Swannies Theorems, Art in Maths and a follow-up to Pythagoras, De Tinseau and De Gua De Malves.

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Above: Some Pythagorean type Art; a "4:7:11 Eu du Cologne art for the eye" equivalent of the one for the nose used by your granny. You have to know how to add/mix colours and the areas before you attempt such paintings! The one on the left is not really true, too much blue! The better mix one on the right also provides a clue to perhaps how, without further ado, you can see that the painting is mathematically true. Note: BCA is a right angle and D is halfway 'up' from BA to C. The ratios between the relevant areas appears above. However, can you prove it, i.e. that the ratios between the areas  of the inside triangles is the same as that between the outside semicircles. Clue: Re-draw the figure with squares replacing the semi-circles on the sides and extend CD all the way through the square with base AB (such as used in the conventional proof of the Pythagoras theorem).    
Now further more have a good look at the trirectangular tetrahedron 'roof' of the central  prism of the model in the photograph below (or any other trirectangular  tetrahedron).  How would you subdivide it into four tetrahedrons so that the volumes  of three ad up to that of the biggest one, their bases being the  faces of the original  tetrahedron? You should then have a little roof for each of the prisms. Would the volume ratio between these tetrahedrons be the same as that between the prisms?
By the way - carry on with the Eu du Cologne sequence, 4, 7, 11, ---, and the 5, 8,13,21, --- of the Fibonacci sequence till you have at least 10 elements of each. Compare the ratios  found between the last two elements of each of the two sets. Do the same with your day and month of birth set e.g. 18, March, 21, 24, 45,  etc. to see if you conform/are normal!  

Museum: The arrangement in the photograph on the left is what Swannie has in mind for his little science museum. It would have a 4,4m square base with the heights and areas in a 1:2:2:3 proportion and total height of 7,26m.
--- Perhaps you'l find an elegant way of turning a cube in to three cubes and so really line up with Pythagoras's all square classical set up?  Theorem V for cuboids gets nearer there. Likewise, semi spheres into three smaller semi spheres, see Theorem II and III page, bottom right, that also gets near there.
And then there is the quadro-pentahedron, which could be a pyramid with one square horizontal base and four similar slanting faces with  x° angles at the vertex, you could decorate with attachments for the sake of art. Perhaps the only model possible?  May be the multi model magic for Swannie type models is reserved for  trirectangular hedrons only?

By the way and to lead you off the track: Actual values were used on a previous page to convey the workings of the First theorem as it is appreciated that the conversions can be confusing. The theorems of the Frenchmen deal with
areas squared / fourth powers , fairy stuff to some of us, that Einstein and mathematicians play(ed) with. It required somewhat of a witches wand to turn it into third powers for volume. That is at least how Swannie experienced  it and what side tracked him from 'getting there' for some months.
Back to where were: Model on the left: What would you call the pyramid at the centre of the yellow model  to which the prisms are attached?  Just use x° for those vertex angles.

March 2012 addition  of new theorems - See next page

Photographs down below:  Views of a twin model. If crystals can have twins, why not models?                                                       

ADD-ONN for first seven pages: October 2012:  For all the models of the First Theorem shown the objects/shapes on the faces make full contact with the original and the projected areas on the faces of the tetrahedrons and have either vertical or slanting sides. However, is seems an object of any shape can be accomodated by the theorem. If its a ball it can obviously only touch the base in one spot and its projected outline on the base is used to, in turn, find the projected outlines on the other faces. Deviding the original outlined  area into smaller sections/areas (What do the mathematicians call this kind of doing?) and "going vertical through the object", manipulating somewhat with ratios in accordance with the theorem allows of the same type of shapes to be fitted on the rectangular faces of the tetrahedron. For example starting off with a round football/sphere you end up with three sort of rugby balls/ellipsoids. See bottom of page "II and III", the next page. It is not recommended that one uses a model of a live object for such an exercise; depending on how you orientate it on the base you will either end up with very fat or very lean shapes, perhaps also grotesque, on the other faces!

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