- Dear Antreas,

I wrote> in the equation

Not always.

> 3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2

> the four variables are equivalent

> and we can interchange them.

Consider the trivial case when the point P

with distances a, b, c from the vertices of the

equilateral triangle, lies on one side of the

equilateral triangle that has side length d.

A general solution of this Diophantine equation

is given by the formulas

a = m^2 + 3n^2

b = Abs((m + n)(m - 3n))

c = Abs(4mn)

d = Abs((m + 3n)(m - n))

For m = 2, n = 1 we get

a = 7, b = 3, c = 8, d = 5

Which means that the side of the equilateral

triangle can be 5, 3, 8 but it can't be 7.

It is interesting the figure.

Construct the equilateral triangle A1P2P3 with side 8

On P2P3 take P1 such that P2P1 = 5 and P1P3 = 3

and on the equilateral triangle A1P2P3 construct

two equilateral triangles A3P2P1 and A2P1P3.

We have A1P1 = A2P2 = A3P3 = 7 and the points

P1, P2, P3 are the required points for the 3 equilateral

triangles. The conclusion is that there is not fourth

equilateral triangle with side 7. The number 7 in this

tetrad is always the distance of P from the opposite vertex.

Best regards

Nikos - Dear Nikos,

El 17/08/2012 17:22, Nikolaos Dergiades escribió:> Dear Antreas,

there are equilateral tringle with side 7 and points that distfrom its

> I wrote

>> in the equation

>> 3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2

>> the four variables are equivalent

>> and we can interchange them.

> Not always.

> Consider the trivial case when the point P

> with distances a, b, c from the vertices of the

> equilateral triangle, lies on one side of the

> equilateral triangle that has side length d.

> A general solution of this Diophantine equation

> is given by the formulas

> a = m^2 + 3n^2

> b = Abs((m + n)(m - 3n))

> c = Abs(4mn)

> d = Abs((m + 3n)(m - n))

> For m = 2, n = 1 we get

> a = 7, b = 3, c = 8, d = 5

> Which means that the side of the equilateral

> triangle can be 5, 3, 8 but it can't be 7.

> It is interesting the figure.

> Construct the equilateral triangle A1P2P3 with side 8

> On P2P3 take P1 such that P2P1 = 5 and P1P3 = 3

> and on the equilateral triangle A1P2P3 construct

> two equilateral triangles A3P2P1 and A2P1P3.

> We have A1P1 = A2P2 = A3P3 = 7 and the points

> P1, P2, P3 are the required points for the 3 equilateral

> triangles. The conclusion is that there is not fourth

> equilateral triangle with side 7. The number 7 in this

> tetrad is always the distance of P from the opposite vertex.

> Best regards

> Nikos

>

>

vertices 3, 4 and 5. See:

http://www.xente.mundo-r.com/ilarrosa/GeoGebra/DistEnterasTriangEquil.html

From a equilateral triangle with side 8, you can construc 12 of thats

triangles of side 7.

--

Best regards,

Ignacio Larrosa Cañestro

A Coruña (España)

ilarrosa@...

http://www.xente.mundo-r.com/ilarrosa/GeoGebra/ - Dear Ignacio,

thank you very much. You are right.

What I said is nonsense.

I was deceived thinking that the fourth

case of equilateral triangle with side 7

would be a trivial case as the other three.

Best regards

Nikos Dergiades

> there are equilateral tringle with side 7 and points that

> distfrom its

> vertices 3, 4 and 5. See:

>

> http://www.xente.mundo-r.com/ilarrosa/GeoGebra/DistEnterasTriangEquil.html

>

> From a equilateral triangle with side 8, you can construc

> 12 of thats

> triangles of side 7.

> Best regards,

>

> Ignacio Larrosa Cañestro

> A Coruña (España)

> ilarrosa@...

> http://www.xente.mundo-r.com/ilarrosa/GeoGebra/

>

>

>

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