SWANNIES FIRST THEOREM -
25 Centuries after Pythagoras
The photograph above on the left shows the five components of a model. Below on the left a trirectangular tetrahedron, the basic shape of the central component of all models conforming to Swannies Theorem is shown. The four prisms are made to fit onto the faces of this tetrahedron.
The photograph above on the right shows an assembled stainless steel prismatic model (see later: De Gua De Malves). A piece of art by Pat Blaauw of Elitsha Holdings in Port Elizabeth who did this, modeled on my cardboard example, in his spare time for fun* without me having accepted any quote as yet and offered it to me, take it or leave, at an obviously very reasonable price; me knowing how many hours of work/play goes into making models. * Obviously his first 'simply Maths' model.
The photograph on the left shows a cardboard model of the same proportions as the prismatic model above but of cylindrical shape (see later: De Tinseau). A model of a trirectangular tetrahedron of the same size and shape as the tetrahedron at its centre, as it would appear before it is shaped to blend (truncated to fit in?) with the cylinders, is shown in front of it. The projected ellipses of the circle on its base appears on the rectangular faces.. Creating this model consumed about two weeks: Only cardboard it takes but many days to make.
How a model can be made:
1. Mark out in straight lines joining at the edges any fair sized corner on an ordinary (cereal) box. 2. Neatly cut this corner from the box. This gives three rectangular triangles with a common vertex. 3. Cover up the triangular opening with cardboard and glue to form the triangular base of your trirectangular tetrahedron. 4. By measurement and calculation find the areas* of each of the four faces of the tetrahedron using a unit of choice, eg. cm. 5. From cardboard build pyramids to fit the faces of the tetrahedron with height in each case equal to the numerical value of the calculated area of the face and the same length unit type. 6. Stick the prisms onto the faces. *You may wish to use Herons formula for the finding the area of the non rectangular face:
Make the sides of the triangle a, b and c and let s = (a + b + c)/2. Then its Area wil be = square root of [s(s-
The following is the well-
PYTHAGORAS'S THEOREM (± 500 BC)
3^2 + 4^2 = 5^2 (before Pythagoras)
with its triangle and squares.
The fo llowing is the prime example for
SWANNIES THEOREM (2006 AC) :
Formula: x3 + y3 + z3 = w3
3^3 + 4^3 + 5^3 = 6^3 (PLATO 400 BC)
A Projection type, after DE TINSEAU (1774 AC)
and A Special case, after DE MALVES (1783 AC)
as shown above in the photographs.
Note x, y, z and w do not represent any linear dimensions of a model. Specified models such as (3units)3 + (4units)3 + (5units)3 = (6units)3 needs the set up of three equations (see below) and calculation to find the lengths of the three rectangular sides of the tetrahedron required. The results of such calculation is used in an illustration that follows later. Note that a model with equal rectangular prisms e.g. (4.16u)3 + (4.16u)3 + (4.16u)3 = (6,00u)3 would obviously be an easier first attempt (see the star and others under More Models).
Setting up the equations for calculation of tetrahedron dimensions for a model with given volumes for its prisms, should anyone wish to go there:
Given the cube values for a De Malves type Swannie' s Theorem Model, how to set up the equations to find the
dimensions of the tetrahedron at its centre. Units not shown.
Formula: x^3 + y^3 + z^3 = w^3
Say given: 3^3 + 4^3 + 5 ^3 = 6^3 (PLATO)
i.e. 27 + 64 + 125 = 216
Draw a trirectangular tetrahedron and mark the sides meeting at right angles a, b and c, in this order, from short to long. Therefore the areas of the three rectangular triangles are equal to ½ ab, ½ ac and ½ bc, from small to large. Now square these and set each equal to the volume of its prism.
Therefore to the theorem: ¼ a^2 b^2 = 27 -
¼ a^2 c^ 2 = 64 -
Projection and Reflection: For a model starting off as a shape marked out on the triangular 'hypotenuse' face of its tetrahedron, such as the cylindrical model in a photograph on the left, the clue comes from De Tinseau -
Return to Horizontal Index at top of Page