The tie up with Pythagoras:
The illustration above on the left shows box dimensions as vectors while the model on the right shows vectors mounted on the vertex of its tetrahedron that sits atop its base prism. The photograph at the centre shows Plato's values vectors representing both area and height for prisms: Yellow and red results in orange; orange is replicated at the vertex where with the blue vector the resultant is the black one; and pointing downward is its negative the really relevant one as that is where the base prism of a model will go. The square of the diagonal d of a rectangular box with sides/dimensions a, b and c is given by Pythagoras as
d2 = a2 + b2 + c2 . However one of each of a,b,c and d can be construed as the area vectors A,B,C and D of the faces of a trirectangular tetrahedron, one a representing for example say 2,279^2 cm2 (or 5,2 in the box above) for the area of face A and the other a, 2,279^2 cm for the height of a prism on face A, thus giving the volume of the prism on A as 2,279^4 cm3 = 27cm3 = (3cm)3 as .fits in with x3 of the equation of Swannies Theorem. Likewise done with b, c and d gives the set of Plato's values: 3^3 + 4^3 + 5^3 = 6^3 . Note: A length dimension in the Pythagoras equation above squared is equivalent to an area vector times a height vector in Swannies theorem. Thus the line-
The tie up with De Tinseau/De Gua De Malves :
*According to De Gua the sum of the squares of the areas of the three rectangular triangles of a trirectangular tetrahedron add up to the square of the area of its 'hypotenuse'/large/remaining face.* That is like a 4 + b 4 + c 4 = d 4 and with length unit, say cm, also to the power of four, gives for example a 4 = 2,279^4 cm4 = 27cm4 , whatever cm4 means. However if a2 of the a4 is taken for the area of a face, area = a2 = 2,279^2 cm2 = 5,194cm2 , which it is, and the remaining a2 of the a4 is used for the height of a prism on that face, ie. height = 2,279^2 cm2 + assistance of the maths fairy that makes the one cm dissappear = 5,194 cm, the volume of the prism becomes 27 cm3 . De Tinseau/De Malves theorem therefore also serves as a proof for Swannies theorem.
In this model the prisms are stacked next to each other with its tetrahedron capping the large/base prism. On top 'vectors' representing both the areas of its faces and the heights of the pyramids are arranged. More about this set up as a possible model for a museum under Arty Models.
The Photographs below (non theorem-
The photograph on the left: Just another way of showing "three vectors" at right angles and their resultant. The areas of the smaller semicircles also do add up to that of the one on the resultant.
The "Arrow head" at the centre is a re-
To fill this space: On the right in 2006: Maybe? but No -