Aren't you just looking at a three-dimensional surface from above?

I don't think so, projections of 3d coordinated to 2d don't change lengths from real to complex.

But that's what's needed: we need a coordinate system inwhich we have

1) a "triangle" OAB and

2) apoint C on AB with two lengths OC=6, AB=10 and

3)two 90 degree angles between the lines OA,OB and AB,OC

With the complex segments of t4a we will need some definition of angle between lines of complex length and check if they are 90 deg.

But that's what's needed: we need a coordinate system inwhich we have

1) a "triangle" OAB and

2) apoint C on AB with two lengths OC=6, AB=10 and

3)two 90 degree angles between the lines OA,OB and AB,OC

With the complex segments of t4a we will need some definition of angle between lines of complex length and check if they are 90 deg.

Maybe this can be used: right angles in 3d don't need to be right when projected to 2d?

The extra degree of freedom in 3d might solve it?

The extra degree of freedom in 3d might solve it?

You just need to find the projection from 3D to 2D that works, with suitable interpretation for the 6 and 10. If it exists I don't expect it to be unique because of your >=12 etc. thing.

In 3d it won't work -i.e. there doesn't exists a triangle in 3d with the given properties either-.

The triangle needs to lie in a 2d surface in 3d and so you can *rotate* the surface to the regular 2d surface without changing lengths or angles in the triangle. For it to exist in 3d (or 4d,..) it would also have to exist in 2d!

The triangle needs to lie in a 2d surface in 3d and so you can *rotate* the surface to the regular 2d surface without changing lengths or angles in the triangle. For it to exist in 3d (or 4d,..) it would also have to exist in 2d!

It depends what the 6 and the 10 refer to. Lift the point where the 6 and 10 lines meet to create an irregular tetrahedron. The 6 refers to one edge, the 10 refers to the edge that is still horizontal (onto which two edges have collapsed in the projection). Or something.

That would make the triangle non flat in 3d. Lifting and projecting is a form of using a non-euclidian geometry. The points inside the triangle that form the area wont be spanned by the corner points. If we allow for that then another option might be to draw it on a sphere, and there might be many more.

The question then is: is the area always the same for all solutions that can be found in various non-euclidean geometries?

If it is then you can write don't the answer, if not then you need to ask the teacher to specify the specific non-euclidean geometry.

The question then is: is the area always the same for all solutions that can be found in various non-euclidean geometries?

If it is then you can write don't the answer, if not then you need to ask the teacher to specify the specific non-euclidean geometry.

There are clues in the image that haven't been mentioned. There are hexagonal 'tiles' everywhere. I think this is part of an architect's plan. The tiling is a typical cosmetic addition. We need to see the views from other directions to know what is going on

Also: you would need to compute the areas of the triangle in the non-euclidean space and not the 2d shadow of it (i.e. compute the sum of areas of the faces of the tetrahedron). The triangle only exist in non-euclidean space, the 2d projection is not a valid solution an not unique either.

Last edited by outrun on May 15th, 2017, 8:16 am, edited 1 time in total.

The probability of Collector going to solve this with spacetime relativistic coordinates and write a paper about it is negative.

- Traden4Alpha
**Posts:**23951**Joined:**

Who ever said the solution had to be unique? There's an entire category of brainteasers intended to elicit as many alternative answers as possible.

And even if the answer must be unique, who says it's a number and not an analytic expression? "The area of the triangle" could be a number, an equation, or the text from a thread on an internet forum.

What's also interesting is the presumption that the numbers are base 10. What if the numbers are base 12 which makes the triangle "valid" in our normal 2-D Euclidean space? The answer in base-10 math would be 6*12/2 = 36 which converted back to base 12 is "30". And what if the numbers are base 16? The answer in base-10 math would be 6*16/2 = 48 which in base 16 is "30". For any base of 12 or greater, the triangle exists and the answer is "30".

P.S. The hexagonal tiles are clearly projections of cubes which implies the triangles are projections on some 3-D space.

And even if the answer must be unique, who says it's a number and not an analytic expression? "The area of the triangle" could be a number, an equation, or the text from a thread on an internet forum.

What's also interesting is the presumption that the numbers are base 10. What if the numbers are base 12 which makes the triangle "valid" in our normal 2-D Euclidean space? The answer in base-10 math would be 6*12/2 = 36 which converted back to base 12 is "30". And what if the numbers are base 16? The answer in base-10 math would be 6*16/2 = 48 which in base 16 is "30". For any base of 12 or greater, the triangle exists and the answer is "30".

P.S. The hexagonal tiles are clearly projections of cubes which implies the triangles are projections on some 3-D space.

Last edited by Traden4Alpha on May 15th, 2017, 8:57 pm, edited 1 time in total.

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

And ever since I meet this man

My life is not the same

And Nicolai Ivanovich Lobachevsky is his name

Nicolai Ivanovich Lobach...

I am never forget the day

I am given first original paper to write

It was on analytical algebraic topology

Of locally Euclidean metrization

Of infinitely differentiable Riemannian manifold

My life is not the same

And Nicolai Ivanovich Lobachevsky is his name

Nicolai Ivanovich Lobach...

I am never forget the day

I am given first original paper to write

It was on analytical algebraic topology

Of locally Euclidean metrization

Of infinitely differentiable Riemannian manifold

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

So the answer to the question is 3 x Ingrid Bergman. Who'd have guessed?

- Cuchulainn
**Posts:**61185**Joined:****Location:**Amsterdam-
**Contact:**

This thread a bit like a multiple choice test. The new Project Maths?

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

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