A lgebraic solution to a geometric 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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A lgebraic solution to a geometric problem

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 – (moons) because of their similarity with the visible phases of the moon, the moon of earth.using only straightedge and compass – of a square having area equal to that of the original plane figure.


Since (moons) because of their similarity with the visible phases of the moon, the moon of earth.OC is a radius of the semicircle =>OC=r

By the Pythagorean theorem


Construct the midpoint (moons) because of their similarity with the visible phases of the moon, the moon of earth.D of AC.

Construct the

Semicircle

midpoint D of AC.


Our goal is to show that the purple lune (moons) because of their similarity with the visible phases of the moon, the moon of earth.AECF is squarable.

area of

AEC=

ACB=


So it terms of areas
So, it terms of areas (moons) because of their similarity with the visible phases of the moon, the moon of earth.

=1/2


In terms of the first octant of our shaded figure, this says that:

Small semicircle = 1/2 large semicircle.


= that:

=


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 of Hios but expresses the hypothesis that freckles can be square in the following 5 cases: = 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 of Hios but expresses the hypothesis that freckles can be square in the following 5 cases: (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 = + of Hios but expresses the hypothesis that freckles can be square in the following 5 cases:

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


= 0 of Hios but expresses the hypothesis that freckles can be square in the following 5 cases:

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 = only when the numbers generated by the sum of its roots are roots of the equation by Grade 4.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, 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

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.


Tomas klausen
Tomas Klausen solved by algebraic methods by mathematicians’ researches from different nationalities.


Hippocrates of chios about 470 bc about 410 bc
Hippocrates of Chios solved by algebraic methods by mathematicians’ researches from different nationalities.about 470 BC - about 410 BC


Tomas klausen1
Tomas Klausen solved by algebraic methods by mathematicians’ researches from different nationalities.


Edmund landau
Edmund Landau solved by algebraic methods by mathematicians’ researches from different nationalities.


Любомир Чакалов solved by algebraic methods by mathematicians’ researches from different nationalities.


Чеботарьов solved by algebraic methods by mathematicians’ researches from different nationalities.


Анатолий Дороднов solved by algebraic methods by mathematicians’ researches from different nationalities.



Made by
Made by: Chobanov, P. Roussev

Kalina Taneva

Ivelina Georgieva

Stella Todorova

Ioana Dineva

SOU “Zheleznik” Bulgaria


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