Most 3 arbitrary non-collinear points in Plane produce a unique ellipse--AP's 240th book of science -- New True Geometry starting with cycloid correction and Geometry-of-Motion // math research Amazon's Kindle

AP's 240th published book
New True Geometry starting with cycloid correction and Geometry-of-Motion // math research

by Archimedes Plutonium

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 a parallelogram that is not square and not a rectangle. If I can find a unique ellipse inside a unique circle produced by 3 arbitrary points non-collinear, then I can say in general all 3 arbitrary non-collinear points produces a unique 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.

Product details
• ASIN ‏ : ‎ B0C5VL38ZK
• Publication date ‏ : ‎ May 20, 2023
• Language ‏ : ‎ English
• File size ‏ : ‎ 625 KB
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Sticky notes ‏ : ‎ On Kindle Scribe
• Print length ‏ : ‎ 76 pages

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.

AP

On Tuesday, May 23, 2023 at 12:31:33 AM UTC+2, Archimedes Plutonium wrote:

> 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.

3 arbitrary non-collinear points in the plane define a circle - which is (a special case of) an ellipse. (See: https://en.wikipedia.org/wiki/Conic_section)

So this circle ist your unique ellipse?

Hint: Not even 4 points suffice. See: https://math.stackexchange.com/questions/3063610/how-many-points-are-needed-to-uniquely-define-an-ellipse

On Tuesday, May 23, 2023 at 12:43:05 AM UTC+2, Fritz Feldhase wrote:

> Not even 4 points suffice.

Hint: "Since two [different] ellipses can intersect at four points, these 4 points cannot determine a unique one."

So 4 points to not in generall suffice to determine a unique ellipse.

On 5/22/2023 6:47 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 12:43:05 AM UTC+2, Fritz Feldhase wrote:
>
>> Not even 4 points suffice.
>
> Hint: "Since two [different] ellipses can intersect at four points, these 4 points cannot determine a unique one."
>
> So 4 points to not in generall suffice to determine a unique ellipse.

That AP still defends his claim that four (now three!) points define a
unique ellipse shows me he's truly delusional. If you look at the images
on that stackexchange page, it's clear that multiple ellipses can go
through four points.

For example: https://i.stack.imgur.com/Nd9Px.png shows two ellipses
passing through four points, so those four points cannot define a unique
ellipse.

https://i.stack.imgur.com/IIEPx.png shows multiple ellipses passing
through 4 points.

It's one thing if AP was too dumb to understand any of the several
ellipse-is-a-conic-section proofs, but it's another thing entirely to
look at a picture of two or more ellipses passing through a set of four
(or even three) points and seeing that it is obviously so, but still
claim that 3 or 4 points define a unique ellipse. The only possibility
is AP is delusional, he believes something which is clearly false.

On 5/22/2023 4:11 PM, Volney wrote:
> On 5/22/2023 6:47 PM, Fritz Feldhase wrote:
>> On Tuesday, May 23, 2023 at 12:43:05 AM UTC+2, Fritz Feldhase wrote:
>>
>>> Not even 4 points suffice.
>>
>> Hint: "Since two [different] ellipses can intersect at four points,
>> these 4 points cannot determine a unique one."
>>
>> So 4 points to not in generall suffice to determine a unique ellipse.
>
> That AP still defends his claim that four (now three!) points define a
> unique ellipse shows me he's truly delusional. If you look at the images
> on that stackexchange page, it's clear that multiple ellipses can go
> through four points.
>
> For example: https://i.stack.imgur.com/Nd9Px.png shows two ellipses
> passing through four points, so those four points cannot define a unique
> ellipse.
>
> https://i.stack.imgur.com/IIEPx.png shows multiple ellipses passing
> through 4 points.

An ellipse can be defined with two arbitrary points. The fun part is
having to create the ellipse in normalized coordinates, then rotate and
translate it back into the two arbitrary points.

The normal ellipse is colored red, and the original line is in purple:

https://i.ibb.co/K7PNBHq/image.png

The target ellipse is cyan and the target line is yellow.

I can do this for any ellipse. Give me a point, and two radii, this can
work. Give me two points, this can work.

>
> It's one thing if AP was too dumb to understand any of the several
> ellipse-is-a-conic-section proofs, but it's another thing entirely to
> look at a picture of two or more ellipses passing through a set of four
> (or even three) points and seeing that it is obviously so, but still
> claim that 3 or 4 points define a unique ellipse. The only possibility
> is AP is delusional, he believes something which is clearly false.

On Tuesday, May 23, 2023 at 1:13:27 AM UTC+2, Volney wrote:
> On 5/22/2023 6:47 PM, Fritz Feldhase wrote:
> > On Tuesday, May 23, 2023 at 12:43:05 AM UTC+2, Fritz Feldhase wrote:
> > >
> > > Not even 4 points suffice.
> > >
> > Hint: "Since two [different] ellipses can intersect at four points, these 4 points cannot determine a unique one."
> >
> > So 4 points to not in generall suffice to determine a unique ellipse.
> >
> That AP still defends his claim that four (now three!) points define a
> unique ellipse shows me he's truly delusional. If you look at the images
> on that stackexchange page, it's clear that multiple ellipses can go
> through four points.
>
> For example: https://i.stack.imgur.com/Nd9Px.png shows two ellipses
> passing through four points, so those four points cannot define a unique
> ellipse.
>
> https://i.stack.imgur.com/IIEPx.png shows multiple ellipses passing
> through 4 points.
>
> It's one thing if AP was too dumb to understand any of the several
> ellipse-is-a-conic-section proofs, but it's another thing entirely to
> look at a picture of two or more ellipses passing through a set of four
> (or even three) points and seeing that it is obviously so, but still
> claim that 3 or 4 points define a unique ellipse. The only possibility
> is AP is delusional, he believes something which is clearly false.

Absolutely. (100% ACK.)

On 5/22/2023 4:51 PM, Chris M. Thomasson wrote:
> On 5/22/2023 4:11 PM, Volney wrote:
>> On 5/22/2023 6:47 PM, Fritz Feldhase wrote:
>>> On Tuesday, May 23, 2023 at 12:43:05 AM UTC+2, Fritz Feldhase wrote:
>>>
>>>> Not even 4 points suffice.
>>>
>>> Hint: "Since two [different] ellipses can intersect at four points,
>>> these 4 points cannot determine a unique one."
>>>
>>> So 4 points to not in generall suffice to determine a unique ellipse.
>>
>> That AP still defends his claim that four (now three!) points define a
>> unique ellipse shows me he's truly delusional. If you look at the
>> images on that stackexchange page, it's clear that multiple ellipses
>> can go through four points.
>>
>> For example: https://i.stack.imgur.com/Nd9Px.png shows two ellipses
>> passing through four points, so those four points cannot define a
>> unique ellipse.
>>
>> https://i.stack.imgur.com/IIEPx.png shows multiple ellipses passing
>> through 4 points.
>
> An ellipse can be defined with two arbitrary points. The fun part is
> having to create the ellipse in normalized coordinates, then rotate and
> translate it back into the two arbitrary points.
>
> The normal ellipse is colored red, and the original line is in purple:
>
> https://i.ibb.co/K7PNBHq/image.png
>
> The target ellipse is cyan and the target line is yellow.
>
> I can do this for any ellipse. Give me a point, and two radii, this can
> work. Give me two points, this can work.
>
>>
>> It's one thing if AP was too dumb to understand any of the several
>> ellipse-is-a-conic-section proofs, but it's another thing entirely to
>> look at a picture of two or more ellipses passing through a set of
>> four (or even three) points and seeing that it is obviously so, but
>> still claim that 3 or 4 points define a unique ellipse. The only
>> possibility is AP is delusional, he believes something which is
>> clearly false.
>

One reason why I normalize the target ellipse into the unit circle is
that it makes the math a heck of a lot "easier", so to speak...

On Tuesday, May 23, 2023 at 1:53:28 AM UTC+2, Chris M. Thomasson wrote:

> An ellipse can be defined with two arbitrary points.

I don't think, it can, dumbo.

Hint: Not even a circle can be defined with two arbitrary points.

On the other hand, if we allow for two _ordered_ points, it's possible.

The first point may define the center of the circle and the second point a point on its perimeter (for example).

So the ordered pair (m, x) defines a circle, where m is its center (point) and |x - m| is its radius (by convention).

Just given a set of two points {a, b} we do not know which of these points is the center. (You see: {a, b} = {b, a}.)

On Tuesday, May 23, 2023 at 1:53:41 AM UTC+2, Fritz Feldhase wrote:

> > The only possibility is AP is delusional, he believes something which is clearly false.
> >
> Absolutely. (100% ACK.)

We all know that Archie is a nutcase. Though a "special one". Really.

There even are songs in his honor!

See: https://www.youtube.com/watch?v=Jox5GBUAMUY

This man (nuts or not) is a legend!

(I have to admit, I like him too.)

On 5/22/2023 5:07 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 1:53:28 AM UTC+2, Chris M. Thomasson wrote:
>
>> An ellipse can be defined with two arbitrary points.
>
> I don't think, it can, dumbo.
>
> Hint: Not even a circle can be defined with two arbitrary points.

Really? I can do it. Did you read my code?

>
> On the other hand, if we allow for two _ordered_ points, it's possible.
>
> The first point may define the center of the circle and the second point a point on its perimeter (for example).
>
> So the ordered pair (m, x) defines a circle, where m is its center (point) and |x - m| is its radius (by convention).
>
> Just given a set of two points {a, b} we do not know which of these points is the center. (You see: {a, b} = {b, a}.)
>

On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:

> > Hint: Not even a circle can be defined with two arbitrary points.
> >
> Really? I can do it. Did you read my code?

Yes, you "can do it", because you are a math moron.

Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)

On 5/22/2023 5:24 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
>
>>> Hint: Not even a circle can be defined with two arbitrary points.
>>>
>> Really? I can do it. Did you read my code?
>
> Yes, you "can do it", because you are a math moron.

Huh? Did you read my code? Heck, if you are having trouble reading it,
ask a god damn ai for help.

>
> Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)

Really? Wow. You just might be an ahole?

On Tuesday, May 23, 2023 at 2:24:14 AM UTC+2, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
> > >
> > > Hint: Not even a circle can be defined with two arbitrary points.
> > >
> > Really? I can do it. Did you read my code?
> >
> Yes, you "can do it", because you are a math moron.
>
> Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)

Hint: I can read and write code.

On 5/22/2023 5:25 PM, Chris M. Thomasson wrote:
> On 5/22/2023 5:24 PM, Fritz Feldhase wrote:
>> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
>>
>>>> Hint: Not even a circle can be defined with two arbitrary points.
>>>>
>>> Really? I can do it. Did you read my code?
>>
>> Yes, you "can do it", because you are a math moron.
>
> Huh? Did you read my code? Heck, if you are having trouble reading it,
> ask a god damn ai for help.

Sorry for that evil comment. Ask me about it. Fuck the ai.

>
>
>>
>> Seems to me that you are a lost case (even if you appear to be a quite
>> intelligent guy). I'm sorry for you. (Never tried to read some math
>> textbooks?)
>
> Really? Wow. You just might be an ahole?
>

On 5/22/2023 5:25 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:24:14 AM UTC+2, Fritz Feldhase wrote:
>> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
>>>>
>>>> Hint: Not even a circle can be defined with two arbitrary points.
>>>>
>>> Really? I can do it. Did you read my code?
>>>
>> Yes, you "can do it", because you are a math moron.
>>
>> Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)
>
> Hint: I can read and write code.

Well, I can help you.

On Tuesday, May 23, 2023 at 2:25:23 AM UTC+2, Chris M. Thomasson wrote:
> On 5/22/2023 5:24 PM, Fritz Feldhase wrote:
> > On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
> >
> >>> Hint: Not even a circle can be defined with two arbitrary points.
> >>>
> >> Really? I can do it. Did you read my code?
> > >
> > Yes, you "can do it", because you are a math moron.
> >
> Huh? Did you read my code? Heck, if you are having trouble reading it,
> ask a god damn ai for help.

lol. Nice try.

Hint: I'm a programmer by profession.

On 5/22/2023 5:26 PM, Chris M. Thomasson wrote:
> On 5/22/2023 5:25 PM, Fritz Feldhase wrote:
>> On Tuesday, May 23, 2023 at 2:24:14 AM UTC+2, Fritz Feldhase wrote:
>>> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
>>>>>
>>>>> Hint: Not even a circle can be defined with two arbitrary points.
>>>>>
>>>> Really? I can do it. Did you read my code?
>>>>
>>> Yes, you "can do it", because you are a math moron.
>>>
>>> Seems to me that you are a lost case (even if you appear to be a
>>> quite intelligent guy). I'm sorry for you. (Never tried to read some
>>> math textbooks?)
>>
>> Hint: I can read and write code.
>
> Well, I can help you.

Think of a vec2 as being a 2-ary vector. This has two components x and y
such that

vec2 a;

a.x is the x component

a.y is the y component

Okay, lets start small! ;^)

On 5/22/2023 5:27 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:25:23 AM UTC+2, Chris M. Thomasson wrote:
>> On 5/22/2023 5:24 PM, Fritz Feldhase wrote:
>>> On Tuesday, May 23, 2023 at 2:20:45 AM UTC+2, Chris M. Thomasson wrote:
>>>
>>>>> Hint: Not even a circle can be defined with two arbitrary points.
>>>>>
>>>> Really? I can do it. Did you read my code?
>>>>
>>> Yes, you "can do it", because you are a math moron.
>>>
>> Huh? Did you read my code? Heck, if you are having trouble reading it,
>> ask a god damn ai for help.
>
> lol. Nice try.
>
> Hint: I'm a programmer by profession.

Are you? What does this do:

https://pastebin.com/raw/uD7s0B2e

Smarty pants!

On Tuesday, May 23, 2023 at 2:25:23 AM UTC+2, Chris M. Thomasson wrote:
> >
> > Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)
> >
> Really? Wow. You just might be an asshole?

That's for sure. :-)

So please try to educate yourself. Ok?

Maybe you aren't a lost case, after all. :-)

On 5/22/2023 5:30 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:25:23 AM UTC+2, Chris M. Thomasson wrote:
>>>
>>> Seems to me that you are a lost case (even if you appear to be a quite intelligent guy). I'm sorry for you. (Never tried to read some math textbooks?)
>>>
>> Really? Wow. You just might be an asshole?
>
> That's for sure. :-)

So, a true troll to the end. Ahhhhhh..... I understand now. Thanks.

>
> So please try to educate yourself. Ok?
>
> Maybe you aren't a lost case, after all. :-)

Lol. Can you understand my code I did for AP?

On Tuesday, May 23, 2023 at 2:27:46 AM UTC+2, Chris M. Thomasson wrote:

> Sorry for that evil comment.

N/p.

You know the saying: "If you can't stand the heat, get out of the kitchen."

> Ask me about it. Fuck the ai.

Agree. I'm sure we DO have a problem here (to say the least).

On 5/22/2023 5:34 PM, Fritz Feldhase wrote:
> On Tuesday, May 23, 2023 at 2:27:46 AM UTC+2, Chris M. Thomasson wrote:
>
>
>> Sorry for that evil comment.
>
> N/p.
>
> You know the saying: "If you can't stand the heat, get out of the kitchen."
>
>> Ask me about it. Fuck the ai.
>
> Agree. I'm sure we DO have a problem here (to say the least).

Well, My code demonstrates a way to create an ellipse using two points,
p0 and p1. Humm...

How much information can you gleam from this: Any questions? I will be
happy to answer. No problem.

my quick and crude C++ code:
_________________________________________
namespace ct_ellipsoid
{ glm::vec2
ct_rotate(
glm::vec2 p,
float angle
) {
glm::vec2 ret = {
p.x * glm::cos(angle) - p.y * glm::sin(angle),
p.y * glm::cos(angle) + p.x * glm::sin(angle)
};

return ret;
}

void
test_0(
ct::plot::cairo::plot_2d& scene,
glm::vec2 p0,
glm::vec2 p1,
unsigned long n
) {
glm::vec2 dif = p1 - p0;
glm::vec2 dif_mid = p0 + dif * .5f;
float dif_diameter = glm::length(dif);
float dif_radius = dif_diameter / 2.f;
float dif_angle = glm::atan(dif.y, dif.x);
float angle_base = CT_PI2 / n;

glm::vec2 n0 = { -dif_radius, 0 };
glm::vec2 n1 = { dif_radius, 0 };
glm::vec2 radii = { dif_radius, dif_radius / 2 };

scene.line(p0, p1, CT_RGBF(1, 1, 0), 5);
scene.line(n0, n1, CT_RGBF(1, 0, 1), 5);

for (unsigned long i = 0; i < n; ++i)
{
float angle = angle_base * i;
glm::vec2 e0_normal = { glm::cos(angle), glm::sin(angle) };
glm::vec2 e0 = e0_normal * radii;
glm::vec2 e1 = dif_mid + ct_rotate(e0, dif_angle);
scene.circle_filled(e0, .005f, CT_RGBF(1, 0, 0));
scene.circle_filled(e1, .005f, CT_RGBF(0, 1, 1));
}
}

void
manifest(
ct::plot::cairo::plot_2d& scene
) {
std::cout << "ct_ellipsoid::manifest()\n" << std::endl;

{
glm::vec2 p0 = { .5, .5 };
glm::vec2 p1 = { .75, -.8 };

test_0(scene, p0, p1, 1024);
}
}
} _________________________________________

On Tuesday, May 23, 2023 at 2:33:59 AM UTC+2, Chris M. Thomasson wrote:

> So, a true troll to the end.

No. I'm serious, really.

<Over and out>

_____________________

Fascinating:

- (procedure word, idiomatic) Used in films and TV to signal the end of a conversation, especially one conducted by CB radio or the like.

- Not used in real-life radio communications, since /over/ means you want a reply and /out/ means you do not expect a reply.

On Tuesday, May 23, 2023 at 2:37:30 AM UTC+2, Chris M. Thomasson wrote:

> Well, My code demonstrates a way to create an ellipse using two points,

I'm sure you know the saying: "A fool with a tool is still a fool."

:-)