Five points in space with rational distances that are not co-linear

[Trysse]

TL;DR Summary

Is it possible to define five points in Euclidean space so that the distance between all points is rational?

Hi there, experts on three-D space!

while thinking about (physical) space, I have come up with the following (geometry) question: Is it possible to define five points (A, B, C, D, E) in Euclidian space, so that all distances (AB, AC, AD, AE, BC, BD, BE, CD, CE, DE) can be expressed in rational numbers? The five points must not be co-linear.

My intuition says that it is not possible, but I have not found a mathematical proof of why this is so (or not).

These are my considerations so far:
First I have reformulated the problem of the problem:
Instead of rational numbers, I rather think of distances expressed in natural numbers. If five points with all rational distances exist, then I could scale the distances between all points so that the distances have numerical values that are natural numbers (i.e. expand the rational fractions). So now I need to define five points, so that all distances can be expressed in natural numbers.

My guess is, that the problem has something to do with degrees of freedom in space.

I can arrange four points (A, B, C, D) to describe a regular tetrahedron where the edges have a length of one. If I now add a fifth point E, which has a distance of one length unit to the points A, B, and C, the distance between the points D and E is irrational $2* \sqrt{2/3}$ . I guess, no matter in which direction and by which distance I move any of the points, I can never get rid of this one irrational distance.

When I take four points that are not co-planar, there is always one sphere, so that all four points are on the surface of this sphere. The sphere can be described by those four points. If I add a fifth point, the sphere would be overdetermined.

To describe a coordinate system in space I need a set of four points {(000), (100), (010), (001)}.

What is your take on this? Is it possible or not? Is this a known geometry problem, if yes: Do you know of any proofs.Stefan

 

[anuttarasammyak]

Interesting question. I tried to find four points
(3,0,0),(-3,0.0),(4,0,0),(-4,0,0)
but the fifth is difficult to find.

 

[mfb]

I found a solution in 3D, is that what you are looking for?

Make a triangle with side length 3. The center is sqrt(3) away from each point.
Add the two points 1 below and above the center. They are 2 away from each other and sqrt(1+3)=2 away from each of the three vertices of the triangle.

 

[Trysse]

Hi mfb,

Thanks for the response. However, I think your solution doesn't work out.

The distance between the center point of an equilateral triangle and a corner is: 1 divided by the square root of 3 times the edge of the triangle. So in your case: 3 divided by the square root of 3.

Please check.

 

[Trysse]

Interesting question. I tried to find four points
(3,0,0),(-3,0.0),(4,0,0),(-4,0,0)
but the fifth is difficult to find.

Isn't it? I believe it might be a feature of space that it is not possible.

 

[jedishrfu]

This is not unlike the perfect Euler Brick where all corner distances make Pythagorean triples.

https://en.wikipedia.org/wiki/Euler_brick?wprov=sfti1

which has no known solution and I think no known proof either.

 

[mfb]

Hi mfb,

Thanks for the response. However, I think your solution doesn't work out.

The distance between the center point of an equilateral triangle and a corner is: 1 divided by the square root of 3 times the edge of the triangle. So in your case: 3 divided by the square root of 3.

Please check.

3/sqrt(3)=sqrt(3), that's what I wrote. None of the five points are at that spot, so it's not an issue that this length is irrational.

 

[PeroK]

I found a solution in 3D, is that what you are looking for?

Make a triangle with side length 3. The center is sqrt(3) away from each point.
Add the two points 1 below and above the center. They are 2 away from each other and sqrt(1+3)=2 away from each of the three vertices of the triangle.

We can extend this solution to seven points and beyond. We have an equilateral triangle of side length $9$. The distance from each vertex to the centre is $3\sqrt 3$. Then we look for points a height $h$ about the centre, where $h$ is a natural number, and we want the distance to each vertex also to be natural. I.e.
$$h^2 + 27$$ must be a perfect square. This does indeed have two solutions:
$$h = 3, h = 13$$
In general, if we have an equilateral triangle of side length $3l$, then we need to find $h$ such that $h^2 + 3l^2$ is a perfect square.

If we take $l = 4$, then we have three solutions: $h = 1, 4$ and $11$. That gives us nine points.

 

[PeroK]

Isn't it? I believe it might be a feature of space that it is not possible.

The maximum I have is 45 points, using an equilateral triangle of side 240 and points above and below the centre at height $h$, with the distance to each vertex $z$.

l = 80, h = 11, z = 139
l = 80, h = 20, z = 140
l = 80, h = 46, z = 146
l = 80, h = 52, z = 148
l = 80, h = 80, z = 160
l = 80, h = 118, z = 182
l = 80, h = 130, z = 190
l = 80, h = 167, z = 217
l = 80, h = 176, z = 224
l = 80, h = 220, z = 260
l = 80, h = 284, z = 316
l = 80, h = 305, z = 335
l = 80, h = 388, z = 412
l = 80, h = 470, z = 490
l = 80, h = 592, z = 608
l = 80, h = 794, z = 806
l = 80, h = 955, z = 965
l = 80, h = 1196, z = 1204
l = 80, h = 1597, z = 1603
l = 80, h = 2398, z = 2402
l = 80, h = 4799, z = 4801

If you want an exercise, prove that give any number $N$ you can find at least $N$ points in 3D space that are all rational distances apart. Although, in this case, all but three are colinear.

 

[Trysse]

Thanks mfb,
I needed that extra "3/sqrt(3)=sqrt(3)" explanation. It wasn't obvious to me at first sight.
Did you "know" this answer instantly when you read the question or did you have to think about it?

Thank you Perok for the long explanation.

Regarding the "exercise": From your explanation I assume, that all other points besides those of the original equilateral triangle lie on a straight line that goes perpendicular through the center of the triangle. And if I choose the triangle large enough there should be enough solutions with natural numbers. However, I guess that does not qualify as proof. Would you mind to show me the proof?

Can there be other combinations of points without an equilateral triangle?

And what about my other consideration? Is it possible to have five points that all lie on the surface of one sphere with all rational distances between the five points? Six hours ago I would have been pretty sure that it is not possible. However, I am getting more cautious with trusting my intuition...

 

[PeroK]

I haven't looked for a proof myself.

You could look for a rectangular solution. E.g. side lengths of 6 and 8 and 10 on the diagonals. Then look for points above and below the centre. That works with a height of 12 I think.

I haven't thought about your sphere question.

 

[anuttarasammyak]

Summary:: Is it possible to define five points in Euclidean space so that the distance between all points is rational?

The five points must not be co-linear.

I find the solution that not all but one point is out of the line. Showing it In 2D case of (x,y) plane, point (0,1) and points ( $\frac{1}{2}(r-\frac{1}{r}$), 0) have rational number distance between where r is any rational number since
$$\sqrt{1^2+[\frac{1}{2}(r-\frac{1}{r})]^2}= |\frac{1}{2}(r+\frac{1}{r})|$$
a rational number. Thus we have countable infinite points on x-axis satisfying the condition.

 

[mfb]

Did you "know" this answer instantly when you read the question or did you have to think about it?

Three points are a plane, so I looked at the simplest case first where the other two points are symmetric above and below that plane. If we start with an equilateral triangle with side length q (=$3l$ in PeroK's notation) then we get to the equation $\frac{q^2}{3} + h^2 = z^2$ where h (height) and z (edges of the figure) should be integers. q=3 is the obvious first number to check and it leads to a solution.

Five points on a sphere would surprise me. The most obvious symmetric option doesn't work: If you use an equilateral triangle and put two points above/below the center (asymmetric but on that line) then you get a contradiction trying to make every distance rational.

 

[Trysse]

This is not unlike the perfect Euler Brick where all corner distances make Pythagorean triples.

https://en.wikipedia.org/wiki/Euler_brick?wprov=sfti1

which has no known solution and I think no known proof either.

Yes, I agree. I looked at the Euler Brick and I think so too. Only that my hunch was disproven within hours.

However, I wonder if the Perfect Euler Brick issue could be reformulated. If I look at a brick, I can also consider the sphere that circumscribes the brick. The eight apices of the brick all lie on this sphere. The space diagonal of the brick is the diameter of the sphere.
So instead of looking at a brick and the diagonals, I only look at five of the points on a sphere and the distance between those points. If I was able to arrange five points in a way, that all distances between them are rational, I could fill in the missing three points and get a perfect Euler Brick. Oh, and two of the points must be antipodal.

Five points on a sphere would surprise me. The most obvious symmetric option doesn't work: If you use an equilateral triangle and put two points above/below the center (asymmetric but on that line) then you get a contradiction trying to make every distance rational.

So maybe the way to prove, that perfect Euler Bricks cannot exist would be to prove, that there cannot be five points on the surface of a sphere that have all rational distances between them.