycloid
figure some other figure to obtain in the end a circle (semicircle). Almost
useless to Google search for their search cannot understand I seek the
reverse.
This is probably the very same flaws in all robots, AI or not AI, they
cannot understand the concept of reverse.
Archimedes Plutonium
May 1, 2023, 1:48:32 PM
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Alright, some good signs in all of this mess. I am looking for rolling of
closed curves that recovers or retrieves the Semicircle. While the rolling
of circle on flat surface gives the cycloid. How to recover a semicircle is
what I want.
Looking at Roulette Curves, I see the rolling of a circle upon another
circle the center of rolling circle etches out a Trochoid.
Now we use an off-center to the rolling circle upon another circle and we
get a Limacon Roulette curve.
So almost instantly I realize if we use a point on the rolling ellipse
(replace rolling circle) we get a new circle.
This is the answer I was seeking. To have a given cycloid curve, how does
one restore a Semicircle from that cycloid. And the answer appears to be
that you roll a specific ellipse upon the cycloid curve.
Archimedes Plutonium
May 1, 2023, 2:00:38 PM
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Now looking at the Limacon, if we stipulate that the off-center point has
to always be as a perpendicular to the tangent of surface of the circle it
is rolling over, that we generate a semicircle.
This gives me the idea that if we stipulate a ellipse rolling over a
identical ellipse and have this marker always perpendicular to the surface
rolled over- it generates a semicircle.
Also, if we stipulate a cycloid rolling over a identical cycloid with
off-center marker always perpendicular to surface of cycloid rolled over,
generates a semiellipse.
I am surprised no-one in math or physics did this before AP did it in May
of 2023, fresh new mathematics on the world scene.
Now the title of that MathStackexchange is "Rolling ellipses" question
asked 9 years, 8 months ago.
And watching the animation the one ellipse traces out almost, say 99.9% of
a semicircle.
But I wonder, with my keen math intuition, that if the center point was
Free to rotate and if that center point were a marker so that it could be
perpendicular to the surface of the second stationary ellipse, that the two
ellipses indeed, trace out a circle.
Just wondering. And I need to make cut-outs of cycloids and ellipses and
hand roll them. We cannot trust computer animations.
Archimedes Plutonium
May 1, 2023, 9:53:05 PM
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So, well, Wikipedia has a animation of a Limacon and the Convex Limacon is
almost a circle but a part of it looks flattened.
My objection is that the point-marker is allowed to rotate. If the point
marker was always staring of the circle it was revolving around at a
perpendicular, then that etches out another circle.
In the MathStackExchange of a ellipse moving around the same ellipse, they
almost get a circle. But I wonder if they programmed the computer animation
to not be perpendicular pointer marker and thus be of the path of a circle
ever so slightly.
Now it is intuitive that if the pointer-marker in Limacon was fixed
stationary and perpendicular to other circle at all times that we etch out
concentric circles depending on where from the center the pointer-marker is
placed. As if the pointer-marker was caught in a magnetic field that did
not allow it to move from being perpendicular.
Now if the pointer-marker were directly on the rim, the circumference
starting out, can we expect the result to be a circle, only a circle with
one point intersection with the circle it is rotating around in 360 degrees.
These figures are actually far more complicated and complex than what
Wikipedia and Websites make them out to be.
Archimedes Plutonium
May 1, 2023, 10:57:40 PM
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Now I am going to go contrary to the description of the Dimpled Limacon and
say that the Pointer-Marker needs to stay Perpendicular to the tangent of
the surface of stationary circle. In this manner, the dimple and the
flattening disappears and what we end up with is a perfect circle, larger
than stationary circle and intersection of both circles at one point.
Archimedes Plutonium
May 1, 2023, 11:23:28 PM
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Now I am pretty sure this is a flaw in Old Math Geometry of their Limacon
pointer marker going around like a clock as the motion of the moving circle
goes around. The AP Limacon is where the pointer marker needs to stay
perpendicular to the tangent of the stationary circle, thus creating a new
larger circle with one intersection point. I say this because there is no
other mechanical means of producing concentric circles or intersection at
one point of two circles.
Physics needs the **perpendicular** for the pointer-marker, for electricity
is always perpendicular to magnetism.
The Old Math Geometry Limacon is actually 2 rollings, the rolling of one
circle over another and the independent rolling of a pointer marker. In the
AP Limacon, there is but one rolling the circle over the other circle,
while the pointer-marker is a fixed perpendicular point.
This messy obfuscation of what is moving also can solve the two identical
ellipses rolling one over the other as tracing out a perfect circle. It is
almost producing a perfect circle in the MathStackExchange animation but
slightly off. And I am saying it is only slightly off because the animation
was not programmed to allow the pointer marker, the center of the rolling
ellipse to always have a perpendicular to the tangent of the stationary
ellipse. You see the programmers of that animation, are like the Limacon
animation and not restricting the pointer-marker to be always a
perpendicular, and that is the reason it is not a perfect circle. This
means having the pointer marker be a tiny tiny length, the infinitesimal of
mathematics, a length of 1*10^-604, a small positive number and length that
needs to stay perpendicular to stationary figure surface tangent. But in
small applications such as the animation for humans to see figures probably
the 10,000 Decimal Grid System or even the 1,000 Decimal Grid System was
used as our eyes see objects well in 1,000 or 10,000 Decimal Grid System.
And in those two grids the smallest length is 0.001 for 1,000 Grid and
0.0001 for 10,000 Decimal Grid System. So how much off being a perfect
circle in that StackExchange animation of ellipse rolling over identical
ellipse. If we can get how much off, we can likely predict what decimal
grid system the animation was working in.
Now I need to do the same thing with the cycloid, see if I can roll one
cycloid over a stationary cycloid with a pointer-marker and see if I trace
out a semicircle.
Yes, I believe that returns two cycloids, one rolling on the other, to be a
semicircle. And later in my research I come to realize the cycloid is a
semiellipse, and rolling a ellipse over a identical ellipse yields a
semicircle.
*5) Old Math's Limacon, Cycloid and similar curves in error.*
Flaws and error in Old Math's Limacon, in Cycloid, and in ellipses rolling
on identical ellipse-- Their unwarranted rotation of their Pointer-Marker.
The worst of these animations is Wikipedia's convex Limacon where they show
the pointer marker as rotating also, when in truth, that Pointer-Marker
should be at all times perpendicular to a tangent line of the stationary
circle. In this way, the trace curve is another circle, larger than either
the stationary or orbiting circle.
This second motion mistake when only one circle should be in motion of
rolling is also seen in a answer reply on Math Stack Exchange
--- quoting MathStackExchange "Rolling ellipses" 9 years, 8 months ago ---
> I'm struggling to prove the following.
>
> Set one ellipse in contact with a congruent one so that the minor axis of
one is aligned with the major axis of the other. Now roll one round the
other. The locus of the centre of the rolling ellipse is a circle centre
the centre of the other, radius a + b.
>
> Is there an obvious line of attack?
>
> (Counterexamples in replies.)
> --- end quoting MathStackExchange ---
So now, in this video animation has these dimensions-- parameters a = 2, b
= 1 and shows a gray circle with radius 3. and the author claims the Locus
has some deviation from perfect circle.
What AP claims in that animation, is why in the world would you have a
arbitrary rotation of the Point-Marker. It is because of this arbitrary
rotation that you do not trace a perfect circle, for if the Point marker
were kept at a constant perpendicular to the stationary ellipse tangent,
then the trace is indeed a perfect circle.
And, if AP's conjecture is true that the only smooth curves in existence in
all the world are conic or cylinder sections, anything else described as a
curve, has vertices.
It is the animation in Wikipedia's limacon entry that alerted me to what
the error of Old Math's cycloid curve, especially the cycloid curve, and
all other rolling curves mechanisms was. Why Old Math was making
error-filled geometry with arbitrary motions. Allowing the pointer-marker
to move arbitrarily. The Pointer Marker has to be a fixed constant
perpendicular to the surface it is rolling over. Thus, the Cycloid curve is
truly a Ellipse curve. And this only presses more in favor of AP's
conjecture-- all the curves of math that exist are in the conic and
cylinder sections, and any other curves have at least one vertex in their
"body curve". The Cycloid being truly a ellipse means it has no vertex
point. But Old Math's cycloid where the pointer-marker is arbitrarily in
motion results in geometry figures with at least 1 or more vertices, and
thus, not a smooth curve. The AP conjecture, once proven true if true, says
only conic and cylinder curves are smooth curves having no vertex point.
--- quoting Wikipedia's limacon entry with animation in full display ---
In geometry, a limaçon or limacon /ˈlɪməsɒn/, also known as a limaçon of
Pascal or Pascal's Snail, is defined as a roulette curve formed by the path
of a point fixed to a circle when that circle rolls around the outside of a
circle of equal radius. It can also be defined as the roulette formed when
a circle rolls around a circle with half its radius so that the smaller
circle is inside the larger circle. Thus, they belong to the family of
curves called centered trochoids; more specifically, they are epitrochoids.
The cardioid is the special case in which the point generating the roulette
lies on the rolling circle; the resulting curve has a cusp.
Depending on the position of the point generating the curve, it may have
inner and outer loops (giving the family its name), it may be heart-shaped,
or it may be oval.
--- end quoting Wikipedia's limacon entry with animation in full display in
a rightward box ---
Archimedes Plutonium
May 2, 2023, 7:45:23 AM
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On Monday, May 1, 2023 at 11:23:28 PM UTC-5 Archimedes Plutonium wrote:
Now I am pretty sure this is a flaw in Old Math Geometry of their Limacon
pointer marker going around like a clock as the motion of the moving circle
goes around. The AP Limacon is where the pointer marker needs to stay
perpendicular to the tangent of the stationary circle, thus creating a new
larger circle with one intersection point. I say this because there is no
other mechanical means of producing concentric circles or intersection at
one point of two circles.
It is understandable that people would be interested in the 2-motions of
Old Math Limacon, the motion of rolling one circle over another circle, but
also the motion of the Pointer-Marker, for the Limacon of Old Math is
obfuscating but yields figures for Art and Artwork, but not figures for
Physics of electricity and magnetism.
In Physics we are more precise than Old Math with their double motions. In
New Math we have only one motion, the rolling of one circle over another.
Or the rolling of 1 ellipse over another and producing a new circle-- once
the mess up of the motion of the pointer-marker is corrected.
In New Math, in rolling figures we allow only one motion. And we
Well-Define the Pointer Marker as not having motion and has a Infinitesimal
length of at least 1*10^-604 distance length. In 10 Grid the length is 0.1,
in 100 Grid the length is 0.01, and not accounting for the length of the
Pointer Marker causes the mistake of thinking rolling a ellipse over
identical ellipse is not a semicircle when it surely is. And this Pointer
Marker must be perpendicular at all times to the rolled-over tangent of
stationary figure. So in the MathStackExchange figure of ellipse rolling
over identical ellipse, it actually is a Circle once the Pointer Marker
mistakes are corrected.
Now it took me all of just 2 minutes today to get a sheet of tracing paper
and trace out the cycloid on page 293 of Jacobs "Mathematics: A Human
Endeavor". And then roll my second cycloid over the perimeter of Jacobs
cycloid. Using the pointed tip (not the center of second cycloid but the
pointed tip as Pointer-Marker. Now keeping the pointed tip always with its
length as a perpendicular to the tangent of the stationary cycloid what
that rolling traces out is indeed a Semicircle.
So here we can start some Conjectures:
Conjecture 1, rolling a ellipse over a identical ellipse, keeping
pointer-marker perpendicular produces a circle.
Conjecture 2, rolling a circle over another circle keeping pointer-marker
perpendicular produces a circle, a larger circle than either the stationary
and rolling circle.
Conjecture 3, rolling a cycloid over identical cycloid, keeping
pointer-marker perpendicular produces a semicircle. And here I found the
cycloid itself is a semiellipse once the perpendicular conditional is
applied.
Physics needs the **perpendicular** for the pointer-marker, for electricity
is always perpendicular to magnetism.
The Old Math Geometry Limacon is actually 2 rollings, the rolling of one
circle over another and the independent rolling of a pointer marker. In the
AP Limacon, there is but one rolling the circle over the other circle,
while the pointer-marker is a fixed perpendicular point with tangent of
surface rolled over.
Old Math Geometry of its roulette math and Limacon are ill-defined in their
pointer-marker and theirs is mostly math for Art work of a computer
animation not programmed to account for a perpendicular conditional, not
physics electricity and magnetism.
This messy obfuscation of what is moving also can solve the two identical
ellipses rolling one over the other as tracing out a perfect circle. It is
almost producing a perfect circle in the MathStackExchange animation but
slightly off. And I am saying it is only slightly off because the animation
was not programmed to allow the pointer marker, the center of the rolling
ellipse to always have a perpendicular to the tangent of the stationary
ellipse. You see the programmers of that animation, are like the Limacon
animation and not restricting the pointer-marker to be always a
perpendicular, and that is the reason it is not a perfect circle. This
means having the pointer marker be a tiny tiny length, the infinitesimal of
mathematics in each Decimal Grid System, for it is 0.1 in 10 Grid, and 0.01
in 100 Grid, and at infinity borderline the length is 1*10^-604, a small
positive number and length that needs to stay perpendicular to stationary
figure surface tangent.
Yes, in geometry if you are not in the habit of first doing a problem with
hand cut outs and seeing for yourself the outcome, then you are not doing
geometry correctly. Those that think they can apply algebra without seeing
a hand on demonstration are prone to much error in math geometry.
Archimedes Plutonium
May 2, 2023, 3:23:05 PM
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Alright, good on MathStackExchange for they have something on cycloid
rolling. Titled "Cycloid rolls on another identical Cycloid" 6years-4months
ago. They have a animation and the figure it traces is looking like half a
stadium geometry shape. However, they use as Point-Marker the center of the
straightline segment of the cycloid. And I suspect they do not keep their
point-marker always at a perpendicular to the surface of stationary
cycloid. If they keep their point-marker perpendicular to tangent of
surface, I suspect it becomes a semi-ellipse rather than what I see as a
partial-stadium figure, a flattened top.
What AP uses as Point-Marker is either one of the corners and keeping this
point-marker perpendicular, I trace out a Semicircle.
Archimedes Plutonium
May 2, 2023, 5:59:55 PM
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Major, major Geometry in Motion discovery by Archimedes Plutonium. I
discovered this in the writing of my 238th book. The discovery I made is
that the cycloid Wave is an Ellipse wave. The error of Old Math Geometry is
that they had 2 motions in a rolling circle on the straightline. They had
the motion of the circle which is wanted, but they also had a 360 degree
motion of the Pointer-Marker on the surface of the rolling circle. What
this causes is the end product of a cycloid with 2 vertices if we put half
and half together.
When we stipulate only one motion and not two motions, and when we
stipulate the Point-marker is always perpendicular to the surface of
straightline, that the Point-Marker is actually a infinitesimal distance
length that is constantly perpendicular to the straightline, that the final
curve is a ellipse and has no vertices.
A cycloid with vertices is actually a fiction geometry produced by a circle
with two motions.
This chimes back to one of my conjectures a long time ago, that the only
smooth curves produced by circle rolling is either ellipses or circles
(ovals are two dissimilar ellipses joined together at midsection) or New
Math conic sections. There is a quantization of smooth curves as produced
by conic sections and so the Old Math Geometry of Motion Geometry erred in
motion by a ill-defined notion of the Point-Marker, and that this
Point-Marker is ill-defined in Old Math and it has a arbitrary motion
itself.
This is a huge major discovery for Geometry in Motion for it directly
impacts Physics electromagnetism.
The cycloid curve of Old Math is a error filled fictional curve. Much like
the slant cut of right-circular cone is a Oval, never the ellipse, and the
oval broadly defined is the adding of two different ellipses to form a
oval. Still within the AP conjecture-- the only smooth curves in geometry
(have no vertices) are New Math conic sections.
This deserves a separate math book for it is a huge major change in
Geometry of Motion.
*6) Proof the Cycloid is a ellipse and smooth curve.*
Archimedes Plutonium
May 2, 2023, 6:15:44 PM
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Now I have a easy intuitive proof that I am correct the cycloid is actually
a ellipse. Some geometers call this a synthetic proof where we argue a
simple insight. In this case we focus upon the midpoint of the cycloid
wave-- 1 cusp. This midpoint is rather flat and it is flat because the
Pointer-Marker is parallel to the straightline it is rolling upon. That
means the points on both sides of the midpoint are also parallel and thus a
straightline ridge. Now for AP's Pointer Marker that is always
perpendicular to straightline causes the midpoint ridge of cycloid to be
higher than the points to the right and left of the midpoint ridge. This is
the description of the ellipse. While the cycloid with its flat ridge
actually is not a smooth curve at the ridge but contains vertices as it
transits from straight line ridge to being a curve again. This constitutes
a Proof that Cycloids are actually Ellipses. And also I need to show that
at the ends of the cycloid cusp that those angles are different in the AP
cycloid than the ill-defined Old Math cycloid.
This is again, a huge major discovery, not so much for Mathematics which
used this stuff like cycloid and Limacon for Art and Art Work, but for
Quantum Electrodynamics, for EM of physics is all about Particle and Wave,
and the cycloid is the particle of the rolling circle upon the straightline
which is the wave of Physics. The Faraday law of a Torus being the Proton
is all about geometry of smooth curves. And geometry is basic and
fundamental to physics.
Archimedes Plutonium
May 2, 2023, 8:04:56 PM
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This makes sense that circle rolled on straightline is ellipse. Ellipse
rolled on identical ellipse makes circle. But it leaves out what happens
when circle is rolled on ellipse or ellipse rolled on circle all of which
the Pointer-Marker is kept perpendicular to rolled surface.
In the Limacon's of circle rolled over another circle produces circles when
the Pointer Marker is kept perpendicular.
So what is a ellipse rolled over a circle? I hazard to guess another
ellipse. And for circle rolled over ellipse, I guess another ellipse. What
about two different sized ellipses? Here the problem is going to be if
there is a fractional rolling? By fractional I mean both do not come to a
smooth finish of 360 degrees and have left-over arc-length. In those cases
I would say look and examine the curve to see if it is going to be a circle
or becoming a ellipse.
Archimedes Plutonium
May 2, 2023, 9:54:19 PM
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This deserves a new separate math book on the massive corrections and
changes to Geometry in Motion. The rolling requires only 1 motion as a
category of figures. If we have two items in motion such as the circle then
the Marker-Point in motion also, then different geometry figures result.
So I have the Conjecture that all and only smooth curves in mathematics can
all be got from cone, and cylinder cuts. Anything that is not a cone and
cylinder cut is not a Smooth Curve.
And let me get on what that Well-Defined Smooth Curve is. I said that the
well defined smooth curve has no vertices and in discrete math since all of
math geometry is discrete with its Decimal Grid Number System-- there
exists empty space between any two given numbers in mathematics. So a well
defined smooth curve, each point and its successor point and predecessor
point is related by the number constant of pi=3.14159.... If a point is not
related to the next point on a curve (discrete geometry) by 3.1415... then
the curve is __not smooth___ and we call that point a vertex.
In line geometry we had no trouble in defining a vertex as where a angle
that is not 180 degrees from one point to the next occurs.
But in curve geometry of Old Math, they never well-defined a vertex on a
curve. I define it as a breakdown in pi = 3.14..... of a point and its
successor point.
*7) Defining what a Smooth Curve is in New Geometry.*
While doing my 238th book of science, I ran into a huge error of Old Math
Geometry of their geometry in motion such as the cycloid curve. So huge and
important of an error that this subject needs its own separate book of
science.
This book should start with the animation of the Limacon by Wikipedia for
it alerted me to the huge mistake made in Old Math Geometry. And this is
Geometry in Motion, far more important than stationary geometry which we
study in High School, stationary figures, for geometry in motion is the
calculus.
Watch that Limacon animation and see for yourself the arbitrariness of the
motion of the Pointer-Marker.
If the Pointer-Marker had been kept a Perpendicular to the tangent of the
circle rolled-over, the trace result would be another larger circle.
And Old Math Geometry was deficient to even ask the question-- how do you
get a new circle by rolling two identical circles-- one upon the other?
In this book I need to make a final proof of my Conjecture poised in my
238th book. That Conjecture said that the Only Smooth Curves existing in
all the world come from conic or cylinder sections. Every other made up
curve is not a smooth curve. Now that is a mighty big conjecture, for it
speaks of all smooth curves. So I need to define what a smooth curve is.
And I define smooth curve as not having any vertices, not one single vertex
plus an additional feature as each point on the curve surface is related to
the next successor point by a number factor of pi = 3.14159.... In New True
Math we have only Discrete Geometry with Decimal Grid Numbers. We thrown
out the pollution of Old Math with their silly continuum and their cesspool
Reals.
I am writing this book while the events of discovery in my 238th book are
still very fresh on my mind.
I had always loved the cycloid and knew that it was vastly more important
to physics than the present state of knowledge of the cycloid. So keen was
I on the cycloid that several years back, I started to write that the Light
Wave--our most important wave of all, is likely to be cycloid and the
sinusoid wave was a silly error of Old Math and Old Physics. For the error
is plain to see, in Old Math they allowed the x-axis in graphing to be
different than the y-axis. So they thought it alright to have the x-axis be
angles while the y-axis was numbers. This is a logical no. For when you
make one axis different from the other axis, you no longer have
mathematics. Perhaps you have commercial sales of bar graphs for selling
fruit as the x-axis is a fruit such as watermelon, lemon, orange, apple and
the y-axis be numbers of sales. This is no longer mathematics or physics.
*8) Researching the Cycloid and Geometry of Motion, rolling one figure over
another.*
And so I headed into writing my 238th book of science on proving the
Maximum Electricity Production Principle. And I knew I would likely have to
use cycloid in the proof. Turns out I did not need the cycloid in the
proof, but I kept digging deeper and deeper into the research. For I needed
to know about what happens when a cycloid wave rolls over on top of another
cycloid wave, much as what happens when you roll a circle over an identical
circle. And from previous glimpses into the math literature, it was barren
of what happens when you roll a ellipse over an identical ellipse. A circle
rolled over another circle. And worst of all, a cycloid rolled over a
identical cycloid. Mostly barren of literature.
And so I dug deep and was able to find examples of all of these, thanks to
modern day computers, people can do animations of this rolling. But be
wary, for computers in geometry often only display the bias and prejudice
of the person who programs the computer and not the actual true geometry
that is underlying.
And so, as I was searching I ran into a animation of the Limacon on
Wikipedia, and saw that the Pointer Marker was moving as well as the circle
itself was moving. So here I see 2 Motions, when I expected only one
motion, the circle rolling over a stationary circle.
And I had seen in Stackexchange a argument that a ellipse rolled over a
identical ellipse was ___ not a circle__. And looking at the Wikipedia
Limacon, the arbitrary motion of that Pointer-Marker, the bells & whistles
& lights went sound and flashing on. What if we placed a Electromagnetic
restriction on the Pointer Marker, my mind said to me. What if we
constrained the Pointer-Marker to be always at a perpendicular to the
tangent on the surface it is rolling over.
Well, beautifully the question of what is the trace figure of a ellipse
rolling over an identical ellipse-- it is a perfect Circle. And the only
reason that StackExchange answers said "no" to being a circle, is that
their Pointer Marker was arbitrarily in motion.
Same as Limacon, if the Pointer Marker is arbitrary you can get all sorts
of goofball figures, but if constrained to being perpendicular the circle
rolled over another circle is a new and larger circle.
But the biggest gain in this is the Cycloid, because before this discovery,
it was thought the cycloid is not a half ellipse. But with the
perpendicular constraint placed on the Pointer Marker, the cycloid is
indeed a Half Ellipse. And rolling a cycloid over a identical cycloid
retrieves back a Circle.
But also important is that Geometry is Discrete and the Pointer Marker has
to have some Length. The largest length is the 10 Grid with length of 0.1
for the Pointer Marker. The smallest possible length for pointer marker is
the 10^604 Grid with its length of 1*10^-604 length. Since every Pointer
Marker has length, then the constraint of being Perpendicular to surface
tangent makes a huge difference on the trace.
As for oddball and goofy figures such as Limacon, well those are just
wastrel figures, for we can mess around with their motion and make all
sorts of silly traces. When something is arbitrary, you get all sorts of
goofy figures.
Now the hard part of this book is going to prove several of the Conjectures
made about smooth curves, namely-- the only smooth curves in all of
geometry are circle, ellipse, and oval where oval is defined as the joining
of two different ellipses at what can be called a midsection, and are conic
or cylinder sections. So the proof of this conjecture is a difficult proof.
But vastly important. It throws out all the oddball and silly curves as
they have at least one vertex and thus, not smooth.
And a second conjecture is the idea that the circle, ellipse, oval,
parabola and hyperbola are all Polynomials, where the formula of Parabola
is Y = x^2 which is a polynomial. In Old Math, they made the mistake of
thinking conic sections involve a orientation of ><, while AP says the
correct orientation is base to base as <>. I have no idea why Ancient Greek
math wanted the >< apex to apex. Kind of silly, really, but perhaps they
were enamored with a open ended curve, not closed.
In the AP configuration, a <> begs the question of the hyperbola and
parabola are now closed. They still are half parabola with formula Y = x^2.
But the conjecture here is whether the Parabola is still a curve without
vertices, same for hyperbola. And the importance of this conjecture, is
that if I can prove the Parabola is indeed a "smooth curve" without
vertices, then I can apply that beautiful polynomial formula to circle and
to ellipse. That circle and ellipse and also oval are in line with a
polynomial formula of Y = x^2.
This 2nd conjecture, I feel is much more tough than the first, unless I
find it does have vertices.
The first conjecture, I have a feeling of proving it from that of
application of Rolling Motion itself. That the rolling of a circle upon
another circle yields a larger circle; the rolling of a ellipse upon
identical ellipse yields a circle. And the rolling of a cycloid which is a
half ellipse yields a circle. There is a proof trick in which, if rolling
of smooth curves yields only other smooth curves, then we can finalize that
data to say these are the only smooth curves in existence.
So in this book I aim to prove the conjectures I left behind in book 238th
for that book was focused on maximum electricity production in a torus
geometry. And these conjecture proofs will be one of my most important
geometry math contributions, for that the cycloid is close and dear to
Light Waves of Physics.
Conjecture #2 is tricky, and depends on whether two cones, base to base <>
forms a vertex at the base. If a Vertex then there is no chance of this
Conjecture being true that a Parabola is a ellipse with equation Y= x^2.
I vaguely remember my study of this question in years past.
The trouble is that in Reals and continuum, those bases form a ellipse
provided we think of the rim of base being absorbed by the rim of the other
base in <>. But in New Math with Discrete Geometry, we can take a base of
one ellipse and be a single number short of the other ellipse base, thus
when we conjoin the bases it remains a Smooth Curve.
What the benefits of a true Conjecture 2 are-- are magnificent. In that all
the smooth curves as listed.
1) Circle
2) Ellipse
3) Oval is just two different ellipses joined together
4) Parabola
5) Hyperbola
The benefits of conjecture #2 is that all five smooth curves can be
translated into a Polynomial of Y= x^2. No longer do we have square roots
involved as non-polynomials.
Does it make sense in a larger view?? I think so, in the idea that a
parabola of Old Math just did not go far enough out there in space to see
it be a ellipse.
And how about the hyperbola? Well, far easier to see it is a ellipse and
possible one of them or one of the parabolas as forming a circle.
But the proof of Conjecture #1-- all smooth curves that exist in the world
must come from conic or cylinder sections, I believe the key to that proof
is the rolling of identical smooth curves one on the other, and any other
curve figure cannot retrieve any of the 5 listed above.
So, yes I do believe I have both Conjectures true and a proof for each.
With the Cones base to base <> what saves me is that I take two identical
cones, and thus I snip away one number layer on the rim of one base in
Discrete Geometry and then can attach both cones base to base and that no
longer is a vertex, but a smooth curve for all sections intercepting the
new conjoined bases. Or possibly instead of snipping away a circle layer
since both have equal circles at the base, is superimpose the equal layers
and the upper and lower make the conjoined figure smooth, as per definition
of smooth being related by the number 3.1415....
A Parabola is much like in Physics of their Linear Momentum in astronomy.
Until someone says to you-- you need to take the circle out far enough such
as the Milky Way galaxy and you no longer see the linear momentum.
So I am ready to write this 240th book of mine, with proofs of two
important conjectures.
These proofs possible because Geometry is Discrete with holes in between
successor numbers. You can say that 20th century mathematics was a cesspool
with their Reals, their continuum, that completely held them back from the
truth of mathematics. For what should have happened in the year 1900
January, was a smart mathematician saying-- our most important math, our
most important job is to solve a geometry proof of Fundamental Theorem of
Calculus. That proof would have thrown out Reals, thrown out continuum.
Because only a geometry proof of FTC is possible in discrete geometry. The
Reals and continuum to math of the 20th century is what Flat Earth was to
science in past centuries.
What is interesting about Conjecture 2-- The Parabola has formula Y = x^2 a
pure polynomial and when bases of cones are oriented <> the parabola is a
closed loop circuit-- a ellipse. This means we can attribute all the other
smooth curves as being Polynomials instead of their circle like formula.
Now yesterday I was trying to get the Pencil Ellipse from conics <>. Long
long long pencil ellipses. And the way to do that is to make the cones
super thin and narrow. And then, either a hyperbola on the <> or a parabola
on the <>, both yield pencil ellipses.
If I had a normal looking cone the only pencil ellipses I can obtain is
very near the apex. But that is a tiny tiny pencil ellipse in length. I
want a huge long pencil ellipse and that is obtained from making the cones
themselves almost be pencil thin.
Now in this discussion, one can begin to ask what is the difference really
of a Parabola and a Hyperbola since both form Ellipses in true conic
sections of <>. Both form ellipses. And the answer appears to be that ---
Some parabolas form Ovals rather than ellipses as they intercept only a
small portion of the second cone where the two cones are joined <>. While
the Hyperbolas can also form ovals, they are not as convergence-limiting as
the Parabola oval. It is the slant of the cones walls that determines the
parabola and thus a convergence-limit on ovals. The Hyperbolas have a
convergence-limit also of a figure that comes closest to the parallelogram
formed from perpendicular apex to apex intersection <> with the bases. The
parabola converge on side wall angle. The hyperbolas converge on
parallelogram <>.
Speaking of the convergence-limit of the Hyperbola to the slant side of
cones of <>. So, picture a Perpendicular to bases conjoinment and directly
through single point apexes at both ends of <>.
Now offset the Hyperbola perpendicular of the Apexes by having the cut be
through 2 points and we end up with a remarkable figure of a Ellipse
resembling almost the parallelogram that was the apexes of <>.
*9) Conic sections with base to base <>, rather than apex to apex ><.*
Of course, AP's conics with base to base rewrites the entire subject of
Conic Sections for <> is the true schematic and not the silly apex to apex
><. The only reason I can think of for apex to apex is the Ancient Greeks
had only 1 cone, not 2, and the Ancients thought of the parabola as open as
well as the hyperbola. And only in modern times was the stupid apex to apex
set up to talk about branches of the hyperbola.
What AP does to conics is makes the entire subject new and very different.
All smooth-curves of geometry comes from the conics base to base and no
smooth curves exist if not found in the conics. With base to base, the
hyperbola and parabola are also closed curves.
My work so far finds the Hyperbola can be a oval, ellipse but no circles
emerge.
However the Parabola has ovals, ellipses and I suspect one circle. That
circle comes from the parabola angle equal to the slant wall angle and
forming a radius of equidistance from the center of the two cones <>. Since
the hyperbolas do not have a angle equal to slant wall of cones, I suspect
they cannot generate a circle. Let me call it the Parabolic Circle, and it
is not 1/2 distance down the slant wall for if the cones are long and thin,
this Parabolic Circle is nearer to the center of the 2 cones.
Is there any saving feature of the Old Math conics of apex to apex?? I
would say no. A total trip down wastrel lane is the apex to apex conics.
This is such a massive overhaul of Old Math Geometry, especially its
geometry in motion. It is far beyond the boundaries of correcting the
Cycloid curve, but in correcting all of Old Math Geometry of its most
important feature-- Geometry calculus, for calculus is geometry in motion.
Perhaps this as title: New Math Geometry of Motion, correcting Conic
sections and cycloid and rolling one figure over another// math research by
Archimedes Plutonium. No, I decided upon this title: New True Geometry
starting with cycloid correction and Geometry-of-Motion // math research.
Now an easy proof of Cycloid being not what Old Math had, but being a Half
Ellipse is for me to show that given any Half Ellipse, and drawing a circle
inside at midpoint, that exactly 3 of those circles is the area of that
Half Ellipse. At least that is my preliminary results of taking ellipses in
Jacobs book of Mathematics a Human Endeavor and testing out with compass
and straightedge. It maybe the case that it is not 3 circle area but
perhaps 2.71.... or something else. But preliminarily it looks to be 3. And
I am assuming the Old Math mistake is that of saying the parametric of
cycloid is the same as parametric of ellipse.
Now Old Math gives the cycloid the parametric equations of:
x = r(t-sin(t))
y = r(1-cos(t)) where t is an angle.
Yet Old Math had for the ellipse the parametric equations far different
than their parametric for cycloid as that of:
(x,y) = (a(cos(t)), b(sin(t)), 0<= t < 2pi. (Source Wikipedia)
AP no longer bothers with trigonometry functions for in New Math, we have
only polynomials to deal with. In New Math we convert everything that is
not polynomial into being polynomial by the Lagrange transform. Only then
do we respond to a function.
Kind of fascinating that Old Math comes up with the correct numerics of
circle rolling is 1/3 of the area inside the cycloid generated, yet Old
Math could not come to grips with the fact that the curve traced is 1/2 of
a Ellipse, not a brand new curve different from a ellipse.
So a nice proof that the Cycloid is a Half Ellipse is this area proof.
So, then the question comes to be, how could Old Math--- blindly not see
that the cycloid was a Half Ellipse yet come up with the correct answer of
circle is 1/3 the area of the cycloid area? How could they do that? And a
reasonable explanation is that Old Math mathematicians lacking logical
intelligence often fell back on a excuse-- they would say-- that cycloid
curve is not a half ellipse but close to it, and then mix and match by
saying ellipse-cycloid, not understanding that it is actually and exactly a
Half ellipse.
Again, how did Old Math mathematicians screw up badly on the Cycloid?? They
screwed up by having a Continuum and their numbers as Reals, when geometry
is Discrete with numbers being Decimal Grid Numbers. So that as they rolled
the circle on a line, they thought the Pointer-Marker also should have a
second motion along with the circle having motion. They never understood
that the Pointer Marker has a small finite Length. And this Length can be
fixed to be perpendicular (like electromagnetism of physics) be
perpendicular at all times with the tangent of the surface rolled over.
Thus, the cycloid curve that is traced is Half a Ellipse.
This Pointer-Marker error is seen in all of Old Math geometry motions of
rolling as a grave error with such ludicrous results of a ellipse rolling
over an identical ellipse and the trace __not being a circle__ when in
truth the trace is a circle.
*10) Proof that all Smooth Curves are obtained from conic sections base to
base <>.*
The proof that all smooth curves as defined smoothness has no vertex and
all the points are governed by the number 3.14159.... one point to the next
point, that all smooth curves can be got from a Conic Sections base to base
<>, that proof is obtained by showing the rolling circle can obtain only
another circle or a ellipse. Then, a rolling ellipse obtains only a circle
or another ellipse. And that is all that is obtainable when the demand on
the Pointer-Marker is Perpendicular to the tangent of surface rolled over.
This even involves the Parabola and Hyperbola curves for when conics are
base to base, the parabola and hyperbola are most often either a ellipse or
a oval.
Reflection: That makes commonsense in the idea that pi = 3.14159... is a
constant and forces a curve to be closed. This forcing determines the true
Conics to be base to base as <> and not as apex to apex with its open
parabola and branched hyperbola.
It has always bothered me to think the parabola would go to infinity as I
walk the x axis. So let me put that notion to rest once and for always.
Say I have the x-axis and feeding the function Y -> x^2 a parabola. First
quadrant only, mind you.
It has been extremely difficult for me to see this function be "open" and
not closed like a ellipse.
So in 10 Grid, the function stops at square root of 10 is 3.16... In 100
Grid the function stops at 10. Because in Grid Systems we pretend the
largest number is the infinity borderline. So in 10 Grid, our parabola
stops at 3.16... and we form a quarter of a ellipse. In 100 Grid, our
parabola stops at 10 forming a quarter of a parabola a closed loop ellipse
that is cut when we cut a parabola of Conic Sections <>.
So my first and mighty proof of Geometry, geometry in motion is to prove
all Smooth Curves come from Conic Sections, and that proof is obtained from
noting the rolling of circle produces only a circle or ellipse and the
rolling of ellipse produces only a circle or ellipse. The proof is that
rolling a circle on another circle with Pointer-Marker always perpendicular
to tangent of surface rolled over is another circle. Same goes for rolling
a ellipse over identical ellipse is traced out a circle. And rolling a
cycloid over identical cycloid, since cycloids are ellipses is a circle.
The proof of Conjecture 1 Statement: All smooth curves are begot from conic
sections base to base. Proof: The rolling of identical figures, smooth
curves over one another yields only circles and ellipses where oval is two
dissimilar ellipses joined at midsection. If a curve has a vertex, then
rolling one over another, those vertex points will make a bump in the
rolling and a bump in the trace remaining.
*11) The oval in conic sections, base to base <>.*
Now I patch over the Oval for most conic cuts are ovals because parabolas
and hyperbolas and slant cuts in the <>, many to most are Ovals as a trace.
But not to worry, because the definition of Oval itself is that of joining
2 ellipses of different sizes at what I call the oval midsection. Most
parabolas and hyperbolas are ovals with two joined different sized ellipses
keeping the final figure a Smooth Curve.
Now, a problematic question arises here. And perhaps needing a proof in
itself. The question of Pi = 3.14159.... is the Smoothness number. Joining
one point in Discrete Geometry to the next point by that number 3.14159....
The question is sort of philosophical in nature, that is, it may not be
provable but has to be a axiom or principle.
The question is -- does a curve out of necessity by a closed loop circuit
to be smooth and obey pi smoothness. So as I started this lesson on saying
the parabola and hyperbola must be a closed loop. Here I end this lesson
with the question that pi= 3.14159... governs Smoothness-- no vertex, and
questioning that smoothness has to be out of necessity a closed loop
circuit? I suspect I cannot prove it as a theorem and must accept it as a
axiom of mathematics geometry.
The beauty of Conjecture #2:: all parabolas and hyperbolas are ellipses or
ovals is that it allows us to define the Function of Circles, Ellipses,
Parabolas and Hyperbolas and Ovals define them all as scalars of the
Polynomial function Y -> x^2. In other words, no more is the circle
equation, the ellipse equation a square root or higher roots. All of these
closed curves are a scalar of Y = kx^2.
All of the circles, ellipses are polynomials and polynomial functions. And
working with them as functions for Calculus, we do the calculus on a "half
of the function" so that each x value has a unique y-value.
*12) Proof the Parabola is a conic ellipse or oval.*
So in my earlier posts I reflected back to the time I was a freshman or
sophomore in college and coming across the Conic Sections and having to
spend time on the Parabola curve, the Y = x^2. And the page 185, of Fisher
& Ziebur, Calculus & Analytical Geometry, 2nd edition, 1965 shows the
function Y = x^2 and it haunted me. Haunted me but not able to do anything
about it, until now.
What haunted me was my mind could not grasp that the curve would follow me
as I walked down the x-axis. So I reach 100 and sure enough 100^2 is y.
Somehow, my mind told me that at a distance from 0 on the x-axis, if I
looked up, there would no longer be the parabola curve. That was what
haunted me.
And now I can fully answer that the haunting was legitimate, in that the
true numbers of mathematics are not the Reals, but instead, Decimal Grid
Systems. So in the 10 Decimal Grid System, I need to walk only to 3.1 and
look up perpendicular and still see the parabola, but step on x-axis 3.2,
and look up perpendicular, and my parabola no longer exists. If I go to the
100 Grid System, my parabola disappears at 10. And if I wanted to attach
another parabola, identical to this one on its loose open ends, I form a
ellipse. And I use the formula of Y = x^2 to characterize the half ellipse
or the full ellipse.
Now the same fate happens to a circle, or hyperbola or a ellipse, as I walk
on the x-axis, at some point on the x-axis, the curve ceases to have a
y-point. In Old Math, they thought a parabola and hyperbola can follow you
forever on the x-axis and deliver you a y-point. Not true, for when
Geometry is all Discrete and the numbers are Decimal Grid Systems, then the
hyperbola, parabola, circle, ellipse, oval, all end at some x-axis point
where no longer there is a y-axis point.
And this is all due to the fact that Conics, true conics are base to base
<> and not the silly apex to apex >< or the single gone infinite in one
direction.
Old Math never understood the Conics started by the Ancient Greeks, and in
fact, Old Math was so so stupid on Conics that by 2016, when AP discovered
the slant cut of right circular cone is a Oval, not the ellipse, it is
really remarkable of the stupidity of Old Math Geometry, that no noticed it
is a oval, not an ellipse. I suppose before 2016, every working
mathematician was too eager to get to the donuts and coffee in the lounge
room rather than fix geometry mistakes.
Here again, the proof the Parabola and even Hyperbola are smooth curves and
are ellipses is the observation of conics base to base, yields a ellipse or
oval for both Parabola and Hyperbola.
*13) Pi is actually square root of 10 = 3.16... when factoring in
Geometry-of-Motion.*
Exquisite Conjecture #3:: Pi is actually 3.16.. the square root of 10 when
factoring in Geometry of Motion.
Alright, experiments in physics places the neutron at 940MeV, and places
the proton at 938MeV and the muon at 105MeV. The true proton is actually
945MeV as being 8 muons that compose a proton with an additional muon
inside the other 8 doing the Faraday law.
So experimentally we have Sigma Error for proton as 945/938 as 0.7% Sigma
Error. For the neutron which is a capacitor particle, a parallel capacitor
that is the outer coating of a proton and collects the electricity produced
by proton + muon and is a growing particle from 1 eV all the way to 945MeV,
for neutron the Sigma Error is 945/940, sigma error of experiments
measuring the neutron is 0.5%.
Now previously in Conjecture #2 I wrote that the parabola was actually a
closed loop circuit curve once we have Conics base to base <> and not the
stupid apex to apex ><. I also wrote that because the true numbers of
mathematics are __not the Reals__ but are the Decimal Grid Systems of 10
Grid, then 100 Grid then 1000 Grid etc etc.
True Mathematics is Discrete Geometry, just as true physics is Quantum
Physics of holes and gaps in between one number and the next true number.
In true physics and true mathematics, there is ____ no continuum ____
instead, there is discreteness.
The Sigma Error if pi were really 3.16... the square root of 10 rather than
3.14.... would be 3.16/3.14 = 0.6%.
Recapping: sigma error in proton measure is 0.7%, in neutron measure is
0.5% and in pi measure is 0.6%. In between proton and neutron sigma error
is pi sigma error.
Sigma Error is simply the measurement error due to biology trying to
measure a piece of the Universe and the error of measurement that creeps
into the measurement. When we measure we try to isolate the entire rest of
the Universe, and we can come close to the true value but some of the
Universe background creeps into the measurement process. For example, the
true muon rest mass is exactly 105MeV but as we measure it, we get outside
0.5MeV Dirac magnetic monopoles that impinge upon the measuring processes
and end up with 105.5MeV. Sigma Error is a outside influences on the
measurement itself.
So in this conjecture of mine, Conjecture #3, I am saying the true Pi of
mathematics is actually square root of 10 as 3.16... and not 3.14....
And what I need to prove this conjecture is the idea that Geometry is in
Motion with a Pointer Marker. Just like the Cycloid has a Pointer-Marker on
the edge (rim) of the circle. But the mistake of Old Math is that their
Pointer-Marker was allowed to be in motion also, along with the circle
being in motion. And because they allowed these 2 motions, Old Math cycloid
wave turns out ____ not to be a half ellipse___.
In new true geometry, our Pointer Marker is always a Finite Length. In 10
Grid that length is 0.1 length. In 100 Grid that finite length of the
Pointer Marker is 0.01, in 1000 Grid it is 0.001, etc etc.
And the Pointer Marker for Geometry-in-Motion must always be at a
Perpendicular to the surface on which the circle is moving over. The
Perpendicular requirement comes from the Physics of Electromagnetism.
Magnetic field is always perpendicular to Electric field.
So, well, say we are in 10 Grid with Conics and dealing with the Parabola Y
= x^2 in 1st Quadrant Only. As we move from 0 to 0.1 to 0.2,... out to 3.0
we still have a y value for our boundary is 10. Now we move to 3.1 on
x-axis and we still have a y-value for our parabola still exists. But now
we move to 3.2 on x-axis, and loo and behold, our parabola no longer exists
in mathematics, for it ran into a number that no longer exists in 10 Grid.
To prove Conjecture #3, I need to show that motion in Decimal Grid Systems
stops the existence of the Parabola, not at 3.14.... but at 3.16.... the
square root of 10. All because of perpendicular requirement for Geometry in
Motion.
I believe I see that Motion as being the motion of 10 Grid goes to 100 Grid
and the walking of the X-axis and the y-value is the perpendicular
component of x value.
Does this make sense, commonsense, that the number 10 as square root is
actually pi? Yes it makes common sense for the probability function of
physics is based upon square root. What is called psi and psi squared.
At one time I collected the digits of square root of 10 and am curious
where the first time it has three zero digits in a row. For 3.14159.... its
first time is the 604th place value to the right of the decimal point.
I spot a row of three zeros 85000 and estimate the place value to be 80 x 7
= 560 place value estimate. Not far off from 3.14.... of its 604th place
value.
3.
16227766016837933199889354443271853371955513932521
68268575048527925944386392382213442481083793002951
87347284152840055148548856030453880014690519596700
15390334492165717925994065915015347411333948412408
53169295770904715764610443692578790620378086099418
28371711548406328552999118596824564203326961604691
31433612894979189026652954361267617878135006138818
62785804636831349524780311437693346719738195131856
78403231241795402218308045872844614600253577579702
82864402902440797789603454398916334922265261206779
26516760310484366977937569261557205003698949094694
21850007358348844643882731109289109042348054235653
40390727401978654372593964172600130699000095578446
31096267906944183361301813028945417033158077316263
86395193793704654765220632063686587197822049312426
05345411160935697982813245229700079888352375958532
85792513629646865114976752171234595592380393756251
25369855194955325099947038843990336466165470647234
99979613234340302185705218783667634578951073298287
51579452157716521396263244383990184845609357626020
31676804240795894693424781414580651430453325889714
46769311137592404705077018546043927212835894192143
79843263432294100698417738335607269111071255492745
61841707758654442076025678341820374148294554615347
20993410591702356226115911404732754291627011270301
Now, do not get me wrong, New Math does not throw out 3.14... but keeps it
along with 3.16.... and keeps them both as where pi hovers as a quantum
duality. Sometimes you use 3.14... other times you use 3.16... Much like
particle wave duality, sometimes the wave is better, sometimes the particle
is better.
So I trimmed and cleaned up the square root of 10 with 50 digits per row.
The 602th digit is a 0 and the 1208th digit is a 0. For pi as 3.14... the
602th, 603rd, and 604th digit were all zeros as well as the 1208th digit
was a 0.
Now in pi=3.14... it was evenly divisible by 120 = 5! at the 604th and the
1208th place value. And stunningly remarkable that pi= 3.14... is evenly
divisible by 6! at 1208th with its solo 0 digit.
So here I have to ask computers to check and see whether pi = 3.16.... if
it is evenly divisible by 120 at its 602th digit of a solo 0 and at 1208th
digit of its solo 0 ??? And also, see if either 602th and 1208th digit of
solo zero is evenly divisible by 6! = 720 ?? If so, that would be
astoundingly exquisite, for it would be proof that pi is really both
3.14... and simultaneously 3.16.... as quantum duality pi.
3.
16227766016837933199889354443271853371955513932521
68268575048527925944386392382213442481083793002951
87347284152840055148548856030453880014690519596700
15390334492165717925994065915015347411333948412408
53169295770904715764610443692578790620378086099418
28371711548406328552999118596824564203326961604691
31433612894979189026652954361267617878135006138818
62785804636831349524780311437693346719738195131856
78403231241795402218308045872844614600253577579702
82864402902440797789603454398916334922265261206779
26516760310484366977937569261557205003698949094694
21850007358348844643882731109289109042348054235653
40390727401978654372593964172600130699000095578446
31096267906944183361301813028945417033158077316263
86395193793704654765220632063686587197822049312426
05345411160935697982813245229700079888352375958532
85792513629646865114976752171234595592380393756251
25369855194955325099947038843990336466165470647234
99979613234340302185705218783667634578951073298287
51579452157716521396263244383990184845609357626020
31676804240795894693424781414580651430453325889714
46769311137592404705077018546043927212835894192143
79843263432294100698417738335607269111071255492745
61841707758654442076025678341820374148294554615347
20993410591702356226115911404732754291627011270301
I had better show the pi= 3.14.... digit sequence and correct my mistake
the three zeroes are in the 601st, 602nd, and 603rd place value. The 1208th
is a zero digit.
And pi= 3.16.... we have the 602nd and 1208th place value is a 0 digit.
3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510
58209 74944 59230 78164 06286 20899 86280 34825 34211 70679
82148 08651 32823 06647 09384 46095 50582 23172 53594 08128
48111 74502 84102 70193 85211 05559 64462 29489 54930 38196
44288 10975 66593 34461 28475 64823 37867 83165 27120 19091
45648 56692 34603 48610 45432 66482 13393 60726 02491 41273
72458 70066 06315 58817 48815 20920 96282 92540 91715 36436
78925 90360 01133 05305 48820 46652 13841 46951 94151 16094
33057 27036 57595 91953 09218 61173 81932 61179 31051 18548
07446 23799 62749 56735 18857 52724 89122 79381 83011 94912
98336 73362 44065 66430 86021 39494 63952 24737 19070 21798
60943 70277 05392 17176 29317 67523 84674 81846 76694 05132
00056 81271 45263 56082 77857 71342 75778 96091 73637 17872
14684 40901 22495 34301 46549 58537 10507 92279 68925 89235
42019 95611 21290 21960 86403 44181 59813 62977 47713 09960
51870 72113 49999 99837 29780 49951 05973 17328 16096 31859
50244 59455 34690 83026 42522 30825 33446 85035 26193 11881
71010 00313 78387 52886 58753 32083 81420 61717 76691 47303
59825 34904 28755 46873 11595 62863 88235 37875 93751 95778
18577 80532 17122 68066 13001 92787 66111 95909 21642 01989
38095 25720 10654 85863 27886 59361 53381 82796 82303 01952
03530 18529 68995 77362 25994 13891 24972 17752 83479 13151
55748 57242 45415 06959 50829 53311 68617 27855 88907 50983
81754 63746 49393 19255 06040 09277 01671 13900 98488 24012
85836 160
Maybe the title of this book should be New True Geometry starting with
cycloid correction and Geometry of Motion// math research by Archimedes
Plutonium
3.
16227766016837933199889354443271853371955513932521
68268575048527925944386392382213442481083793002951
87347284152840055148548856030453880014690519596700
15390334492165717925994065915015347411333948412408
53169295770904715764610443692578790620378086099418
28371711548406328552999118596824564203326961604691
31433612894979189026652954361267617878135006138818
62785804636831349524780311437693346719738195131856
78403231241795402218308045872844614600253577579702
82864402902440797789603454398916334922265261206779
26516760310484366977937569261557205003698949094694
21850007358348844643882731109289109042348054235653
40390727401978654372593964172600130699000095578446
31096267906944183361301813028945417033158077316263
86395193793704654765220632063686587197822049312426
05345411160935697982813245229700079888352375958532
85792513629646865114976752171234595592380393756251
25369855194955325099947038843990336466165470647234
99979613234340302185705218783667634578951073298287
51579452157716521396263244383990184845609357626020
31676804240795894693424781414580651430453325889714
46769311137592404705077018546043927212835894192143
79843263432294100698417738335607269111071255492745
61841707758654442076025678341820374148294554615347
20993410591702356226115911404732754291627011270301
So here I am asking the computers to see if the 602nd place value digit of
0 is evenly divisible by 5! = 120 and 6! = 720.
Then, whether the 1208th place value 0 is evenly divisible by 120 and 720.
For Pi the 601th, 602nd, 603rd place value was evenly divisible only by 120
not 720, yet amazingly the solo 0 in the 1208th place value of pi was
evenly divisible both by 120 and 720.
I need to know if square root of 10 follows a similar characteristic.
Also while I have a computer working on this division
> 3.
> 16227766016837933199889354443271853371955513932521
> 68268575048527925944386392382213442481083793002951
> 87347284152840055148548856030453880014690519596700
> 15390334492165717925994065915015347411333948412408
> 53169295770904715764610443692578790620378086099418
> 28371711548406328552999118596824564203326961604691
> 31433612894979189026652954361267617878135006138818
> 62785804636831349524780311437693346719738195131856
> 78403231241795402218308045872844614600253577579702
> 82864402902440797789603454398916334922265261206779
> 26516760310484366977937569261557205003698949094694
> 21850007358348844643882731109289109042348054235653
>
See if those first three 0 digits of sqrt10 at the 555th, 556th and 557th
place value are evenly divisible by 120 and 720?
Why am I so concerned about being divisible by 120 and then 720? I am
concerned that at infinity borderline we need the existence of Regular
Polyhedra-- the tetrahedron, cube, octahedron, dodecahedron and
icosahedron. At infinity we must have the angles of these figures existing
and those angles exist if at the infinity borderline it is evenly divisible
by 5 factorial = 120 and 6 factorial = 720.
I know pi = 3.14159.... is evenly divisible by 120 at 1*10^-604 place value
and that pi is evenly divisible by both 120 and 720 at the 1*10^-1208 place
value, the algebraic completeness of infinity. I have no idea if 3.16...
follows a similar pattern and need computers to do the divisions. Instead
of computers messing up geometry, computers can shine in algebra and
numbers with their fast speed division.
*14) Conic Sections in 2nd dimension.*
I am still in awe over Conjecture #2 about the Parabola is actually a
closed loop circuit when Conic Sections are base to base <> for all
parabolas are either a ellipse or oval but one case is a circle in base to
base <>. All hyperbolas are either ellipse or oval.
There is one and only one Euclidean straightline geometry figure, the
parallelogram, a cut from apex to apex perpendicular with base.
The drawings alone offer the proof.
And so far, I have just offered identical cones put base to base, with one
layer of circle removed from one cone base so the configuration remains a
Smooth curve.
But I can also apply two comes that are dissimilar and connect them base to
base, much like what the Oval in 2D is, two dissimilar ellipses joined at
the midsection.
But the conic sections base to base that I think are the most important of
all are those that are very long and super thin for they are pencil
ellipses that which Light Waves of physics are pencil ellipses.
We do math, not so much for math sake, but for what it gains in physics.
Conic Sections in 2D with 3D being base to base <> cones.
Alright, there exists a theory of Conic Sections in 2D as there is in 3D
with bases of cones joined. And the result of 3D conic sections is to give
all possible Smooth curves that exist-- the circle, the ellipse, the oval=
2 dissimilar ellipses joined at a midsection. Notice I have no parabola nor
hyperbola for they are in base to base conics either a circle, ellipse or
oval.
The silhouette of Conics base to base is a quadrilateral, a parallelogram.
When the cones are of a equilateral right triangle base to base on
hypotenuse forms a square. When cones are right triangles with base to base
on hypotenuse and not equilateral are rectangles. When cones are right
triangles and base to base not on hypotenuse can be a triangle or
quadrilateral parallelogram.
Now a conic section in 2D is a straightline cut. And here a cut leaves two
figures whose summation of sides of these two figures can be no more than
8. Some sums are 7 or less.
In 3D we do not have sides to cones. And the cut is a planar cut. We
inspect and study the after-result of the planar cut and see if it is a
circle, ellipse, or oval. Again, a parabola and hyperbola no longer exists
in cones base to base.
In 2D conics, is it possible to leave a remnant that is a six sided figure?
I see plenty of triangles 3 sided figures and the other remnant as 5 sided
figure. But I see no triangle as one remnant and the other as 6 sided. This
again supports the idea that the sum of the two remnants cannot exceed 8.
Now if we had a single solo triangle in 2D conics, the two remnants cannot
exceed 7.
Alright, there is no good reason to bring up this subject of 2D conic
sections, unless they are useful to physics or hard science.
Here I believe the use of 2D conic sections is essential to Light Waves of
Physics as electricity magnetism is a surface phenomenon-- 2D, for electric
current moves along the surface. Pencil ellipses are Light Waves and these
are 2D geometry.
The triangle is the cone analog from 2D to 3D. And apparently the triangle
is the most fundamental and primitive closed figure. So we can expect the
cone be the most fundamental and primitive 3D closed figure.
Alright I have a collection of Staedtler math tool products of triangles,
compass, protractor etc. And the right triangles made of plastic are
rounded on the vertices.
So the question becomes how long and thin of a right triangle can we
become. Just in the 100 Grid System what sort of length of a right triangle
can we get whose leg is 0.01 and 0.02 vertex a distance length apart of
0.01.
This immediately leads into the question does every parallelogram determine
a unique ellipse? Then we construct the parallelogram first whose minor
axis has a length of 0.01. And see how far the major axis can be in 100
Grid. If we allow 1000 Grid or higher Grids we see how long the major axis
can go and determine unique pencil ellipses.
So yes, 2D conic sections are important.
*15) Three non-collinear points determine a unique plane in space.*
Now I pride myself on logic above all else in science. And so the next big
topic of conversation is that 3 arbitrary non-collinear points in the plane
form a unique circle, and from those 3 arbitrary points forming a unique
circle, also forms a unique ellipse. This is proven true in the next
several chapters.
But before I get to those proofs, I need to, as a logician start with the
ultimate first discussion of 3 arbitrary points in space and what can be
said of their forming some unique figure of geometry?
The answer here is that 3 arbitrary points in space, non-collinear forms a
line segment with 2 of those arbitrary points leaving the 3rd point
sticking out from the line segment. And now a unique plane in this space is
formed that contains both the line segment of the 2 points and the 3rd
point sticking out.
And one can picture the 2 points forming a line segment then the end points
of the line segment connecting up with the 3rd point forming a triangle.
And this triangle determining a unique plane in which it lies in that plane.
Now I need not prove that, nor anyone else need not prove that 3 arbitrary
points not-collinear form a unique plane. Because that is a axiom or
postulate of mathematics. A axiom is a statement that is fundamentally true
and obvious and apparent that no proof is necessary.
So I just wanted to write this as the starting point of ideas concerning
how many points are necessary to form a unique circle and then how many
necessary to form a unique ellipse in the plane.
Starting with the idea that 3 arbitrary points, not collinear in Space
determine a unique plane.
*16) Takes only 3 arbitrary points to determine a unique ellipse, not Old
Math's 5.*
The math literature says five points are needed to uniquely define a
ellipse. If true then 4 points of the vertices of a parallelogram
determines a ellipse, several ellipses but not a unique one.
I changed my mind. I am going to wade into this debate of how many points
determine a unique ellipse, seeing that 3 non-collinear points determine a
unique circle.
--- quoting a Google Search of the math literature---
Introduction. Five points are required to define a unique ellipse.
5-Point Ellipse | Lee Mac Programming
Lee Mac Programming
http://www.lee-mac.com › 5pointellipse
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Stack Exchange
Question
I recently asked a question on this forum regarding why 3 points guaranteed
the presence or absence of a unique equation representing a specific
circle. (link here What do "3 different points" have to do with linear
dependence in determining a unique circle?) Shortly after this, I came
across a question in my book that provided a picture of 4 red dots (image
below) and asked, "How many ellipses do these 4 red points define". Having
read the comments on my post with the circle, I thought that this was
fairly straight forward. I chose " 1 ". This was wrong. The answer was
infinite. This caught me as surprising as I didn't think of the equations
for a circle and an ellipse as differing by much beyond a scaling factor
for each quadratic term. I know that the general equation for an ellipse is
as follows: (x−ha)2+(y−kb)2=1 The only thing I can think of is that because
of the added scaling factors, there are now technically two additional
unknowns (for a total of 4 different unknowns... h, a, k, and b), and
therefore I need 4 points to specify an unique ellipse. However, I thought
to myself again, even if the ellipse is not centered at the origin, if all
4 given points happened to coincide with the intersection between the major
axis and the ellipse and the minor axis and the ellipse, then certainly
that would specify an unique ellipse. If this is true, then why does the
arrangement of the points matter in determining whether or not an unique
ellipse is specified? Visual explanations would be greatly appreciated!
Answer · 26 votes
The equation (x−ha)2+(y−kb)2=1 is the equation for an ellipse with major
and minor axes parallel to the coordinate axes. We expect such ellipses to
be unchanged under horizontal reflection and under vertical reflection
through their axes. In this equation, these reflections are effected by x↦2h−x
and y↦2k−y. This means, if all you have is one point on the ellipse and the
three reflected images of this point, you do not have 8 independent
coordinates; you have 2 and uninformative reflections forced by the
equation. We can see this by plotting two ellipses at the same center (same
h and k), intersecting at 4 points, with, say, semiaxes of length 1 and 2.
These clearly have four points of intersection. But as soon as you know an
ellipse is centered at the origin and contains any one of the four points
of intersection, by the major and minor axis reflection symmetries, it
contains all four. This is still true if you use generic ellipses, which
can be rotated. Remember that the reflecti…
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Studysmarter.us
Question
Three non-collinear points determine a unique circle. Do three
non-collinear points determine a unique ellipse? If so, explain why. If
not, provide three non-collinear points that are on two distinct ellipses.
Answer · 0 votes
No, three non collinear points does not determine a unique ellipse.For
example :-Graph the following two ellipses :-x24+y216=1 and x216+y+224=1 as
following :-There are three non collinear points 0,-4,-1.996,-0.267 and
1.996,-0.267 and these points are on two distinct ellipses. So these three
points 0,-4,-1.996,-0.267 and 1.996,-0.267 are required points show that
three non collinear points does not determine a unique ellipse as these are
lies on two different ellipses.
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Quora
Question
Is it possible to define an ellipse (and only one) from only two points and
its center (or one focal point)? How can it be done?
Answer · 6 votes
Assume (for simplicity) that the center is the origin. Any conic (ellipse
or otherwise)* centered at the origin must satisfy the equation
[math]Ax^2+Bxy+Cy^2=1\tag*{}[/math] for some choice of A, B, C. With two
given points—say, [math]P(x_1,y_1)[/math] and [math]Q(x_2,y_2)[/math]—that
leaves one degree of freedom; that is, we must make one additional “free”
(arbitrary) choice before we can determine the coefficients. For example,
if we decide to choose a specific value of B, we can then solve for A and C
to satisfy the system [math]\begin{bmatrix} x_1^2 & y_1^2\ x_2^2 & y_2^2
\end{bmatrix} \cdot \begin{bmatrix} A\ C \end{bmatrix} =\begin{bmatrix}
1-Bx_1y_1\ 1-Bx_2y_2 \end{bmatrix}\tag*{}[/math] for which the solution is
[math]\begin{bmatrix} A\ C \end{bmatrix}
=\dfrac{1}{x_1^2y_2^2-x_2^2y_1^2}\begin{bmatrix}y_2^2 & -y_1^2\ -x_2^2 &
x_1^2\end{bmatrix} \cdot \begin{bmatrix} 1-Bx_1y_1\ 1-Bx_2y_2
\end{bmatrix}\tag*{}[/math] We can see that A and C depend on B, so in
general, the degr…
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Quora
Question
What are the conditions that determine whether a point lies inside, outside
or on an ellipse?
Answer · 1 vote
An ellipse in the plane, orthogonal to the axis, looks like a(x-h)^2 +
b(y-k)^2 = 1. If you pick some point (h,k), plug it into the left
expression. If you get 1 it's on the ellipse, if you get less than 1 it's
inside, and if you get greater than 1 it's outside.
More
Study.com
Question
Any three distinct points A,B,CA,B,C in space determine a unique plane.
True False Explain.
Answer · 0 votes
Consider three points that all lie on the same line, so that the points are
distinct. Note that there are an infinite number of planes that pass
through these three points, i.e. that contain the line. We can draw any
one, then spin it around the line (which basically acts like an axis) to
create the others. Thus any plane determined by these three points is not
at all unique, and the statement is clearly false.
More
Stack Overflow
Question
How to find the equation for an ellipse from 3 points using python
Answer · 4 votes
Sounds like a fun problem! If your 3 click points are in the same quadrant
then one of the angles of the triangle defined by those points must be
obtuse. Call that B and the other two vertices A and C. The general
equation for an x-y oriented ellipse has 4 parameters in it. Substituting
the x and y coordinates of A, B and C into the ellipse equation will give
you three equations. You need to come up with a 4th. You have the freedom
to pick that 4th point in some way that makes sense in the context of your
game. Here's a SymPy code snippet that assigns an x value to the center of
the ellipse based on A, B and C and a parameter 'f'. def e3(p1, p2, p3, f):
t = Triangle(p1, p2, p3) a = t.angles for p in a: if a[p] > pi/2: break
else: return while p != p2: p1, p2, p3 = p2, p3, p1 pts = A, B, C = p1, p2,
p3 # perpendicular line to AB passing through A perp = Segment(A,
B).perpendicular_line(A) # intersection between that and the perpendicular
to that # which passes through C m = perp.perpen…
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Study.com
Question
Determine the points at which dy/dx is zero or does not exist to locate the
endpoints of the...
Answer · 0 votes
Differentiating both sides of the equation
x2+4y2+6x−16y+9=0x2+4y2+6x−16y+9=0 with respect to xx gives
2x+8ydydx+6−16dydx+0=0⟹dydx=2x+616−8y.2x+8ydydx+6−16dydx+0=0⟹dydx=2x+616−8y.
When x=−3x=−3, the slope dydxdydx of the tangent line is 00, namely, the
tangent line is horizontal. Plugging x=−3x=−3 into the equation of the
ellipse gives y2−4y=y(y−4)=0y2−4y=y(y−4)=0, so we have co-vertices
(−3,0),(−3,4).(−3,0),(−3,4). When y=2y=2, the slope dydxdydx of the tangent
line is ∞∞, namely, the tangent line is vertical. Plugging y=2y=2 into the
equation of the ellipse gives x2+4x−21=(x+7)(x−3)=0x2+4x−21=(x+7)(x−3)=0,
so we have vertices (−7,2),(3,2).(−7,2),(3,2).
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Stack Overflow
Question
Error: Too few points to calculate an ellipse with 3 points? - R
Answer · 5 votes
I've faced the same problem. The solution is to use geom_mark_ellipse from
ggforce package. It's possible to create an ellipse around 3 points (even
around 1 point). So, the workflow should be as follows: library(factoextra)
library(ggforce) data(iris) iris2<-iris[c(1:3,51:53,101:103),] # 3 points
for each factor res.pca <- prcomp(iris2[, -5], scale = TRUE)
fviz_pca_ind(res.pca, label='none',alpha.ind = 1, habillage=iris2$Species,
repel = TRUE, # Don't use default Ellipses!!!! # addEllipses = TRUE,
invisible='quali') + # ADD ggforce's ellipses
ggforce::geom_mark_ellipse(aes(fill = Groups, color = Groups)) +
theme(legend.position = 'bottom') + coord_equal()
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MathOverflow
Question
Can an ellipse's center be determined from a perimeter point's coordinates?
Answer · 3 votes
To elaborate on J.M.'s comment, your parameters d and ϵ determine the size
and shape of an ellipse, and ψ determines its orientation in the plane, but
once those are set you're still free to rigidly translate the ellipse
(without rotating it), which means the point (x1,y1) can be anywhere along
its perimeter. If you replace ψ with θ, you're now allowing yourself to
rotate the ellipse and then place it up against a specified tangent line.
(If you keep ψ and add θ, you've got it down to two possibilities, one on
either side of the tangent line, but at this point you've got six
parameters instead of five.) So all in all, the answer to your question is
no, there's no way to express (x0,y0) uniquely in terms of d, ϵ, ψ (or θ),
x1 and y1. You might, however, be able to solve for the ellipse whose
center is closest to the origin. But that's a different question.
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Answers.com
Question
What two points define the shape of the ellipse?
Answer · 0 votes
The two foci are necessary to define the location of an ellipse, but the
shape depends on the eccentricity, which is related to the lengths of the
two axes.
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How many points are needed to uniquely define an ellipse?
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Alright, I suspect, that the math literature on how many points
(non-collinear) in the plane determine a Unique Ellipse is 4 and not the
reported 5. And I further suspect the 5 number was a "so called computer
generated proof" and not a human generated proof. So much of Old Math is
tarnished by having computers involved in the "proof process". Beyond the
initial excitement of the computer proof by Appel & Haken 1976 in 4 Color
Mapping that computer proof is a silly absurd fakery and not a proof at
all. It is touted and hyped as a proof only because too many ignorants
accepted it and they accepted it because it meant growth in money paying
colleges to install a computer in their math department--- money money
money and not truth of math.
I suspect this claim of 5 non-collinear points yet a circle needs just 3
non-collinear points to determine a unique circle (a human generated
proof), is a mistake on relying upon computers to give a geometry proof.
Here is where I think the Computers went wrong. They were fed a square or
rectangle vertex points. Then asked the computer to generate ellipses upon
those 4 vertex points. Of course in the mind's eye one can easily see that
the Computer would make the mistake of having the 4 points have two
different ellipses depending on which axes-- minor or major the ellipse was
running so that two different ellipses perpendicular to one another and
only if a 5th point was introduced do you escape 2 different ellipses.
So AP thinks this 5 points to determine a unique ellipse is a Computer
generated proof in terrible error.
To correct that idea. Instead of using square or rectangle vertices as 4
points, say we tried out using only parallelograms that are not square or
rectangle. Now, ask the computer how many points using a parallelogram do
we need for a unique ellipse? And the answer is just the 4 vertices of the
parallelogram produces a unique ellipse.
But I have to let that simmer on my mind for a few days to be sure I myself
did not make a mistake.
And computers in math, are good for calculation, such as is 3.16.... or
3.14.... evenly divisible by 120 or 720 in the nth place digital place
value, is what computers are good for. As for genuine proofs of
mathematics, all computers are horrible at that task, horrible and
laughable as likely the points needed to determine a unique ellipse are the
4 points as vertices of a parallelogram.
Computers have no mind, hence, no logic, hence are never able to do a proof
of mathematics.
And the silliness of AI talk that they will overrun humanity is just
another idiocy pushed by people who want to make more money money money by
selling these machines that fail to do science or math proofs.
The only danger that AI poses to humanity, is when humanity is dumb enough
to install AI into operating its ICBM missiles, and no longer run by humans
in a protocol of countdown. If we install AI to have a key role in
launching ICBM, then we will have a accident where AI initiates a Atomic
War.
*17) AP's proof that 3 arbitrary points determine a unique circle, and the
3 same arbitrary points determine a unique ellipse.*
On Friday, May 19, 2023 at 4:34:47 PM UTC-5, Archimedes Plutonium wrote:
> But I have to let that simmer on my mind for a few days to be sure I
myself did not make a mistake.
>
I think I already have a proof.
A major problem in this geometry is that scatterbrained minds have a
problem with the statement. They realize that if you had a given point and
string of given length they can construct a circle. Likewise a given two
points and length of string they can rotate the string about the 2 points
and generate a ellipse, a unique ellipse.
So this is where bad logic in Statements can fool a person in thinking that
this statement 1 point generates a unique circle, 2 points generate a
unique ellipse. Is the same Statement as 3 arbitrary points in plane, not
collinear produces a unique circle.
A person needs logic training in separating out the difference in Statement
of 1 point and a distance length as opposed to the Statement: 3 arbitrary
non-collinear points determine a unique circle.
But I need to prove that 4 non-collinear points determines a unique
ellipse, and not what the Old Math Literature says it requires 5 points.
I suspect the error in Old Math is they focused on 4 vertices of rectangle
or square and realized they had at least 2 ellipses from those 4 points,
two ellipses, one perpendicular to the other as trading off the major with
the minor axis. Seeing this, they then upped the ante to saying it took 5
points for a unique ellipse. But AP says that if you chose the
parallelogram 4 vertices, you end up with a unique ellipse.
So let me see if I can prove that Statement: Every ellipse is uniquely
determined by the 4 vertex points of a Parallelogram.
Makes me rather want to look up the proof by humans, not silly computers
that 3 arbitrary points determine a unique circle.
Statement: Every ellipse is uniquely determined by the 4 vertex points of a
Parallelogram.
PROOF: Start in the plane with 3 arbitrary points determine a circle. Draw
that circle. Using 2 of those 3 arbitrary points to construct the side of a
rectangle that is inscribed in the circle. 1 of those 2 points will end up
being a vertex of a unique Parallelogram. Now construct the Unique
Parallelogram into being a Unique Ellipse of the 4 vertices of the
Parallelogram.
QED
Comments: I am satisfied that all ellipses are uniquely determined by just
4 points.
Alright the unique circle which generates a unique rectangle in turn
generating a unique parallelogram inside the rectangle with one vertex
point being one of the original arbitrary 3 points that made the circle,
shows us that a cylinder is uniquely determined by 4 points-- the vertices
of a parallelogram.
So far I am using just 1 of the 3 arbitrary points of the generating a
unique circle by those 3 points, using 1 of those points as a vertex of a
unique parallelogram which delivers a unique ellipse from the 4 vertices of
the parallelogram, one of the vertex being a original point of the 3
arbitrary points.
I am looking to see if I can use 2 of the 3 arbitrary points to form the
parallelogram. So far I do not see it. As the 3 arbitrary points that forms
the unique circle can be configured that all 3 points be on one 1/2 circle
or two of the 3 be on 1/2 circle while third on other 1/2 circle. So there
are 2 options for the 3 arbitrary points.
The parallelogram formed inside the rectangle which is inside the circle is
shaped somewhat like this <> as parallelogram inside the rectangle. The
outer vertices, only one has to touch the circle rim.
Now in this rectangle formed from the 3 arbitrary points that constructed a
unique circle, I am going to build a parallelogram inside the rectangle
composed of 2 right triangles and joined at one leg. And then I conjecture
that every ellipse, every single ellipse that exists has a inscribed two
right triangles where the 4 vertices determine a Unique Ellipse.
And this also shows the ignorance and fault and error of thinking ellipse
required 5 points to determine a unique ellipse. Because I can go both ways.
1st Way:
4 vertices of parallelogram not a square or rectangle forms a unique
ellipse.
2nd Way:
Given a ellipse a unique parallelogram inside that ellipse exist.
With the idea that 5 points determine a unique ellipse, there is no 2nd
way, only 1 way.
And the more and more I think of this, I am convinced the error in Old Math
was that they used Computers to hand them a proof of 5 points. They fed
their computer with a square or rectangle and of course, the fools got a
computer answer that the 4 vertices were not enough, so they elevated it to
needing 5 points.
But if the fools had instead, insisted on using only parallelograms not
square nor ellipse, they would have obtained the true proof-- 4 points
necessary for a unique ellipse.
This is the price mathematicians pay when they rely ever more on computers,
geometry mistakes here there and everywhere.
Near the beginning of this book I defined the Smooth Curve as a curve
without vertices, further adding the idea that one point to the next
successor point is arithmetically related to pi = 3.14.... in Discrete
Geometry.
So here I end up with a Ellipse being unique from 4 points as the vertices
of a parallelogram (not a square, not a rectangle).
In essence I end up with a Smooth Curve as ellipse being the joining of 2
right-triangles to form that parallelogram.
So this harkens back to a question of can right-triangles be directly
related to having one number be related to its successor number by pi =
3.14..... Which is asking whether right-triangles have a "pi in them"?
And we see in trigonometry, that the semicircle wave is formed from the
movement of different right-triangles in succession motion of varying
angles. Trigonometry right-triangles build the circle. And now we see that
right-triangles by joining together into a parallelogram build the ellipse.
Every true mathematician in the world, is able to do a proof of a
statement, without looking it up. Without having to go to the literature
and find what the proof was.
In keeping with that idea. Let me prove that Statement: Given 3 arbitrary
non-collinear points in the Plane, prove that a unique circle is obtained.
Alright, mind is blank on how Old Math proved that statement and frankly I
am excited in proving it myself and do not want to look up how Old Math
proved it. I am going to guess that my proof is going to be radically
different from that which is already existing in the literature.
And the reason I am proving it, because I will flip it around, once proven
and show that 4 non-collinear points determines a unique ellipse, counter
to Old Math claim that you need 5.
Statement: Given 3 arbitrary non-collinear points in the Plane, prove that
a unique circle is obtained.
Proof: Well, in an earlier post I broke this down to two cases. One case is
that all 3 points are on a semicircle of the circle and the other case is
where 2 points are on one semicircle and the other on the other semicircle.
I reckon that to prove any 3 arbitrary points determines a unique circle,
is best done by a synthetic geometry proof and not analytic. So I reach for
a compass and take the two cases. What I search for is a center of what is
to be my unique circle. In case one, the 3 points form a curve for the
circle. And if the points are close together it is obvious all 3 are in one
semicircle, and forming a arc of the circle which is going to be a big
circle. But if the 3 points are widely separated it is obvious the center
is between them and drawing a small circle.
So now, in this proof I drop down to looking at just 2 of the 3 arbitrary
points. Looking at just 2 of the arbitrary points and asking myself how
many circles can these two arbitrary point form? Now some will say infinite
number off the bat, immediately. But I think this line of questioning has
never been done before. And so let me draw a picture of the two arbitrary
points.
* *
And let me call the first North and the second South. And immediately I see
one circle for those two points with a center half way between them. Next,
I call the first point pie slice corner 1 and the second pie slice corner 2
and see a center of a huge circle upward from the two points, remembering
that they are not collinear but arced.
Then I see there is another large circle from these two non-collinear but
arced points downward from the picture.
So how many circles do I have from just 2 arbitrary points?? I have 3 of
them, certainly not a unique circle. Are there more than 3 that I can draw?
No, there are only 3 as given a arc between the 2 points.
Now, if I throw a 3rd arbitrary point into that above 2 does the 3rd point
eliminate 2 of the possible circles leaving behind just one Unique circle?
Yes, and depending where the 3rd point is, if in-between the two as forming
the point in the other semicircle, eliminates the two large circles, and if
the 3rd point is a continuation of the arc, eliminates one of the huge
circles and the small circle, leaving only the huge circle as unique.
High School geometry never taught us Synthetic Geometry proofs and I was
able to learn them from a book that heavily discussed synthetic geometry
proofs.
Summary: We look at 2 arbitrary points and they have three possibilities.
*
*
1) A circle between the two points.
2) The two points create a arc in this direction (
3) The two points create a arc in this direction )
Since there are three possibilites, we need a extra third point to
eliminate possibilities and make a unique circle. By adding a 3rd arbitrary
point, it eliminates two of those possibilities leaving a unique circle for
3 arbitrary non-collinear points.
QED
Now I use the above proof that 3 arbitrary non-collinear points in the
plane determines a unique circle to prove that 4 non-collinear points of
the plane determines a unique ellipse. But I must use the already given
proof of 3 arbitrary points produces a Unique Circle and use that unique
circle to render a unique ellipse.
So in the ellipse, the 4 points are the 4 vertices of a parallelogram. And
synthetically let me describe them as North, South, equator_1 and equator_2.
Now suppose I had just 3 of these 4 vertices. And the next question would
be, how many ellipses can be drawn if just 3 of those points were given??
Say I was missing equator_2 point. Then with the remaining 3 that can form
a unique circle. And with those 3 remaining points they are all on the one
side of a semicircle. So I can add on an infinite number of half-ellipses
to the 3 points that form a half-ellipse.
So I need a 4th point to eliminate all the infinite number of half-ellipses
forming a oval. And that 4th point forms a 4th vertex of a parallelogram.
If I add a 4th point, that narrows down all the add on half ellipses to be
a unique ellipse formed from the 4 points. And this is a synthetic Geometry
proof that 4 points non-collinear determine a unique ellipse.
The Old Math literature says 5 points are needed to form the unique
ellipse, and AP says only 4 are needed.
Summary for the 4 points that uniquely determine a ellipse, not 5.
Let us start with 3 arbitrary points and find the unique circle. Then using
2 of the 3 points construct a interior rectangle in the circle. From that
rectangle construct a parallelogram inside the rectangle and using the 1
arbitrary point as a vertex of the unique ellipse. The other 3 vertices of
the parallelogram finish off the unique ellipse. The appearance of the
parallelogram inside the rectangle looks like this <>.
QED
Now I could not leave this topic without looking at what the Old Math proof
was of 3 non-collinear points in plane determines a unique circle. So I had
a look. And the mechanism of proving it is use of the perpendicular
bisector. So you have 3 arbitrary points in plane. You immediately connect
the three points with a straight line. Then take the perpendicular bisector
of the two line segments. Now draw the two perpendicular bisectors so they
meet and this point is the center of your unique circle.
How does it compare with my proof? I would say it is better, provided I can
utilize this same argument on the ellipse. For my proof was designed to
facilitate the proof of unique ellipse is determined by 4 points, not 5.
And my proof was designed to then aid the ellipse proof.
But maybe Old Math's proof of 3 points determine circle aids and
facilitates the 4 points also. So I have any two identical right-triangles
joined to make a parallelogram. Now I find the perpendicular bisectors of
the 4 straightline segments, and sure enough they meet in the 2 foci.
I am proud of myself for coming up with an independent proof, never known
before, which is the mark of every good mathematician-- tell him there is a
theorem proven true and without looking at it-- come up with a self
independent proof.
Comments on the toughness of seeing the difference between 3 arbitrary
points determine a unique circle which is not the same as 4 arbitrary
points determine a unique ellipse. We do not pluck 4 arbitrary points on a
graph paper and proceed in the same manner as we did for circle. Because
for the Ellipse, we need to construct a Parallelogram first, then we say 4
points determine a unique ellipse. So this is a problem of Logical
soundness and consistency, and many a student and math professor will be
annoyed with this soundness and consistency of Logic, that the pattern for
circle uniqueness is not followed for ellipse uniqueness.
On Sunday, May 21, 2023 at 3:42:08 AM UTC-5, Archimedes Plutonium wrote in
sci.math, sci.physics and plutonium atom universe newsgroups:
Huge surprise, huge surprise, 3 arbitrary points in the plane produce a
unique Ellipse.
So I read the Old Math literature on how they prove 3 arbitrary
non-collinear points in plane yield a unique circle from those 3 points
being the circumference points of that circle. The trick there was connect
by straightline and perpendicular bisectors yield the center of the circle.
I know the Parallelogram is the key to the Ellipse formation.
So let us take any 3 arbitrary non-collinear points in plane, just like we
do for the unique circle.
Say we picked these three points arbitrarily
*
*
*
Now we connect them with straightline and have 2 straightline segments
attached.
Now, we construct a Parallelogram from that arrangement. There is a Unique
Parallelogram using those two segments.
And from that unique Parallelogram we easily find the 2 foci as
perpendicular bisectors. And our ellipse is Unique to those two line
segments from 3 arbitrary points of the plane.
I should have realized this before.
Because the question goes for the Unique square given arbitrary points,
unique regular hexagon given arbitrary points. All of them fall back upon
unique circle for you can fit the regular polygons inside a circle.
So, well I complained of Old Maths needing 5 points, and reduced it to 4
points. Now I reduced it further to 3 arbitrary points in the Plane yields
a unique Ellipse.
There is a lot of Logic involved in this proof. A lot of logic to unpack as
to what we mean by "uniquely determined". Especially the Statement needs
crystal clarity of its mechanics. There cannot be any sloppiness. For
instance, if we do not state that the 3 arbitrary points in the plane and
non-collinear, we still have to remove the idea that 2 arbitrary points,
the center of circle and a single point on circumference yields a unique
circle from those two points.
Mathematicians usually do not get this fine grained precision into their
statements and thus have logical loopholes.
And just as the perpendicular bisectors reveal the center of circle of 3
arbitrary points on the circumference of circle, so too does the building
of the other 2 line segments to construct a unique full parallelogram which
contains the 3 arbitrary points on the circumference of the ellipse.
Old Math says we need 5 points to determine a unique ellipse. I initially
thought it was 4 points. Now I see it is really just 3, same as circle.
Alright, I am going to test drive my proof that the unique Ellipse comes
from 3 arbitrary non-collinear points in the Plane. This exercise requires
good graph paper. I like the ten by ten blocks for easy counting for
coordinate points, pencil and slide-rule to make parallelogram of 3
arbitrary coordinate points.
The way the unique circle works from 3 arbitrary points non-collinear, is
the perpendicular bisector of the two line segments the 3 points generate.
The way the unique ellipse works, is that the 3 arbitrary points
non-collinear generates a unique parallelogram, the vertices of which
determine a unique ellipse. But careful in stating that for you have to
remember that a square or rectangle, although a parallelogram, cannot
provide a unique ellipse. So one must state that precisely in the Statement
and in the Proof thereof.
So, you have 2 straightline segments from the 3 arbitrary points, and what
is crucial and key is that those two line segments determine a Unique
Parallelogram.
I select for 3 Arbitrary Non-collinear coordinate points as (0,0), (22,44)
and (70,75) and these are exact points, but as I produce the ellipse and
its foci, I shall be reading from the graph paper and the future points
maybe not precise, but approximates. I hope the center of unique circle
falls also in the 1st Quadrant Only, but if not, not to worry for I just
extend the 1st Quadrant. Also I hope the 2 foci of unique ellipse falls in
1st Quadrant Only, but if not, again, I extend the 1st Quadrant and get rid
of the negative numbers. This is what we do throughout New Math, whenever
negative numbers crop-up, we extend the 1st Quadrant Only to rid ourselves
of negative numbers.
So, using a Slide Rule, I get the 4th coordinate point of a unique
Parallelogram which is (48,31) approximately as all further points are
approximates since I read them from the graph paper.
Now the distance from (0,0) to (22,44) is 48 unit distance.
And the distance from (22,44) to (70,75) is 57 unit distance.
I need that for perpendicular bisectors.
Unfortunately the perpendicular bisectors for the circle center is way in
the negative numbers quadrant and my graph paper ran out to determine that
center. But we can see that a unique circle center is obtained.
But I was able to use the perpendicular bisectors of Parallelogram to
determine the Unique Ellipse and its two focal points. The foci are (21,18)
and (45,60) approximately, ignoring measurement precision.
Splendid, splendid result, and makes commonsense, that a arbitrary 3 points
in plane, non-collinear should determine uniquely both the Circle and the
Ellipse all at once. Because the ellipse is determined uniquely from the 4
vertices of a parallelogram (not a square, not a rectangle).
And so given two straightline segments joined, they produce a unique
parallelogram which in turn yields a unique ellipse. All from starting with
3 arbitrary points in plane, not collinear. And where the parallelogram is
not a square nor a rectangle. The 3 arbitrary points do not form a 90
degree angle, is basically the 3 point proof.
So there may still be "some life" in the Old Math 5-point requirement for a
unique ellipse.
And as for my analysis so far, if the 3 arbitrary points has a 90 degree
angle, I can only envision 2 ellipses formed, one I call a North-South
ellipse and the other I call a East- West ellipse, two different ellipses
not unique. And it is this two different ellipses that I believe propelled
Old Math to say it takes 5 points.
So what can be done to make a unique ellipse for *all* 3 arbitrary points
regardless if it is a square or rectangle, if anything can be done? I am
tempted to define Parallelogram such that 2 identical right-triangles
joined at one of the 2 sides that is not the hypotenuse. This construction
defines the Parallelogram as never being a square nor a rectangle. Or I
could say all arbitrary 3 points not forming a 90 degree angle. But, better
yet, I look for a unique ellipse inside the unique circle formed from the
square or rectangle.
It is notable, that a unique circle from 3 arbitrary points is not hindered
or hampered if the angle is 90 degrees. But an ellipse formed from square
or rectangle can be either North-South ellipse or East-West ellipse.
Another way out is to see if a circle formed from square or rectangle
posses a unique ellipse to that circle formed by square or rectangle.
So here I see where Old Math came up with their 5 point requirement, for
they were stymied and set off course by the square and rectangle.
Still some life left in Old Math's 5 point requirement for unique ellipse.
If the 3 Arbitrary points do not form a 90 degree angle then a unique
ellipse is formed by the Parallelogram (a parallelogram that is not a
square nor rectangle). But if the 3 arbitrary points do form a square or
rectangle then we no longer have a unique ellipse for we can have a
north-south ellipse on the 4 vertices or a east-west ellipse on the 4
vertices.
So unless I can find a unique ellipse to that square or rectangle forming a
unique circle, my 3 arbitrary points have to be conditional on no right
triangle.
So, well, if I define a Parallelogram as every 2 joined identical
right-triangles into a 4 sided figure, joined not at the hypotenuse, and
joined not to form a larger triangle, but joined at the side (the leg of
right-triangle) in which you have 4 vertices that forms a parallelogram
___without a 90 degree angle___. Then all of these figures form a unique
ellipse in the plane from starting out with 3 arbitrary non-collinear
points.
That is the stipulation. But, if the 3 arbitrary points when connected does
equal a 90 degree angle, thus forming a square or rectangle, then, either I
find a unique ellipse from the unique circle, or forced to use Old Math's 5
points to obtain a unique ellipse, a 5th point to distinguish between
north-south ellipse or east-west ellipse.
At this moment in time, I suspect I can find a unique ellipse from the
unique circle formed from square or rectangle, and if so, then in general
__All arbitrary 3 non-collinear points in plane__ form a unique ellipse
regardless of what angle the 3 points deliver. If not, the theorem needs to
restrict the definition of Parallelogram to not include a 90 degree angle.
So I can see why Old Math thought 5 points was necessary to handle 90
degree angles in their analysis. For a square and rectangle, all of them
produce two different ellipses, what I described as a North-South ellipse
and a East-West ellipse from the 4 vertices of square or rectangle. But I
suspect I can utilize the unique circle of those 4 vertices to uncover a
unique ellipse that causes me to say All Parallelograms, including squares
and rectangles provide a unique ellipse from starting with just 3 arbitrary
points.
If not, I restrict the definition of Parallelogram to exclude square and
rectangle.
I wrote on 22May2023, a universal proof as given below.
Alright, so far, I have a proof of a unique ellipse when given 3 arbitrary
non-collinear points in the plane, provided those 3 points do not form a 90
degree angle resulting in a square or rectangle. All other parallelograms
provide a unique ellipse.
So what I am going to do in the case of square or rectangle is look for a
unique ellipse in the circle formed. If I find one, means I can elevate
this theorem to include all 3 arbitrary points in the Plane, non-collinear.
I believe I found the solution. I have four plastic right triangles at my
table. 2 of them 60-30-90 and 2 of them 45-45-90.
One forms a rectangle, the other forms a square.
If I start with the square formed by 3 arbitrary non-collinear and draw the
unique circle which is 3 vertices of the square, now I scoot the one
right-triangle from its joining at the hypotenuse, which does not have all
3 vertices as the 3 arbitrary points, and place a new joining on the leg
which forms a parallelogram, I have a unique parallelogram.
Same is done for rectangle.
Thus, I have found a unique parallelogram that is not a square and not a
rectangle from starting 3 arbitrary non-collinear points in Plane.
Question arises as to whether the unique parallelogram in a circle
constructed from square or rectangle is a north-south ellipse or a
east-west ellipse, is a remaining question. And the answer is, it is
neither. For depending on which of the 4 points is the 3 arbitrary points,
overrides the orientation of north-south or east-west.
Thus I have proven that the same as a circle,-- all 3 arbitrary
non-collinear points in plane yield a unique circle, and those 3 arbitrary
points also yield a unique ellipse.
So, finally, the Statement and Proof are in order.
Statement:: Given any 3 arbitrary non-collinear points in the Plane, they
form a unique parallelogram (including the square and rectangle) and this
unique parallelogram forms a unique ellipse. Same as those 3 points form a
unique circle.
Proof:: Straightforward for 3 arbitrary points forming parallelogram that
has no 90 degree angle and where the foci are perpendicular bisectors
meeting. As for when the 3 arbitrary points form a 90 degree angle we have
to unscramble the two right triangles of square or rectangle and rejoin
them leg to leg to make the unique parallelogram, all from the starting 3
arbitrary points. QED
Comments:: So glad I could make this a Universal Theorem, and eliminate the
square and rectangle encumbrance.
Archimedes Plutonium
May 22, 2023, 5:32:05 PM
to Plutonium Atom Universe
Alright, as it stands I have a proof of a unique ellipse when given 3
arbitrary non-collinear points in the plane, provided those 3 points do not
form a 90 degree angle resulting in a square or rectangle. All other
parallelograms provide a unique ellipse.
So what I am going to do in the case of square or rectangle is look for a
unique ellipse in the circle formed. If I find one, means I can elevate
this theorem to include all 3 arbitrary points in the Plane, non-collinear.
I believe I found the solution. I have four plastic right triangles at my
table. 2 of them 60-30-90 and 2 of them 45-45-90.
One forms a rectangle, the other forms a square.
If I start with the square formed by 3 arbitrary non-collinear and draw the
unique circle which is 3 vertices of the square, now I scoot the one
right-triangle from its joining at the hypotenuse, which does not have all
3 vertices as the 3 arbitrary points, and place a new joining on the leg
which forms a parallelogram, I have a unique parallelogram.
Same is done for rectangle.
Thus, I have found a unique parallelogram that is not a square and not a
rectangle from starting 3 arbitrary noncollinear points in Plane.
Question arises as to whether the unique parallelogram in a circle
constructed from square or rectangle is a north-south ellipse or a
east-west ellipse, is a remaining question. And the answer is, it is
neither. For depending on which of the 4 points is the 3 arbitrary points,
overrides the orientation of north-south or east-west.
Thus I have proven that the same as a circle,-- all 3 arbitrary
non-collinear points in plane yield a unique circle, and those 3 arbitrary
points also yield a unique ellipse.
So, finally, the Statement and Proof are in order.
Statement:: Given any 3 arbitrary non-collinear points in the Plane, they
form a unique parallelogram (including the square and rectangle) and this
unique parallelogram forms a unique ellipse. Same as those 3 points form a
unique circle.
Proof:: straightforward for 3 arbitrary points forming parallelogram that
has no 90 degree angle and where the foci are perpendicular bisectors
meeting. As for when the 3 arbitrary points form a 90 degree angle we have
to unscramble the two right triangles of square or rectangle and rejoin
them leg to leg to make the unique parallelogram, all from the starting 3
arbitrary points. QED
Comments:: So glad I could make this a Universal Theorem, and eliminate the
square and rectangle encumbrance.
Archimedes Plutonium
May 23, 2023, 12:45:54 AM
to Plutonium Atom Universe
Could I have anticipated this final result-- both the ellipse and circle
require just 3 arbitrary points non-collinear to be a unique ellipse and
unique circle?
No, not unless I knew beforehand that the Parallelogram is unique from 3
arbitrary points and a unique parallelogram provides a unique ellipse.
Alright, on 22May2023, I thought I was finished with this proof, closed and
wrapped up. Turns out I had a huge gap in reasoning, for my analysis did
not cover all types of angles for the 3 arbitrary points. And what happened
was that on 26May2023, someone sent me a message showing 4 vertices of a
parallelogram for which it was not a parallelogram of 2 identical
right-triangles joined together leg to leg and thus a unique ellipse
formed. And showing a sketch of 2 ellipses covering the 4 vertices. So this
was a counterexample to my proof unless I filled the gap.
So there were 3 possibilities for sure. The messenger was correct and the
unique ellipse needed 5 points, not AP's 3 points. Or, the messenger had an
error in that one of the two shown ellipses was not an ellipse. Or, AP just
simply missed a whole new class of parallelograms and AP needed to patch
that gap in his proof so far. (Looking ahead, it turns out that AP missed a
whole class of parallelograms and needed to patch his gap.)
Filled the gap I did in the subsequent day, as can be read in this chain of
correspondence.
Archimedes Plutonium
May 26, 2023, 7:26:27 PM
to Plutonium Atom Universe
Alright, I am not done with this book yet for a message sent to me with
diagram that a parallelogram of vertices (-2, -.5), (2, -.5), (-1.5, .5)
and (2.5, .5) create two dissimilar ellipses, one ellipse I call a
north-south ellipse using minor axis and another different ellipse, I call
east-west using the major axis.
Now if this person is correct, then my claim of 3 arbitrary points
determines a unique ellipse is wrong, and Old Math with its 5 points is
needed.
So, this is serious business if my proof is to survive and be the truth, I
must overcome this problem.
So, well, this message sends me the news that I really have not proven the
theorem that all parallelograms not a square, and not a rectangle create a
unique ellipse from their 4 vertex points. I have not proven that, but
hinted of the proof as coming from all Parallelograms are formed by 2
identical right-triangles joined not at hypotenuse, not at legs forming a
square or rectangle but formed from two legs that does create a
parallelogram.
And here I have a parallelogram not formed from 2 identical
right-triangles. Clearly I have a huge gap in my proof and need to patch
it, or my proof sinks.
If the author of this message is correct, then some parallelograms are _not
built_ from 2 identical right-triangles, and these parallelograms have at
least 2 ellipses formed.
So upon seeing this message, my first questions were whether the two
ellipses are actually ellipses. Granted one is certainly an ellipse but is
the second one an ellipse, which the author of the message needs to prove.
So I need to engage and challenge the author of this message, or else the
counterexample ruins my claim and proof.
What I have going for me, is that if I make the right triangles so steep
and thin, that it is impossible to have a ellipse for the minor axis. And
if one such pair of right triangles delivers only one unique ellipse then
all such parallelograms built of 2 identical right-triangles deliver a
unique ellipse.
If that is true, then the mistake made by the author of this message is
that his second ellipse is not actually a ellipse at all, and the burden on
him is to prove it is a ellipse and not from his eyeballing it as a ellipse.
So my burden is to prove a parallelogram not a square not a rectangle
yields a unique ellipse but that some parallelograms are not built from
joining 2 identical right triangles conform to proof. His burden is to
prove his second ellipse is actually a ellipse, not from "I say so".
But, thanks for the challenge. For I see I have more proving to do.
I am confident I will win this battle for I cannot envision a very steep
right triangle say a 89-1-90 can have both a north-south ellipse and also a
east-west ellipse.
The author of this message, his/her parallelogram, is too much shaped like
a rectangle, and the vertices do not mark out the major axis nor minor
axis. In my 2-right triangle parallelograms the 4 vertices mark the major
and minor axes. I suppose we can do some property analysis of some number
factor as to how close a parallelogram is shaped like a rectangle, and if
so, well the parallelograms I am concerned with are looking more like two
needles. And whence it is impossible to form a ellipse to capture both the
minor axis along with major axis. Easy to capture the major axis,
impossible for the minor axes, or vice versa.
So more kilometers to go on this book.
So, yes, AP insists that the author of this message has to prove both are
indeed ellipses. His/her proof that the second figure is a ellipse is
_perhaps_ in error, for he uses focal points. Focal point 1 he describes as
(-1.4898424...., - 0.751094....)
and focal point 2 he describes (1.9898424...., -0.7510948...) with not a
hope of ever proving the second figure is actually a ellipse.
AP's proofs in this book all along is the use of Synthetic Geometry, which
reminds me I need to cite that book where I learned the techniques of
synthetic geometry proofs.
Anyway, AP's proof relies on the idea of making huge needle like
parallelograms, for which one simply sees it is impossible for a second
ellipse to capture 2 of the 4 vertex points.
Archimedes Plutonium
May 26, 2023, 10:19:16 PM
to Plutonium Atom Universe
Alright, I was not prepared for this. The idea that there is a whole new
class of Parallelograms.
I took the square to be 2 right triangles of 45-45-90. Then I take any
rectangle and divide it into 2 right triangles of for example 60-30-90.
(Here the question arises, is the rectangle in general vary through all
angles, and not stick to only 60-30-90?. The answer is definitely varying
in angles. Whereas all squares are 45-45-90 right triangles.)
But there are a entire class of parallelograms which are not formed by
right triangles.
The Rhombus is a parallelogram and what I call generic parallelograms such
as this /____/ which is not two right triangles joined.
But not to worry about AP's 3 point theorem for a Generic Parallelogram, we
take 3 of the 4 vertices. We draw the unique circle around these 3 points.
Next we form a unique isosceles trapezoid from the 3 points inside the
circle, now we cut a piece off one end and transfer to other end of
trapezoid and flip around forming a rectangle. And now we have 2 right
triangles of those starting 3 points of 60-30-90 and now we have a unique
ellipse.
So AP's proof of 3 points still stands true but becoming more complicated.
More complicated as to the angle of the 3 arbitrary points formed and
whether it produces a isosceles trapezoid inside unique circle.
So we start with 3 arbitrary non-collinear points in the plane. We connect
those 3 points by straightline segments. We inspect the angle between the
two straightline segments. If the angle is 90 degrees we have either a
square or rectangle when we draw the unique circle. And both a square or
rectangle yield a 2 identical right-triangles that is transformed into a
parallelogram joined at leg to leg. This parallelogram forms a unique
ellipse. But if the angle of the 2 straightline segments is smaller or
larger than 90 degrees, the figure formed from the unique circle is a
isosceles trapezoid. And here we have to perform more transformations of
turning the isosceles trapezoid into a rectangle or square, then finally
proceed to form the unique ellipse on the starting 3 arbitrary points.
(Actually one of the 3 points is buried in the snipped triangle of
trapezoid to make it a rectangle.)
All because 3 arbitrary points determines a unique circle, and this unique
circle yields a unique rectangle inside the circle, or a piece outside the
circle if isosceles trapezoid, which yields a 2 right-triangle
parallelogram which finally yields the unique ellipse.
So the proof really all hinges on the fact that 2 right triangles when
joined leg to leg that forms a parallelogram, unique parallelogram can only
form 1 unique ellipse.
I am staring at these 2 right triangles on the table and can see no way of
forming more than 1 ellipse. All because 2 of the 4 points are far away
from one another.
In the case of the (-2, -.5), (2, -.5), (-1.5, .5) and (2.5, .5)
parallelogram sent to me by the author-messenger. which is shaped much like
this /____/ one can see a north-south ellipse and also a east-west ellipse
as being an actual ellipse. Much like the two ellipses constructed on any
rectangle, a north-south ellipse as well as a east-west ellipse. But when
you construct the parallelogram with the 4 vertices, two forming the major
axis and other two forming the minor axis, you deny one of the two ellipses
and end up with only a unique ellipse to cover the 4 vertices.
So in that example if we started with just the three points (-2, -.5), (2,
-.5), (-1.5, .5) we construct the unique circle. Now construct the unique
isosceles trapezoid for the angle is less than 90 degrees inside that
circle. Then snip off one end of trapezoid and attach to other end forming
a rectangle. Part of this rectangle sticks outside the circle, but no
problem. Finally, split the rectangle into 2 right triangles and join them
leg to leg to form a parallelogram. And now you have a unique ellipse from
starting 3 points (third point buried in forming rectangle).
So basically, all that AP needs to prove is that any 3 arbitrary point
yields unique circle and if not already a rectangle or square, can fetch a
isosceles trapezoid that in turn fetches a rectangle or square that fetches
2 identical right triangles which then determines a unique ellipse when the
2 right triangles are joined at leg to leg forming a unique parallelogram.
Pretty pretty cool proof.
Now, do all rectangles form a unique ellipse? None do, for all rectangles
have a north-south ellipse and a east-west ellipse. It is when we carve out
the rectangle of 2 right-triangles and join them together, that forms a
unique ellipse.
And we can see why, for as we form the north-south ellipse we capture all 4
vertices, but in the east-west direction of ellipse we cannot capture the
other 2 vertex points.
It is because the 4 vertices of this 2-right-triangle formation, those 4
vertices define the major axis and the minor axis and thus forcing
uniqueness.
The messenger's parallelogram as (-2, -.5), (2, -.5), (-1.5, .5) and (2.5,
.5) is not a parallelogram formed by 2 right triangles, and so it has to
transformed of its three points (-2, -.5), (2, -.5), (-1.5, .5) to a
isosceles trapezoid inside that unique circle, first.
And the above is Analytic Geometry, distinguished by coordinate points,
formulas, functions, and distances using Reals. As compared to Synthetic
Geometry, the axioms of Euclid and no numbers.
It was the book Elementary Geometry from an Advanced Standpoint, Edwin
Moise, 3rd ed, 1989 that I learned these two different ways of doing
geometry.
So what is the status of the claim that (-2, -.5), (2, -.5), (-1.5, .5)
and (2.5, .5) has both a north-south ellipse and a east-west ellipse? It
could well be it has 2 ellipses, but the starting premiss is the 3 points
of (-2, -.5), (2, -.5), (-1.5, .5) which produces unique circle, which in
turn produces unique isosceles trapezoid inside circle, that in turn
produces a parallelogram of 2 right triangles which joined at leg to leg
yields a unique ellipse.
Alright, I was not prepared for this message of a counterexample. The idea
that there is a whole new class of Parallelograms. And exposing a huge gap
in my proof.
I took the square to be 2 right triangles of 45-45-90. Then I take any
rectangle and divide it into 2 right triangles of 60-30-90 and varying
angles such as 50-40-90, etc. Only from these parallelograms built of 2
right triangles is it true that 3 arbitrary points yields a unique ellipse.
Now I have to go back and review this. In the idea that all squares are
45-45-90. But any rectangle not a square can vary, and not all be 60-30-90.
And if the three starting out arbitrary points not-collinear are not
forming a 90 degree angle, I must contend with a isosceles trapezoid and
convert it into 2-right-triangles to obtain that unique ellipse.
Now the angle posed by the author of the message with (-2, -.5), (2, -.5),
(-1.5, .5) and using a protractor, I measure to be 80 degrees
approximately, and whose unique circle has center approx (0, -.8) and
constructing approximately the isosceles trapezoid as its 4th point of
(1.85, -1.6) using just compass.
Now with that isosceles trapezoid I need to snip one end off of a right
triangle and upside down attach to the other end to make a rectangle and
get my 2 identical right-triangles.
Alright, I am grateful of the unnamed author who sent me a diagram of
his/her 3 arbitrary points, then showing the parallelogram fits at least 2
different ellipses. His/her (-2, -.5), (2, -.5), (-1.5, .5) fetching a
parallelogram of vertices (-2, -.5), (2, -.5), (-1.5, .5) and (2.5, .5).
Grateful for this message exposes a gap in my proof.
And this diagram is not what AP says has to be done. For AP says the three
arbitrary points need to go through the Unique Circle construction. And
once through that construction, the 3 arbitrary points construct a Isoceles
Trapezoid using (-2, -.5), (2, -.5), (-1.5, .5). And the 4th point in the
isosceles trapezoid construction is approximately (compass instrument use
to find this 4th point, is (1.85, -1.6) from circle center (0, -.8).
Remember the center of circle is obtained from perpendicular bisectors of
the two line segments from (-2, -.5), (2, -.5), (-1.5, .5) intersecting at
the center.
So I was far far from complete in my proof of 3 arbitrary points in plane,
not-collinear yield not only a unique Circle but yield a Unique Ellipse, by
the use of that unique circle.
Also, another item came up concerning the variation of right triangles for
a rectangle. The idea that different sizes of rectangles offers more than
just a 60-30-90 dissection then joining at leg to leg. So if I take one
unit block 1 by 1, it is a 45-45-90 dissection. A 2 units block of 1 by 2
yields what rectangle, a 1 by 3 units joined together yields what rectangle
that yields what dissected two right triangles? And this would be
legitimate work for Old Math's trigonometry.
As of 27May2023, I am sure a isosceles trapezoid like the 2 identical
60-30-90 right triangles joined at leg to leg yield a unique ellipse. And I
now wonder as I can always take a isosceles trapezoid, cut off one end and
attach to other end to form the rectangle which provides for the 2 right
triangles of the unique ellipse.
So, thanks to the unnamed messenger for I had missing essentials in my
proof.
So I have before me on my math table two identical 60-30-90 right triangles
put leg to leg forming a parallelogram that has only one unique ellipse to
circumscribe the 4 vertices.
And looking at that configuration it is easy to see it can have only one
ellipse.
Now next to the 2 right triangles sits a Isosceles Trapezoid and the only
difference between the trapezoid and parallelogram is the opposite sides of
trapezoid are not parallel but oriented to meet with the same complimentary
angles. And this isosceles trapezoid, once transformed into a rectangle by
lopping off one end and attaching to the other end, preserving our 3
starting points like the parallelogram allows only 1 unique ellipse to
circumscribe the 4 vertices.
Does this make commonsense? Yes of course because I can snip one end of the
isosceles trapezoid and join to the other end and form a rectangle which
thence possesses the 2 right triangles. The 2 right triangles that forms a
unique ellipse.
So, now my proof that 3 arbitrary non-collinear points of plane suffice to
produce a unique ellipse is complete.
And this in turn makes common sense for if 3 points produce a unique
circle, it is common sense that 3 points should produce unique ellipse. For
a circle is a special type of ellipse.
And this ends and fulfills my proof that unlike Old Math with their silly 5
points needed for unique ellipse, AP proves 3 points are all that is needed
for unique ellipse.
Alright, I revise my recent published book 240th by adding more material on
the proof that 3 arbitrary non-collinear points determine a unique ellipse,
much the same as those 3 points determine a unique circle. And here is the
logical deficiency of Old Math professors doing math, little to no logical
marbles. They all recognize 3 points determines the unique circle but lack
the where-with-all to understand if circle is represented by 3 points and
since the ellipse is merely a special type of a circle, that stands to
reason 3 points, not 5 would define a unique ellipse. When you have little
to no logical marbles in the head, you come up with 5 points needed.
My proof relies on the idea that some 4 sided figures, parallelograms
mostly but isosceles trapezoids also, allow for only one of two types of
ellipses to form. What I called north-south ellipse and distinguished from
east-west ellipse.
In my proof, we start with 3 arbitrary non-collinear points and those
points form one of three possibilities. A 90 degree angle formed by the two
line segments, a less than 90 degrees and a greater than 90 degrees. If 90
degrees the 3 points form either a rectangle or square. If less than 90
degrees we have the three points form concave inward (, and if greater we
have concave outward ) for the unique circle. So we draw the unique circle.
And if not 90 degrees we end up constructing a isosceles trapezoid unique
to those 3 points and the unique circle. Here we have to snip off the one
end of the isosceles trapezoid and paste onto the other end leaving us with
a rectangle and yet, still the 3 starting points. From this rectangle we
proceed to dissect the rectangle into 2 identical right triangles and draw
the unique ellipse.
Archimedes Plutonium
8:24 PM, 27May2023
to Plutonium Atom Universe
So we have to ask, how is it that the 2 identical right-triangles from
rectangle or square placed leg to leg forming parallelogram and in addition
the isosceles trapezoid have this ability to have only one unique ellipse
cover its 4 vertices?
The rectangle and square have no such ability for both allow at least 2
ellipses to cover their 4 vertices. The two ellipses I call a north-south
ellipse and a east-west ellipse. So what is the underlying reason for this
property of only one ellipse??
From observation it seems as though the positioning in a 2-right-triangle
parallelogram and in a isosceles trapezoid which is transformed to
rectangle then a 2-right-triangle transform, the positioning of the 4
vertices allows one of the ellipses by negating the existence of the other
kind of ellipse. And this is why the rectangle and square have the two
types of ellipse. For the 2-right-triangles, leg to leg, still keeping the
3 arbitrary points (except in isosceles trapezoid) that the vertices are so
to speak 4 pointed to ensnare the major and minor axis of the unique
ellipse.
When you have the 4 points ensnare other than the major and minor axes,
then you allow for other ellipses to exist on those 4 points.
And this is why the requirement of getting the 3 arbitrary, not-collinear
points to form a unique circle that in turn forms a isosceles trapezoid
which in turn forms a rectangle (or square) is so important. Because once
we fetch that rectangle or square it is turned into 2 identical right
triangles that form 4 vertices of a parallelogram that forms the major axis
and minor axis of a unique ellipse. So so many math persons in favor of the
5 points to form a unique ellipse are empty headed in the realization that
2 identical right triangles placed leg to leg forming a parallelogram forms
a Unique Ellipse, yes, unique ellipse for it forms a unique set of major
and minor axis. When you form a major and minor axis, you cannot cover
those 4 vertices with 2 or more ellipses. You can only have one ellipse, a
unique ellipse from those constraints.
And yet, still, those that believe in 5 points to gain a unique ellipse
cannot understand the logic of 2 identical right-triangles set leg to leg
yields a unique ellipse.
Now, out of fun, let me go in the exact opposite direction of trying to
seek a unique ellipse or unique circle but destroying the circle or
ellipse. I am trying to help this author who sent me this parallelogram.
> > Alright, I am not done with this book yet for a message sent to me with
diagram that a parallelogram of vertices (-2, -.5), (2, -.5), (-1.5, .5)
and (2.5, .5) create two dissimilar ellipses, one ellipse I call a
north-south ellipse using minor axis and another different ellipse I call
east-west using the major axis.
So now let me start out with 3 points of (0,0), (0,5) and (10,0). And let
me pretend they are arbitrary just for illustration sake. We pretend those
are 3 arbitrary points of the plane, and obviously not collinear. And we
know for a fact that as we connect our 3 arbitrary points as line segments
they have a 90 degree angle between them, telling us that the unique circle
formed will complete the 3 points with a 4th point that forms a rectangle
and be the point (10,5) as we obtained the unique circle center from
perpendicular bisectors 2.5 on one line segment and perpendicular bisector
5 meeting at center (5,2.5).
Now here is a beautiful question. Let us say we have our starting arbitrary
3 points of (0,0), (0,5) and (10,0) and say we throw into that a 4th point
of say (10,8) or perhaps a 4th point of (7,7) or something else. The
question is-- does that 4th point destroy the chances of that 4th point and
the 3 starting points from forming a circle, or a ellipse.
Here I am using a 4th point to ask if that 4th point destroys the formation
of a circle or ellipse. And keeping with the non-collinear edict.
So the answer to this question-- the question of destruction of forming a
circle or ellipse, lends insight on our other focus of attention--
sufficient arbitrary points to form a circle or ellipse. Here we want to
know how many arbitrary points destroys the formation of a circle or
ellipse.
Now one could say they are so in tune in picking points that are on the
circle formed from 3 arbitrary points, and so they pick a 4th point and it
happens to fall on the circle, pick a 5 and again happens to fall on that
same circle.
And this is getting into Probability theory. For the question now becomes,
what is the probability of 4 arbitrary points forming a circle? What is the
probability of 3 arbitrary points forming a ellipse and according to AP
theorem it is 100% provided not collinear points. But then the question is
4 or 5 non-collinear points arbitrary what is the chances of forming a
ellipse?
Now I raise these questions to help those that are still confused about
uniqueness and about necessary and sufficient conditions. For these
problems are often mind bending, especially those novices of logic.
*18) Synthetic proof that 3 arbitrary non-collinear points forms unique
ellipse since those points form a unique circle.*
Let me review again what a synthetic proof is. It is a proof without
numbers involved. If numbers are involved it is called analytic proof.
Numbers such as distances, and formulas is analytic. In analytic proofs we
use coordinate systems and points as numbers. In synthetic proof we rely on
axioms and theorems and constructions, and the use of compass, ruler,
protractor.
So the 3 arbitrary non-collinear points is plane produces a unique circle,
because three points determine two connected straightline segments, and
from those two segments we take the perpendicular bisectors until they
intersect. They intersect in a unique point which is the center of a circle
for, from that center all three arbitrary points are on that circle of that
center. A unique intersection as center, a unique circle of radius.
Before the end of May 2023, I thought I had the complete proof of 3
arbitrary points, not collinear in plane produces a unique ellipse. But
with the help of discord and complaint in sci.math, AP begins to notice a
gap in this proof.
The 3 arbitrary points formed a unique circle, and those 3 points formed 2
line segments that now forms a unique square or rectangle or isosceles
trapezoid inside the circle. From those possible 4 sided quadrilaterals all
can be turned into a square or rectangle, and finally 2 identical
right-triangles emerge inside that unique circle. And as we reassemble
those 2 right triangles to form a parallelogram, for which 2 of the
vertices are an ellipse major axis and the other 2 are the ellipse minor
axis. Those 4 vertices cannot be the major and minor axis endpoints of any
other ellipse except for this unique ellipse.
Then by June 3, 2023, AP acknowledges the gap and fixes his proof with
another new figure-- the HourGlass figure, from a further movement of the 2
right triangles.
2 right triangles 45-45-90 in HourGlass figure forms unique circle// 2
right triangles 60-30-90 in HourGlass figure forms unique ellipse. 3
arbitrary non-collinear points in plane forms unique ellipse-- not Old
Math's 5 points
3:10 PM 3May2023
to sci.math
On Saturday, June 3, 2023 at 2:21:39 PM UTC-5, Archimedes Plutonium wrote:
> On Saturday, June 3, 2023 at 5:10:46 AM UTC-5, Archimedes Plutonium
wrote:
> > Alright, I thought I was completely finished with this 3 points
determines a unique ellipse. Turns out I have quite a more distance to run
with it. I had the oversight, or missed-sight of thinking I had both the
major and minor axes when both identical right triangles are placed leg to
leg into a parallelogram. I was thinking both axes surely determines a
unique ellipse, that was an oversight for the 2 right triangles only
determines the major axis and not the minor axis.
> >
> > That means more distance to run on this proof, and the possibility of
failure.
> >
> > But there are strong signs that I will be triumphant because the square
and rectangle and rhombus and then all sorts of parallelograms not 2 right
triangles --- they do not mark out the major axis. Only the 2 right
triangles mark out the major axis. And the others that have 2 or more
ellipses (perhaps that can even be sharpened up to say all parallelograms
have only 2 ellipses to cover their 4 vertices), have no axes involved in
their ellipses.
> >
> > What I am considering is whether the perpendicular bisectors of these 2
right triangles are indeed the foci of the unique ellipse. If that is true,
then indeed the 2 right triangles determine a unique ellipse.
> >
> > Everyone that is thinking of disputing or attacking AP on this issue,
ask yourself first, how much are you relying on Computer Graphics? A math
geometry proof never has a computer enter the picture for the proof itself.
And I am pretty sure the Old Math claim of 5 points required for unique
ellipse in the plane is a computer graphics mindless idiocy.
> >
> Sorry, I made a mistake here, in thinking the 2 right triangles captured
both the Major and Minor axis, but it turns out the 2 right triangles
captured only the Major axis. The minor axis is in-between, straddled on
both sides by the 2 vertex of short legs joined together.
>
> However, all is saved, capturing both Major and Minor axis if I join the
2 right triangles, first by leg to leg, then I slide the one right triangle
down the leg of the other forming a hour-glass
>
> /|
> ..|/
>
> I still have 4 vertices but no longer parallelogram, instead a hourglass
especially with the 45-45-90, not so much with the 60-30-90.
>
> So, many more kilometers to run with this proof.
>
A glitch has turned up in AP's proof, for he thought he captured both Major
and Minor axis. Turns out I have to do a final movement to obtain the
unique ellipse to capture both Major and Minor axes. The final figure is
much like a hourglass, or two right triangles joined vertex to vertex.
Caution, as you make the final movement, the center of the unique ellipse
also appears.
And it appears that the hourglass for the 45-45-90 right triangles ends up
being a Circle, then, all other 2 right-triangles in the Hourglass final
movement ends up being ellipses, unique ellipses because they carry both
the Major and the Minor axes.
P.S. thanks to all those in Germany, and West Germany for trying to
convince AP he was wrong and had nothing, which only forced AP to make a
final HourGlass movement. Thanks. Your strident hatred only helps AP to
further and with higher success. As the late .... always- used to try to
say here in sci.math, but was too shy "eram semper", and .... had to say it
for him.
P.P.S. The only thing left for me is to count how many movements of figures
to form the unique ellipse. Let me see, here, I, start with 3 points to
form a circle. If the circle 3 points has no 90 degrees but is a isosceles
trapezoid, I need to cut and paste the trapezoid into a rectangle. Then a
second movement of cutting rectangle into 2 right triangles leg to leg.
Then a third movement to where the 2 right triangles are a HourGlass
forming the unique ellipse. Counting 3 movements, but see if my starting 3
points is saved and preserved in all these 3 movements???
Alright the glaring mistake I made was after finishing the proof, I did not
go backwards. In math you often can go backwards, not just forwards. So in
this proof, I failed to start with a ellipse then work backwards to see the
2 right triangles that form the unique ellipse.
For the best drawing of this there is the book by Harold Jacobs,
"Mathematics A Human Endeavor", 1970, page 305 of a graphed ellipse. His is
in 4 quadrants which I try to avoid since in AP mathematics there are no
negative numbers, but his coordinate points of major and minor axes are
(0,9), (0,-9), (0,6), (0,-6).
Now I bring this to your attention because that drawing of the ellipse
shows 2 sets of right triangles in a HourGlass pattern.
/|
..|/
And
.....|\
...\|
What I failed to do before now was to keep the 2 right triangles leg to leg
like this
/|
|/
Keeping them as a parallelogram composed of 2 right triangles leg to leg.
When I should have moved those 2 right-triangles to form a HourGlass
geometry figure. For the Hourglass makes the major and minor axes unique.
I should have reviewed the proof and gone backwards. Start with a Ellipse
and look for the 2 right triangles.
And now I see there is a 2, a second Set of 2 right triangles in Jacob's
page 305.
So I had another Movement step of moving the 2 right triangles as a
Parallelogram into becoming a HourGlass figure.
And that then forces the appearance of a Major axis and Minor axis and thus
a Unique ellipse.
Not only forcing the appearance of the Major and Minor axis but clearly
showing the center of the ellipse, a unique ellipse.
One last thing I have to do is see how many of the original 3 arbitrary
non-collinear points of the Plane determine a Unique Ellipse are retained.
Not that missing one of the three points is at all damaging to the proof,
for when I have to cut and paste the isosceles trapezoid when the unique
circle does not give me 90 degrees, in the cut and paste, we move one of
the original 3 arbitrary points and is lost.
But not to worry because in 3 arbitrary points forming a unique circle, we
have to do some movement in perpendicular bisectors to obtain the unique
circle center. And so movement of points is not damaging to the proof, for
the statement of these proofs is that 3 arbitrary points **determine or
force** a unique circle, force a unique ellipse.
So my error over these last few weeks is an error of not going backwards in
the proof argument. Starting with a ellipse and looking for 2 right
triangles that determine the unique ellipse. A mistake I will not make
again.
AP
*19) Computers in geometry often just mess things up and can never be
trusted for geometry proofs, for they are "artwork" in geometry, not
mathematics.*
I would like to end this book on reflections of where computers have
muddied up the waters of geometry and not helped geometry. Where computers
are "artwork" and not math truth.
Old Math guided by computers says 5 points are required for a unique
ellipse, while AP proves 3 points are all that is needed.
This is not the first time I ran into a computer error in geometry. And
this is the big danger of using computers in geometry. Because computers
are programmed to the bias and prejudice and ignorance of its programmers.
We see this in tiling a sphere surface and in 5 points needed for unique
ellipse. What I recommend is get rid of computers out of geometry math.
Computers are art in math geometry, not science. Three big cases I ran into
so far. There are the computers that simulate a slant cut in right circular
cone and come up with a ellipse when in reality it is a oval. All because
of the program is programmed bias, prejudice and outright ignorance. Then
there was the time computers fooled me in the tiling of a sphere, and now
computers making a mockery of unique ellipse needs 5 points.
--- quoting my 152nd published book ---
The 6th Regular Polyhedron-- hexagonal faces at infinity is nonexistent //
Math proof series, book 12
by Archimedes Plutonium (Author) Format: Kindle Edition
Preface: I started this book in September 2021, and not until July 2022,
did I uncover my gross error-- the nonexistence of the 6th Regular
Polyhedron. I so much wanted there to be a 6th regular polyhedron and
looking in the Internet, the world wide web, are many images of a cell of 7
regular hexagons, a central hexagon surrounded by 6 more regular hexagons
tiling a sphere surface. Plenty of these images, but the tipping point for
me is the Goldberg polyhedron, here again the cell of 7 regular hexagons
tiling a sphere surface. And so, using that 7 cell as supporting evidence
of the existence of a 6th Regular Polyhedron, AP proceeds to publish such.
Even though I knew of the University of Utah beware caution web page
stating that a vertex of 3 regular polygons is an angle of 120 +120+120=
360 degrees and thus laying flat as a plane, no bending, hence no tiling a
sphere.
So I published this book in Sept2021, and not until July2022, needing a
coordinate system of points on a sphere for my Ecology book "_Complete
Ecology_ with Generalized Faraday Law and revised food chain // Ecology
science". That I finally realize my mistake-- Uof U completely correct, and
why on Earth did I want to believe Goldberg polyhedron and all those fake
geometry images of regular hexagons tiling a sphere surface. This is a
massive computer problem of our times, in that it is super easy to make
optical illusions in geometry and filling web sites with fake geometry
images.
Well, AP was fooled and fell victim to computer graphics showing where a
sphere surface tiling of a central regular hexagon and surrounded by 6 more
regular hexagons. There are many pictures and images of a sphere tiling on
the Internet of 7 regular hexagons, a central one and surrounded and
encircled by 6 more regular hexagons. There is even geometry of what is
called Goldberg polyhedron with more pictures and images, all deceptive,
all wrong. So this book ends up about the theme of how deceptive computer
imaging can be, and not what AP hoped for-- the existence of a regular
polyhedra with regular hexagon faces.
If it were true that a cell of 7 regular polygons has a bend to it, so that
it can eventually circle around a sphere surface, then my first publication
of this book would have been true. But instead, the truth is the
nonexistence of the 6th Regular Polyhedron.
Cover Picture: is my iphone photograph of a soccer ball of 20 hexagons, 12
pentagons; and a glass ball covered by netting of tiny hexagons. Both
objects I use in experiments of trying to prove the 6th Regular Polyhedron
only it is nonexistent as I eventually found in July 2022.
--- end quoting my 152nd book of science ---
Also I should include the Dandelin fake geometry spheres, computer
animation. The trouble here is that a fake proof of mathematics, for the
slant cut of cone is not a ellipse but a oval, and as to whether a oval has
foci is unknown. In AP's definition of Oval is two dissimilar ellipses cut
and joined at a midsection.
My complaint is that a fakery in mathematics is easily inputed into a
computer that will then further entrench the fake math.
Computers have a tendency to entrench, entrench errors in geometry.
Computers in geometry are more artwork than science of truth. And computers
in geometry are programmed such that bias and prejudice and sheer idiocy
are programmed into what the computer gives as output.
Computers debuted in math in the Appel & Haken fake 4 Color Mapping, 1976
and as the years rolled by, computers have become more unreliable, so much
that they cannot be trusted in math proofs. Another atrocious computer in
geometry, the Hales silly allegation of proving Kepler Packing by
computers. It is that Hales himself programmed his silly bias, prejudices
and flawed ideas into the computer and that is what the computer gave him,
back.
Computers can be stimulus of thought, but cannot be bearers of truth in
math geometry. Computers are quick calculators, not proving machines.
--- quoting my 24th published book ---
World's First Proof of Kepler Packing Problem KPP // Math proof series,
book 3
by Archimedes Plutonium (Author) (Amazon's Kindle)
There has been a alleged proof of KPP by Thomas Hales, but his is a fakery
because he does not define what infinity actually means, for it means a
borderline between finite and infinite numbers. Thus, KPP was never going
to be proven until a well-defined infinity borderline was addressed within
the proof. And because infinity has a borderline means that in free space
with no borderlines to tackle and contend with, the 12 kissing point
density that is the hexagonal close packed is the maximum density. But the
truth and reality of Kepler Packing is asking for maximum packing out to
infinity. That means you have to contend and fight with the packing of
identical spheres up against a wall or border. And so, in tackling that
wall, we can shift the hexagonal closed pack to another type of packing, a
hybrid type of packing in order to get "maximum packing". So no proof ever
of KPP is going to happen unless the proof tackles a infinity border wall.
In free-space, a far distance away from a wall barrier of infinity border,
then, hexagonal closed pack reigns and is the packing in all of free
space-- but, the moment the packing gets nearby the walls of infinity
border, then, we re-arrange the hexagonal closed pack to fit in more
spheres. Not unlike us packing a suitcase and then rearranging to fit in
more.
Cover picture: is a container and so the closed packing must be modified
once the border is nearly reached to maximize the number of spheres.
--- end quoting my 24th book of science ---
--- quoting my 28th published book ---
World's First Valid Proof of 4 Color Mapping Problem// Math proof series,
book 4
by Archimedes Plutonium (Author) (Amazon's Kindle)
Now in the math literature it is alleged that Appel & Haken proved this
conjecture that 4 colors are sufficient to color all planar maps such that
no two adjacent countries have the same color. Appel & Haken's fake proof
was a computer proof and it is fake because their method is Indirect
Nonexistence method. Unfortunately in the time of Appel & Haken few in
mathematics had a firm grip on true Logic, where they did not even know
that Boole's logic is fakery with his 3 OR 2 = 5 with 3 AND 2 = 1, when
even the local village idiot knows that 3 AND 2 = 5 with 3 OR 2 = either 3
or 2 depending on which is subtracted. But the grave error in logic of
Appel & Haken is their use of a utterly fake method of proof-- indirect
nonexistence (see my textbook on Reductio Ad Absurdum). Wiles with his
alleged proof of Fermat's Last Theorem is another indirect nonexistence as
well as Hales's fake proof of Kepler Packing is indirect nonexistence.
Appel & Haken were in a time period when computers used in mathematics was
a novelty, and instead of focusing on whether their proof was sound,
everyone was dazzled not with the logic argument but the fact of using
computers to generate a proof. And of course big big money was attached to
this event and so, math is stuck with a fake proof of 4-Color-Mapping. And
so, AP starting in around 1993, eventually gives the World's first valid
proof of 4-Color-Mapping. Sorry, no computer fanfare, but just strict
logical and sound argument.
Cover picture: Shows four countries colored yellow, red, green, purple and
all four are mutually adjacent. And where the Purple colored country is
landlocked, so that if it were considered that a 5th color is needed, that
5th color should be purple, hence, 4 colors are sufficient.
--- end quoting my 28th book of science ---
Of course I cannot blame computers for screwing-up trigonometry sine and
cosine et al for computers were mostly a 1900s onward phenomenon. But I can
blame computers from 1900 to 2023 for **entrenching that b.s. fake math**.
For trigonometry was never designed for anything more than figuring out all
aspects of a triangle given angles along with sides of that triangle in
question. When you have "kook mathematicians" extending something that
cannot be extended, you end up with this cesspool of incomprehensible
gibberish pretending to be mathematics.
Usually Kook mathematics starts out with being kookish. But trigonometry
history shows that it was legitimate math working to solve triangles in
Islamic math, true good use, but then along comes some kook mathematicians
like Euler who bends trigonometry into being wastrel cesspool kook math.
And I do not think a single student of math or physics or science can ever
forgive Euler for how he dragged a simple piece of math-- trigonometry to
be this modern day ugly cesspool kook math.
Trigonometry was never meant to have the Y-axis be numbers, while the
X-axis be angles. Why that turns math into being bar graphs and pie charts
with fruit and lollipops on x-axis and gummy bears on y-axis.
And, no human mind, not even the famous Euler, could think in a medium of
numbers along with angles and then throw in the silly "i" imaginary number,
square-root of -1, along with Euler's constant "e = 2.71828....". No, Euler
was in late stages of kook math when he dabbled with trigonometry, and the
world of mathematics has suffered ever since Euler ruined trigonometry. I
am attempting to revive and resuscitate what Euler mangled and destroyed.
If my mind cannot operate smoothly in trigonometry that which Euler
wrought, then no mind can operate with this insanity of Euler trigonometry.
And part of the blame is that Euler never had the true numbers of
mathematics-- Decimal Grid Number Systems. No, Euler had the delusion of
continuum with Reals.
--- quoting Wikipedia on trigonometry history ---
All six trigonometric functions in current use were known in Islamic
mathematics by the 9th century, as was the law of sines, used in solving
triangles. With the exception of the sine (which was adopted from Indian
mathematics), the other five modern trigonometric functions were discovered
by Arabic mathematicians, including the cosine, tangent, cotangent, secant
and cosecant. Al-Khwārizmī (c. 780–850) produced tables of sines, cosines
and tangents. Muhammad ibn Jābir al-Harrānī al-Battānī (853–929) discovered
the reciprocal functions of secant and cosecant, and produced the first
table of cosecants for each degree from 1° to 90°.
The first published use of the abbreviations sin, cos, and tan is by the
16th-century French mathematician Albert Girard; these were further
promulgated by Euler (see below). The Opus palatinum de triangulis of Georg
Joachim Rheticus, a student of Copernicus, was probably the first in Europe
to define trigonometric functions directly in terms of right triangles
instead of circles, with tables for all six trigonometric functions; this
work was finished by Rheticus' student Valentin Otho in 1596.
In a paper published in 1682, Leibniz proved that sin x is not an algebraic
function of x. Roger Cotes computed the derivative of sine in his Harmonia
Mensurarum (1722). Leonhard Euler's Introductio in analys in infinitorum
(1748) was mostly responsible for establishing the analytic treatment of
trigonometric functions in Europe, also defining them as infinite series
and presenting "Euler's formula", as well as the near-modern abbreviations
sin., cos., tang., cot., sec., and cosec.
--- end quoting Wikipedia ---
I needed to include this chapter on how computers have messed up geometry,
badly. For computers are simply automatons which regurgitate back to the
author what his bias and prejudices and ignorance are, repeat it back, and
the author thinks he has a computer proof.
Computers in geometry are "art" not truth of science. For it is really
silly to think that a man on a computer pressing in keys and a image forms
on the screen, to think that pressing keys is geometry truth is very silly.
It is art, not math.
AP
zzzzzzzzzz
This is an Ebook textbook, meaning that I am constantly adding new
material, amplifying and detailing points of interest. And I am constantly
editing and correcting and revising.
Archimedes Plutonium
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