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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

But to get that you had to introduce new points.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

x0 x1 y0 y1 are the same intersection points. the other one for each polynomial will be the known point.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Yes, but that polynomial needs 3 known points to be a unique quadratic. You have 2 and one unknown point. You can not determine that quadratic knowing only two.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi Herc11

Do you have a test example we could try?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

What is wrong with post #9? Can anyone get A and B?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I need a of four quadratics.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

Ok. You are right. But I am confused with the equation of defining a2.

In equation of a2 only the 4 variables x0 x1 y0 y1 exist.

And

if I have four equations for a2 (I do not know the 4 polynomials, only that they have the same 2 inters. points x0 y0 x1 y1 (unknowns),

I know

each polynomial's leading coef. i.e. a2

and one point of each of the polynomials.

Then from the set of 4 equations of a2, x0..y1 can be defined?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Me too, and putting more points on the drawing does not give them to me.

I need more info about those quadratics!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I need you to give me the leading coefficient of two more quadratics through those two unknown points and a point on each of those.

*Last edited by anonimnystefy (2013-06-20 01:50:27)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

anonimnystefy wrote:

I need a of four quadratics.

So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

For post #9? Yes any additional info would help. I have been asking the OP to state the full problem. It might be possible to get more info out of it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Herc11 wrote:

anonimnystefy wrote:

I need a of four quadratics.

So, if you have 4 quadratics plus a point of each of them you can solve it? i.e you can find x0...y1?

I think so.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

From just the first coefficient?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

First coefficient of four quadratics and a point on each.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

anonimnystefy

I think so.

I have the same opinion.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

It is more information than you provided in post #1 so I am willing to post the challenge. It is worth a shot.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

bobbym wrote

Want to use post #9 and I will provide the two more quadratics since I know what A and B is?

What do you mean exactly?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

I know the answer for post #9 only because I created it. No one else can know it from that information ( maybe ). I can give 2 more quads.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi bobbym

Where does it say that we have only 2 quadratics in the first post??

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Herc11****Member**- Registered: 2013-06-19
- Posts: 169

It does not, we just went from there. But so far no one has proved it is possible with n quadratics.

Thats why I am asking.

It seems to me that If there are 4 quad. you can define the 2 interesection points.

Similarly, if the polynomials were of degree 3, 6 polynomials of 3 degree might be needed...and so on...

But I am not sure

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

bobbym wrote:

I am going to say that with 4 quadratics or 176 of them you still can not define the two points. That is my opinion.

Okay, then give me the first coefficient of four quadratics and a point on each of them and I will try finding the 2 intersections.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Yes, give some time. I am very interested in your result. Please hold.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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