Pi OF THE CIRCLE VOL-IV Squaring an Arbelos and Circle

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

mailto:[email protected]

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



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



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



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



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



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



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



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

http://www.rsjreddy.webnode.com/

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)

20

[Reddy*, 4.(10): October, 2015] ISSN: 2277-9655

(I2OR), Publication Impact Factor: 3.785

http: // www.ijesrt.com © International Journal of Engineering Sciences & Research Technology

[39]

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

21

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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

22





[41]

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

23





[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”.

24


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[43]

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

25





[44]

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

26





[45]

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

27





[46]

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

28





[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

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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


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.

39



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



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



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



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



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



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

http://www.rsjreddy.webnode.come/


e-ISSN: 2278-5728, p-ISSN:2319-765X. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 14-15



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


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


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. III (May-Jun. 2014), PP 17-20



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

48

Pythagorean way of Proof for the segmental areas of one square with that of rectangles of …..


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

49




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



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

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