AP's 240th book of science// Correcting Cycloid Wave of Old Math-- Geometry in Motion// math research;; by Archimedes Plutonium
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 14, 2023, 10:45:34 AM (5 days ago)
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AP's 240th book of science// Correcting Cycloid Wave of Old Math-- Geometry in Motion// math research;; by Archimedes Plutonium

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 a error that this subject needs its own separate book of science.

Principle of Maximum Electricity Production by muon-as-electron inside proton toruses// physics research

by Archimedes Plutonium (Amazon's Kindle)

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.

-----------------------------
Table of Contents
-----------------------------

1) My own history on this subject matter.

2) Explore some math data.

3) Looking for unique features of Torus geometry.

4) The speed of Rest Mass particles.

5) Working on Reincarnation science.

6) Preliminary-Proof of Maximum Electricity Production Principle.

7) Is Maximum Electricity Production Principle a reverse of Least Action Principle?

8) Nature is never half way in doing something, never halfway in producing electricity.

9) Efficiency of electric generator is e = B x L x v.

10) Photons and Neutrinos as the wiring system of Atoms.

11) All Atomic structure is composition of magnetic monopoles.

12) Proton, muon and neutron all composed of photons and neutrinos.

13) I need semicircle and semi ellipse cycloid waves -- so intense research into this.

14) AP uses wood blocks to better understand Torus geometry.

15) How much more surface area for torus over volume?

16) Proof that the proton is a torus with muon inside doing Faraday law for hydrogen and for higher atomic number, the geometry is a Mainframe torus with smaller toruses of S, P, D, F orbitals inside doing the Faraday law.

17) Proof of Maximum Electricity Production Principle inside Atom toruses.

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Archimedes Plutonium
May 14, 2023, 3:03:31 PM (5 days ago)
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This book should start with the animation of the Limacon by Wikipedia for it alerted me to the huge
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Archimedes Plutonium
May 14, 2023, 8:52:32 PM (5 days ago)
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2-AP's 240th book of science// Correcting Cycloid Wave of Old Math--It is a Half-Ellipse//
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Archimedes Plutonium
May 14, 2023, 9:30:07 PM (5 days ago)
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I am writing this book while the events of discovery are still very fresh on my mind. I had always
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Archimedes Plutonium
May 14, 2023, 10:42:34 PM (5 days ago)
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Now the hard part of this book is going to prove several of the Conjectures made about smooth curves,
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Archimedes Plutonium
May 15, 2023, 12:42:36 AM (5 days ago)
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Conjecture #2 is tricky, and depends on whether two cones, base to base <> forms a vertex at
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Archimedes Plutonium
May 15, 2023, 1:20:31 AM (5 days ago)
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So, yes I do believe I have both Conjectures true and a proof for each. With the Cones base to base
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Archimedes Plutonium
May 15, 2023, 11:19:06 AM (4 days ago)
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What is interesting about Conjecture 2-- The Parabola has formula Y = x^2 a pure polynomial and when
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Archimedes Plutonium
May 15, 2023, 3:56:51 PM (4 days ago)
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Speaking of this convergence-limit of the Parabola to the slant side of cones of <>. So,
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Archimedes Plutonium
May 15, 2023, 4:50:50 PM (4 days ago)
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Speaking of the convergence-limit of the Hyperbola to the slant side of cones of <>. So,
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Archimedes Plutonium
May 16, 2023, 1:01:36 AM (4 days ago)
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Of course, AP's conics with base to base rewrites the entire subject of Conic Sections for <
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Archimedes Plutonium
May 16, 2023, 9:00:37 AM (3 days ago)
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I think I need to alter the Title more. For this is such a massive overhaul of Old Math Geometry,
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Archimedes Plutonium
May 16, 2023, 3:57:45 PM (3 days ago)
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The proof that all smooth curves as defined smoothness has no vertex and all the points are governed
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Archimedes Plutonium
May 16, 2023, 8:21:05 PM (3 days ago)
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The beauty of Conjecture #2:: all parabolas and hyperbolas are ellipses or ovals is that it allows us
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Archimedes Plutonium
May 17, 2023, 12:11:39 AM (3 days ago)
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The title of this book seems to change by the day, even twice a day. My newest title to reflect the
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Archimedes Plutonium
May 17, 2023, 12:40:13 PM (2 days ago)
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Exquisite Conjecture #3:: Pi is actually 3.16.. the square root of 10 when factoring in Geometry of
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Archimedes Plutonium
May 17, 2023, 1:45:07 PM (2 days ago)
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At one time I collected the digits of square root of 10 and am curious where the first time it has
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Archimedes Plutonium
May 17, 2023, 2:15:53 PM (2 days ago)
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Now, do not get me wrong, New Math does not throw out 3.14... but keeps it along with 3.16.... and
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Archimedes Plutonium
May 17, 2023, 2:29:49 PM (2 days ago)
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I had better show the pi= 3.14.... digit sequence and correct my mistake the three zeroes are in the
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Archimedes Plutonium
May 17, 2023, 6:52:36 PM (2 days ago)
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Maybe the title of this book should be New True Geometry starting with cycloid correction and
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Archimedes Plutonium
May 18, 2023, 12:02:11 AM (2 days ago)
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3. 16227766016837933199889354443271853371955513932521
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Archimedes Plutonium
May 18, 2023, 12:21:34 PM (yesterday)
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Conic Sections in 2D with 3D being base to base <> cones. Alright, there exists a theory of
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Archimedes Plutonium
May 18, 2023, 3:07:07 PM (yesterday)
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In 2D conics, is it possible to leave a remnant that is a six sided figure? I see plenty of triangles
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Archimedes Plutonium
May 18, 2023, 11:32:39 PM (yesterday)
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Alright I have a collection of Staedtler math tool products of triangles, compass, protractor etc.
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Archimedes Plutonium
May 18, 2023, 11:55:30 PM (yesterday)
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Also while I have a computer working on this division > 3. >
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Archimedes Plutonium
May 19, 2023, 12:29:07 AM (yesterday)
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The math literature says five points are needed to uniquely define a ellipse. If true then 4 points
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Archimedes Plutonium
May 19, 2023, 1:24:38 PM (19 hours ago)
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I changed my mind. I am going to wade into this debate of how many points determine a unique ellipse,
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Archimedes Plutonium
May 19, 2023, 4:36:43 PM (16 hours ago)
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Alright, I suspect, not sure yet, that the math literature on how many points (noncollinear) in the
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Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 19, 2023, 6:23:11 PM (14 hours ago)
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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.
>

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 noncollinear 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 noncollinear points determines a unique ellipse, counter to Old Math claim that you need 5.

Statement: Given 3 arbitrary noncollinear 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 noncollinear 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.

QED

Now I use the above proof that 3 arbitrary noncollinear points in the plane determines a unique circle to prove that 4 noncollinear 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.

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 noncollinear 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.

QED

AP

On Saturday, May 20, 2023 at 9:28:44 AM UTC-5, Archimedes Plutonium wrote:
> 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 noncollinear 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 noncollinear points determines a unique ellipse, counter to Old Math claim that you need 5.
>
> Statement: Given 3 arbitrary noncollinear 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 noncollinear 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.
>
> QED
>
> Now I use the above proof that 3 arbitrary noncollinear points in the plane determines a unique circle to prove that 4 noncollinear 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.

And adding on these infinite number of half ellipses to an existing half ellipse is a Oval. As a Oval is defined as a joining of 2 dissimilar ellipses at a midsection.

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 noncollinear 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.
>
> QED
>
> AP

Let me summarize these two proofs.

On Saturday, May 20, 2023 at 10:06:13 AM UTC-5, Archimedes Plutonium wrote:

> > Statement: Given 3 arbitrary noncollinear 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 noncollinear 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 noncollinear points.

> >
> > QED
> >
> > Now I use the above proof that 3 arbitrary noncollinear points in the plane determines a unique circle to prove that 4 noncollinear 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 noncollinear 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.
> >
> > QED

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 <>.

AP