Regular Tetrahedron and 4 Parallel Lines?

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Duke, in the example you gave about 3 g-parallel lines in which to find an equilateral triangle, that determines the 4th vertex the tetrahedron, and therefore the 4th point on the plane, given 3 others. But I'm sure you want some kind of a property that all four points, in symmetrical fashion, must meet for a solution to exist. I'll get around to that. Euler has already given a good answer, so let's see if that can be improved on.

Edit: Given 4 points on the plane, with all 6 lines drawn between the 4 points, is it an orthogonal projection of a regular tetrahedron? One way, which is not pretty, is to find the lengths of those lines to each vertex, say, a, b, c, terminating in vertex A. Then a, b, c, must satisfy the equation:

3(a² + b² + c²) + 2(√((1-a²)(1-b²)) + √((1-b²)(1-c²)) + √((1-c²)(1-a²)) ) = 7

If this is true for all 4 vertices, then it's a orthogonal projection of a regular tetrahedron with edge length = 1.

As an example, for an equilateral triangle on the plane, where a = b = 1, and c = 1/√3, then this formula results in 7, as it should.

I'll see if I can find a more elegant condition, using all 6 lengths, a, b, c, d, e, f.

Edit 2: Okay, Duke, what you are looking for is this: Let a = (x1 + iy1), b = (x2 + iy2), c = (x3 + iy3), and d = (x4 + iy4), where {x1,y1}, {x2,y2}, {x3,y3}, {x4,y4} are the coordinates of the 4 given points. Then if

(a + b + c + d)² = 4(a² + b² + c² + d²)

then the 4 points are projected points of a regular tetrahedron. I think this can be generalized.

Edit 3: This relationship, when slightly rearranged, bears an uncanny similarity to the well known inequality for real numbers:

(a + b + c + d)/4 ≤ √( (a² + b² + c² + d²)/4 )

even though inequalities have no meaning for complex numbers. Now this has got my interest. Is there somehow a connection between the two?

Edit 4: Duke, I'd like to be able to give you an elegant method for how I've come to this, but that's not how I've come to it. What I can do is to offer you a sort of a proof why this condition works. First, note that for any {a, b, c, d}, complex or real, the value of this expression is unchanged when any number e, complex or real, is added to all a, b, c, d. So, now we only have to prove that a² + b² + c² + d² = 0 for any tetrahedron where its centroid is at (0,0,0). This simplifies things. Starting with {{1,1,1},{1,-1,-1},{-1,1,-1},{-1,-1,1}}, apply 2 successive rotations (one about the z-axis and the other about the y-axis, for example) at arbitrary angles. (a+b+c+d) will still be 0, since rotations are done about its centroid, but you'll find that a² + b² + c² + d² = 0 for any arbitrary angles of rotations. This line of work is just a spin-off of my earlier effort looking at lengths of the lines between the points on the planes, I just thought maybe putting them the complex plane could yield something.

Maybe someone can offer a elegant proof of this?...Show more

Revised for clarity:

Let a, b, c, d ∈ ℝ² be the respective projections onto the first two coordinates of the vertices
α, β, γ, δ of a regular tetrahedron sitting in ℝ³. Note this projection is orthogonal.
By translating by their mean, we can assume
α + β + γ + δ = (0,0,0). It follows that a+b+c+d = (0,0).

Denote their negatives as
-a = a', -b = b', -c = c', -d = d',
which are projections of -α, -β, -γ, -δ, respectively.

Now all eight points together are the orthogonally projected vertices of a cube sitting in ℝ³.
We can express these eight points as all ± combinations

±u ± v ±w

where
u = (a-b')/2 = (a+b)/2,
v = (a-c')/2 = (a+c)/2,
w = (a-d')/2 = (a+d)/2.

Note that u, v, and w are projections of
(α + β)/2, (α + γ)/2, and (α + δ)/2,
which are mutually orthogonal and all the same length (Use α + β + γ + δ = (0,0,0)).
Let D be the 3-by-3 matrix whose rows are these row vectors (in that order).
Note D is an orthogonal matrix (in the sense that DD⁺ is a scalar multiple of I).

Now let C be the 3-by-2 matrix whose rows are the row vectors u, v, and w, in that order.
Note C is just the first two columns of D.
It follows that the columns of C are orthogonal (as column vectors) and have the same square modulus.

It is not hard to see that this necessary condition (that the columns of C are orthogonal as column vectors and have the same square modulus) is also a sufficient condition for there to exist a regular tetrahedron of whose vertices a, b, c, and d are the projections: indeed, the third column of D can be taken to be the normalized cross product of the columns of C. The vectors a, b, c, and d can be recovered from the rows of C as the combinations
±u ± v ± w
having an even number (0 or 2) of minus signs....Show more