Algebraic solution to 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.

Since OC is a radius of the semicircle => OC=r

By the Pythagorean theorem

2 .AC r

Construct the midpoint D of AC.

2 ,AC r

/ 2.( 2 )AD r

Construct the

Semicircle

midpoint D of AC.

Our goal is to show that the purple lune AECF is squarable.

area of

AEC= 2( ) / 2.r

ACB= 2.r

So, it terms of areasSo, it terms of areas

=1/2

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

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= 1m= 3 n= 1m= 3 n= 2m= 5 n= 1m= 5 n= 3

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

, b = sin 1а = sin

C = 122

1

SINSIN

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.

n Sin(m) = m Sin (n 1 )

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.

1

2k

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.

Х8 + Х 7 - 7Х6 + 15Х4 +10 Х3 – 10Х2 – 4Х + 1 - 17 = 0

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.

Chakalov use and another equation Х = cos2 + sin 2

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:

22 )1

1()

1

1()(

Xm

XXP XXX n

nmm

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.

Tomas Klausen

Hippocrates of Chiosabout 470 BC - about 410 BC

Tomas Klausen

Edmund Landau

Любомир Чакалов

Чеботарьов

Анатолий Дороднов

References: “Bulgarian mathematicians” Sofia, avt.Ivan Chobanov, P. Roussev

Made by:

Kalina Taneva

Ivelina Georgieva

Stella Todorova

Ioana Dineva

SOU “Zheleznik” Bulgaria