Regular Tetrahedron and 4 Parallel Planes?

Question

Do You remember the old problem about the equilateral triangle on the rails? Given are 3 co-planar parallel lines, the distance between 1st and 2nd is a, between 2nd and 3rd - b (and a + b between 1st and 3rd of course).
Please follow the link below to see a picture:

http://i219.photobucket.com/albums/cc101/Andrey_Dyukmedzhiev/2D3D.gif

Easily follows that for all a ≥ 0 and b ≥ 0 there exists an equilateral triangle, having one vertex on each line with side length:
(2/√3)*√(a² + ab + b²)

Now the 3D analog: imagine 4 parallel planes at distances a, b and c between them. Does a regular tetrahedron exist, having one vertex on each plane? If so, find its edge length.

Answer 1

EDIT:

I noticed that I use different notation.
In my notation, a, b, and c are distances
between the lowest plane and the other planes -
not the distances between neighboring planes.

To make transition to the original notation
one should make the substitutions: a=a, b = a+b, c = a+b+c
(left sides are my a, b and c).

Then the answer will be

X = √ (3*a^2/2+2*b^2+3*c^2/2+2ab+ac+2bc).

One more remark. Once we know that the required construction exists, it is easy to calculate the polyhedron edge X. X^2 is a quadratic function of distances a, b, c (in my notation). The expression should be symmetric to arbitrary permutatons of [a, b, c]. Then X^2 = s1 (a^2+b^2+c^2) + s2 (ab+ac+bc), where s1 and s2 are some constants. In two limits a=0, b=0, and a=b=c, c is equal to the height of a regular polyhedron: c/X=√ (2/3). Applying the above expression to these two limits we have s1=3/2, 3(s1+s2) = 3/2, and hence s2 = -1.

*****************************************************

Choose the coordinate system, such as the lowest plane passes through the cooridnate origin (0,0,0). Put the vertex v_0 of the tetrahedron in the coordinate origin. The other 3 vertices have coordinates:

Eq. (1)

v_1=(0, L, L),
v_2=(L, 0, L),
v_3=(L, L, 0).

Here, L is some number. The distance between any two vertices v_i and v_j is equal to L √ 2, so this is a regular tetrahedron.

Let P=(P_x,P_y,P_z) is the unit vector normal to the planes. The vertices of the tetrahedron lie on planes, if and only if v_j have the form v_j = d_j P + Q_j, where d_j =a, b, c, and Q_j are some vectors perpendicular to P. Taking scalar products of these relations with the vector P, we have v_j*P_j = d_j, or

y + z = a,
x + z = b,
x + y = c,

where x= L*P_x, y=L*P_y, z=L*P_z. The solution is

Eq. (2)

x = (-a+b+c)/2,
y = (a-b+c)/2,
z = (a+b-c)/2.

This means that by choosing L = √(x^2+y^2+z^2); and P = (x,y,z)/L we solve the problem.

The edge length X of the tetrahedron is equal to
X = L √ 2=
√ (3a^2/2+3b^2/2+3c^2/2-ab-ac-bc).


For explicit construction we start with an arbitrary coordinate system. For given a, b, and c we calculate x,y,z, according to Eq. (2) and build the tetrahedron according to Eq. (1). Then draw the vector P and build the planes. After that, we can rotate and shift this construction to transform the planes into the desired position.