pi of the circle vol-iv squaring an arbelos and circle
TRANSCRIPT
Pi OF THE CIRCLE
VOL-IV
Squaring an Arbelos and Circle
(A Compendium of articles)
By
R.D. Sarva Jagannadha Reddy
Retired Zoology Lecturer
19-42-S7-374, S.T.V. Nagar, Tirupati – 517 501, India.
Ph: +91 9494724403, E-mail: [email protected]
December-2015
Dedicated humbly
to
HIPPOCRATES OF CHIOS
(450 B.C.)
Founding Father of Mathematics
An Appeal
Dear Professor,
Namasthe !
I request your valuable comments, please, and it will be
included as and when they are received in the book. I spent
43 years to author this manuscript. I am now 70. Can’t you spend
some hours to read and comment, Sir ?
RSJ Reddy Author India
Comments
1
2
Preface
The widely accepted view on the “Squaring a circle” is that it is
an unsolved geometrical problem, leave alone arbelos of Archimedes.
Inspite of this view, mathematicians of the world have been successful
but their works were not recognized and on the other hand such people
were ridiculed as “Circle Squarers”. This concept is divine. The fault
lies in choosing a polygon’s number 3.1415926… of Exhaustion method.
Due to improper understanding of the circle and square this number
has ruled the world as Pi of the Circle. In this small book different
articles pertaining to the squaring of circle, circling a square and
squaring an arbelos are found. The readers are requested their
comments and suggestions for the improvement of this book.
Author
CONTENTS
S.No. Title Page No.
Preface
1. Hippocratean Squaring Of Lunes, Semicircle and Circle 1-8
2. Hippocrates’s Squaring of A Semi Circle (RSJ Reddy’s
Apology to CLF Lindemann)
9-12
3. One More Evidence for the Squaring of Circle from the
Proof of Howard Eves for the Cavalieri’s Principle
13-20
4. Leonardo Da Vinci’s Ingenious Way of Carving One-
fourth Area of A Segment in a Circle
21-29
5. A Durga Labarum 30
6. Circle A Square 31-37
7. Squaring of circle and arbelos and the judgment of
arbelos in choosing the real Pi value (Bhagavan Kaasi
Visweswar method)
38-45
8. Durga Method of Squaring A Circle 46-47
9. Pythagorean way of Proof for the segmental areas of one
square with that of rectangles of adjoining square
48-51
10. Squaring of Circle, Squaring of Semi Circle and Semi
Circle Equated to Rectangle and Triangle (Euclid’s
Knowledge of Algebraic Nature of Circle and its Pi)
52-60
APPENDIX 61
11. Trisection of an Angle of 900 62
12. Duplication of the Cube 63-64
IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. II (May-Jun. 2014), PP 39-46
www.iosrjournals.org
www.iosrjournals.org 39 | Page
Hippocratean Squaring Of Lunes, Semicircle and Circle
R. D. Sarva Jagannadha Reddy 19-9-73/D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati – 517 501, A.P., India
Abstract: Hippocrates has squared lunes, circle and a semicircle. He is the first man and a last man. S.
Ramanujan is the second mathematician who has squared a circle upto a few decimals of equal to
3.1415926… The squaring of curvature entities implies that lune, circle are finite entities having a finite
magnitude to be represented by a finite number.
Keywords: Squaring, lune, circle, Hippocrates, S. Ramanujan, , algebraic number.
I. Introduction Hippocrates of Chios was a Greek mathematician, geometer and astronomer, who lived from 470 until
410 BC. He wrote a systematically organized geometry text book Stoicheia Elements. It is the first book. And
hence he is called the Founding Father of Mathematics. This book was the basis for Euclid’s Elements.
In his days the value was 3 of the Holy Bible. He is famous for squaring of lunes. The lunes are
called Hippocratic lunes, or the lune of Hippocrates, which was part of a research project on the calculation of
the area of a circle, referred to as the „quadrature of the circle‟. What is a lune ? It is the area present between
two intersecting circles. It is based on the theorem that the areas of two circles have the same ratio as the
squares of their radii.
His work is written by Eudemus of Rhodes (335 BC) with elaborate proofs and has been preserved by
Simplicius.
Some believe he has not squared a circle. This view has become very strong with the number
3.1415926… a polygon‟s value attributed to circle, arrived at, from the Exhaustion method (EM) prevailing
before Archimedes (240 BC) of Syracuse, Greece, and refined it by him, hence the EM is also known as
Archimedean method. This number 3.1415926… has become much stronger as value, and has been
dissociated from circle-polygon composite construction, with the introduction of infinite series of Madhavan
(1450) of South India, and independently by later mathematicians John Wallis (1660) of England, James
Gregory (1660) of Scotland.
With the progressive gaining of the importance of 3.1415926… as value from infinite series, the
work of „squaring of circle‟ of Hippocrates has gone into oblivion. When the prevailing situation is so, in the
mean time, a great mathematician Leonhard Euler (1707-1783) of Switzerland has come. His record-setting
output is about 530 books and articles during his lifetime, and many more manuscripts are left to posterity. He
had created an interesting formula ei
+1 = 0 and based on his formula, Carl Louis Ferdinand Lindemann
(1852-1939) of Germany proved in 1882 that was a type of nonrational number called a transcendental
number. (It means, it is one that is not the root of a polynomial equation with rational coefficients. Another
way of saying this is that it is a number that cannot be expressed as a combination of the four basic arithmetic
operations and root extraction. In other words, it is a number that cannot be expressed algebraically).
Interestingly, the term transcendental number is introduced by Euler.
When all these happened, naturally, the work on the Squaring of circle by Hippocrates was almost
buried permanently.
This author with his discovery in March 1998 of a number 14 2
4
= 3.1464466… from Gayatri
method, and its confirmation as value, from Siva method, Jesus proof etc. later, has made the revival of the
work of Hippocrates. Hence, this submission of this paper and restoring the golden throne of greatness to
Hippocrates has become all the more a bounden duty of this author and the mathematics community.
II. Procedure I. Squaring of Lunes-(1)
Hippocrates has squared many types of lunes. In this paper four types of lunes are studied.
“Consider a semi-circle ACB with diameter AB. Let us inscribe in this semi-circle an isosceles triangle
ACB, and then draw the circular are AMB which touches the lines CA and CB at A and B respectively. The
segments ANC, CPB and AMB are similar. Their areas are therefore proportional to the squares of AC, CB and
1
Hippocratean Squaring Of Lunes, Semicircle and Circle
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AB respectively, and from Pythagora‟s theorem the greater segment is equivalent to the sum of the other two.
Therefore the lune ACBMA is equivalent to the triangle ACB. It can therefore be squared.”
The circular arc AMB which touches the lines CA and CB at A and B respectively
can be drawn by taking E as the centre and radius equal to EA or EB.
1. AB = diameter = d
2. DE = DC = radius = d/2
3. F = midpoint of AC
4. N = midpoint of arc AC.
5. NF = 2d d
2 2
, DM =
2d d
2
, MC =
2d d
2
6. Area of ANC = Area of CPB = 2d
216
7. Area of AMB = Areas of ANC + CPB = 2d
28
8. Area of ACM = Area of BCM = 2d
416
9. Area of ACB triangle =
21 d dd
2 2 4
10. According to Hippocrates the area of the lune ACBMA is equal to the area of the triangle ACB.
Area of Lune ACBMA = Area of triangle ACB
(ANC + CPB + ACM + BCM)
2 2 2d d d
2 2 2 416 16 4
Squaring of Lunes-(2)
11. “Let ABC be an isosceles right angled triangle
inscribed in the semicircle ABOC, whose centre is O. On
AB and AC as diameters described semicircles as in the
figure. Then, since by Ecu. I, 47,
Sq. on BC = Sq. on AC + Sq on AB.
Therefore, by Euc. XII, 2,
Area semicircle on BC = Area semicircle on AC + Area semicircle on AB.
Take away the common parts
Area triangle ABC = Sum of areas of lunes AECD and AFBG.
Hence the area of the lune AECD is equal to half that of the triangle ABC”.
12. BC = diameter = d,
13. OB = OC = radius = d/2
14. AB = AC = 2d
2 = diameter of the semicircle ABF = ACD
15. GI = EJ = 2d 2d
4
16. Sq. on BC = Sq. on AC + Sq. on AB
17. Area of the larger semicircle = BAC =
2d
8
18. Area of the smaller semicircle = ABF = ACD
Diameter = 2d
2
2
Hippocratean Squaring Of Lunes, Semicircle and Circle
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Area =
2
2 2
2d
2d d
8 8 16
18. a) Areas of two smaller semicircles =
2 2d d2
16 8
19. Area of the triangle ABC = 1
base altitude2
Base = BC = d, Altitude = OA = d
2
Area = 21 d d
d2 2 4
20. Segment AIBG = Segment AJCE
Areas of AIBG + AJCE = 2 2 2d d d
2 2 216 16 8
21. Lune AGBF = lune AECD
22. Area of the lune (AGBF or AECD)
= Semicircles (ABF & ACD) – Segments (AIBG & AJCE) 2
2 22 dd d
8 8 4
Squaring of lunes-(3)
“There are also some famous moonshaped figures. The best known
of these are the crescents (or lunulae) of Hippocrates. By the
theorem of Thales the triangle ABC in the first figure is right
angled: Thus p2 = m
2 + n
2. The semicircle on AB = p has the area
AAB = p2/8; the sum of the areas of the semicircles on AC
and BC is AAC+ABC= (n2 + m
2)/8 and is thus equal to AAB.
From this it follows that:
The sum of the areas of the two crescents is the area of the
triangle.”
23. AB = diameter = d
24. BC = radius = d/2
25. AC = 3d
2
26. DF = d/4
27. DE = 2d 3d
4
28. EF = 3d d
4
29. GH = d/4
30. GJ = 3d
4
31. HJ = GJ – GH = 3d d 3d d
4 4 4
32. Area of the semicircle BDCF
= 2d d 1d
2 2 8 32
where BC = diameter =
d
2
33. Area of the semicircle AGCJ
= 23d 3d 1 3
d2 2 8 32
where AC = diameter =
3d
2
3
Hippocratean Squaring Of Lunes, Semicircle and Circle
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34. Area of the triangle ABC = 1 1 3d d
AC BC2 2 2 2 = 23
d8
35. Area of the curvature entity BDCE = 22 3 3d
48
36. Area of the curvature entity AGCH = 1
Circle AGCH3
22d 3 3 1
d4 16 3
= 24 3 3d
48
Area of the triangle = 23 3d
16
, where side = AC = 3
d2
37. Area of the lune BECF =
Semicircle BDCF – BDCE segment = 2 23 3d d
32 48
= 26 3d
96
(S.No. 32) (S.No. 35)
38. Area of the lune AHCJ
Semicircle AGCJ – AGCH segment = 2 23 4 3 3d d
32 48
= 26 3d
96
(S.No. 33) (S.No. 36)
39. Sum of the areas of two lunes = area of the triangle
(S.No. 37) + (S.No. 38) (S.No. 34)
= 2 26 3 6 3d d
96 96
= 23d
8
Squaring of lunes – (4)
The sum of the areas of the lunes is eqal to the area of the square.
40. AB = side = d
41. DE = EC = d/2
42. AO = OC = 2d
2
43. EF = 2d d
2
44. FG = 2d d
2
45. Area of the circle =
2d
4
Where diameter = 2d
12d 2d
4 =
22d
d2 2
46. Area of the semicircle DECG
Where DC = diameter = d = 2
2dd
8 8
47. Area of the curvature entity DECF = 22d
8
48. Area of the lune DFCG
Semicircle DECG – Curvature entity DECF = 2
2 22 dd d
8 8 4
49. The sum of the areas of 4 lunes = the area of the square
4
Hippocratean Squaring Of Lunes, Semicircle and Circle
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2 2 224 d d d
8 8
III. Squaring of a semicircle
“Hippocrates next inscribed half a regular hexagon ABCD in a
semicircle whose centre was O, and on OA, AB, BC, and CD as
diameters described semicircles. The AD is double any of the lines
OA, AB, BC and CD,
Sq. on AD = Sum of sqs. On OA, AB, BC and CD,
Area semicircle ABCD = sum of areas of semicircles on OA,
AB, BC and CD.
Take away the common parts.
Area trapezium ABCD = 3 lune AEFB + Semicircle on OA”.
50. DA = diameter = d
51. Area of the semicircle DABC = 2
2dd
8 8
52. DA
2 radius of larger semi circle =
dAB
2
53. AB = d
2 = diameter of smaller semi circle ABE
22d d d 1
d8 2 2 8 32
54. Areas of semicircle on OA, AB, BC and CD = 2d32
55. Area of sector OAFB = 2
2d 1d
4 6 24
56. Area of the triangle OAB = 23d
16
57. Area of the segment AFB = Sector – Triangle
2 2 23 2 3 3d d d
24 16 48
58. Area of lune AEBF = Semicircle on AB – AFB segment
=
2 2 26 3 3 42 3 3
d d d x32 48 96
Area of one lune = x
59. Area of 3 lunes =
26 3 3 4
3 d96
=
26 3 3 4
d32
60. Area of 3 lunes + semicircle on OA
=
2 26 3 3 4
d d32 32
=
2 2 26 3 3 4 6 3 3 3
d d d32 32 16
61. Area of trapezium = 3 x OA triangle
(S.No. 56)
= 2 23 3 33 d d
16 16
62. Area of 3 lunes + semicircle on OA =Area of trapezium = 2 23 3 3 3d d
16 16
(S.No. 60)
5
Hippocratean Squaring Of Lunes, Semicircle and Circle
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IV. Squaring of circle
“Consider two concentric circles with common centre O and radii such that the
square of the radius of the larger circle is six times the square of the radius of
the smaller one. Let us inscribe in the smaller circle the regular hexagon
ABCDEF. Let OA cut the larger circle in G, the line OB in H and the line OC
in I. On the line GI we construct a circular segment GNI similar to the segment
GH. Hippocrates shows that the lune GHIN plus the smaller circle is
equivalent to the triangle GHI plus t he hexagon.”
63. OA = radius of the smaller circle = d
2
64. OH = radius of the larger circle =
2d
62
= 6
d2
65. Third circle: GI = radius = GK + KI
66. OH = OI = 6
d2
67. OK = OH 6
d2 4
68. KI = 2 2 3 2
OI OK d4
69. Radius of the third circle
= GI = 2 x KI = 3 2
d2
70. Area of the GHI triangle = 1
GI HK2
OH 6HK d
2 4
= 1 3 2 6
d d2 2 4
= 23 3d
8
71. Area of the AOB triangle
OA = AB = d
2; AP =
OA d
2 4
PB = 2 2
AB AP = 3
d4
;
Area =
21 1 d 3 3OA PB d d
2 2 2 4 16
72. Area of the hexagon = Area of the triangle AOB x 6
= 2 2 23 6 3 3 3d 6 d d
16 16 8
73. Area of the smaller circle = 2
2dd
4 4
74. Area of the segment GH = Segment HI
75. Area of the larger circle =
2d
4
Where d = 6
d 22
= 6d = 21 66d 6d d
4 4
76. Area of the larger circle is divided into 6 sectors = 2 26 1d d
4 6 4
Hippocratean Squaring Of Lunes, Semicircle and Circle
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77. Area of the triangle OGH = OHI = GHI = 23 3d
8
(S.No. 70)
78. Area of the GH segment = HI segment
= Sector – Triangle (OGH) = 2 23 3d d
4 8
= 23 3d
4 8
(S.No. 76) (S.No. 77)
There are two segments GH and HI = 2 23 3 2 3 32 d d
4 8 4
79. Similarly, GNIK is also another segment which is the part of the sector GNIQ. It consists of the triangle
GIQ and GNIK segment.
80. To find out the area of the sector GNIQ, let us first find out the area of the circle whose diameter is equal
to that of the third circle.
Diameter of the third circle = riadus x 2 = 3 2
d 2 3 2 d2
(S.No. 69)
Area = 2d
4
=
13 2 d 3 2d
4 = 2 21 9
18 d d4 2
81. Then let us find out the area of the sector = 1
th6
= 2 2 29 1 9 3
d d d2 6 12 4
82. Now let us find out the area of the triangle GIQ = 1
GI KQ2
Where KI = GI 3 2 1 3 2
d d2 2 2 4
GI = QI = GQ = Radius of the third circle.
2 2
2 2 3 2 3 2KQ QI KI d d
2 4
= 3 6
d4
Area = 1
GI KQ2 = 21 3 2 3 6 9 3
d d d2 2 4 8
83. Area of the segment GNIK = Sector – Triangle
(S.No. 81) (S.No. 82)
2 2 23 9 3 6 9 3d d d
4 8 8
84. Now it has become possible to calculate the area of GHIN segment
= Triangle GHI – Segment GNIK
(S.No. 70) (S.No. 83)
= 2 2 23 3 9 3 6 3 3d d d
8 8 4
Area = 26 3 3d
4
85. Area of the lune GHIN
Segments + Segments + Circle =
GH & HI GHIN
S.No. 78 S.No. 84 S.No. 73
= 2 2 2 22 3 3 6 3 3 3 3d d d d
4 4 4 4
86. Area of the triangle GHI + Area of the hexagon ABCDEF
(S.No. 70) (S.No. 72)
7
Hippocratean Squaring Of Lunes, Semicircle and Circle
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2 2 23 3 3 3 3 3d d d
8 8 4
87. Area of lune + Circle = 23 3d
4
= Area of triangle+ hexagon = 23 3d
4
(S.No. 85) (S.No. 86)
So, the sum of the areas of lune and circle is equal to the sum of the areas of triangle and hexagon.
V. Post Script The following are the points on which some thinking is necessary:
1. 3.14159265358… is accepted as value.
2. 3.14159265358… is a transcendental number.
3. As this polygon‟s value is accepted as of the circle, circle and its value have become transcendental
entities.
4. The concept of transcendental number vehemently opposes squaring of circle.
Latest developments
5. 14 2
4
= 3.14644660942… is the new value.
6. 14 2
4
is the exact value.
7. This number is an algebraic number, being the root of x2 – 56x + 97 = 0
8. Squaring of circle is done with this number.
Conclusion
9. Hippocrates did square the circle.
10. 3.14159265358… is a transcendental number – it is correct.
11. 3.14159265358… can not square a circle, - is also correct.
Final verdict
12. As Hippocrates did the squaring a circle, it amounts to confirming that circle and its value are algebraic
entities. It implies that as 3.14159265358… is a borrowed number from polygon and attributed to circle,
called a transcendental number, said squaring a circle an unsolved geometrical problem, the final verdict
is, all are correct, except one, i.e. attributing 3.14159265358 of polygon to circle. Hence
3.14159265358… is not a value at all.
References [1]. T. Dantzig (1955), The Bequest of the Greeks, George Allen & Unwin Ltd., London.
[2]. P. Dedron and J. Itard (1973) Mathematics and Mathematicians, Vol.2, translated from French, by J.V. Field, The Open University
Press, England. [3]. W.W. Rouse Ball (1960), A short Account of the History of Mathematics, Dover Publications, New York.
[4]. W.G.H. Kustner and M.H.H. Kastner (1975). The VNR Concise Encyclopedia of Mathematics, Van Nostrand Rusinhold Company.
[5]. R.D. Sarva Jagannadha Reddy (2014). Pi of the Circle at www.rsjreddy.webnode.com
8
1
HIPPOCRATES’S SQUARING OF A SEMI CIRCLE
(RSJ Reddy’s Apology to CLF Lindemann)
Introduction
It has been said that is a transcendental number and squaring of
circle is not possible. However, Hippocrates had squared a semicircle
and squared a full circle. Both semi circle and full circle were combined
with lunes and equated to straight-lined geometrical constructions.
Semicircle + 3 lunes = Trapezium
Full circle + lune = Triangle + Hexagon
This paper is clear that is not a transcendental number and
squaring a circle is not an impossible concept. 3.14159265358… is a
transcendental number. Infact, it is a polygon’s number, attributed to
circle, as its value. If the above proof is right, becomes an algebraic
number. And thus Hippocrates had foresaw that was an algebraic
number.
Here is a very very interesting and an eye opener observation
about the work of C.L.F. Lindemann by Petr Beckmann in his book A
History of in Page numbers 40 to 52.
“The fourth man of interest in this early Greek period is Hippias of Elis,
who came to Athens in the second half of the 5th century B.C., and who was the
first man on record to define a curve beyond the straight line and circle. It is
perhaps ironic that the next curve on the list should be a transcendental one,
skipping infinitely many algebraic curves, but the Greeks did not yet know
about degrees of a curve, and so they ate fruit from all orchards.” (Page No.40)
“A rectangle with sides u/2 and r then has area
½ u r = r2,
9
2
i.e., the same area as a circle with radius r, and since a rectangle is easily
squared by an elementary construction, the circle is squared by the use of
compasses and straightedge alone. Nevertheless, as we shall see, this
construction failed to qualify under the implicit rules of Greek geometry.”
(Page No. 42)
“What Lindemann proved in 1882 was not that the squaring of
the circle (or its rectification, or a geometrical construction of the number )
was impossible; what he proved (in effect) was that it could not be reduced to
the five Euclidean axioms.” (Page No. 52)
The Five Euclidean axioms are
“Euclid’s five foundation stones, or “obvious” axioms, were the
following:
I. A straight line may be drawn from any point to any other point.
II. A finite straight line may be extended continuously in a straight
line.
III. A circle may be described with any center and any radius.
IV. All right angles are equal to one another.
The fifth postulate is unpleasantly complicated, but the general idea
is also conveyed by the following formulation (not used by Euclid,
whose axioms do not contain the concept parallel).
V. Given a line and a point not on that line, there is not more than one
line which can be drawn through the point parallel to the original
line.” (Page No. 50)
Are we to blame for misunderstating C.L.F. Lindemann in saying
that he meant squaring a circle impossible by his proof of 1882 ? Yes.
This author is the first man who expresses his fault and requests late
C.L.F. Lindemann to excuse this humble worker. The polygon’s number
3.14159265358 is / may be transcendental but not constant definitely.
10
3
Because, this number is not of the circle. He differs still with C.L.F.
Lindemann even now (This author is fortunate that he could buy
Beckmann’s book on 14-06-2015 which he has been trying for many
years, but obtained through flipkart.com online and this author is
highly indebted and grateful to Beckmann for this author has been
blind to the reality all these 17 years seeing the literature –
misrepresented-about squaring of circle by CLF Lindemann.)
Procedure:
1. Square ABCD
2. Side = AB = Diameter
= EF = a
3. Draw two arcs, with
centres E and F and
with radius OE and
OF equal to a
2.
4. Draw a semicircle
with centre G and
radius DG= GC equal
to a
2.
5. Two quadrants OED
and OCF
Area of each quadrant = 2 2a 1 a
4 4 16
Areas of two quadrants = 2 2a a
216 8
11
4
6. Rectangle DEFC
DE = a
2, EF = a
Area of DEFC rectangle = DE x EF = 2a a
a2 2
7. Area of S segment (shaded area)
Rectangle DEFC – Two quadrants = S segment
S segment = 2 2a a
2 8 =
2 24a a
8 = 24
a8
8. Semi circle on DC diameter = a
Area of semi circle = 2a
8
9. According to Hippocrates of Chios (450 B.C)
Semi circle + S segment = Rectangle DEFC = EABF
Area of Semicircle = 2a
8
Area of S segment = 24a
8
Area of rectangle DEFC = 2a
2
Semicircle + S segment = Rectangle
2 24a a
8 8 =
2a
2
12
1
ONE MORE EVIDENCE FOR THE SQUARING OF CIRCLE FROM
THE PROOF OF HOWARD EVES FOR THE CAVALIERI’S
PRINCIPLE
Introduction
Hippocretes of Chios (450 B.C) has squared a semi circle and a
full circle along with a lune. He has equated the sum of the areas of
three lunes and a semicircle with the area of a trapezium. Similarly, he
has equated the sum of the areas of a lune and a full circle with the sum
of the areas of a triangle and a hexagon.
We have one more evidence for the squaring of a circle. Here, a
circle in area is equated to the area of a rectangle. This evidence is
obtained from the classic on .
: A Biography of the World’s
Most Mysterious Number
By
Alfred S. Posamentier & Ingmar Lehman
Page 293, 2004
Prometheus Books
59, John Glen Drive
Amherst, New York, 14228-2197
This author, humbly expresses his grateful thanks to the authors, Prof.
Alfred S. Posamentier and Prof. Ingmar Lehman and the Publisher
M/s. Pormetheus Books, New York for letting the world of
mathematics to have a great opportunity to see that squaring of circle is
not impossible, though the view about it is otherwise in the
mathematical establishment. I thank them again and again. This
author believes people will feel very happy that there is a clear evidence
13
2
of squaring of circle, questioning the false idea of calling as a
transcendental number. A latest evidence, indeed. Two great
mathematicians have been seen by the world of mathematics again,
when some thing different is going on in the Pi world. We are very
fortunate to see Francesco Bonaventura Cavalieri (1598-1647) and
Howard Eves, our legendary. F.B. Cavalieri is an Italian mathematician
and is famous for his Cavalieri’s principle. It states that
“Two solid figures are equal in volume if a randomly selected plane cuts
both figures in equal areas”.
Our mathematics historian professor Howard Eves developed
highly ingenious and very simple proof for the Cavalieri’s principle.
For this proof he was awarded, 1992 “George Polya Award”. His proof
says that “there exists a tetrahedron which has the same volume as a given
sphere”, or, as he says, where the two solids are “Cavalieri Congruent”.
Thus, we have two very recent evidences questioning 1. The
validity of upper limit of , i.e. less than 1/7 of Archimedes and 2. The
impossibility of squaring a circle of James Gregory (1660) of Scotland
and C.L.F. Lindemann (1882) of Germany.
Professor C.H. Edwards Jr and Professor David E. Penny of the
University of Georgia, Athens, have given in Page No. 295, of their very
voluminous classic.
Calculus and Analytic Geometry
2nd Edition, 1986
and published by Prentice-Hall International that “ lies between
3.133259323 3.133 and 3.14659265 3.147”.
14
3
Archimedean upper value is 3 1/7 = 22/7 = 3.142. The official
value 3.141592653 is in total agreement with the upper limit 3.142 of
Archimedes.
But, the latest finding for upper limit of is 3.1465… There is a
clear cut opinion on the value of and the concept of squaring a circle.
Let us see the proof of our Professor Howard Eves.
(By courtesy of authors and publisher)
UV = (r + x) (r – x) = (r2 – x2)
“Thus the area of circle H and the area of rectangle LSTK are
equal. So by Cavalieris’ theorem, the two volumes must be the same”.
For details, the above book on and / or
2. Howard Eves “Two supporting Theorems on Cavalieri
Congruence’s”. College Mathematics Journal 22, No. 2 (March 1991):
123-124.
I beg of you Sirs, the above two are crucial evidences for the
probable accurate value of and a clear evidence for the squaring of
circle. They imply further, that 3.14159265358… is a far lower value
and is not a transcendental number.
–Author
15
4
We call 3.14159265358…. and , a transcendental number. Is it apt
to call this number a transcendental ?
We inscribe a regular polygon in a circle say with 6 sides. We
double the number of sides and continue it, till the gap between the
inscribed polygon and the circumference disappears, leaving no gap
what so ever. We imagine the inscribed polygon ultimately becomes a
circle. Ludolph Van Ceulen (1610) of Germany has obtained 35
decimals of 3.14159265358… from the inscribed polygon having 262
sides. It means 4, 611, 686, 018, 427, 387, 904 sides.
However, according to J. Houston Banks et al of the book
Geometry (1972), in Page 409,
“…. Yet no matter how many times we repeat this process the perimeter
of the polygon will never actually reach the circumference.”
Oxford dictionary says
Transcendental = going beyond human knowledge
16
5
In the process we know the perimeter of the inscribed polygon
grows and grows, but it reaches an ending because the circle obstructs
the growth of the polygon. In otherwords, it is limited and not
limitless. Further, the surd 3 is used to calculate the perimeter of the
polygon. Number of decimals do not change from the beginning of 6
sides to 1062 sides. Only value changes. The number of decimals
remain unchanged, because 3 stands till the end. The Japan
mathematical society in its Encyclopaedic Dictionary of Mathematics,
in Page 1310, says “The theory of transcendental members is, however, far
from complete. There is no general criterion that can be utilized to characterize
transcendental numbers”.
Secondly, which one increases its extent in the above diagram ? Is
it inscribed polygon or circle ? Which one really a transcendental entity
? Is it polygon or circle ? The circle remains static. It observes silently
the coming of smaller polygon slowly towards it i.e. circle. The one
(circle) that is obstructing the continuous growth of the polygon is being
called a transcendental !
By the by, is there any entity that can be called, transcendental ?
Yes, there is, i.e. Space and nothing else. How ? Let us imagine. The
Cosmos consists of two entities. They are, physical universe comprising
of planets, stars, solar families, galaxies, clusters of galaxies; and two:
radiation. The radiation is often called electromagnetic radiation. Let
us do this in imagination. Let us go and close the eyes Sir, to our
imagination, very slowly. Let us think our Sun, other planets have
disappeared leaving Earth untouched. We stand in our imagination.
Next, other solar families, galaxies, clusters of galaxies have
disappeared one by one. And finally, our Earth has also disappeared
leaving us intact. What exists ? It is Space. Space has no physical
17
6
quality. So, no concept of “distance”. The space between two bodies is
distance. No two bodies and no distance. Here, now, we are in space.
Yes, in space, if we say so, we are wrong. How ? We are not in space. If
we say, we are “in space”, it implies that universe is a large container or
bowl of astronomical magnitude. Then, a question comes. What is
beyond the container ? No answer Sir. Albert Einstein has rightly said
that
“Physical objects are not in space, but these objects are spatially
extended. In this way the concept, ‘empty space’ loses its meaning.”
There is thus a clarity on “space” in saying not “in space” but
“spatially extended” of planets, stars etc. Let us come back Sir, from
our imagination.
“Science” is called a self correcting subject. It appears that it is
not the case with mathematics. Tomorrow science may correct the
following hypothesis. Mathematics, ultimately certifies, if correct.
What is that ? Here is a brief “thought experiment”. Either
“imagination” or “intuition”, both are necessary in mathematics just
like, they in science. Did anybody think of, how God controls the
World ? The universe is very big. Its distance is measured in “light
years”.
If God, say is with us. He gets a message that some accident has
happened on the other side/ boundary of the physical world, say
distance is trillions of kilometers or one thousand light years away. If
He starts instantly to go to the accidental spot with the speed of an
electromagnetic ray, He takes more time. By the time He goes, nothing
He can do there. In this imagination, our boy was standing in an open
ground, thinking about His administrative capability looking high
above into the sky.
18
7
A divine form suddenly appeared before this boy and enquired
about his problem. The boy narrated what was in his mind.
Immediately the divine form carried the boy on His shoulders, went
with such a high speed, to a far of place, say to the distance of one
thousand light years, and returned, leaving the boy and disappeared.
The boy looked at his watch and found, he took just less than a minute
for his up and down journey with the divine form. The boy thought and
thought for many years and came to the following conclusion. What is
the conclusion ?
1. The velocity of the light may not be fixed / constant.
2. Or, there must be a non-electromagnetic radiation whose
velocity of travel doubles successively in its onward passage
along the distance, for every unit of distance of velocity.
i.e. 1 2 4 8 16 32 …….
In other words, the speed of travel by the God, doubles successively, for
every unit of distance travelled in the first second of the first unit. For
example
A B C D E |--------------|--------------|--------------|--------------| 186,000 186000 186000 186000
Miles
For light ray to travel from A to E, it takes four seconds. But God
travels this distance between A and E in less than two seconds.
How ? First unit of distance (186000 miles) takes one second,
second unit half a second, third unit quarter second and fourth unit one
8th second.s
19
8
To end the discussion, the surd 3 which is used in arriving at
3.14159265358… is itself never ending in its decimal form. Inscribed
polygon is a finite entity drawn in another finite entity : circle. So,
naturally, it has to be represented by a finite number. As 3 is an exact
number in its surd form, 3 is suffice. Calling either 3.14159265358 of
polygon or circumference or area of circle or circle constant , the term
“transcendental” becomes redundant. And, to be accurate nothing in
the world is transcendental except Space.
Finally, as the mathematical giant Prof. Howard Eves has
equated the area of a circle to the area of a rectangle or in otherwords,
called, squaring a circle, the number can no more be called a
transcendental number. Further, the concept of “Squaring of Circle”
tells us that is a finite number, representing either area or
circumference of a circle, and can be equated with that of either square
or rectangle or triangle or trapezium. The new value 14 2
4 has
squared exactly a circle, it has circled a square exactly, it has
constructed a triangle whose area is equal to that of a given circle, it has
squared arbelos of Archimedes. 14 2
4 stands at every step for
exactness with all the straight-lined geometrical constructions. Though
all the constructions are very elementary in nature, that does not mean
the quality of this work is substandard.
“medeis ageometrtos eisito” - Plato
(“Let no one without geometry enter here” is at the portals of the
academy of Plato)
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IJESRT INTERNATIONAL JOURNAL OF ENGINEERING SCIENCES & RESEARCH
TECHNOLOGY
LEONARDO DA VINCI’S INGENIOUS WAY OF CARVING ONE-FOURTH AREA OF
A SEGMENT IN A CIRCLE R.D. Sarva Jagannadha Reddy*
* 19-42-S7-374, STV Nagar, Tirupati-517501, INDIA
ABSTRACT Hippocrates of Chios (450 BC) has squared lunes, semicircle and full circle. Forgetting his squaring of circle, and 1.
believing that 3.1415926 of polygon as Pi of the circle, 2. accepting wrong interpretation of Leonard Euler’s equation
of Pi radians and calling Pi constant as transcendental number by C.L.F. Lindemann, squaring of circle has become
as an unsolved geometrical problem. S. Ramanujan has partially succeeded in squaring a circle. Leonardo da Vinci
has carved out quarter area from a circle with the help of other circles. This paper is yet another attempt to prove that
circle and its Pi are algebraic entities.
KEYWORDS: Algebraic number, circle, diameter, squaring.
INTRODUCTIONCircle and square are basic entities of geometry. Straightedge is associated with square. Whereas, circle is associated
with compass, in addition to straightedge for its existence. From the beginning of human civilization till today, the
circle has not been understood properly. The reason is that the circumference of circle is a curvature. The instrument
compass can draw a circle but it can not measure its length. Unfortunately, we are yet to find out an instrument that
can measure the length of a circumference, like the straightedge for the length of the side, diagonal of a square.
The age old method called Exhaustion method helps us to understand the circle only partially. Inspite of this
deficiency, Hippocrates of Chios (450 B.C) has squared lunes, semi circle and full circle (Ref.9). The work that has
been based on the value 3.1415926 of polygon of the Exhaustion method has misled the world. Thus, nearly 3000
years have gone by. The new branch Calculus too has supported polygon’s value 3.1415926… as Pi of the circle
(Ref. 11). It is too has failed to reveal the true Pi value of circle. In March 1998 Nature had been kind and revealed
the true Pi value as 14 2
4 = 3.1464466… This value has survived next 17 years, fighting non-violently,
submissively and politely and single handedly against the army of supporters of 3.1415926… of polygon, calling itself
as Pi of the circle.
The new Pi value 14 2
4 has succeeded in convincing almost all members of the mathematics community and
that Hippocrates had squared circle. Surprisingly, no Professor came out openly in opposing either the value
3.1415926… is Pi value and in opposing that squaring of circle is not an impossible geometrical problem and come
out boldly in restoring the golden throne to the deserving Hippocrates of Chios because he was the first and the last
mathematician who had squared a circle (Ref. 9).
Prof. Robert Burn of Exeter, U.K. has sent this author on 19.09.2015, ingenious construction of Leonardo da Vinci
in carving out
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[40]
Diagram sent by Prof. Robert Burn
one-fourth area of a segment in a circle. It is explained and submitted to the world of mathematics that yet another
evidence is here to show a circle or a group of circles demarcate a certain area equal to quarter magnitude in a circle.
Nobody has done before this great construction. Now the mathematics community openly can come out in support
of Hippocrates and Leonardo da Vinci that squaring of circle was done. And believing polygon’s number 3.1415926…
still as Pi of the circle will definitely can be taken as a disservice to mathematics and misleading the world and the
innocent student community of the world in particular.
PROCEDURE
Diagram explained
Explanation of the diagram
1) AB = diameter of the larger circle = 1
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2) AD = diameter of the smaller circle = 2
2
3) DE = EB =
2 2
4
4) From Siva method (Ref. 4)
Area of FED =
22d
32 where d = 1
Area of 4 segments =
22 24 d
32 8 where d = 1
5) Area of larger circle =
2d
4 4 where d = 1
6) Area of smaller circle =
2d 2 2
4 4 2 2 8 where d =
2
2
7) Area of shaded area =
Area of larger circle – (Area of smaller circle + Areas of 4 segments)
=
2 1
4 8 8 4
So, the shaded area is equal to 1
4. It means this area is squared.
CONCLUSION Leonardo da Vinci has squared a segment of a circle with the help of other circles.
ACKNOWLEDGEMENTS Prof. Robert Burn of Sunnyside, Exeter, U.K. has been arguing with this author for the last more than a decade that
3.1415926… is the value of Pi implying further that squaring of circle is impossible too. This day i.e. Saturday, 19-9-
2015 he has sent this diagram of Leonardo da Vinci to this author. This author was shocked at first but felt very happy
that a great soul like Leonardo da Vinci (1452-1519), an Italian Renaissance Painter, sculptor, draftsman, architect,
engineer and scientist could alone change Prof. Robert Burn’s opinion that what is the true nature of Pi value. This
author congratulates this Professor of U.K. and is highly indebted to him for bringing this highly valuable
construction of Leonardo da Vinci to the world unknown till now.
REFERENCES [1] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, 2nd edition, Springer-
Verlag Ney York Berlin Heidelberg SPIN 10746250.
[2] Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the World’s Most Mysterious
Number, Prometheus Books, New York 14228-2197.
[3] David Blatner, The Joy of Pi (Walker/Bloomsbury, 1997).
[4] RD Sarva Jagannada Reddy (2014), New Method of Computing Pi value (Siva Method). IOSR Journal of
Mathematics, e-ISSN: 2278-3008, p-ISSN: 2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 48-49.
[5] RD Sarva Jagannada Reddy (2014), Jesus Method to Compute the Circumference of A Circle and Exact
Pi Value. IOSR Journal of Mathematics, e-ISSN: 2278-3008, p-ISSN: 2319-7676. Volume 10, Issue 1 Ver.
I. (Jan. 2014), PP 58-59.
[6] RD Sarva Jagannada Reddy (2014), Supporting Evidences To the Exact Pi Value from the Works Of
Hippocrates Of Chios, Alfred S. Posamentier And Ingmar Lehmann. IOSR Journal of Mathematics, e-ISSN:
2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. II (Mar-Apr. 2014), PP 09-12
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[42]
[7] RD Sarva Jagannada Reddy (2014), New Pi Value: Its Derivation and Demarcation of an Area of Circle
Equal to Pi/4 in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: 2321 –
4767 P-ISSN: 2321 - 4759. Volume 2 Issue 5, May. 2014, PP-33-38.
[8] RD Sarva Jagannada Reddy (2014), Pythagorean way of Proof for the segmental areas of one square with
that of rectangles of adjoining square. IOSR Journal of Mathematics, e-ISSN: 2278-3008, p-ISSN:2319-
7676. Volume 10, Issue 3 Ver. III (May-Jun. 2014), PP 17-20.
[9] RD Sarva Jagannada Reddy (2014), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal
of Mathematics, e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. II (May-Jun. 2014), PP
39-46
[10] RD Sarva Jagannada Reddy (2014), Durga Method of Squaring A Circle. IOSR Journal of Mathematics,
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 14-15
[11] RD Sarva Jagannada Reddy (2014), The unsuitability of the application of Pythagorean Theorem of
Exhaustion Method, in finding the actual length of the circumference of the circle and Pi. International
Journal of Engineering Inventions. e-ISSN: 2278-7461, p-ISSN: 2319-6491, Volume 3, Issue 11 (June 2014)
PP: 29-35.
[12] R.D. Sarva Jagannadha Reddy (2014). Pi treatment for the constituent rectangles of the superscribed square
in the study of exact area of the inscribed circle and its value of Pi (SV University Method*). IOSR Journal
of Mathematics (IOSR-JM), e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. I (Jul-Aug.
2014), PP 44-48.
[13] RD Sarva Jagannada Reddy (2014), To Judge the Correct-Ness of the New Pi Value of Circle By Deriving
The Exact Diagonal Length Of The Inscribed Square. International Journal of Mathematics and Statistics
Invention, E-ISSN: 2321 – 4767 P-ISSN: 2321 – 4759, Volume 2 Issue 7, July. 2014, PP-01-04.
[14] RD Sarva Jagannadha Reddy (2014) The Natural Selection Mode To Choose The Real Pi Value Based On
The Resurrection Of The Decimal Part Over And Above 3 Of Pi (St. John's Medical College Method).
International Journal of Engineering Inventions e-ISSN: 2278-7461, p-ISSN: 2319-6491 Volume 4, Issue 1
(July 2014) PP: 34-37
[15] R.D. Sarva Jagannadha Reddy (2014). An Alternate Formula in terms of Pi to find the Area of a Triangle
and a Test to decide the True Pi value (Atomic Energy Commission Method) IOSR Journal of Mathematics
(IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug. 2014), PP 13-
17
[16] RD Sarva Jagannadha Reddy (2014) Aberystwyth University Method for derivation of the exact value.
International Journal of Latest Trends in Engineering and Technology (IJLTET) Vol. 4 Issue 2 July 2014,
ISSN: 2278-621X, PP: 133-136.
[17] R.D. Sarva Jagannadha Reddy (2014). A study that shows the existence of a simple relationship among
square, circle, Golden Ratio and arbelos of Archimedes and from which to identify the real Pi value (Mother
Goddess Kaali Maata Unified method). IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-
ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug. 2014), PP 33-37
[18] RD Sarva Jagannadha Reddy (2015). The New Theory of the Oneness of Square and Circle. International
Journal of Engineering Sciences & Research Technology, 4.(8): August, 2015, ISSN: 2277-9655, PP: 901-
909.
[19] RD Sarva Jagannadha Reddy (2015), Pi of the Circle (III Volumes), at www.rsjreddy.webnode.com.
APPENDIX THE SECOND GREAT CONSTRUCTION OF LEONARDO DA VINCI
A Study That Shows The Oneness Of Square And Circle From The Mathematical Configuration Of The
Human Body
INTRODUCTION Prof. Mario Livio (of Is God a Mathematician, a book of him, published by Simon & Schuster Paperbacks, New
York, 2009, purchased by this author on 25-05-2015 and the next day this paper of Pi of the Circle (Page No. 195)
Ref. www.rsjreddy.webnode.com has acquired a new life this way) says in Preface “In this book I humbly try to clarify
both some aspects of the essence of mathematics and, in particular the nature of the relation between mathematics
and the world we observe”.
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In page 1 James Jeans (1877-1946) a British Physicist quoted as saying by him “The universe appears to have been
designed by a pure mathematician”.
So, it is very clear that here is one evidence, for the idea, of God as a mathematician, from ancient times, and as this
figure has its origin in the intelligence of the past and understanding, of the right relation between this Creation and
Mathematics.
This author, in one of his papers also said that “The one that converts “Nothing” into Everything is Mathematics”.
If one starts at this angle, searching, innumerable evidences will be our mathematical truths.
This author has been struggling to drive to a point that there is no difference between square and circle. Both are one.
Both are reversible in their origin. So, naturally, it implies that exists both in circle and square. This paper shows
that Human body also reflects the above concept of oneness of square and circle.
PROCEDURE 1. Square = ABCD, Side = AB = a
2. Inscribe a circle. Diameter = GF = Side = AB = a
3. Diagonals = AC = BD = 2a
Fig-1
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4. Two diagonals intersect the circumference at four points E, F, G and H, creating smaller square EGHF,
whose side is equal to 2a
2 = OE = OF = Radius =
a2
2 =
2a
2.
5. As a result of the formation of smaller EGHF square, four more third squares have formed. They are
QAKG, LHMB, FJCN and DIEP.
6. Let us find out the side of four third squares. For example
IJ = Parallel side = a
EF side = 2a
2 (S.No. 4)
IE = IJside EF side
2
= FJ
= 2a 1 2 2
a a2 2 4
= IE = CJ = 2 2
a4
= ID = DP = PE = NF = NC = FJ = CJ
7. AB + AD + DC + CJ
= a + a + a + 2 2
a4
= 2 2
3a a4
= 14 2
a4
8. 14 2
a4
is equal to the circumference of the inscribed circle.
9. The length of the circumference can be obtained from the following way also
14 AB sides ACdiagonal
4
= 14a 2a
4
= 14 2
a4
10. One fourth of the circumference of the inscribed circle can be obtained by the following process.
11. HF arc is equal to one fourth of the inscribed circle.
BC side of ABCD square also gives the exact length of HF arc. How?
12. BC = side = a
CJ = side = 2 2
a4
JB = BC – CJ = 2 2
a a4
= 2 2
a4
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Fig-2
13. Bisect JB twice
2 2 2 2 2 2a a a
4 8 16
14. Now deduct one fourth of JB side of 2 2
a4
from the side BC=a
JB JR + RB RS + SB
1
4 of JB side =
2 2 2 2a a
4 16
So, SB = 2 2
a16
2 2 14 2a a a
16 16
which is equal to quarter length of the circumference = FRH arc.
15. This way the full length of the circumference of the inscribed circle can be earmarked in the perimeter of
the ABCD square as BA + AD + DC + CJ = 3a + 2 2
a4
= 14 2
a4
of S.No. 7 and an
Arc of circumference say one quarter of it FRH can also be earmarked as follows
2 2CJ a
4
2 2JR a BR
8
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2 2SB a RS
16
Let us verify Side CB = CJ + JR + RS + SB = a
2 2 2 2 2 2 2 2a a a a a
4 8 16 16
1
4 of circumference = arc FRH which is equal to = CS =
= CJ+JR+RS = 2 2 2 2 2 2
a a a4 8 16
= 14 2
a16
i.e., CS = 14 2
a16
of S.No. 14
16. So, arc FRH = CS
Part-II
Square – Circle – Human Body
LEONARDO DA VAINCI’S CONSTRUCTION
Fig-3: Human body in the circle – square nexus
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[47]
Of Fig-1 = Of Fig.3 17. Centre O = Belly-button of the Human body.
18. E = Tip of the Right Fore Limb
19. F = Tip of the Left Fore Limb
G = Tip of the Right Hind Limb
H = Tip of the Left Hind Limb
20. Human body has a bilateral symmetry (which means “A type of arrangement of the parts and organs of
an animal in which the body can be divided into two halves that are mirror images of each other along one
plane only (usually passing through the midline at right angles to the dorsal and ventral surfaces). Bi-
laterally symmetrical animals are characterized by a type of movement in which one end of the body always
leads”).
21. In the human body when both the limbs are stretched they create a square equal to EGHF of Fig.1.
22. This EGHF square can be an inscribed square of ABCD larger square, when a circle of diameter equal to
side AB is inscribed in it.
23. The EGHF square has formed because of intersection of two diagonals of the larger ABCD square, at four
points of the inscribed circle.
24. Further, four tips of both the fore limbs and the hind limbs have created the four corners of EGHF.
25. Thus, it is clear that the smaller square EGHF in which human body’s configuration fits in cent per cent
exactly.
26. When four equi-distant tangents are drawn it creates ABCD square.
27. To conclude, the human body has mathematically been designed by The Nature. This is one clear example
for the belief that
GOD IS A MATHEMATICIAN And Mathematics is not thus a human creation, as Everything is God & God is
Everything.
29
30
31
32
33
34
35
36
37
IOSR Journal of Engineering (IOSRJEN) www.iosrjen.org
ISSN (e): 2250-3021, ISSN (p): 2278-8719
Vol. 04, Issue 07 (July. 2014), ||V3|| PP 63-70
International organization of Scientific Research 63 | P a g e
Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan Kaasi Visweswar method)
R.D. Sarva Jagannadha Reddy
Abstract: - value 3.14159265358… is an approximate number. It is a transcendental number. This number
says firmly, that the squaring of a circle is impossible. New value was discovered in March 1998, and it is
14 2
4
= 3.14644660942…….. It is an algebraic number. Squaring of a circle is done in this paper. With
this new value, exact area of the arbelos is calculated and squaring of arbelos is also done. Arbelos of
Archimedes chooses the real value.
Keywords: - Arbelos, area, circle, diameter, squaring, side
I. INTRODUCTION Circle and square are two important geometrical entities. Square is straight lined entity, and circle is a
curvature. Perimeter and area of a square can be calculated easily with a2 and 4a, where ‘a’ is the side of the
square. A circle can be inscribed in a square. The diameter ‘d’ of the inscribed circle is equal to the side ‘a’ of
the superscribed square. To find out the area and circumference of the circle, there are two formulae r2 and
2r, where ‘r’ is radius and is a constant. constant is defined as “the ratio of circumference and diameter of
its circle. So, to obtain the value for , one must necessarily know the exact length of the circumference of the circle. As the circumference of the circle is a curvature it has become a very tough job to know the exact value
of circumference. Hence, a regular polygon is inscribed in a circle. The sides of the inscribed polygon doubled
many times, until, the inscribed polygon reaches, such that, no gap can be seen between the perimeter of the
polygon and the circumference of the circle. The value of polygon is taken as the value of circumference of
the circle. This value is 3.14159265358…
In March 1998, it was discovered the exact value from Gayatri method. This new value is 14 2
4
=
3.14644660942.
In 1882, C.L.F. Lindemann and subsequently, Vow. K. Weirstrass and David Hilbert (1893) said that
3.14159265358… was a transcendental number. A transcendental number cannot square a circle. What is
squaring of a circle ? One has to find a side of the square, geometrically, whose area is equal to the area of a
circle. Even then, mathematicians have been trying, for many centuries, for the squaring of circle. No body could succeed except S. Ramanjan of India. He did it for some decimals of 3.14159265358… His diagram is
shown below.
Then the square on BX is very nearly equal to the area of the circle, the error being less than a tenth of an inch when the diameter is 40 miles long.
– S. Ramanujan
38
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 64 | P a g e
With the discovery of 14 2
4
= 3.14644660942… squaring of circle has become very easy and is done
here. Archiemedes (240 BC) of Syracuse, Greece, has given us a geometrical entity called arbelos. The shaded area
is called arbelos. It is present inside a larger semicircle but outside the two smaller semicircles having two
different diameters.
In this paper squaring of circle and squaring of arbelos are done and are as follows.
Squaring of inscribed circle
QD is the required side of square
Squaring of arbelos YB is the required side of square
II. PROCEDURE 1. Draw a square and inscribe a circle.
Square = ABCD, AB = a = side = 1
Circle. EF = diameter = d = side = a = 1
2. Semicircle on EF
EF = diameter = d = side = a = 1
Semicircle on EG
EG = diameter = 4a
5 =
4
5
Semicircle on GF = EF – EG = 4
15
= 1
5
GF = diameter = a
5=
1
5
3. Arbelos is the shaded region.
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Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 65 | P a g e
Draw a perpendicular line at G on EF diameter, which meets circumference at H. Apply Altitude theorem to
obtain the length of GH.
GH = EG GF = 4 1
5 5 =
2
5
4. Draw a circle with diameter GH = 2
5 = d
Area of the G.H. circle =
2d
4
=
2 2
4 5 5 25
5. Area of the G.H. circle = Area of the arbelos
So, area of the arbelos = 14 2 1
25 4 25
= 14 2
100
Part II: Squaring of circle present in the ABCD square 6. Diameter = EF = d = a = 1
Area of the circle =
2d
4
= 1 1
4 4
7. To square the circle we have to obtain a length equal to 4
. It has been well established by many
methods – more than one hundred different geometrical constructions – that value is 14 2
4
. Let
us find out a length equal to 4
.
8. Triangle KOL
OK = OL = radius = d
2 =
a
2 =
1
2
KL = hypotenuse = 2d
2 =
2a
2 =
2
2
DJ = JK = LM = MC = Side hypotenuse
2
= 2a 1
a2 2
=
2 2a
4
= 2 2
4
So, DJ = 2 2
4
9. JA = DA – DJ = 2 2
a a4
= 2 2
a4
. So, JA = 2 2
4
Bisect JA twice
JA JN + NA NP + PA
= 2 2
4
2 2
8
2 2
16
So, PA = 2 2
16
10. DP = DA side – AP = 2 2
116
= 14 2
16
40
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 66 | P a g e
11. 14 2
DP4 16
(As per S.No. 7)
12. Draw a semicircle on AD = diameter = 1
AP = 2 2
16
, DP =
14 2
16
13. Draw a perpendicular line on AD at P, which meets semicircle at Q. Apply Altitude theorem to obtain
PQ length
PQ AP DP = 2 2 14 2
16 16
= 26 12 2
16
14. Join QD
Now we have a triangle QPD
26 12 2PQ
16
,
14 2PD
16
Apply Pythagorean theorem to obtain QD length
QD = 2 2
PQ PD =
2 2
26 12 2 14 2
16 16
= 14 2
4
15. 14 2
4
is the length of the side of a square whose area is equal to the area of the inscribed circle
4
, where
14 2
4
,
14 2
4 16
Side = 14 2
a4
Area of the square = a2
2
14 2
4
= 14 2
16
Thus squaring of circle is done.
Part III: Squaring of arbelos
The procedure that has been adopted for squaring of circle is also adopted here. Here also the new value alone
does the squaring of arbelos, because, the derivation of the new value 14 2
4
= 3.14644660942… is based
on the concerned line-segments of the geometrical constructions.
16. Arbelos = EKHLFG shaded area. GH = Diameter (perpendicular line on EF diameter drawn from G
upto H which meets the circumference of the circle.
Area of the arbelos = Area of the circle with diameter GH = 25
of S.No.4
So, 25
14 2 1
4 25
= 14 2
100
, where
14 2
4
17. To square the arbelos, we have to obtain a length of the side of the square whose area is equal to area
of the arbelos 14 2
100
.
18. EG = diameter = 4
5. I is the mid of EG.
41
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 67 | P a g e
EI + IG = EG = 2 2 4
5 5 5
So EI = 2
5
19. Small square = STBR
Side = RB = EI = 2
5
Inscribe a circle with diameter 2
5 = side, and with Centre Z. The circle intersects RT and SB diagonals at K’
and L’. Draw a parallel line connecting RS side and BT side passing through K’ and L’.
20. Triangle K’ZL’
ZK’ = ZL’ = radius = 1
5
K’L’ = hypotenuse = 1
25 =
2
5
RB = 2
5
21. L’U = Side hypotenuse
2
=
2 2 1
5 5 2
=
2 2
10
22. So, L’U = 2 2
10
= BU
BT = Side of the square = 2
5
UT = BT – BU = 2 2 2 2 2
5 10 10
So, UT = 2 2
10
23. Bisect UT twice
UT UV + VT VX + XT
2 2 2 2 2 2
10 20 40
So, XT = 2 2
40
24. BT = 2
5; XT =
2 2
40
BX = BT – XT = 2 2 2
5 40
BX = 14 2
40
25. Draw a semi circle on BT with 2
5 as its diameter.
26. Draw a perpendicular line on BT at X which meets semicircle at Y.
42
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 68 | P a g e
XY length can be obtained by applying Altitude theorem
14 2 2 2XY BX XT
40 40
=
26 12 2XY
40
27. Triangle BXY
14 2BX
40
,
26 12 2XY
40
BY can be obtained by applying Pythagorean Theorem
2 2BY BX XY =
22
14 2 26 12 2
40 40
= 14 2
10
BY is the required side of the square whose area is equal to the area of the arbelos of Archimedes.
Side = 14 2
10
= a
Area of the square on BY = a2 =
2
14 2
10
= 14 2
100
of S.No. 16
= Area of arbelos
Part-IV (The Judgment on the Real Pi value)
In this paper, the correctness of the area of the arbelos of Archimedes can be confirmed. How ? Here are the
following steps.
28. New value 14 2
4
gives area of the arbelos as
14 2
100
= 0.12585786437. Whereas the official
value 3.14159265358… gives the area of the arbelos as
2d
4
= 3.14159265358 x d x d x
1
4
d = GH = 2
5 of S.No. 3
3.14159265358 2 2 1
5 5 4 = 0.12566370614
Thus, the following are the two different values for the same area of the arbelos.
Official value gives = 0.12566370614
New value gives = 0.12585786437
29. Diameter of the arbelos circle GH = d = 2
5
Square of the diameter = d2 = 2 2
5 5
= 4
25
Reciprocal of the square of the diameter = 2
1 1 25
4d 4
25
30. Area of arbelos, if multiplied with 25
4 we get the area of the inscribed circle in the ABCD square
Area of the circle =
2d
4
43
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 69 | P a g e
d = a = 1, 14 2
4
= 14 2 1
1 14 4
=
14 2
16
31. Area of the arbelos reciprocal of the square of the arbelos circle’s diameter = Area of the
inscribed circle in ABCD square
14 2 25
100 4
= 14 2
16
S. No. 16 S.No.29 S.No. 30
32. Let us derive the following formula from the dimensions of square ABCD
ABCD square, AB = side = a = 1
AC = BD = diagonal = 2a = 2 , Perimeter of of ABCD square = 4a
Perimeter of ABCDsquare
1Half of 7 timesof ABsideof square th of diagonal
4
= 4a 4
7a 2a 7 2
2 4 2 4
= 4 16
14 2 14 2
4
33. In this step, above 2 steps (S.No. 29 and 32) are brought in.
Arbelos area x 25 16
4 14 2
= Area of the ABCD square, equal to 1.
As there are two values representing for the same area of the arbelos, let us verify, with the both the values, which is ultimately the correct one.
Arbelos area of official value 3.14159265358
25 160.12566370614
4 14 2
= 0.99845732137 and
Arbelos area of new value 14 2
4
14 2 25 161
100 4 14 2
This process is done by understanding the actual and exact interrelationship among, 1. area of the ABCD
square, 2. area of the inscribed circle in ABCD square and, 3. area of the arbelos of Archimedes.
34. For questions “why”, “what” and “how” of each step, the known mathematical principles are
insufficient, unfortunately.
So, as the exact area of ABCD square equal to 1 is obtained finally with new value. The new value equal to
14 2
4
is confirmed as the real value. This is the Final Judgment of arbelos of Archimedes.
III. CONCLUSION This study, proves, that squaring of a circle is not impossible, and no more an unsolved geometrical problem.
The belief in its (squaring of circle) impossibility is due to choosing the wrong number 3.14159265358… as
value. The new value 14 2
4
has done it. The arbelos of Archimedes has also chosen the real value in
association with the inscribed circle and the ABCD superscribed square.
44
Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan
International organization of Scientific Research 70 | P a g e
REFERENCES
[1] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, 2nd edition,
Springer-Verlag Ney York Berlin Heidelberg SPIN 10746250.
[2] Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the World’s Most Mysterious Number, Prometheus Books, New York 14228-2197.
[3] David Blatner, The Joy of Pi (Walker/Bloomsbury, 1997).
[4] RD Sarva Jagannada Reddy (2014), New Method of Computing Pi value (Siva Method). IOSR Journal
of Mathematics, e-ISSN: 2278-3008, p-ISSN: 2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP
48-49.
[5] RD Sarva Jagannada Reddy (2014), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-ISSN: 2278-3008, p-ISSN: 2319-7676. Volume 10,
Issue 1 Ver. I. (Jan. 2014), PP 58-59.
[6] RD Sarva Jagannada Reddy (2014), Supporting Evidences To the Exact Pi Value from the Works Of
Hippocrates Of Chios, Alfred S. Posamentier And Ingmar Lehmann. IOSR Journal of Mathematics, e-
ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. II (Mar-Apr. 2014), PP 09-12
[7] RD Sarva Jagannada Reddy (2014), New Pi Value: Its Derivation and Demarcation of an Area of
Circle Equal to Pi/4 in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN:
2321 – 4767 P-ISSN: 2321 - 4759. Volume 2 Issue 5, May. 2014, PP-33-38.
[8] RD Sarva Jagannada Reddy (2014), Pythagorean way of Proof for the segmental areas of one square
with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-ISSN: 2278-3008, p-
ISSN:2319-7676. Volume 10, Issue 3 Ver. III (May-Jun. 2014), PP 17-20. [9] RD Sarva Jagannada Reddy (2014), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR
Journal of Mathematics, e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. II (May-Jun.
2014), PP 39-46
[10] RD Sarva Jagannada Reddy (2014), Durga Method of Squaring A Circle. IOSR Journal of
Mathematics, e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 14-
15
[11] RD Sarva Jagannada Reddy (2014), The unsuitability of the application of Pythagorean Theorem of
Exhaustion Method, in finding the actual length of the circumference of the circle and Pi. International
Journal of Engineering Inventions. e-ISSN: 2278-7461, p-ISSN: 2319-6491, Volume 3, Issue 11 (June
2014) PP: 29-35.
[12] R.D. Sarva Jagannadha Reddy (2014). Pi treatment for the constituent rectangles of the superscribed
square in the study of exact area of the inscribed circle and its value of Pi (SV University Method*). IOSR Journal of Mathematics (IOSR-JM), e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4
Ver. I (Jul-Aug. 2014), PP 44-48.
[13] RD Sarva Jagannada Reddy (2014), To Judge the Correct-Ness of the New Pi Value of Circle By
Deriving The Exact Diagonal Length Of The Inscribed Square. International Journal of Mathematics and
Statistics Invention, E-ISSN: 2321 – 4767 P-ISSN: 2321 – 4759, Volume 2 Issue 7, July. 2014, PP-01-04.
[14] RD Sarva Jagannadha Reddy (2014) The Natural Selection Mode To Choose The Real Pi Value Based
On The Resurrection Of The Decimal Part Over And Above 3 Of Pi (St. John's Medical College
Method). International Journal of Engineering Inventions e-ISSN: 2278-7461, p-ISSN: 2319-6491
Volume 4, Issue 1 (July 2014) PP: 34-37
[15] R.D. Sarva Jagannadha Reddy (2014). An Alternate Formula in terms of Pi to find the Area of a
Triangle and a Test to decide the True Pi value (Atomic Energy Commission Method) IOSR Journal of Mathematics (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug.
2014), PP 13-17
[16] RD Sarva Jagannadha Reddy (2014) Aberystwyth University Method for derivation of the exact value. International Journal of Latest Trends in Engineering and Technology (IJLTET) Vol. 4 Issue 2
July 2014, ISSN: 2278-621X, PP: 133-136.
[17] R.D. Sarva Jagannadha Reddy (2014). A study that shows the existence of a simple relationship among
square, circle, Golden Ratio and arbelos of Archimedes and from which to identify the real Pi value
(Mother Goddess Kaali Maata Unified method). IOSR Journal of Mathematics (IOSR-JM) e-ISSN:
2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug. 2014), PP 33-37
[18] RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com.
45
IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728, p-ISSN:2319-765X. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 14-15
www.iosrjournals.org
www.iosrjournals.org 14 | Page
Durga Method of Squaring A Circle
RD Sarva Jagannadha Reddy
Abstract: Squaring of circle is an unsolved problem with the official value 3.1415926… with the new value
1/4 (14- 2 ) it is done in this paper.
Keywords: Exact Pi value = 1/4 (14- 2 ), Squaring of circle, Hippocrates squaring of lunes.
I. Introduction Squaring a circle is defined as constructing a square having an area equal to that of a given circle. It is
also called as quadrature of the circle.
This concept has been there from the days of Rhind Papyrus (1800 B.C) written by a scribe named
Ahmes. Hippocrates of Chios (450 B.C) has squared lunes, full circle and semicircle along with lunes. He
fore saw the algebraic nature of the value. value 3.1415926… has failed to find a place for it in the squaring
of lunes. Though the World of Mathematics has accepted his squaring of lunes, they became silent for why
3.1415926… is a misfit in his constructions. Further, there is a false opinion that Hippocrates could not square
a circle. However, Hippocrates did square a full circle and a semicircle along with a lune. In both the cases –
squaring a lune, squaring a circle along with a lune – the new value, 14 2
4
has explained perfectly well the
constructions of Hippocrates. Thus the propositions of Hippocrates which remained theoretical all these 2400
years, have become practical constructions with the discovery of 14 2
4
. It is clear therefore, we have
misunderstood Hippocrates because, we believed 3.1415926… as the value of . I therefore apologize to
Hippocrates on behalf of mathematics community for the past mistake done by us. And to atone the
academic sin committed by us, I bow my head and dedicate the explained parts (for details: Pi of the
Circle, last chapter: Latest work, Pages from 273 to 281) to Hippocrates, in www.rsjreddy.webnode.com
James Gregory (1660) has said squaring of circle is impossible. His view has been confirmed by
C.L.F. Lindemann (1882) based on Euler’s formula ei
+1 = 0. Von K. Weiertrass (1815-1897) and David
Hilbert (1893) have supported the proof of Lindemann by their proofs.
S. Ramanujan (1913) has squared a circle upto some decimals of 3.1415926… Prof. Underwood
Dudley doesn’t accept Lindemann’s proof because this is based on numbers which are approximate in
themselves.
Now, the exact value is discovered. It is 1
14 24
. It is an algebraic number. The following is
the procedure how to square a circle.
II. Procedure We have to obtain a side of the square
whose value is 1
2 ; when
14 2
4
, then
1 1 14 2 14 2
2 2 4 4
CD = a, OK = OF = radius = 2
a, FK =
2
2
a, JK =
FG = GC,
GC = 1
JG KF2
= 2a 1
a2 2
=
46
Durga Method Of Squaring A Circle
www.iosrjournals.org 15 | Page
2 2a
4
; GB = BC – GC = 2 2
a a4
= 2 2
a4
;
Bisect GB. GH = HB = 2 2
a8
. Bisect HB. 2 2
a16
= HI = BI
CI = BC – BI = a – 2 2
a16
= 14 2
a16
= 4
; Area of the circle =
214 2a
16
CB = diameter = a;
Draw a semicircle on CB, with radius a
2 and center O’; CO’ = O’B =
1
2 where a = 1
Draw a perpendicular line on CB at I, which meets semicircle at Y. Apply altitude theorem to obtain IY length.
IY = 14 2 2 2 26 12 2
CI IB16 16 16
Connect YC which is the side of the square CYUT whose area is equal to that of the inscribed circle in the
square ABCD.
Apply Pythagorean theorem to get CY from the triangle CIY.
Side of the square CY =
22
2 2 14 2 26 12 2 14 2CI IY
16 16 4
Area of the square CYUT =
2
14 2 14 2
4 16
= area of the inscribed circle in the square ABCD.
III. Conclusion
14 2
4
is the exact value of circle. Hence, squaring of circle is done now. The misnomer “Circle
squarer” will sink into oblivion. Hippocrates will now gets his deserving throne of greatness though delayed
unfortunately for 2400 years.
47
IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. III (May-Jun. 2014), PP 17-20
www.iosrjournals.org
www.iosrjournals.org 17 | Page
Pythagorean way of Proof for the segmental areas of one square
with that of rectangles of adjoining square
R. D. Sarva Jagannadha Reddy 19-9-73/D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati – 517 501, A.P., India
Abstract: It is universally accepted that 3.14159265358… as the value of . It is thought an approximation, at
its last decimal place. It is a transcendental number and squaring a circle is an unsolved problem with this
number. A new, exact, algebraic number 14 2
4
= 3.14644660942… is derived and verified with a proof that
is followed for Pythagorean theorem. It is proved here squaring of circle and rectification of circumference of a
circle are possible too.
Keywords: Pythagorean thorem, circle, square, .
I. Introduction
Official value is 3.14159265358… It is obtained from the Exhaustion method, which is a geometrical
method. This method involves the inscription of a polygon in a circle and increased the sides of the polygon,
until the inscribed polygon touches the circle, leaving no gap between them. The value 3.14159265358… is
actually the length of the perimeter of the inscribed polygon. And it is not the value of circle. There was no
method till yesterday to measure the circumference of a circle, directly or indirectly.
3.14159265358… has four characteristics: 1) It represents polygon, 2) It is an approximation, 3) It is a
transcendental number and 4) It says squaring a circle is impossible. And such a number is attributed and
followed as of the circle based on limitation principle, because of the impossibility of calculating, the length
of the circumference of circle and in such a situation this field of mathematics has been thriving for the last 2000
years.
From 1450 Madhavan of South India and a galaxy of later generations of mathematicians have
discarded geometrical construction and have introduced newly, the concept of infinite series.
In this paper geometrical constructions are approached again, for the derivation of value. New value has
been derived. It is 14 2
4
= 3.14644660942… It is an exact value, an algebraic number and makes squaring
of circle possible and done too.
II. Procedure
Siva method for the area of the circle of 1st square ABCD
Construction procedure
Draw a square ABCD. Draw two diagonals. ‘O’ is the centre.
Inscribe a circle with centre ‘O’ and radius ½. E, F, H and J are
the mid points of four sides. Join EH, FJ, FH, HJ, JE and EF.
Draw four arcs taking A, B, C and D as centres and radius ½.
Now the circle square nexus is divided into 32 segments. Number
them 1 to 32. 1 to 16 segments are called S1 segments. 17 to 32
segments are called S2 segments. 17 to 24, S2 segments are
outside the circle. 25 to 32, S2 segments are inside the circle.
Draw KP, a parallel line to the side DC which intersects diagonals
at M and N.
Square = ABCD
Side = AB = 1 = EH = diameter
Areas of S1 and S2 segments S1 = 6 2
128
; S2 =
2 2
128
16S1 + 16S2 = Area of square 6 2 2 2
16 16 1128 128
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Pythagorean way of Proof for the segmental areas of one square with that of rectangles of …..
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16S1 + 8S2 = Area of circle 6 2 2 2 14 2
16 8128 128 16
Square has 32 constituent parts
Fig-1: Segmental areas calculated; Fig-2: Areas of Rectangles are calculated
Both values are same
This method is taken from the book Pi of the Circle of this author (available at
www.rsjreddy.webnode.com).
In this method there are two squares of same sides. First square has an inscribed circle divided into 32
segments of two dimensions called S1 and S2 segments, each category of 16 in number. And areas of these
segments are calculated using the following two formulas
2
1
aS 2
32 and
2
2
aS 4
32
which are obtained by solving two equations (in Square 1)
16 S1 + 16 S2 = a2 = Area of the square (Eq.1)
16 S1 + 8 S2 = a2/4= Area of the inscribed circle (Eq.2)
This method is called as Siva method. In the present method: Siva Kesava method, second square is
joined to the 1st square. One side CB is common to both the squares.
The second square is similarly divided, as in the case of 1st square, into 32 rectangles. Rectangles are also of
two dimensions each category of 16 numbers. The areas of each type of rectangle is equal to S1 and S2 segments
of the 1st square. These rectangles are formed, based on the division of common side of the both the
squares. The areas of rectangles agree cent percent with the above two formulas of Siva method, where
value is 14 2
4
. Thus, the division of 1
st square is exactly duplicated in the second square, except for the
difference, in the 1st square, 32 segments are curvy linear, and in the 2
nd square, 32 segments are rectangles,
naturally, of straight lines.
Now let us see how the common side CB is divided.
1. Squares 1 = ABCD, 2 = BZTC
2. Side = diameter of the inscribed circle = 1
3. KP = Parallel side to the side DC
4. OM = ON = radius ½
5. MON = triangle; MN = hypotenuse = 2
2
6. DK = KM = NP = PC = KP MN
2
=2 1 2 2
12 2 4
7. So, CP = 2 2
4
, PB = CB – CP =
2 2 2 21
4 4
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Pythagorean way of Proof for the segmental areas of one square with that of rectangles of …..
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8. Bisect PB. PB PQ + QB QR + RB = 2 2 2 2 2 2
4 8 16
9. QB = 2 2
8
, CB = 1, CQ = CB – QB = CQ =
2 2 6 21
8 8
10. Bisect CQ CS + SQ = 6 2 6 2
8 16
11. We have started with side = 1, divided next, into CP = 2 2
4
, and PB =
2 2
4
. In the second step
PB is bisected into PQ = 2 2
8
and QB =
2 2
8
. In the third step CQ =
6 2
8
is bisected into CS =
6 2
16
and SQ =
6 2
16
.
12. After divisions, finally we have CB Side divided into 4 parts
CS =6 2
16
, SQ =
6 2
16
, QR =
2 2
16
and RB =
2 2
16
13. 2nd
Square BZTC is divided horizontally into four parts: CS, SQ, QR and RB.
14. Now BZ side of 2nd
square is divided into 8 parts. So, each length is 1/8.
15. Finally, the 2nd
square is divided into 16 rectangles of one dimension equal in area to S1 segments of 1st
square and 16 rectangles of 2nd
dimension, equal in area to S2 segments of 1st square.
16. Square BZTC consists of first two rows are of S1 and 3rd
& 4th
rows are of S2 segments.
17. Area of each rectangle = S1 segments of 1st square =
6 2 1 6 2
16 8 128
Sides of rectangles of 1st & 2
nd rows=TW=WX=
6 2
16
and other side =
1
8
18. Area of each rectangle = S2 segment of 1st square =
2 2 1 2 2
16 8 128
Sides of rectangles of 3rd
& 4th
rows=XY=YZ= 2 2
16
and other side =
1
8
19. The areas of 16S1 and 16S2 segments of 1st square
2 2a a
16 2 16 432 32
= 1 = Area of the 1
st square; where a = 1
20. The area of all the 32 rectangles.
6 2 2 216 16
128 128
= 1 = Area of the 2
nd square
21. The area of the inscribed circle in the 1st square = 16S1 segments + 8S2 segments
= 6 2 2 2 14 2
16 8128 128 16
= Area of the circle in the 1
st square
22. The areas of the 1st, 2
nd and 3
rd rows of rectangles of 2
nd square.
6 2 6 2 2 2 14 28 8 8
128 128 128 16
23. Thus area of the circle from 1st and 2
nd squares is =
214 2 d
16 4
14 2
4
where side = diameter = d = 1
50
Pythagorean way of Proof for the segmental areas of one square with that of rectangles of …..
www.iosrjournals.org 20 | Page
24. When is equal to 14 2
4
, the length of the inscribed circle in the 1
st square is = d = a =
14 2 14 21
4 4
where side = diameter = 1
25. Perimeter of the rectangle QXWTCS is equal to the circumference of the inscribed circle in the 1st
square.
QX = 1 = TC; XW = WT = CS = SQ = 6 2
16
QX + XW + WT + TC + CS + SQ = 6 2 6 2 6 2 6 2 14 2
1 116 16 16 16 4
In the first square we have seen that the length of the circumference of the inscribed circle is the outer edge of
the 16 S1 segments. In the 2nd
square also the outer edges of the 1st and 2
nd rows of 16 rectangles are equal to
14 2
4
.
26. Thus, Siva Kesava Method supports the value 14 2
4
obtained by earlier Gayatri, Siva, Jesus
methods.
27. And also, the curvy linear 16S1 and 16S2 segments of 1st square are all squared in the 2
nd square.
III. Conclusion
Two squares of same sides are drawn with one common side. Circle is inscribed in one square. Areas
of square and its inscribed circle are calculated from their constituent curvy linear segments. The correctness of
areas of constituent segments are verified with that of the areas of rectangles of the adjoining square. All the
values thus are proved correct.
51
1
SQUARING OF CIRCLE, SQUARING OF SEMI CIRCLE AND SEMI
CIRCLE EQUATED TO RECTANGLE AND TRIANGLE
(Euclid’s Knowledge of Algebraic Nature of Circle and its Pi)
Introduction
Till 1450 of Madhavan of Kerala, India, number had its base in
the Exhaustion method of Archimedes. With Madhavan and
independently, John Wallis (1660) of England and James Gregory
(1660) of Scotland, number has become a special number, dissociated
from geometrical constructions almost. Radius or diameter or
circumference has become nothing in its relation with constant then
onwards. From here, constant has travelled choosing a new path of its
further journey.
Even in the Exhaustion method, the derivation of value
3.1415926358… is only partial. Here, the perimeter of the polygon is
divided by the diameter of the circle. In other words, 3.14159265358 is
not a pure one. It is a hybrid value. Its outcome is based on the
perimeter of the polygon and the diameter of the circle.
Perimeter of thepolygon
Diameter of the circle
From 1450 onwards, even this partial association with polygon
has been slowly cut off permanently. Ultimately, 3.14159265358 has
gained wings of infinite series and has grown to the magnitude of
trillions of its decimals, sitting on the shoulders of Super Computers.
Till 1882 of C.L.F. Lindemann, the nature of 3.14159265358 or
constant was not definite i.e., is an algebraic number or a
transcendental number ?
52
2
When the situation was like this
“In 1873, Charles Hermite (1882-1901) proved that the number e is
transcendental; from this it follow that the finite equation
a e r + b e s + c e t + … = 0 (6)
cannot be satisfied if r, s, t … are natural numbers and a, b, c, … are rational
numbers not all equal to zero.
In 1882, F. Lindemann finally succeeded in extending Hermite’s
theorem to the case when r, s, t, … and a, b, c, … are algebraic numbers, not
necessarily real. Lindemann’s theorem can therefore be stated as follows:
If r, s, t, …, z, are distinct real or complex algebraic numbers, and a, b,
c, … n are real or complex algebraic numbers, at least one of which differs from
zero, then the finite sum
a e r + b e s + c e t + … + n e z (7)
cannot equal zero.
From this the transcendence of follows quickly. Using Euler’s
Theorem in the form
e i + 1 = 0, (8)
we have an expression of the form (7) with a = b = 1 algebraic, and c and all
further coefficients equal to zero; s = 0 is algebraic, leaving r = i as the only
cause why (8) should vanish. Thus, i must be transcendental, and since i is
algebraic, must be transcendental.” (Page No.172) Petr Beckmann
“A History of ”
What is the base for Euler’s formula e i + 1 = 0. Let us see the
following Sir,
53
3
“But if Euler finished off one chapter in the history of , he also started
another. What kind of number was ? Rational or irrational ? With each new
decimal digit the hope that it might be rational faded, for no period could be
found in the digits. There was no proof as yet, but most investigators sensed
that it was irrational. However, Euler asked a new question: Could be the
root of an algebraic equation of finite degree with rational coefficients ? By
merely asking the question, Euler opened a new chapter in the history of , and
a very important one, as we shall see. He was also the one who started writing
it, for later investigations were based on one of Euler’s greatest discoveries, the
connection between exponential and trigonometric functions,
e ix = cos x + i sin x (16)
Euler discovered a long, long list of theorems. They are known as “Euler’s
theorem on…” and “Euler’s theorem of …” But this one is simply known as
Euler’s Theorem.” (Page No. 156 of A History of )
So, it was settled in 1882 that constant was a transcendental
number. Let us observe here Sir, in Euler’s Theorem,
e i + 1 = 0
refers to radians 1800. constant 3.14… has no right of its
participation in theorem. When it participates the theorem itself
becomes wrong.
We know Archimedes, Isaac Newton are the two greatest
mathematicians. To the place of third position, there is a doubt. Who ?
Is Carl Friedrich Gauss (1777-1855) or Leonhard Euler (1707-1783)? So,
the theorem is from such a greatest mathematician. He can’t be wrong.
Here, now comes a question, how did C.L.F. Lindemann chose
Euler’s theorem to prove that constant was a transcendental number?
54
4
Because, constant has no place in the Theorem. But it was
called transcendental, showing Euler’s theorem.
Are we to accept that
Pi radians 1800 = Pi constant 3.14 ?
Does Mathematics accept this ?
There is no iota of doubt if Lindemann calls that polygon number
3.14159265358… is transcendental. He is cent percent right. But he
becomes wrong if he calls constant as transcendental based on Euler’s
Theorem.
To sum up thus, constant was attached to polygon based on
Exhaustion principle of Eudoxus is the beginning. Then it left the
polygon in 1450 with Madavan of India. C.L.F. Lindemann wrongly
interpreted Euler’s Theorem and called constant as a transcendental
number.
This paper goes back turning the wheel of Time and even before
Egyptian mathematics, and starts returning in a forward journey and
stops at Hippocrates of Chios (450 B.C.), gets the support of squaring of
lunes, squaring of semicircle and squaring of full circle and submits
itself humbly before the World of Mathematics now, that the real
value is 14 2
4 = 3.14644660941… and is an algebraic number, and it
merges with triangle, rectangle, square etc. revealing, that circle –
square – triangle – rectangle stands as one whole geometrical entity.
Sir, In Figure 5 we see a semi circle on PT converting a rectangle
CJQP into a square of its side as QU. How ? Is it possible, if circle were
55
5
to be a transcendental entity, to convert rational entities into one form to
another form ? Think over Sir, please !
Procedure:
Draw a square ABCD. Side = AB = 1 Inscribe a circle with centre
‘O’ and radius 1/2. So, Diameter = side = 1
This combined geometrical construction consists of
1. Square
2. Inscribed circle
3. Left side semicircle
4. Triangle
5. Rectangle
6. Right side semicircle (shaded area)
7. Larger semicircle
8. DL Line-segment as the side of the square whose area is equal to
the area of the inscribed circle in the ABCD square and
9. QU line-segment as the side of the square whose area is equal to
the area of the shaded semi circle.
Let us study one by one
I. Figure 1
10. To obtain a length equal to 4
, where is equal to 14 2
4 =
3.14644660941… the following steps are followed:
4 = Quarter circumference of the inscribed circle =
14 2
16
11. EH = side = 1
12. OF = OG = radius = 1
2
56
6
57
7
13. FOG = triangle
FG = Hypotenuse = OF x 2 = 1
22
= 2
2
14. EF = GH = HC = Side hypotenuse
2 =
2 11
2 2 =
2 2
4
15. HB = CB – CH = Side – CH = 2 2
14
= 2 2
4
16. Bisect HB twice
HB HI + IB IB IJ + JB
= 2 2 2 2 2 2
4 8 16
17. CB – JB = 2 2
116
= 14 2
16 4
18. 14 2
4 16 = CJ
II Figure 2
19. AB = Diameter. Draw a semi circle with radius 1
2.
CJ4
of Fig.1 = DK of Fig.2
20. 2 2AK
16,
14 2DK
16
Apply altitude theorem to obtain KL
KL = AK DK = 2 2 14 2
16 16
KL = 26 12 2
16
21. DL is the side of the square whose area is equal to the area of the
inscribed circle in the ABCD square.
22. DK = 14 2
16, KL =
26 12 2
16
58
8
23. Apply Pythagorean theorem to obtain the length of DL
22
2 2 14 2 26 12 2DK KL
16 16
= 14 2
4
Which is the side of the square whose area is equal to the area of
the inscribed circle, which is equal to 1
2. Where
14 2
4
III Figure 3 : Semicircle (Shaded area)
24. PV = Diameter = 1
R = centre. Draw a semi circle
With centre R and radius 1
2.
Circle Area = 2d
4. Semi circle =
2d
8
Where d = 1, then 14 2 1
8 4 8 =
14 2
32
IV Figure 4 : Rectangle CJQP
25. CJ = 14 2
4 16
CP = JQ = 1
2 = radius
Area of the rectangle = CJ x CP
= 14 2 1
16 2 =
14 2
32
26. So, Semicircle on PV = Rectangle CJQP
(S. No. 24)
14 2
32 =
14 2
32
59
9
V Figure 5: Squaring a rectangle
27. Rectangle CJQP = 14 2
32
28. Diameter PT = radius4
= 14 2 1
16 2 =
22 2
16
29. S is the centre of diameter PT.
Draw a semi circle on PT.
30. Draw a perpendicular line at Q on PT which meets circumference
at U. To obtain QU apply altitude theorem.
QJ = 1
2 = QT =
1
2, PQ =
14 2
16
14 2 1PQ QT
16 2
It is a well established fact that a rectangle can be squared
when QT = QJ
Euclid III. 35 (Book III, Theorem 35)
31. So, it implies, that Euclid must have been aware of the algebraic
nature of circle and its , necessarily.
VI Figure 6 : Semicircle = Triangle
32. AB = Base of the triangle ABN = 1
M = Mid point of AB
MN = CJ4
= 14 2
16 = altitude
Area of the triangle ABN
1 1 1 14 2ab MN AB 1
2 2 2 16
= 14 2
32
So, area of the semi circle (shaded area) on PV is equal to that of
ABN triangle.
60
APPENDIX
61
62
63
64