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.
SinceOC is a radius of the semicircle =>OC=r
By the Pythagorean theorem
Construct the midpoint D of AC.
midpoint D of AC.
Our goal is to show that the purple lune AECF is squarable.
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
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= 2n= 1
m= 3n= 1
m= 3n= 2
m= 5n= 1
m= 5n= 3
а = sin
, b = sin 1
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
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
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.
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
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.
References: “Bulgarian mathematicians” Sofia, avt.Ivan Chobanov, P. Roussev
SOU “Zheleznik” Bulgaria