WHETHER it is POSSIBLE to CONSTRUCT “ CIRCLE QUADRATURE “? WHETHERCAN be CONSTRUCTED “ CIRCLE QUADRATURE “?
Construction subject “ not solvable “ tasks it is presented by very large volume of the literature and information which is laid out on the Internet. For example, the verses said by the astronomer Meton are quoted: I will take a ruler, I will draw a straight line,
I in a trice a circle will turn back a square, In the middle the market we will suit
A from it streets will go -
Well, as to the Sun! Though it
I round, and beams direct!.
Verses, say that the task already was known by then in Greece. One of Socrates`s contemporaries - the sophist Antifon considered that the quadrature of a circle can be carried out by means of a circle kvadrirovaniye. However already Aristotle proved that it will be only the approximate, but not exact solution of a task as the polygon can never coincide with a circle.
At the end of the 18th century the German mathematician I. Lambert and the French mathematician A. Legendre established irrationality of number π. In 1882 the German mathematician F. Lingdemang proved that number π (so i) it is transcendental, that is does not satisfy to any algebraic equation with the whole coefficients. Lingdemang`s theorem put an end to attempts of the solution of a task on a circle quadrature by means of compasses and a ruler.
The algorithm of creation of a quadrature of a circle can be calculated, without resorting to plotting as it was done by Hippocrates Hiossky, and to begin with creation of a square: measuring compasses of 10 equal pieces on a straight line. We build a parallel straight line (the second party of a square) which also we break into 10 equal pieces. We connect both parallel on the lines received as a result of a marking. At us the square broken into 10 levels turned out (vertical breakdown in this case it is not obligatory to do).
We carry out square diagonal from the left bottom corner to the right top corner. The diagonal passing through the marked square levels also is marked on 10 parts. we Find
the center of a square and we measure four pieces on diagonal from the center to the right top corner. This radius from the center of a square we draw a circle. The area of the received circle - is equal the areas of a square from the center which the circle is drawn.
Notes: This calculation is based on the principle of system construction when 4 - y (9 - y), the level of system is transforming. The method of this construction corresponds to the method of creation of a quadrature of a circle given above in Meton`s verse. The ancient observer had no modern knowledge and just in case give Thales` theorem: If the parallel direct crossing sides of angle, cut equal pieces on one its party, then they cut equal pieces and on other its party.
Solved and now, apparently, it is possible to go “ stomach forward “ on the contrary, because there is a certain dirty trick as we learned on the Internet that even the tasks solved these will be considered as unresolved. There is a certain cunning when for the solution of tasks the ruler is given without divisions and any tsifira, and work will be taken over as it is necessary “ with scales and " weights;. Any references to the oracle and “ Divine Providence “ will not help.
However the dishonourable relation of science to the solution of ancient tasks forces to draw the corresponding conclusions: Problems of creation of a quadrature of a circle of trisection of a corner and doubling of a cube are solved according to the principle “ Logical reasoning “ when for construction and verification of the solution of a task the same tools have to be used. From here the consequence follows, - the conclusions of scientists based on use of the tools which are not participating in creation of a task are not correct, and cannot approve impossibility of creation of the above-stated tasks.
From here: Lingdemang`s theorem cannot claim that the aforesaid tasks are not solved:
1. It contradicts conditions of tasks and conditions of creation of geometrical figures;
2. In a statement of the problem there are no cubic, and square roots there are no goniometric tools. On a condition of tasks mechanical construction by means of compasses and a line of a certain task, namely an equal square and a circle is necessary;
3. Geometrical figures a circle, a square or a cube are stable and steady figures which under the terms of the construction cannot be calculated by transcendental sizes.
Therefore Lingdemang`s theorem cannot serve as the incontestable proof of impossibility of the solution of the tasks stated above. It became clear that by results of Lingdemang`s theorem, today we have no tool to calculate the area of a circle. Is transcendental π which is used till this day, and there is no constant which has to be the second area of a circle, komponenty for calculation.
We do not even know how to check correctness of performance of these tasks, without applying any tsifira as it is demanded by statements of the problem. Unless only to return on two thousand years ago to Archimedes (who knew how to make it) and his bathtub with water. Here let scientists who cannot deal with the solution of Ancient Greek tasks will also be engaged in it. Therefore time the oracle told it will be necessary to build, so to build.
We measure a hypotenuse of a rectangular triangle with the parties 10 and 10.
So: The root square of 200 which we calculate approximately to 4 - y figures after zero is necessary, using a notebook in a cage and a ruler = 14. 1422.
Further we calculate radius: 14. 1422:10 × 4=5. 65688; then r ² =5. 65688 × 5. 65688=32. We calculate 000291
" From here; constant “: 100:32. 000291=3. 12497
So: The constant necessary for creation of a circle is found =3,12497
Check: 32. 000291 × 3. 12497=99. 9999
we Repeat calculation and we will take other rectangular triangle with the parties 11 × 11; We Calculate a hypotenuse and we find radius: 15. 55:10 × 4=6. 22; 6. 22 × 6. 22 × 3. 12497=120. 900;
at the same time 11 × 11=121.
We will repeat experiment once again and we will take the following rectangular triangle with the parties 12 × 12; We Calculate a hypotenuse and we find radius: 17:10 × 4=6. 8; 6. 8 × 6. 8 × 3. 12497=144. 49;
at the same time 12 × 12=144.
we Remove a formula Skp. = π territory of r ²; where Skp. - area of a circle; π the territory - Pythagoras`s constant (we will pay tribute to the founder of mathematics); r- circle radius.
We check calculations with an old and new formula on a square with the parties 10 × 10
Use of a formula Skp. = π r ²: External radius: 7. 0711 × 7. 0711=50. 00 × 3. 14=157
Apothem: 5 × 5=25. 00 × 3. 14=78. 5
Use of a formula Skp. = π territory of r ²: External radius: 7. 0711 × 7. 0711=50. 00 × 3. 12=156
Apothem: 5 × 5=25. 00 × 3. 12=78. 00
Two systems of the equations are called equivalent if these systems have the same decisions.
Consequence: If to replace each equation of system with the equivalent equation, then the system equivalent will turn out this.
The task was solved with use of a formula Skp. = π r ² where π = 3. 14, but then after calculation “ constant “ it is counted on a formula Skp. = π territory of r ² where π = 3. 12
Conclusion: A problem of construction equal on the area of a square and circle (“ circle quadrature “) by means of a ruler and compasses it is solvable.