Heights in a tetrahedron III

First define the function
U(x,y,z) = (x + y + z)² – 2(x² + y² + z²)
If x, y, z are the squares of the side lengths in a triangle then U(x,y,z) equals the area of triangle, squared, times 16. This follows from Heron’s formula.

In the figure let… a, b, c be the *squares* of the sides of the inner triangle. The other three triangles, drawn in heavy lines, are those which we will try to fit together with the middle one as faces of a tetrahedron with equal faces. The dashed blue lines outline three triangles congruent to the middle one. We know from earlier posts on this subject that they fit together iff they are acute-angled. They can be looked upon as references from which we try deviations, though it is sufficient to really consider those drawn in heavy lines. The Greek letters denote side lengths, and we let x, y, z be the side squares (x = xi², y = eta², z = zeta²).

If all three triangle areas are equal then
U(a,y,z) = U(a,b,c)
U(x,b,z) = U(a,b,c)
U(x,y,c) = U(a,b,c)
or
(a + y + z)² – 2(a² + y² + z²) = (a + b + c)² – 2(a² + b² + c²)
(x + b + z)² – 2(x² + b² + z²) = (a + b + c)² – 2(a² + b² + c²)
(x + y + c)² – 2(x² + y² + c²) = (a + b + c)² – 2(a² + b² + c²)

If the squares are expanded. terms containing x, y, z collected to the left and constants (i e terms containing just just a, b, c) to the right this gives

2a(y + z) – y² – z² + 2yz = 2ab +2bc +2ca – b² – c²
2b(x+ z) – x² – z² + 2xz = 2ab + 2bc +2ca – a² – c²
2c(x + y) – x² – y² + 2xy = 2ab +2bc +2ca – a² – b²

Now, let’s introduce the square differences
δ = x – a, ε = y – b, λ = z – c, from which x = a + δ, y = b + ε, z = c + λ. The right hand sides here are substitute for x, y, z in the above system of equations:

2a(b + c + ε + λ) – (b + ε)² – (c + λ)² + 2(b + ε)(c + λ) = 2ab +2bc +2ca – b² – c²
2b(a + c + δ + λ) – (a + δ)² – (c + λ)² + 2(a + δ)(c + λ) = 2ab + 2bc +2ca – a² – c²
2c(a + b + δ + ε) – (a + δ)² – (b + ε)² + 2(a + δ)(b + ε) = 2ab +2bc +2ca – a² – b²

Here the entire right hand sides cancel terms to the left, giving

2a(ε + λ) – (2bε + ε²) – (2cλ + λ²) + 2εc + 2ελ + 2bλ = 0
2b(δ + λ) – (2aδ + δ²) – (2cλ + λ²) + 2aλ + 2cδ + 2δλ= 0
2c(δ + ε) – (2aδ + δ²) – (2bε + ε²) + 2aε + 2δb + 2δε = 0

from which

(ε + λ)a + (-ε + λ)b + (ε – λ)c = ( ε – λ)²/2
(-δ + λ)a + (δ + λ)b + (δ – λ)c = (δ – λ)²/2
(-δ + ε)a + (δ – ε)b + (δ + ε)c = (δ – ε)²/2

This system is *linear* in a, b, c! Thus is can rather easily be solved for a, b, c expressed in the differences δ, ε, λ. If all of these are non-zero the solution is unique (the determinant is -8δελ) and otherwise the system simplifies so that it can still be used to draw conclusions.

Now we consider in turn the four possibilities
(I) δ = ε = λ = 0
(II) δ ≠ 0, ε = λ = 0
(III) δ ≠ 0, ε ≠ 0, λ = 0
(IV) δ ≠ 0, ε ≠ 0, λ ≠ 0
(For symmetry reasons only the *number* of zero/non-zero differences matters.)

(I) Trivial: The triangles are congruent and as we know from earlier post they fit together iff they are acute-angled.

(II) Two of the equations reduce to -δa + δb + δc = δ²/2 by which δ = 2(-a + b + c) and so x = 2b + 2c – a. Also y = b, and z = c since ε = λ = 0. If these triangles fit together in a tetrahedron then this will have edges squared a, b, c, 2b + 2c – a. But the volume would be zero so no tetrahedron. To see this I shamelessly used the Cayley–Menger determinant:
http://en.wikipedia.org/wiki/Tetrahedron
(By ”shamelessly” I mean that I just relied on it; I would very much like a proof but I haven’t seen one or constructed one myself – yet. 🙂

(III) Two of the equations become εa – εb + εc = ε²/2 and -δa + δb + -δc = δ²/2, which give ε = 2(a – b + c) and δ = 2(-a + b + c), respectively. So δ + ε = 4c and δ – ε = 4(b – a). If this is inserted into the third equation we get (a – b) ² = c². The edges of the tetrahedron squared will be a, b, c, -a + 2b + 2c, 2a – b + 2c, c, and using also (a – b) ² = c² the Cayley–Menger determinant shows that the volume is zero.

(IV) In this case the equation system can be solved uniquely for a, b, c (non-zero determinant):
a = δ(ε – λ)²/(4ελ)
b = ε(δ – λ)²/(4δλ)
c = λ(δ – ε)²/(4δε)

From this it follows that δ ≠ ε ≠ λ ≠ δ, i e all the differences are different since otherwise some side length would have become zero.

The other sids squared:
x = δ(ε – λ)²/(4ελ) + δ
y = ε(δ – λ)²/(4δλ) + ε
z = λ(δ – ε)²/(4δε) + λ

Using once more the Cayley–Menger determinant it is found that the volume becomes zero.

So there are no possibilities to get all face areas and thus heights of a tetrahedron equal unless the faces are congruent triangles.

Conclusion:
*************************************
Any four congruent, acute-angled triangles fit together as faces of a tetrahedron in which then all heights become equal. This is the only possibility to have all heights equal.
*************************************

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