Can "Q" be any eutrigon?
This web page is reproduced from the author's book Principles of nature; towards a new visual language, WA Roberts P/L Canberra. 2003. [Used with permission of the author]- edited and reformatted for the web.
Showing how the encompassing large equilateral triangle can accommodate every possible eutrigon shape
The visual proof as presented in Figures ET2 & ET3 includes no stipulation as to the absolute size of the figure and its component shapes. The geometric relationships hold whether the diagram fills a football field, or whether it fits on the back of an envelope with room to spare. The entire diagram is scaleable: it can be made larger or smaller without affecting the geometrical relations within it. Thus the particular eutrigon in the diagram can be any size at all, and this does not alter the internal relations of the diagram, and therefore our theorem is unaffected by scaling the entire diagram up or down by some factor.
But we need to still check whether or not the geometric construction also accounts for every possible shape of eutrigon . This is a little trickier.
Our definition of a eutrigon is of a triangle with one angle equal to 60-degrees. Its shape is thus solely determined by changing the relative lengths of its legs a and b (the sides adjacent to the 60-degree angle). From this point of view, the hypotenuse length is passively determined to the extent that it merely joins the 'free' end of leg a with that of leg b, and this length is determined by the fact that the angle between legs a and b is fixed at 60-degrees. So the length of leg a relative to the length of leg b is all the information we need to determine the shape of any eutrigon [see diagram below]. This makes our analysis of possible shapes so much easier.
We may combine the two values a and b into a single value reflecting their relative lengths—the ratio, a/b. This ratio thus defines the shape of any eutrigon. If we make a vanishingly small, b (in comparison) 'becomes' immensely large; if the ratio a/b is the same, the shape is the same. The following three possibilities exist: a = b; a > b; or, a < b. When a = b, the eutrigon is the special case of an equilateral triangle. In every other case, one side is smaller than the other. Let us define side 'a' to be always the smaller of the two legs a and b. Hence, a/b is less than or equal to "1" for any eutrigon. So if our geometric construction can accommodate eutrigons of every leg-ratio (a/b) between zero and one, then the proof holds for all eutrigons.
We are now ready to refer back to our visual proof of Figure ET2. The lower half of that diagram has been reproduced in the diagram at the top of this page, except this time, the dynamics of eutrigon-change is hinted at.
We first stipulate that the three corners of the large equilateral triangle always contain three congruent eutrigons (shaded in the figure), whose shape may vary but whose leg-length sum (a + b) is equal to the side-length of the outer equilateral triangle. This latter point means that they will always fit around the edge of the equilateral triangle in 'tip-to-toe' fashion..
It can be seen that each one of the outer-equilateral-triangle's vertices is also the 60-degree angle of a eutrigon, and that the central triangle C (constructed on the eutrigons' hypotenuses) remains equilateral under these transformations although its size and orientation change. It can also be seen that the ratio a/b changes smoothly from one to zero, as side a of the eutrigon begins first of all, equal to b (a/b = 1), but then shrinks through every lesser value towards zero length. Its 'lost length' is added to that of side b which gradually increases to itself approach the limiting fixed length of a+ b [see footnote]. It can only 'reach' this limit when a = 0, giving a/b = 0 (since zero divided by any other number whether it be 2, 3, or 10000000, is still zero). Therefore, the transition from a=b to a = 0, sees the ratio a/b pass seamlessly from one to zero, and therefore the geometric construction of [figures ET2 and ET3] can be made for eutrigons of any possible shape, and so, our proof is complete.
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