Squaring the circle by equalization.
Of ferman: Fernando Mancebo Rodriguez--- Personal page. ----Spanish pages

Email: ferman25@hotmail.com
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Squaring the circle by equalization.
Using inscribed triangle

Page in preparation

Fig. 1

This method of squaring the circle, (through the equalization of inscribed triangle segments) and as it is verified with the numerical adjustment that is exposed, represents a correct method and an easy achievement of the quadrature in a very simple way.
However, and although it will have its detractors due to the personal method that I use of square segment by approximation, I think that it can be considered as the exact square of the circle.
As we can see, it is based on the internal properties of the circumference, which as we will discover, said circumference possesses and numerous segments of interrelation with Pi are extracted from its interior, a question that to date I have not seen reflected in any work in this regard.

These segments embedded in the interior of the circumference range from the cube root of the diameter (2), which I already used to duplicate the cube, to various segments that are products and roots of the number Pi.
Well, these operations that can be done and adjusted numerically (for which I give data) can also be used with a ruler and compass to get the square of the circle.
As I have said before, my method of work is of approximation, confluence and equalization of segments related to each other.
Therefore, instead of making multiple adjustments and compositions of segments, here what is done is several (3-8) approximation measures but only on a pair of segments dependent on each other, until their confluence or equalization is achieved.

The first thing to discover and expose is a triangle inscribed to the circumference, which in turn contains segments that join them to Pi through mathematical operations such as duplication, product of segments, power of segments, roots, etc.
In figure1, we already have the main essence of the quadrature model, consisting of an inscribed triangle, whose base half (a) represents the fourth part of Pi, (pi / 4).
This would be devoid of importance if it were not for the fact that its height (b) to the center of the circumference (O) is in turn the square of this base half (b = a^2)
Thus, having this formula of segments, it is already beginning to suppose that it would be easy to find any of these segments as a function of the other, if only by simple product of segments, (one of the methods to use for quadrature).

Fig. 2

In the second drawing (Fig. 2) this triangle is exposed, but with much more data on segments, their location and properties.
This second drawing and data is very important because the diversity of segments relative to Pi, which the circumference has internally, is already glimpsed, and also, because another method of quadrature can be used (the one I like the most) due to the use of duplicate segments that It exists in this drawing (h1, h2), which added together make us reach the segment of half of Pi (pi / 2) located laterally from the vertex A of diameter to the point B that cuts the circumference.
That is, these segments (h1, h2) added, lead us to the key point to measure (Pi / 2).

Fig. 3

At the moment, and although there are several methods of quadrature with ruler and compass, here I am only going to expose the one that I normally use, that is, the one that I use the segments (h1, h2).

In figure 3 we see the ruler/square is placed on the vertical diameter, and at a distance of approximately 1/3 of the diameter of the circumference.
In this situation we fix the point (rm) and make an arc from the vertex of the vertical diameter (A) to the reference point or situation (rs).
And from this reference point (rs) we mark a perpendicular until we cut the circumference at the point (cs), and then extend that segment to the other side of the circumference (ps), which will help us to see if the measurement taken fits.
Now from the point (C) lower part of the vertical diameter, and with the length of the segment (s) we make an arc towards the point previously marked (ps), and with the same length, another arc from (ps) towards point (C) to check if the two arcs meet in the center.
As it will not be correct the first time, then we repeat the process until we get these arcs to coincide, because then it will be when the quadrature and all its segments are correct.

Fig. 4

Thus, we have made the final measurement (figure 4) in which we check that the segments (m) drawn on the position of Pi/2 (segment P) are coincident, and that therefore as we said, this segment P represents the correct value of Pi / 2.
Now this segment Pi/2, with the compass we move it to the vertical diameter reaching the point (pp), and from here we make its projection on the circumference to the point Sqr.
Well, from the lower vertex of the vertical diameter C, we mark another segment up to Sqr, which will be the square root of Pi, and therefore the side of the square that we are looking for to square the circle.