# Re: non-congruent tetrahedra with shared invariants-

*From*: JEmebius <jemebius@xxxxxxxxx>*Date*: Mon, 02 May 2011 22:18:13 +0100

Han de Bruijn wrote:

On May 2, 2:26 pm, Han de Bruijn <umum...@xxxxxxxxx> wrote:On May 2, 2:10 pm, achille <achille_...@xxxxxxxxxxxx> wrote:

On May 2, 7:23 pm, Han de Bruijn <umum...@xxxxxxxxx> wrote:Okay. Congruent, but apart from mirroring I suppose?On Apr 28, 5:24 pm, achille <achille_...@xxxxxxxxxxxx> wrote:This is for the special case (corresponds to what LeonOn Apr 28, 7:29 pm, Han de Bruijn <umum...@xxxxxxxxx> wrote:I only see three edges in your formula, while the tetrahedron has six.On Apr 26, 9:41 pm, quasi <qu...@xxxxxxxx> wrote:Similar to the area, the volume of the tetrahedron canOn Mon, 25 Apr 2011 19:17:27 -0700 (PDT), tensegriboyThere are six (6) edges with a tetrahedron. If you have the lengths of

<Space...@xxxxxxxxxxx> wrote:

face areas have to conform to A + B + C => D,However based on Leon Aigret's reply, it seems the conditions

and the other three cyclic permutations.

the more "acute" inequality is on the six edges.

I guess, the answer is, no;

those relations are unique, up to reflection.

I specified may not be sufficient to force a unique congruence

type.

these edges then you can construct the tetrahedron, given some obvious

restrictions (triangle inequalities). There are four (4) faces, there

is one (1) volume. It all results in five (5) equations, with six (6)

unknowns. So indeed it is _impossible_ to determine the tetrahedron up

to congruence. Still apart from the fact that the area of each face is

expressed into surrounding lengths (a,b,c) of the edges, by Heron from

Alexandria's formula:

A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2 -- four (4) times

http://en.wikipedia.org/wiki/Heron's_formula

be expressed in terms of the length of the edges using

Cayley Menger determinant:

288 V^2 = | 0 1 1 1 1|

| 1 0 a^2 b^2 c^2|

| 1 a^2 0 c^2 b^2|

| 1 b^2 c^2 0 a^2|

| 1 c^2 b^2 a^2 0|

= 4(a^2 + b^2 - c^2)(a^2 + c^2 - b^2)(b^2 + c^2 - a^2)

so it not very hard to find two non-congruent tetrahedra

with same area and volume.

REF:http://mathworld.wolfram.com/Cayley-MengerDeterminant.html

Thus resulting in a system of fourth degree equations for eventually

solving (a,b,c, .. ) when given the areas of the faces. Go ahead ..

This kind of problems, though interesting, is notoriously difficult.

? Did you copy and paste wrong bits and pieces from the Wolfram site ?

Han de Bruijn

suggest) where all 4 triangular faces of a tetrahedron

are congruent to each other and have sides a, b and c.

But ah, that's nonsense. We are in _space_. Sorry.

Han de Bruijn

MIRROR IMAGE ISSUE NOT ONLY TWO-DIMENSIONAL

Quotations from previous posts:

(A) From: Han de Bruijn 2011-05-02 13:26 CEST

------------------------------------------------------

Okay. Congruent, but apart from mirroring I suppose?

Han de Bruijn

(B) From: Han de Bruijn 2011-05-02 14:15 CEST

------------------------------------------------------

> Okay. Congruent, but apart from mirroring I suppose?

But ah, that's nonsense. We are in _space_. Sorry.

Han de Bruijn

I decided to do some Fröbel ( http://en.wikipedia.org/wiki/Froebel ) paperwork. Here is the recipe:

(1) Draw a pair of congruent scalene triangles with acute angles only.

(2) Take care that you lay them down such that they are directly congruent, not each other's mirror images.

(3) Connect the midpoints of the sides so as to obtain a subdivision into four congruent scalene triangles of half the size.

(4) Find out for yourself where you must provide gluing strips in order to do the folding and gluing of the next and last step.

(5) Finally turn the one triangle into a tetrahedron with congruent faces by folding the three corners upwards; the same for the other one by folding the corners downwards.

One will observe that the two tetrahedrons are each other's mirror images while not being directly congruent; they do not have a plane of symmetry of their own.

Now the math behind all this:

Imagine the still flat large triangles of (1) laid down on top of each other (midpoints of the sides denoted by A, B, C) in the ABC plane.

After folding and gluing we have tetrahedron ABCD with D at the one side of the ABC plane, and tetrahedron ABCE with E at the other side of the ABC plane. Vertices D and E are each other's mirror images in the ABC plane. So ABCD and ABCE are at least indirectly congruent.

How could ABCD and ABCE be directly congruent? Only if there is a second plane of symmetry. Such a plane would intersect the ABC plane along a line of symmetry of triangle ABC. Therefore ABC must be isosceles.

Ciao: Johan E. Mebius

.

**Follow-Ups**:

**References**:**Re: non-congruent tetrahedra with shared invariants***From:*Han de Bruijn

**Re: non-congruent tetrahedra with shared invariants***From:*achille

**Re: non-congruent tetrahedra with shared invariants***From:*Han de Bruijn

**Re: non-congruent tetrahedra with shared invariants***From:*Han de Bruijn

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