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36-sided polygon

36-SIDED POLYGON

Symmetric hexagonal star polygon

Example of angle trisection
on the pattern of 36-sided polygon

Why are we opting for the 36-sided polygon construction in this chapter, and as an example taking every fourth divisional point as our “tool”, is not only because the vertices of the 36-sided polygon are 10° (360°÷36 = 10°), but also since it is an essential element in some other so-called “unknowns” or “impossibilities” of construction only with a compass and unmarked straightedge, such as:

  • diametrical divisions and quadratures,
  • doubling the cube
  • the Pi of ancient peoples
  • pyramidal constructions etc.

There is no doubt that the construction from an aspect of exactness is demanding (the greater number of divisions calls for greater precision), but the constructions up to 99 are relatively easy. On top of this, the 36-sided polygon is based on 6 equilateral triangles of the basic circle’s radius. Someone might think it resembles the Star of David, but that is just seemingly. In this case there are only two of three centerlines of the hexagon triangle, and their intersections are the centers of equilateral triangles, therefore enough to circumscribe described circles from each of the centers (in truncated form). And in order to distinctly show this we will resort to a biggest possible truncation, regardless of a loss of a series of data, so as to come as close as possible to present-day geometry but with the addition of everything else that we have demonstrated in our previous chapters.
Let us recapitulate:

  • the basic circle is a reflection of everything that is ongoing inside and outside of it (intersection points inside and outside are of other radii that create segments of other and different divisions on the basic circle);
  • any division on the basic circle begins from an initial point and finishes in that same point;
  • as regards trisecting, we may ascertain the following: trisecting is never direct, unmediated, hence it requires a “pattern, scheme or tool” to perform its function, but nevertheless we underline that this mode of the 36-sided polygon is not the only mode of construction, as we will see from the artifacts of the pentagram or the yin and yang in some of our upcoming chapters. Finally, it is important to stress that in sacred geometry all of this is performed only with a compass and unmarked straightedge.

* * *

Therefore:  the circle is described as usual and its radius divided into six segments. This time applying the truncated procedure, i.e. only by means of the inner segments of the partitioned circles.

* * *

Inscribe a chord to each segment of the resulting hexangular division.

* * *

Thereafter we delineate the areas with lengths of opposite poles through the center. Thus we get six equilateral triangles. We seek out the center of each.

* * *

If we inscribe a triangle joining en route from 1st pole – 3rd pole – 5th pole – 1st pole we get the centerline of each hexangular triangle.

* * *

If we then inscribe another subtended linkage from the 4th pole -.6th pole – 2nd pole –4th pole, we get two centerlines of each of the six equilateral triangles of the hexagon. Thus we have already found 6 centers of 6 triangles of the hexagon.

* * *

Anyway, trace a third centerline even if it isn’t essential.

* * *

We will join the midpoints with a circle from the center of the basic circle (even though it also isn’t essential) – but, …

* * *

… as we can see, it is of the same magnitude as the circles described around the six equilateral triangles of the hexagon’s basic circle.

* * *

Applying the “truncated method” those circles are divided with their radius, starting from the poles of the basic circle of the hexagon…

* * *

… then from the vertices of the described circles that arose from the division of poles. All six of the described circles of the six triangles of the hexagon’s basic circle have arcs larger (more removed from the center of the basic circle) than the arc of the basic circle.

* * *

If we now enter into the compass the radius length to the center of the basic circle – the exterior poles of the described circles of the equilateral triangles of the hexagon – then circumscribe  the whole „scheme“, we know that said radius of the basic circle will give us the result of the division of its arc.

* * *

If we start the division from the pole of the basic circle until we get back to the starting point, we will get the division of 36 = 360° ÷ 36 = 10°. Thus, thirty-six segments of the arc of the basic circle’s midpoint peak angles are 10°. Therefore, we can apply the 36-sided polygon as a pattern for trisection of angles up to 120°.

* * * *

Example:

Let us now use the 36-sided polygon as a “tool” for trisecting angles greater than 90° and smaller than 120°. We specify an arbitrary angle with an arc and chord, and a centerline.

* * *

The chord of the arbitrary angle is a side of the equilateral triangle and we inscribe it with the bisectors of its sides. (The arbitrary angle is put aside to the very end.)

* * *

We find the center of the equilateral triangle and describe its circle (the scheme’s basic circle).

* * *

We divide it with its radius and get a hexagon, from the vertex of its chord.

* * *

A bisector triangle was inscribed and now we inscribe another one from the subtended pole of the basic circle.

* * *

It is not necessary to show the third centerline, nor the chords of the hexagon segments of the basic circle for easier reference. There is also no need to inscribe the circle radius from the center of the basic circle – center of the segmental triangles.

* * *

From these six centers we circumscribe the segmented circle from the radius center of the basic circle – center of the hexagon segments.

* * *

On basis of pattern shown in scheme of 36-sided polygon, we divide the segments of the circle …

* * *

… with their radius , in which we get the vertex poles.

* * *

We insert the radius between the center and circumference of the basic circle into the compass and the exterior poles of the segments of the described circles and from the peak pole of the basic circle we start the division on its arc.

* * *

In this way we have divided the basic circle with its radius into 36 parts.

* * *

If we join every fourth dividing point on the arc of the basic circle starting from the peak pole which is at the same time the pole of the arbitrary angle (the endpoint of the chord), we get three equal parts on the arc of the basic circle and above the arbitrary angle.

* * *

Therefore, from each fourth point on the basic circle towards the vertex of the randomly determined angle we delineate half-lines. They divide the arc of the arbitrary angle into three equal parts.

* * *

We have thus divided the arbitrary angle into three equal parts using the 36-siced polygon scheme (36 ÷ 4 = 9), of angle greater than 90° and smaller than 120°. As for angles greater than 120° we have sequences of more simplex duplicate “tools”, so we will continue using the nonagon only with the compass.

* * * * * *

4 Responses to “36-sided polygon”

  1. Ricardo says:

    Sorry Mr Tomo, that circle gives me 51 (17*3) and not 36 (its a tiny bit smaller 36) to make sure, draw 4 times the 9 and you can see for yourself. just a bit smaller. 😉 ohh Im back to the beggining … again…

  2. Ricardo says:

    Its impossible not to go bak to the beggining and reconsider everything again and again, but can you be more specific, wich part should I look at, at the Geo for Kids?? cheers.

  3. Brodic says:

    Meni je izašlo 30 dijelov
    a luka :-/

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