Mathematical Innovation The problem that was given by the ancient mathematician (ARCHIMEDES) concerning the squaring of the circle by using ruler and diabetes challenged me to take the time with it, just as many people did in the past.

This known and still (unsolved) problem it’s true that it has bothered for more 2000 years many mathematicians and not only them, and it will remain unsolved for the years to come.

This happens many times; a difficult maths problem ends up in an easy solution and vice versa. This actually means that there is always a solution, unless the clues of a problem are given incorrectly, just like this particular problem (our problem). Let’s see the basic mistake, why the circle be squared from its real dimension.

We know that if take the diameter of a circle 1m. its length is said to be (3,1415…) m. a transcendental number (infinite). This is right, that (3,1415…) is a transcendental and irrational number, and it was proven by Ferdinand Von Lindermann 1882 (a) and Johann Heinrich Lambert 1761 (b).

For this reason and only, because of this (particular) irrational number ( 3,1415…) which was given wrong, the circle cannot be squared. It is logical! Because it is not the real π. number, the QUOTIENT. Since there is no theory or a formula that can prove or check the π (3,1415) the length of the periphery of a circle to its diameter that was given, then there can be a doubt if it is ( correct ).

Theorem and Proof I.E. ( Ioannis Efthimiadis )

In order to square the circle we have to know its two basic factors, the real number π and its radius. The first meaning of squaring the circle means that I check up the number π, in connection with its radius to the basis of the square. The second meaning is that they must have the same area. Therefore, up to today, they haven’t managed to square the circle and to prove number ( 3,1415…).

The ( 3,1415…) is a number approximate to the real number π. The proven number π, with the formula I.E. 2r-(2r/Φ)+r is an Sacred number: π I.E. 3,111...

Confirmation of π I.E. 3,111… with the formula I.E. 2r-(2r/Φ)+r

We take the radius of a circle 1m,
Which covers an area of π I.E. 3,111…

Square meters ( shape 1 ).

We shall square the circle with the formula
I.E. 2r-(2r /Φ)+r and we shall prove the
real number π I.E. 3,111… in the
following shape (2), (3) and (4). Then we take the radius of the circle
1m and we double it 2r = 2m
(shape 2). We set the golden mean in radius AO
2 r = 2m with the golden number Φ
(shape 3).    The formula I.E. 2r-(2r/Φ)+r confirms the π I.E. 3,111…
in relation with the radius of the circle, to the basis of the square,
so that the square has the same area with the circle.

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