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A lgebraicÂ solution to a Â geometricÂ problem

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A lgebraic solution to a  geometric problem. S quaring of a lune. Even in ancient times people have watched and studied the dependence of the moons and their daily lives.

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### Algebraic solutionto a geometric problem

Squaring of a lune

Even in ancient times people have watched and studied the dependence of the moons and their daily lives
Figure enclosed by two arcs of circles are called Freckles (moons) because of their similarity with the visible phases of the moon, the moon of earth.
The squaring of a plane figure is the construction – using only straightedge and compass – of a square having area equal to that of the original plane figure.
Construct the midpoint D of AC.

Construct the

Semicircle

midpoint D of AC.

area of

AEC=

ACB=

Small semicircle = 1/2 large semicircle.

=

=

In many archaeological excavations in the Bulgarian lands have found drawings of the moon in different

Phases.

Excavations in Baylovo

In 1840 T. Clausen (Danish mathematician) raises the question of finding all the freckles with a line and a compass, provided that the central angles of the ridges surrounding are equal. That means that there must be a real positive number Q and positive mutual goals primes m, n, such that the corners are met:

= m.Q 1= n Q

He establishes the same cases to examine and Hippocrates of Hios but expresses the hypothesis that freckles can be square in the following 5 cases:

m= 2 n= 1

m= 3 n= 1

m= 3 n= 2

m= 5 n= 1

m= 5 n= 3

а = sin

, b = sin 1

C =

In 1902 E. Landau deals with the question of squaring a moon. He proves that the moon can be squared of the first kind Numbers

Sin(m

) =

Sin(n

1 )

If the number of c= 0 is squarable moon, and if the corners are not commensurate Moon squarable. It is believed that he used the addiction, which is familiar and

T. Clausen.

Landau considered n = 1, m = p = +

a gaussian number in which the moon is squarable. When k = 1, k = 2 are obtained Hioski cases of Hippocrates.

= 0

Х8 + Х 7 - 7Х6 + 15Х4 +10 Х3 – 10Х2 – 4Х + 1 -

In 1929 Chakalov Lyubomir (Bulgarian mathematician) was interested of tLandau’s work and used algebraic methods to solve geometric problems. Chakalov consider the case p = 17, making X = cos 2 and obtained equation of the eighth grade.

He proves that this equation is solvable by radicals square only when the numbers generated by the sum of its roots are roots of the equation by Grade 4.

Х = cos2

+ sin 2

Chakalov use and another equation

Then:

n Xn ( Xm - 1 )2 – m Xm ( Xn – 1 )2 = 0

X = 1 is the root of the equation

So he gets another equation:

Chakalov consider factoring of this simple polynomial multipliers for different values ​​of m and n. So he found a lot of cases where the freckles are not squrable. It extends the results of Landau for non Gaussian numbers.
In 1934 N.G. Chebotaryov (Russian mathematician) Consider a polynomial Chakalov and proves that if the numbers m, n are odd, freckles squrable is given only in cases of Hippocrates, and in other cases not squaring.
In 1947 a student of Chebotaryov,

A. C. Dorodnov had proven cases in which polynomial of Chakalov is broken into simple factors and summarizes the work of mathematicians who worked on the problem before him. So the case of clauses is proven.

Thus ended the millennial history of a geometrical problem, solved by algebraic methods by mathematicians’ researches from different nationalities.
Made by:

Kalina Taneva

Ivelina Georgieva

Stella Todorova

Ioana Dineva

SOU “Zheleznik” Bulgaria