Principles of Nature: towards a new visual language
© copyright 20032015 Wayne Roberts. All rights reserved.
The area of a new class of semiregular triangle (eutrigons) in etuThis web page is quoted from (or largely based on) the author's book Roberts, W Principles of nature; towards a new visual language, Canberra. 2003. Beginning with our defined unit (etu), and having obtained an expression for the area of the most symmetric of triangles (the equilateral triangle) in terms of our defined unit (p^{2} etu), we are now wellplaced to find an expression for the areas of lessregular triangles [yet also] expressed in etu’s. A new class of triangle defined: the eutrigonWe begin by considering the area of an important semiregular triangle [in fact an important new class of triangle], the analogue of the righttriangle in orthogonal (Cartesian) coordinate geometry. This triangle has, instead of one angle equal to 90 degrees, one angle equal to 60 degrees. Let us call such a triangle, a eutrigon.
Minimovie with narration, 72Kb:
The area of a eutrigon in etuThis neglected, hitherto unnamed, and unclassified triangletype is a very special class of triangle, as we shall see later. But for now we seek to find an expression for the area of any eutrigon given in etu . Let triangle ABC be any eutrigon. We define C to be the 60^{o} angle, a and b to be the ‘legs’ or sides adjacent to C, and c the ‘hypotenuse’ or side opposite angle C (fig. 56).
Figure 56 Example of a eutrigon We shall form a geometric construction around the eutrigon in order to determine its area in [terms of] etu’s. The reasons for this will become apparent, but suffice to say that scalestructure theory hints at relating such a form to a 'resonant' symmetric whole. We proceed as follows: Let the unknown area (in terms of etu) of eutrigon ABC be Q. Make a copy of eutrigon ABC and place this hypotenusetohypotenuse with the first to form a parallelogram ADBC. (Fig. 57) Figure 57 The area of the parallelogram ADBC is simply 2Q since it consists of two identical eutrigons each of defined area Q. The reasoning behind this approach is that it creates a resonant scalar construction of unequal parts which together form a symmetrical whole, or ‘scale’ (much like completing an octave in music). Now construct centrifugal (outward pointing) equilateral triangles on the legs of one of the component eutrigons as in figure 58... Figure 58 It can be seen from an examination of figure 58, that the construction has formed a large equilateral triangle DEF. Also we note from figure 58 that, in terms of areas, DEF = ACE + BCF + 2Q Applying our knowledge that the area of an equilateral triangle in etu is simply the sidelength 'squared', or p^{2}... we observe that ACE and BCF are each equilateral triangles of sidelengths b and a respectively, and therefore their areas, expressed in terms of etu’s, are given by b^{2} and a^{2} respectively. Thus, [the] equation [above], expressed in etu, becomes: Area of DEF = a^{2} + b^{2} + 2Q But DEF is equilateral and therefore its area in etu’s is its sidelength 'squared'. Its sidelength can be seen from an inspection of fig. 58 to be simply (a + b). Thus the area of DEF in etu is (a + b)^{2} . Substitution in the equation [immediately above] gives (a + b)^{2} = a^{2} + b^{2} + 2Q [Scale structure theory implies that there must be resonances or consonances between geometric scales and number scales (including the more generic numbers of algebra). Any student of algebra, learns very early on that the expansion of the expression (a + b)^{2} is equal to a^{2} + 2ab + b^{2}. See footnote for a scalestructural visual proof of this, but this time using the traditional square areas with which we are all so familiar] Thus we accept that the term (a + b)^{2} is again equal to a^{2} + 2ab + b^{2}, including triangular systems of analytic geometry. Thus, upon substitution of the expanded version of (a + b)^{2} in our equation above, we now have, a^{2} + 2ab + b^{2} = a^{2} + b^{2} + 2Q After cancelling corresponding terms on the left and righthand sides of the above equation, we have simply: 2ab = 2Q dividing both sides by 2, this equation reduces to the elegant identity [expressed in terms of relative units, in this case, etu], areaQ = ab Thus, the area of any eutrigon expressed in etu is ab, where a and b are the lengths of its legs (namely those sides adjacent to the 60degree angle, as illustrated in fig. 59), Area_{eutrigon} = ab Figure 59 email author: contact/feedback

