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Pi OF THE CIRCLE

Vol. II (Published papers in the International Journals)

By

R.D. Sarva Jagannadha Reddy

Retired Zoology Lecturer

19-42-S7-374, S.T.V. Nagar, Tirupati – 517 501, India.

August, 2014

(For copies, send email please, to the author: [email protected])

mailto:[email protected]

Dedication

to

SRI GOVINDARAJA SWAMI VARU

(Sri Maha Vishnu of Vaikuntam)

Tirupati Temple, Chittoor District,

Andhra Pradesh, India

Sri Balaji of Tirumala Temple, Sri Govindaraja Swami of Tirupati Temple,

Sri Ranganatha Swami of Sri Rangam Temple, Sri Anantha Padmanabha

Swami of Tiruvanthapuram Temple are one and the same.

i

Preface

3.14159265358… has been used as a value for the last 2000 years. This

number actually represents polygon of Exhaustion Method of Archimedes (240 BC)

of Syracuse, Greece. This is the only one geometrical method available even now.

The concept of limitation principle is applied and thus this number is attributed to

the circle. In other words, 3.14159265358… of polygon is a borrowed number and

attributed/ thrust on circle as its value, as the other ways, to find the length of the

circumference of circle, has become impossible with the known concepts, principles,

statements, theorems, etc.

From 1660 onwards, 3.14159265358… has been derived by infinite series

also, starting with John Wallis of UK and James Gregory of Scotland. This number

was obtained by Madavan of Kerala, India, adopting the same concept of infinite

series even earlier i.e. 1450. The World of Mathematics has recognized very

recently, that Madavan is the first to invent infinite series for the derivation of

3.14159265358. John Wallis and James Gregory too invented the infinite series

independently though later in period (George Gheverghese Joseph of Manchester

University, UK).

C.L.F. Lindemann (1882), Von K. Weirstrass and David Hilbert have

called 3.14159265358… as a transcendental number. The basis for their proof was

Euler’s formula ei

+ 1 = 0 (Leonhard Euler, Swiss Mathematician, 1707-1783).

With their proofs, squaring of circle has become, without any doubt, an

unsolved geometrical problem. Thus, the present thinking on is, 3.14159265358…

is the value which is an approximation and squaring of circle is impossible with the

number.

At this juncture, the true and an exact value equal to 14 2

4 =

3.14644660942…. was discovered by the grace of God in March, 1998 after a

struggle of 26 years (from 1972) adopting Gayatri method. Hence, this value is

called the Gayatri value as the Gayatri method has revealed the true value for the

first time to the World. It was only a suspected value then, and however, it was

ii

not discarded, by this author. He continued and confirmed 14 2

4 as the real

value with Siva method, Jesus method and later with many more methods only.

A dilemma has thus crept into the minds of the people, which number

3.14159265358… or 14 2

4 = 3.14644660942 is the real value. One

responsibility before this author was to clear this dilemma. And, therefore, a book

was written collecting the work done in the past 12 years and titled Pi of the Circle in

2010, and is available in the website www.rsjreddy.webnode.com

The second responsibility before this author was also, to search for any flaw

in the derivation of the present value of equal to 3.14159265358…

As a result of continuous search for 16 years further deep, two errors have

been identified. And one paper has been published. One is, that, 3.14159265358…

belongs to the polygon and not to the circle. The second error is, to call of the circle

as a transcendental number. They (Lindemann, Weirstrass and Hilbert) may be

right in calling 3.14159265358… and not . Why ? It has been shown in earlier

paragraph that Euler’s formula is the basis in calling 3.14159265358… as a

transcendental number. In the formula ei

+ 1 = 0, refers to radians equal to 1800

and not constant 3.14…. constant has no place for it in the above formula.

When 3.14… is involved in the Euler’s formula, the formula becomes wrong. Is it

acceptable then to call constant as transcendental number even though this has no

right of its participation in the above formula ? However, it is acceptable still, if one

agrees that radians 1800 = constant 3.14 or radians 180

0 is identical to

constant 3.14… Mathematics may not accept this howler.

Thus, on two counts i.e., 3.14159265358… is a polygon number and calling it

as a transcendental number, based on radians 1800. The present work on

unfortunately, is confusing. These are the two simple errors to be rectified

immediately. Here, the NATURE has kindly entered and rectified the errors by

revealing Gayatri value. It is exact and is an algebraic number. Squaring of circle

is also done, by IT’s grace.

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

iii

In this book there are two unpublished papers also, one is to show the first

method the base of this work after 26 years of struggle (Gayatri method) and two is

is, not a transcendental number (Arthanaareeswara method).

Fifteen papers on value have been published by the following international

journals. The World and this humble author are grateful, forever, to these Journals.

1. IOSR Journal of Mathematics

2. International Journal of Mathematics and Statistics Invention

3. International Journal of Engineering Inventions

4. International Journal of Latest Trends in Engineering and Technology

5. IOSR Journal of Engineering (IOSRJEN)

While writing Pi of the Circle, Mr. A. Narayanaswamy Naidu, and while

writing these published papers of this book, Mr. M. Poornachandra Reddy, have

helped this author in simplification of formulas. The Editors of above Journals have

published this author’s work after refining the papers, keeping in mind the standards

expected in the original research. This author could complete his University

Education (1963-68) because of his mother only. He led a happy life for 42 years

with his wife. Now he is leading a peaceful life because of his second daughter

R. Sarada out of three children by staying at her house, after the death of this

author’s wife two and half years ago. Mr. Suryanarayana of M/s. Vinay Graphics,

Balaji Colony, Tirupati, has done DTP work perfectly well. This author, therefore, is

greatly indebted to these well-wishers and prays to the God to bless them with good

health. This author requests the readers to send their comments and they will be

gratefully received and acknowledged. To end, the quantum of contribution of this

author in this work is equal to, square root of less than one, in the square of trillions of

trillions, i.e. 2

1

trillions of trillions.

Author

iv

How did this Zoology Lecturer get encircled

himself in 1972 in Mathematics ?

Some people are curious to know, how did this student of Zoology enter

and entrench himself in this field of mathematics. Here is a brief narration:

This author loves book reading very much. One day in 1972, while reading an

encyclopedia he saw square, triangle, trapezium and so on and found constant

for circle alone in r2 and 2 r and such constant was not there for other

constructions. He questioned himself, “Why”. „Why‟ led to “Why not

without ” for circle too. He thought many days. One day he inscribed a

circle in a square and found the diameter of the inscribed circle and the side of

the superscribed square equal. He was surprised and felt happy that he got the

clue to make real, the question “Why not without ”. He thought and thought,

did many things, searched, studied, enquired fellow mathematicians, did

physical experiments, on-and-off, for next 26 years.

Being a government college teacher, he was transferred in the mean

time, from Piler (1971-81) to Kadapa, to Nagari, to Anantapur and finally to

Chittoor (in December 1995). No answer to his question of 1972.

He was 26 when question came, waited another 26 years and lost self

confidence. He was like a man swimming on the surface of the ocean looking

down and striving hard to take hold of the wanted pin with its visible blurred

image lying on the bottom of the ocean. Man looks up when he is helpless.

This was what happened to him also. He went to the nearby temple of Mother

Goddess Durga (at Chittoor) in 1998 and prayed to HER. He gave a word to

the goddess. “When he gets answer and succeed in finding formulas for the

v

area and circumference of circle without equal to 22/7, he will keep

himself away from receiving awards, royalties, positions and avoid

felicitation functions, meetings etc. on account of future discovery”.

Surprisingly, one Mr. Ramesh Prasad a physics teacher, next neighbor to this

author when discussed with him this long pending problem after the promise

to the Goddess before giving a clue he asked this author how did he had been

doing. The answer to him was, as inscribed circle, it is smaller in size

compared to the larger superscribed square – the concept of difference had

been a dominating point. With this answer, Mr. Prasad told this author to

look at the problem, at the concept of “ratio” also and not only the factor of

difference. This author received this idea of Mr. Prasad and that whole night

worked on the problem and prepared an article and was sent to the Indian

Institute of Technology, Kharghpur, Mathematics Department, next morning.

The department was impressed with the paper and cautioned this author, was

not 22/7 and it was 3.1415926… in its reply with encouraging comments.

There, the search did not stop. Second question came anew. The new

question made this author why should there be 22/7, 3.1416, 3.1415926… By

March, 1998, during the rejuvenated search Gayatri Method, Siva method

came successively after many many many failures. Next 16 years,

continuous search has been focused on confirming, the correctness of 14 2

4

= 3.14644660942 as value. This author thanks the reader for this attentive

reading.

CONTENTS

Page No.

1. Preface i-iii

2. How did this Zoology Lecturer get encircled himself

in 1972 in Mathematics ?

iv-v

3. Gayatri Method (unpublished) 1

4. times of area of the circle is equal to area of the

triangle Arthanaareeswara method (unpublished)

2

5. 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)

3

6. 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)

8

7. Squaring of circle and arbelos and the judgment of

arbelos in choosing the real Pi value (Bhagavan Kaasi

Visweswar method)

13

8. Aberystwyth University Method for derivation of the

exact π value

21

9. New Method of Computing value (Siva Method) 25

10. Jesus Method to Compute the Circumference of A

Circle and Exact Value

27

11. Supporting Evidences To the Exact Pi Value from the

Works Of Hippocrates Of Chios, Alfred S.

Posamentier And Ingmar Lehmann

29

12. New Pi Value: Its Derivation and Demarcation of an

Area of Circle Equal to Pi/4 In A Square

33

13. Pythagorean way of Proof for the segmental areas of

one square with that of rectangles of adjoining square

39

14. To Judge the Correct-Ness of the New Pi Value of

Circle By Deriving The Exact Diagonal Length Of

The Inscribed Square

43

15. 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)

47

16. 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)

51

17. Hippocratean Squaring Of Lunes, Semicircle and

Circle

56

18. Durga Method of Squaring A Circle 64

19. The unsuitability of the application of Pythagorean

Theorem of Exhaustion Method, in finding the actual

length of the circumference of the circle and Pi

66

20. Home page of the Author 73

Gayatri Method

ABCD = Square

AB = Side = a = 1

JG = diameter = a= 1

OF = OG = radius = a

2 = 0.5

FG = hypotenuse = 2a

2

DE = EF = GH = CH =

2aa

2EH FG

2 2 =

2a 2a

4

The length of the circumference of the inscribed circle can be earmarked in the

perimeter of the superscribed square.

Circumference of the circle =

BA + AD + DC + CH = a + a + a + 2a 2a

4 =

14a 2a

4

d = a = 14a 2a

4

= 14 2

4

This is the second method which came to life after this author’s struggle for 26

years i.e. in March 1998. This author saw 14 2

4 for the first time by this

method. He just presumed this number might be the value. After many

failures he confirmed this 14 2

4 as the true value with Siva method (Page

No. 25) where the total area of the square and the total area of the inscribed

circle in square were calculated for the first time.

times of area of the circle is equal to area of the triangle

(Arthanaareeswara method)

Square ABCD

Side AB = 1

Diagonal = AC = 2

Take a paper and construct a square whose side is 1 (=10 cm) and diagonal 2 .

Fold the paper along the diagonal AC. Then bring the two points of A and C of

the folded triangle touching each other in the form of a ring, such that AC

becomes the length of the circumference of the circle whose value is 2 . Now

the folded paper finally looks like a paper crown.

Let us find out the area of the circle

Circumference = 2 = d; d =2

; Area = 2

4

d=

2 2 1 2

4 4

Area of the triangle = 1

2; x Area of circle =

2 1

4 2

Second method

This time let us bring A and D or D and C close together, touching just in such

a way they form a ring (= circle)

Side = AD = 1 (=10 cm); Circumference = 1 = d;

d = 1

; 2 1 1 1 1

4 4 4

d; 2 x Area of circle =

1 12

4 2

The interrelationship between two areas of circle and triangle by , shows that

is not a special number called transcendental number.

D 1 C

A 1 B

1 1 2

2

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 www.iosrjournals.org

www.iosrjournals.org 44 | Page

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*)


Abstract: Pi value equal to 3.14159265358… is derived from the Exhaustion method of Archimedes (240 BC)

of Syracuse, Greece. It is the only one geometrical method available even now. The second method to compute

3.14159265358… is the infinite series. These are available in larger numbers. The infinite series which are of

different nature are so complex, they can be understood and used to obtain trillion of decimals to

3.14159265358… with the use of super computers only. One unfortunate thing about this value is, it is still an

approximate value. In the present study, the exact value is obtained. It is 14 2

4

= 3.14644660942… A

different approach is followed here by the blessings of the God. The areas of constituent rectangles of the

superscribed square, are estimated both arithmetically, and in terms of of the inscribed circle. And value thus derived from this study of correct relationship among superscribed square, inscribed circle and constituent

rectangles of the square, is exact.

Keywords: Circle, diagonal, diameter, area, radius, side, square

I. Introduction

Square is an algebraic geometrical entity. It has four sides and two diagonals which are straight lines.

A circle can be inscribed in the square. The side of the square and the diameter of the inscribed circle are same.

This similarity between diameter and side, has made possible to find out the exact length of the circumference

and the exact extent of the area of the circle, when this interrelationship between circle and its superscribed

square, are understood in their right perspective. The difficulty is, the inscribed circle is a curvature, though, its

diameter/ radius is a straight line as in the case of side, diagonal of the square. When we say a different approach is adopted, it means, these are entirely new to the literature of mathematics. The universal acceptance

to the new principles observed in the following method is a tough job and takes time. However, as the

following reasoning ways are cent percent in accordance with the known principles, understanding of the idea is

easy.

To study the different dimensions, such as, circumference and area of circle, constant is inevitable. Similarly, to understand perimeter and area of the square, 4a and a2 are adopted and hence, no constant similar

to circle is necessary in square. In the present study, the area of the square is divided into five rectangles. The

areas of rectangles are calculated in two ways: they are: 1. Arithmetical way and 2. In terms of of the

inscribed circle. Finally, the arithmetical values are equated to formulas having , and the value of is derived ultimately, which is exact.

II. Procedure

Draw a square and its two diagonals. Inscribe a circle in the square.

1. Square = ABCD, AB = Side = a

2. Diagonals = AC = BD = 2a

3. ‘O’ Centre, EF = diameter = side = a

4. The circumference of the circle intersects two diagonals of four points: E, H, F and G. Draw a parallel

line IJ to the sides DC, passing through G and F.

5. OG = OF = radius = a/2

6. Triangle GOF. GF = hypotenuse = OG 2 = a

22 =

2a

2 = GF

* This author studied B.Sc., (Zoology as Major) and M.Sc., (Zoology) during the years 1963-68 in the Sri

Venkateswara University College, Tirupati, Chittoor district, Andhra Pradesh, India. And hence this author

as a mark of his gratitude to the Alma Mater, this method is named after University’s Honour.

3

Pi treatment for the constituent rectangles of the superscribed square in the study of exact area of …..


7. IJ = side = a

8. DI = IG = FJ = JC = Side hypotenuse

2

=

IJ GF

2

= 2a 1

a2 2

=

2 2a

4

= JC

9. JC = 2 2

a4

, CB = side = a

JB = CB – CJ = 2 2

a a4

= 2 2

a4

10. Bisect JB twice of CB side of Fig-2

JB JL + LB JK + KL + LM + MB

= 2 2

a4

2 2

a8

2 2

a16

11. Similarly, bisect IA twice, of AD side of Fig-2

IA IP + PA IQ + QP + PN + NA

12. Join QK, PL, and NM. 13. Finally, the ABCD square is divided into five rectangles.

DIJC, IQKJ, QPLK, PNML and NABM

Out of the five rectangles, the uppermost rectangle DIJC is of different dimension from the other four bottomed

rectangles.

14. Area of DIJC rectangle

= DI x IJ = 2 2

a a4

=

22 2a

4

15. The lower four rectangles are of same area. For example one rectangle

= IQKJ = IQ x QK = 2 2

a a16

=

22 2a

16

16. Area of 4 rectangles

4



= IQKJ + QPLK + PNML + NABM = 22 2

4 a16

= 22 2

a4

17. Area of the square ABCD = DIJC + 4 bottomed rectangles = a2

= 2 2 22 2 2 2

a 4 a a4 16

Part-II

18. Let us repeat that

Area of the ABCD square = a2

Area of the inscribed circle =

2 2d a

4 4

; where diameter = side = a

19. When side = diameter = a = 1

Area of the ABCD square = a2 = 1 x 1 = 1

Area of the inscribed circle =

2 2d a 1 1

4 4 4 4

20. Corner area in the square (of Figs 1, 2, and 3)

= Square area – circle area

= 4

14 4

21. It is true that any bottomed 4 rectangles, is equal to

the corner area of the square of Figs 1, 2 and 3. Thus,

bottomed rectangle = corner area

22 2

a16

= 2 2

1 116

=

2 2

16

Part-III

22. Let us prove it i.e. S. No. 21

23. The inscribed circle is equal to the sum of the areas of upper most rectangle DIJC = 22 2

a4

of

S.No. 14 and next lower 3 rectangles IQJK, QPLK and PNML, and each is equal to 22 2

a16

of S.No. 15

2 22 2 2 2a 3 a

4 16

=

2214 2 a

a16 4

24. Area of the inscribed circle =

2a

4 4

where a = 1

Area of the corner region = 4

4

(S.No. 20)

Area of the inscribed circle + corner area = square area

4

+

4

4

= 1

25. The sum of the areas of 4 bottomed rectangles

= Square area – Uppermost rectangle DIJC

5



= 2 22 2

a a4

= 22 2

a4

and

S. No. 14 this is equal to S.No. 16 26. As the area of the corner region is equal to any one of the 4 bottomed rectangles,

then it is = 24

a4

(S.No. 20 & 21)

27. Then the sum of the areas of 4 bottomed rectangles

= 24

4 a4

= 24 a

28. Finally,

Area of the uppermost rectangle DIJC

= Square area – 4 bottomed rectangles

= 2 2 2a 4 a 3 a

29. CJ length = 3 a

Side = AB = IJ = a

30. Area of the upper most rectangle DIJC

= CJ x IJ = 3 a a = 23 a

31. Thus, the areas of five rectangles which are interpreted in terms of above, are

Uppermost rectangle DIJC = 23 a

4 bottomed rectangles = 24 a

Area of the ABCD square

Uppermost rectangle + 4 bottomed rectangles

= 2 2 23 a 4 a a

Area of the inscribed circle

= Uppermost rectangle DIJC + 3 bottomed rectangles

= 2 2 243 a 3 a a

4 4

This is the end of the process of proof.

32. As the corner area is equal to

1. Arithmetically = 22 2

a16

= 2 2

16

S.No. 21 where a = 1

and 2. in terms of = 4

4

S.No. 20

then 4 2 2

4 16

14 2

4

III. Conclusion

It is well known, that a2 is the formula to find out area of a square or a rectangle. In this paper besides

a2, formulae, in terms of , of the inscribed circle in a square, are obtained, and equated to the classical arithmetical values of a2. One has to admire the Nature, that, a circle’s area can also be represented exactly

equal, by the areas of rectangles, thus, the arithmetical values of these rectangles, are equated to that of a circle,

which thus give rise to new value 14 2

4

=3.14644660942… This author stands and bow down and

6



dedicates this work to the Nature. The Nature is the visible speck of the infinite Cosmos. The Creator exists

in the invisible Energy form of this infinite Cosmos. We call this Creator as GOD and this author offers

himself, surrenders himself totally and prays to THE LORD of the Cosmos of His/ Hers/ It’s infinite goodness, as an infinitesimally, a small living moving body, as a mark of humble gratitude to THE LORD.

References [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, 2

nd 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, Page. 25 prometheus

Books, New York 14228-2197.

[3]. 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.

[4]. 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.

[5]. 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

[6]. 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.

[7]. 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.

[8]. 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

[9]. 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

[10]. 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.

[11]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com

7

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


e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug. 2014), PP 33-37 www.iosrjournals.org


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)

R. D. Sarva Jagannadha Reddy

Abstract: This study unifies square, circle, Golden Ratio, arbelos of Archimedes and value. The final result,

in this unification process, the real value is identified, and is, 14 2

4

= 3.14644660942…

Key words: Arbelos, area, circle, diameter, diagonal, Golden Ratio, Perimeter, value, side, square

I. Introduction

The geometrical entitles and concepts such as circle, square, triangle, Golden Ratio have been studied

extensively. “The Golden Ratio is the ratio of two line segments „a‟ and „b‟ (when a < b) such that a b

b a b

.

The ratio a 5 1

b 2

0.6180339887498948482045868343656, while the reciprocal

b 5 1

a 2

=

1.6180339887498948482045868343656. Notice the relationship between the decimals. It suggests that

11

”. (A.S. Posamentier and I. Lehmann, 2004, : A Biography of the World‟s Most

Mysterious Number, Page 146).

Archimedes (240 BC) of Syracuse, Greece, called the area arbelos that is inside the larger semi circle,

but outside the two smaller semi circles of different diameters. By its shape it is also called as “a shoemaker‟s

knife”. The Golden Ratio and the arbelos of Archimedes are different concepts. But in this paper by the grace

of God, it has become possible to see that these two concepts too have an interesting and unexpected inter

relationship between each other (one). Further, this relationship has an extended relationship also with the circle

(two). It is a well known fact that there exists simple and understable relationship between circle and square

(three). As circle, square are related, their combined interrelationship has been extended to value also (four). There is, thus, a divine chain of bond (of four interconnecting relations) exists, among square,

circle, Golden Ratio, arbelos and value. (Here value means a true/ real/ exact/ line-segment based value.

The stress here, on the adjectives to , has become necessary, because 3.14159265358… of Polygon is attributed or thrust on circle. In other words, this number to circle is a borrowed number from polygon and its

existence thus can not be seen in the radius of the circle, naturally. However, the new value 14 2

4

=

3.14644660942… (unlike with official value 3.14159265358…) is inseparable with radius and is, here, humbly submitted to the World of Mathematics:

Area of the circle = 27r 2r

r r2 4

Circumference of the circle = 2r 2r

6r2

= 2r

In support of the above formulae, this paper also chooses and confirms that the real value is 14 2

4

=

3.14644660942…

8

A study that shows the existence of a simple relationship among square, circle, Golden Ratio …..


II. Procedure

1. Draw a square ABCD. Draw two diagonals. Inscribe a circle with centre „O‟ and with radius 1

2,

equal to half of the side AB of the square, whose length is 1.

AB = Side = EN = diameter = 1

AC = BD = Diagonal = 2

2. E is the mid point of AD

AE = 1

2, AB = 1

Triangle EAB, EB = hypotenuse = 5

4

EH = 1

2; HB = EB – EH =

5 1 5 1

4 2 2

Golden Ratio = HB = 5 1

2

3. EN = Diameter = 1

EJ = HB = 5 1

2

= 0.61803398874…

JN = EN – EJ = 5 1

12

= 3 5

2

= 0.38196601126…

4. Draw two semicircles on EN. And one semicircle with EJ as its diameter, and second semicircle with

JN as its diameter.

5. So, the diameter of the EJ semicircle = Golden Ratio = 5 1

2

and

the diameter of the JN semicircle = 3 5

2

6. The area present (which is shaded) outside the two semicircles (of EJ and JN) and within the larger EN

semi circle, is called arbelos of Archimedes.

7. Draw a perpendicular line on EN at J which meets circumference at K.

KJ = EJ JN (this is Altitude Theorem)

= 5 1 3 5

2 2

= 5 2 = 0.48586827174

9



8. Draw a full circle with diameter KJ. It has already been established that this area of the full circle is

equal to the area of the shaded region called arbelos.

9. To calculate the area of the arbelos we have the following formulas.

q d q

4

and

2h

2

where h = perpendicular line KJ = 5 2 = diameter of the circle LKM

h

2 = radius of LKM circle.

10. Now, let us see the first formula

q d q

4

q = JN = 3 5

2

d = EN = diameter = 1

=

3 5 3 51

2 2

4

= 5 2

4

= 4

0.23606797749

11. The conventional formula is r2.

KJ = diameter = h = 5 2

Radius = diameter h

2 2 =

5 2

2

= 0.24293413587

x 0.24293413587 x 0.24293413587 = x 0.05901699437

Part-II

12. value is known and hence, it is possible to find out the area of the arbelos either from q d q

4

or

2h

2

13. As there are two values now 3.14159265358… and 3.14644660942 =14 2

4

, the time has come,

to find a way to decide which number actually represents value.

14. The following formula helps in deciding the real value. The formula is Side = a = diameter = d = 1

Diagonal = 2 a = 2

Perimeter of thesquare

1Half of 7 timesof sideof square th of diagonal

4

= 4a

7a 2a

2 4

= 4

7 2

2 4

= 4

14 2

4

=

16

14 2

10



(when a circle is inscribed in a square, or when a square is created from the four equidistant tangents on a

circle, the length of the circumference of the inscribed circle can be demarcated in the perimeter of the

superscribed square. It is called rectification of the circumference of the circle).

Part III (Area of the arbelos)

Let us calculate now the area of the arbelos with the known two values, official value and new one called

Gayatri value.

15. With official value

0.236067977494

=

3.141592653580.23606797749

4 = 0.18540735595

(S. No. 10)

x 0.05901699437 = 3.14159265358 x 0.05901699437 = 0.18540735594

(S. No. 11)

16. With Gayatri value

0.236067977494

=

3.146446609420.23606797749

4 = 0.18569382184

(S.No. 10)

0.05901699437 = 3.14644660942 x 0.05901699437 = 0.18569382183

(S.No. 11)

17. Finally, we obtain two different values representing same area of the arbelos of Archimedes.

Official value gives: 0.18540735595 and

Gayatri value gives: 0.18569382184 Which one is the actual value for the area of the arbelos ? The answer can be found in Part IV.

Part IV

18. In the Figure 1 we have Golden Ratio, HB equal to 5 1

2

= 0.61803398874

19. Let us divide area of the arbelos of S.No. 17 with the Cube of Golden Ratio =

3

5 1

2

and

multiply it with 16

14 2 of the formula, derived in the S.No.14, which finally gives the area of the square

ABCD, equal to 1.

The value that gives the exact area of the square equal to 1 is confirmed as the real value. Here, the

Golden Ratio decides, the real value, by choosing the correct area of the arbelos of Archimedes of S.No. 17

20. Area of the arbelos obtained with official value (S.No. 17)

3

0.18540735595 16

14 25 1

2

Let us use simple calculator for the value of cube of Golden Ratio, which gives 0.23606797748

= 0.18540735595 16

0.23606797748 14 2

= 0.18540735595

1.271275345340.23606797748

= 0.99845732139

21. Area of the arbelos obtained with Gayatri value (S.No. 17). Let us repeat steps of S.No. 20 here:

3

0.18569382184 16

14 25 1

2

= 0.18569382184

1.271275345340.23606797748

= 1

11



As the exact area of the superscirbed square is obtained, it is clear, therefore, that, the real value is 14 2

4

= 3.14644660942…

III. Conclusion

It is well known that there exists a simple relationship between circle and square. In the present study,

it is clear such simple relation also exists between Golden Ratio and arbelos of Archimedes. This paper

combines above two kinds of relations and decides the real value, as 14 2

4

= 3.14644660942…




[2]. Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the World‟s Most Mysterious Number, Prometheus Books,

New York 14228-2197.







Ver. II (Mar-Apr. 2014), PP 09-12



2014, PP-33-38.



2014), PP 17-20.








[11]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com.

[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.

12


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)


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

13

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.

14



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

15



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.

16



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.

17



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


2d

4

18



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.

19



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.

20


Aberystwyth University Method for derivation of the exact π value

R.D. Sarva Jagannadha Reddy Abstract - Polygon’s value 3.14159265358… of Exhaustion method has been in vogue as ππππ of the circle for the last 2000 years. An attempt is made in this paper to replace polygon’s approximate value with the exact ππππ value of circle with the help of Prof. C.R. Fletcher’s geometrical construction.

Keywords: Circle, diagonal, diameter, Fletcher, ππππ, polygon, radius, side, square

I. INTRODUCTION

The official π value is 3.14159265358… It is considered as approximate value at its last decimal place, always. It implies that there is an exact value to be found in its place. a2, 4a, ½ab etc are the formulas of square and triangle which are derived based on their respective line-segments. Similarly, radius is a line-segment and a need is there to have a formula with radius alone and without π. The following formulas are discovered (March, 1998) from Gayatri method and Siva method.

1. Area of Circle = 27r 2rr r

2 4

� �− = π� ��

� �

and

2. Circumference of Circle = 2r 2r

6r 2 r2

−+ = π ; where r = radius

2d 1 1 dd d d

2 2 4 4

� � π− − − =� ��

� ��

where d = diameter = side of the superscribed square In the Fletcher’s geometrical construction there are two line-segments. They are radius and corner

length. To find out the area of the shaded region in which corner length is present Professor has given 1

14

− π .

Fig-1: Professor’s Diagram (by courtesy)

II. CONSTRUCTION PROCEDURE OF SIVA METHOD

International Journal of Latest Trends in Engineering and Technology (IJLTET)

Vol. 4 Issue 2 July 2014 133 ISSN: 2278-621X

21

Fig-2: Siva Method

Draw a square ABCD. Draw two diagonals. ‘O’ is the centre. Inscribe a circle with centre ‘O’ and radius ½. Side of the square is 1. E, F, G and H are the midpoints of four sides. Join EG, FH, EF, FG, GH and HE. Draw four arcs with centers A, B, C and D and with radius ½. Now the circle-square composite system is divided into 32 segments of two different dimensions, called S1 segments and S2 segments. Number them from 1 to 32. There are 16 S1 and 16 S2 segments in the square and 16S1 and 8S2 segments in the circle.

Square: ABCD, AB = Side = 1, AC = Diagonal = 2 ; Circle : EFGH,

JK = Diameter = 1 = Side; Corner length = Diagonal diameter AC JK 2 1

2 2 2

− − −= = ; OL =

2

4;

OK= radius = 1

2; LK = OK – OL =

1 2 2 2

2 4 4

−− = =0.14644660942…

From the diagram of Fletcher the area of the shaded segment cannot be calculated arithmetically. The diagram of the Siva method helps in calculating the area of the shaded segment. How ?

Shaded area of Fletcher is equal to two S2 segments 19 and 20 of Siva method.

This author, in his present study, has utilized radius/ diameter as usual, and a corner length, in addition, of the construction to find out the arithmetical value to the shaded area. A different approach is adopted here. What is that ? As a first step the shaded area is calculated using four factors. They are of Fig-2.

AC and BD, 2 diagonals ( )2 2



22

KC corner length = diagonal diameter AC JK

2 2

− −= =

2 1

2

� �−� ��

Area of the square (a2 = 1 x 1 = 1) and

32 constituent segments of the square.

Their relation are represented here in a formula and is equated to

Professor’s formula

2a132 142 1

2 22

= − π� �−� �

� �

(of Fig.1, where radius = 1)

where, 1

14

− π has been derived with radius equal to 1, and naturally, the diameter = side of the square

= 2. With this, the above formula becomes

4132 142 1

2 22

= − π� �−� �

� �

14 2

4

−∴π =

The accepted value for π is 3.14159265358… With this π, the area of shaded region is equal to

11 3.14159265358 0.21460183661...

4− × =

And, with the new π value derived above, the area of the shaded region is equal to

1 14 2 2 21 0.21338834764...

4 4 16

� �− +− = =� ��

� �

So, this method creates a dispute now. Which π value is right i.e. is 3.14159265358… or

14 23.14644660942...?

4

−=

The study of this method is extended further to decide which π value is the real π value ?

To decide which π is real, a simple verification test is followed here. What is that ? We have a line

segment LK = 2 2

0.14644660942...4

−=

LK is part of the diagonal along with the corner length KC.

So, in the Second step, an attempt is made to obtain the LK length, from the area of the shaded region. How ? Let us take the reciprocal of the area of the shaded region;

1 1

Area of theshaded region 0.21460183661of official=

π = 4.65979236616

and with new π



23

1 1 164.68629150101...

Area of theshaded region 2 2 2 216

= = =+ +

Then, this value when divided by 32, we surprisingly get KL length. It may be questioned ‘why’ one should divided that value. The answer is not simple. Certain aspects have to be believed, without raising questions like what, why and how at times.

Official π = 4.65979236616

0.14561851144...32

=

New π = 4.68629150101

0.1464466094...32

=

0.14644660942… of new π value is in total agreement with LK of Fig.2.

i.e. 2 2

4

−=0.14644660942… and differs however with 0.14561851144… of official π from 3rd decimal

onwards. If this argument is accepted, the present π value 3.14159265358… is not approximate value from its last decimal place, but it is an approximate value from the 3rd decimal.

IV. CONCLUSION

From the beginning to the end of this method, various line-segments are involved. Professor Fletcher’s construction is analyzed arithmetically with the line-segments of the Siva method. This arithmetical

interpretation has resulted in the derivation of a new π value, equal to 14 2

4

−. The new value is exact,

algebraic number.

REFERENCES

[1] C.R. Fletcher (1971) The Mathematical Gazettee, December, Page 422, London, UK. [2] R.D. Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com



24


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 48-49

www.iosrjournals.org


New Method of Computing value (Siva Method)

RD Sarva Jagannada Reddy

I. Introduction

equal to 3.1415926… is an approximation. It has ruled the world for 2240 years. There is a necessity

to find out the exact value in the place of this approximate value. The following method givesthe total area of

the square, and also the total area of the inscribed circle. derived from this area is thus exact.

II. Construction procedure Draw a circle with center ‘0’ and radius a/2. Diameter is ‘a’. Draw 4 equidistant tangents on the

circle. They intersect at A, B, C and D resulting in ABCD square. The side of the square is also equal to

diameter ‘a’. Draw two diagonals. E, F, G and H are the mid points of four sides. Join EG, FH, EF, FG, GH

and HE. Draw four arcs with radius a/2 and with centres A, B, C and D. Now the circle square composite

system is divided into 32 segments and number them 1 to 32. 1 to 16 are of one dimension called S1 segments

and 17 to 32 are of different dimension called S2 segments.

III. Calculations: ABCD = Square; Side = a, EFGH = Circle, diameter = a, radius = a/2

Area of the S1 segment =26 2

128a

; Area of the S2 segment = 22 2

128a

;

Area of the square = 16 S1 + 16S2 = 2 2 26 2 2 2

16 16128 128

a a a

Area of the inscribed circle = 16S1 + 8S2 = 2 2 26 2 2 2 14 2

16 8128 128 16

a a a

General formula for the area of the circle

2 2214 2

4 4 16

d aa

; where a= d = side = diameter

14 2

4

IV. How two formulae for S1 and S2 segments are derived ? 16 S1 + 16 S2 = a

2 = area of the Square … Eq. (1)

25

New Method of Computing value (Siva Method)


16 S1 + 8 S2 =

2

4

a= area of the Circle … Eq. (2)

……………..…………………

(1) – (2) 8S2 =

2 2 22 4

4 4

a a aa

= S2 =

2 244

32 32

a a

(2)x 2 32 S1 + 16 S2 =

22

4

a … Eq. (3)

16 S1 + 16 S2 = a2

… Eq. (1)

………………………………

(3) – (1) 16S1 =

22

2

aa

= S1 =

2 222

32 32

a a

V. Both the values appear correct when involved in the two formulae a) Official value = 3.1415926…

b) Proposed value = 3.1464466… = 14 2

4

Hence, another approach is followed here to decide real value.

VI. Involvement of line-segments are chosen to decide real value.

A line-segment equal to the value of ( - 2) in S1 segment’s formula and second line-segment equal to the

value of (4 - ) in S2 segment’s formula are searched in the above construction.

a) Official : - 2 = 3.1415926… - 2 = 1.1415926….

Proposed : - 2 = 14 22

4

= 6 2

4

The following calculation gives a line-segment for 6 2

4

and no line-segment for 1.1415926..

IM and LR two parallel lines to DC and CB; OK = OJ = Radius = 2

a; JOK = triangle

JK = Hypotenuse = 2

2

a

Third square = LKMC; KM = CM = Side = ?

KM = 2 1 2 2

2 2 2 4

IM JK aa a

; Side of first square DC = a

DC + CM = 2 2 6 2

4 4a a a

b) Official = 4 - = 4 – 3.1415926… = 0.8584074….

Proposed = 4 - = 14 2 2 2

44 4

No line-segment for 0.8584074… in this diagram.

MB line-segment is equal to 2 2

4

. How ?

Side of the first square CB = a

MB = CB – CM = 2 2 2 2

4 4a a a

VII. Conclusion: This diagram not only gives two formulae for the areas of S1& S2 segments andalso shows two line-

segments for ( - 2) and (4 - ). Line-segment is the soul of Geometry.

26


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. I. (Jan. 2014), PP 58-59



Jesus Method to Compute the Circumference of A Circle and

Exact Value

RD Sarava Jagannada Reddy

I. Introduction The Holy Bible has said value is equal to 3. Mathematicians were not satisfied with the value. They

thought over. Pythagorean theorem came in the mean time. A regular polygon with known perimeter was

inscribed in a circle and the sides doubled successively until the inscribed polygon touches the circumference,

leaving no gap between them. Hence this method is called Exhaustion method. The value of the perimeter of

the inscribed polygon is calculated applying Pythagorean theorem and is attributed to the circumference of the

circle. This method was interpreted, first time, on scientific lines by Archimedes of Syracuse, Greece. He has

said value is less than 3 1/7.

Later mathematicians have refined the Exhaustion method and found many decimals. The value is

3.1415926… and this value has been made official.

From 15th

century (Madhava (1450) of South India) onwards infinite series has been used for more

decimals to compute 3.1415926 of geometrical method. Notable people are Francois Viete (1579), Van

Ceulen (1596), John Wallis (1655), William Brouncker (1658) James Gregory (1660), G.W. Leibnitz

(1658), Isaac Newton (1666), Machin (1776), Euler (1748), S. Ramanujan (1914), Chudnovsky brothers

(1989). The latest infinite series for the computation of value is that of David Bailey, Peter Borwein and

Simon Plouffe (1996) and is as follows:

0

1 4 2 1 1

16 8 1 8 4 8 5 8 6ii i i i i

Using above formula Yasumasa Kanada of Tokyo University, Japan, calculated trillions of decimals

to 3.1415926….. with the help of super computer.

Mathematics is an exact science. We have compromised with an approximate value. Hence, many

have tried to find exact value. This author is one among the millions. What is ? It is the ratio of

circumference of a circle to its diameter. However, in Exhaustion method, perimeter of the inscribed polygon is

divided by the diameter of the outside circle. Thus 3.1415926…. violates the definition of . This is about the

value of . Next, about the nature of . C.L.F. Lindemann (1882) has said is a transcendental number based

on Euler‟s formula 1 0ie . In Mathematical Cranks, Underwood Dudley has said “‟s only position in

mathematics is its relation to infinite services (and) that has no relation to the circle…. Lindemann proclaimed

the squaring of the circle impossible, but Lindemann‟s proof is misleading for he uses numbers (which are

approximate in themselves) in his proof”.

Hence, pre-infinite series – days of geometrical method is approached again to find out exact value

and squaring of circle. This author has struggled for 26 years (1972 to 1998) and calculated the exact value of

in March, 1998. The following method calculates the total length of circumference and thus the exact value

has been derived from it.

Procedure: Draw a square. Draw two diagonals. Inscribe a circle. Side = a,

Diagonal = 2a , Diameter is also = a = d.

1) Straighten the square. Perimeter = 4a

Perimeter – Sum of the lengths of two diagonals = 4 2 2a a = esp

esp = end segment of the perimeter of the square.

27

Jesus Method To Compute The Circumference Of A Circle And Exact Value


2) Straighten similarly the circumference of the inscribed circle

3 diameters plus some length, is equal to the length of the circumference.

Let us say circumference = x.

Circumference – 3 diameters = x – 3a = esc

esc = end segment of the circumference of the circle.

3) When the side of the square is equal to „a‟, the radius of the inscribed circle is equal to a/2. So, the

radius is 1/8th

of the perimeter of the square.

4) The above relation also exists between the end segment of the circumference of the circle and the end

segment of the perimeter of the square.

Thus as radius 2

a

of the inscribed circle is to the perimeter of the square (4a), i.e., 1/8th

of it,

so also, is the end segment of the circumference of the circle, to the end segment of the perimeter

of the square.

So, the end segment of the circumference = 8

end segment of the perimeter of the square

4 2 23

8 8

esp a aesc x a

14 2

4

a ax

5) Circumference of the circle = d = a (where a = d = diameter)

14 2

4

a aa

14 2

4

II. Conclusion

value, derived from the Jesus proof is algebraic, being a root of 2 56 97 0x x but also that it

differs from the usually accepted value in the third decimal place, being 3.146…..

28


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. II (Mar-Apr. 2014), PP 09-12



Supporting Evidences To the Exact Value from the Works Of

Hippocrates Of Chios, Alfred S. Posamentier And Ingmar

Lehmann

R.D. Sarva Jagannadha Reddy 19-9-73/D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati – 517 501, A.P., India

Abstract: Till very recently we believed 3.1415926… was the final value of .And no body thought exact

value would be seen in future. One drawback with 3.1415926…is, that it is not derived from any line-segment

of the circle. In fact, 3.1415926… is derived from the line-segment of the inscribed/ circumscribed polygon in

and about circle, respectively. Surprisingly, when any line-segment of the circle is involved two things

happened: they are 1. Exact value is derived and 2 that exact value differs from 3.1415926… from its 3rd

decimal onwards, being 3.1464466… Two geometrical constructions of Hippocrates of Chios, Greece (450

B.C.) and Prof. Alfred S. Posamentier of New York, USA, and Prof. Ingmar Lehmann of Berlin, Germany, are

the supporting evidences of the new value. They are detailed below.

Keywords: value, lune, triangle, area of curved regions

I. Introduction In the days of Hippocrates, value 3 of the Holy Bible was followed in mathematical calculations.

He did not evince interest in knowing the correct value of . He wrote a book on Geometry. This was the first

book on Geometry. This book became later, a guiding subject for Euclid’s Elements. He is very famous for

his squaring of lunes. Prof. Alfred S. Posamentier and Prof. Ingmar Lehmann wrote a very fine

collaborative book on . They have chosen two regions and have proved both the regions, though appear very

different in their shapes, still both of them are same in their areas. These areas are represented by a formula

2 12

r

. The symbol „r‟ is radius. , here must be, the universally accepted 3.1415926…

Every subject in Science is based on one important point. It would be its soul. In Geometry, the soul is

a line-segment. The study of right relationship between two or more line-segments help us to find out areas,

circumference of a circle, perimeters of a triangle, polygon etc. For example, we have side in the square, base,

altitudein the triangle. The same concept is extended here, to show its inevitable importance in the study of

two regions of Professors of USA and Germany. The lengths of the concerned line-segments have been arrived

at and associated with 2 12

r

. 3.1415926… does not agree with the value of line-segments of two regions.

However, the new value 3.1464466… = 14 2

4

has agreed in to-to with the line-segments of the two regions

of the Professors. This author does believe this argument involving interpretation of 2 12

r

with the line-

segments, is acceptable to these great professors and the mathematics community. It is only a humble

submission to the World of Mathematics. Judgment is yours. If this argument in associating line-segment with

the formula looks specious or superficial, this author may beexcused.

II. Procedure The two methods are as follows:

1. Hippocrates' Method of Squaring Lunes And Computation of The Exact Value

“Archimedes's procedure for finding approximate numerical values of (without, of course, referring

to as a number), by establishing narrower and narrower limits between which the value must lie, turned out to

be the only practicable way of squaring the circle. But the Greeks also tried to square the circle exactly, that is

they tried to find a method, employing only straight edge and compasses, by which one might construct a square

equivalent to the given circle. All such attempts failed, though Hippocrates of Chios did succeed in squaring

lunes.

Hippocrates begins by noting that the areas of similar segments of circles are proportional to the

squares of the chords which subtend them

29

Supporting Evidences To The Exact Value From The Works Of Hippocrates Of Chios, Alfred S.


Consider a semi-circle ACB with diameter AB. Let us inscribe in this semi-circle an isosceles triangle

ACB, and then draw the circular arc 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

AB respectively, and from Pythagoras'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.

AB = diameter, d. DE = DC = radius, d/2; F = mid point of AC

N = mid point of arc AC

NF = 2

2 2

d d; DM =

2

2

d d; MC =

2

2

d d

With the guidance of the formulae of earlier methods of the author where a

Circle is inscribed with the Square, the formulae for the areas of ANC, CPB, ACM

and BCM are devised.

1. Area of ANC = Area of CPB =

2d 2 12 1

32 2 2

2. Area of AMB = Areas of ANC + CPB (Hippocrates)

3. Area of ACM = Area of BCM =

2

2d

16

2 18

2

4. Area of ACB triangle = 1 d

d2 2

5. According to Hippocrates the area of the lune ACBMA is equivalent to the area of the triangle ACB

Lune ACBMA = triangle ACB

(ANC + ACM + BCM + CPB)

i.e.

2

2

2d

d 2 1 1 d164 1 2 d32 2 22 2 2 1

82

ANC + CPB ACM+BCM ACB

From the above equation it is clear that the devised formulae for the areas of different segments is

exactly correct.

6. Area of AMB = Areas of ANC + CPB

7. Area of the semicircle = 2d

8

= Areas of ANC + CPB + ACM + BCM + AMB

8. 2

8 Area of thesemicircle

d

=

2

2 2

2

2d

8 d 2 1 d 2 1164 1 2 4 132 32d 2 2 2 22 1

82

=

14 2

4

2. Alfred S. Posamentier’s similarity of the two areas

and decimal similarity between an area and its line-segments

Prof. A.S. Posamentier has established that areas of A and B regions are

30



equal. His formula is 2 1

2r

for the above regions. This author is grateful to the professor of New York for

the reason through his idea this author tries to show that his new value equal to 114 2

4 is exactly right.

1. Arc = BCA; O = Centre; OB = OA = OC = Radius = r

2. Semicircles : BFO = AFO; E and D = Centres; OD=DA = BE = OE= radius= 2

r

OF = 2

2

r; FC = OC – OF =

2

2

rr =

2 2

2

r r

3. Petal = OKFH; EK = 2

r; ED =

2

2

r; EJ =

2

4

r; JK = EK – EJ =

2

2 4

r r =

2 2

4

r r;

JK = JH, HK = JH + JK = 2 2

2

r r

4. So, FC of region A = HK of region B = 2 2

2

r r

5. BFAC = OKFH i.e. areas of A and B regions are equal (A.S. Posamentier and I. Lehmann).

(By Courtesy: From their book )

Formula for A and B is 2

2 1 22 2

rr

Here r = radius = 1

From March 1998, there are two values. The official value is 3.1415926… and the new value is

14 2

4

= 3.1464466… and which value is exact and true ?

Let us substitute both the values in 2

22

r , then

Official value = 2

3.1415926 22

r =

2

1.1415926...2

r

(It is universally accepted that 3.1415926… is approximate at its last decimalplace however astronomical

it is in its magnitude.)

New value = 2

3.1464466 22

r =

2

1.1464466...2

r

6. FC = HK (HJ + JK) line segments = 2 2

2

r r

7. Half of HC and HK are same 2 2 1

2 2 2 2

FC HK r r

= 2 2

4

r r = 0.1464466…..

8. Area of A/B region equal to 1.1464466… is similar in decimal value of half of FC/HK line segment i.e.

0.1464466…

9. Formulae a2, 4a of square and ½ab of triangle are based on side of the square and altitude, base of

triangle, respectively. In this construction, FC and HK are the line segments of A and B regions,

respectively.

31



As the value 0.1464466… which is half of FC or HK is in agreement with the area value of A/B region

equal to 1.1464466… in decimal part, it is argued that new value equal to 114 2

4 = 3.1464466…

is exactly correct.

The decimals 0.1415926… of the official value 3.1415926… does not tally beyond 3rd

decimal with the half

the lengths of HK and FC, whose value is 0.1464466, thus, the official value is partially right. Whereas, FC

& HK are incompatible with the areas of A & B calculated using official value. Then, which is real, Sirs?

III. Conclusion 3.1415926… agrees partially (upto two decimals only) with the line-segments of curved geometrical

constructions. When these line-segments agree totally and play a significant role in these constructions a

different value, exact value 14 2

4

= 3.1464466… invariably appears. Hence,

14 2

4

is the true valueof

.

Acknowledgements This author is greatly indebted to Hippocrates of Chios, Prof. Alfred S. Posamentier, and Prof.

Ingmar Lehmann for using their ingenious and intuitive geometrical constructions as a supportive evidence of

the new value of .

Reference [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]. Alfred S. Posamentier&Ingmar Lehmann (2004). A Biography of the World‟s Most Mysterious Number. Prometheus Books,

New York, Pages 178 to 181.

[4]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, a Canto on-line edition, in the free website: www.rsjreddy.webnode.com

32


International Journal of Mathematics and Statistics Invention (IJMSI)

E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759

www.ijmsi.org Volume 2 Issue 5 || May. 2014 || PP-33-38

www.ijmsi.org 33 | P a g e

New Value: Its Derivation and Demarcation of an Area of

Circle Equal to 4

In A Square


ABSTRACT: value is 3.14159265358… it is an approximation, and it implies the exact value is yet to be

found. Here is a new method to find the most sought after exact value. 3.14159265358… is actually is the value of inscribed polygon in a circle. It is a transcendental number. When line-segments of circle are involved

in the derivation process then only the exact value can be found. 14 2

4

= 3.14644660942… thus obtained

is an algebraic number and hence squaring of circle is also done in the second part (method-3) of this paper.

KEYWORDS: Circle, Diagonal, Diameter, value, Radius, Square, Squaring of circle

I. INTRODUCTION

METHOD-1: Computation of tail-end of the length of the circumference over and above three

diameters of the Circle

The Holy Bible has said value is 3. Archimedes (240 BC) of Syracuse, Greece has said value is

less than 3 1/7. He has given us the upper limit of value. In 3 1/7, 3 represents three diameters and 1/7

represents the tail-end of the circumference of the circle (d = circumference)

In March, 1998, Gayatri method said the value as 14 2

4

= 3.14644660942… and its tail-end of the

length of the circumference of a circle over and above its 3 diameters as equal to 1

2 2 4 when the diameter is

equal to 1.

1/7 of Archimedes = 0.142857142857…

1

2 2 4 of Gayatri method =

1

6.82842712474 = 0.14644660942…

In the days of Archimedes there was no decimal system, because there was no zero. Archimedes is

correct in saying the tail-end length of the circumference is less than 1/7. How ? Gayatri method supports

Archimedes’ concept of less than 1/7 by giving 1

6.82842712474. The denominator part of the fraction, is

actually, less than 7 of 1/7. He is a great mathematician. This fraction 1

6.82842712474 has become possible

because of the introduction of zero in the numbers 1 to 9 and further consequential result of decimal system of

his later period. If he comes back alive, with his past memory remain intact, Archimedes would say, what he

had visualized in 240 BC has become real.

II. PROCEDURE

Let us see how this tail-end value of circumference is obtained: Draw a circle with Centre ‘O’ and

radius a/2. Draw four equidistant tangents on the circumference. They intersect at four prints called A, B, C

33

New Value: Its Derivation and Demarcation of…


and D, creating a square ABCD. The diameter of the circle EF is also equal to the AB side of the square. Draw

two diagonals AC and BD. Their values are 2a . Perimeter of the square is equal to 4a.

In this circle-square composite system there are now 3 types of straight lines and one circumference

which is a curvature. The straight lines are 1. Perimeter of the square, 2. Two diagonals and 3. Diameter of the

circle. The values of these straight lines are known and exact. The length of the circumference is unknown and

hence this method is to find out its exact length with the help of known lengths of three types of straight lines.

Let us repeat here again the perimeter (4a) of the square and the sum of the lengths of two diagonals

2 2a are the outcome of the tangents on the circumference. It is clear therefore that the curved

circumference reflects its true length in all the straight lines of square and circle.

We know very well that there are three diameters and some length called tail-end in the circumference.

Circumference = 3 diameters + tail end length

Tail-end length is unknown and hence it is called x.

3d + x = circumference

Diameter is equal to side of the square i.e. d = a

Let us rewrite 3d + x as 3a + x

To find out the length of the x

We take the help of all the straight lines. The reciprocal of the two diagonals plus the perimeter of

the square and when this product is multiplied by the square of the diameter (= side), will give x value.

21 1x a a

2 2a 4a 2 2 4

Circumference = 3 diameters (3a) + tail-end length (x) = 1 14 2

3a a a42 2 4

Circumference of the circle = d = a

So, 14 2

a a4

14 2

4

34



The length of the circumference is obtained by superscribing a square. The correct understanding of

the relationship among perimeter, diagonals of the square and the diameter of the circle (= side of the square)

results in knowing the exact length of the tail-end of the circumference (x), what the great mathematician

Archimedes has said is less than 1/7 is proved now to be 1 1

6.828427124742 2 4

The denominator of

Gayatri method 6.82842712474 … is less than 7 of 1/7 of 3 1/7 of Archimedes.

METHOD-2: Computation of Segmental Areas (An Elementary Approach)

1. BD = 2 2

4

; CD =

2 1

2

;

AB = OB = 2

4

2. Area of ABC = 1

16

3. Area of a + b = 1

16

4. Area of OAC = 1

8; OAB =

1

16; Sector OAD

1

10

5. Area of segment ‘a’ = OAD – OAB = 3

80 = 0.0375

Area of segment ‘b’ = OAC – OAD = 1

40 = 0.025

6. ‘a’ is larger than ‘b’. So, ‘a’ is greater than half of ABC 1

16

.

‘b’ is smaller than ‘a’. So, ‘b’ is lesser than half of ABC 1

16

.

7. Half of ABC = 1

32 = y

8. Let a – y = s, y – b = t; s = t

9. Let us assume BD AC 2 2

s ;16 128

CD AB 2 2t

16 128

35



10. a – y = s; a = s + y = 6 2

128

11. y – b = t; b = y – t = 2 2

128

12. a + b = ABC

13. Inscribed circle consists of 16a and 8b

so, 26 2 2 2 d

16 8128 128 4

Where d = 1

= 14 2

4

METHOD-3: DEMARCATION OF AREA OF CIRCLE WITH ITS STRAIGHT-LINED BOUNDARY

From Gayatri method, the world came to know in March 1998, for the first time, the length of the

circumference of the inscribed circle, demarcated in the perimeter of the square. The demarcated length of the

circumference of the circle is BA + AD + DC + CH = a + a + a + 2 2 14 2

a a4 4

The area of the circle is now bounded by a curvature called the circumference. In this method this area

can have a straight-lined boundary. In this process the value of plays important role. The official value is

3.14159265358… With this value let us locate the area:

1. Square = ABCD, Side = AB = a; AC = BD = diagonals = 2 a ; O = Centre

2. Inscribed a circle with centre O and radius = a

2; side = diameter = a

3. OF = OG = Radius; FOG = triangle; OF = a

2; FG = hypotenuse =

2a

2;

4. EH = Side – Parallel to CD = a

5. DE = EF = GH = CH = EH FG 2a 1

a2 2 2

= 2 2

a4

6. BA + AD + DC + CH = a + a + a + 2 2

a4

= 14 2

a4

= circumference of the inscribed circle

(from Gayatri method).

7. Let us try to demarcate the extent of area of the inscribed circle with the help of official value

3.14159265358… = 3.14159265358…

3.141592653580.78539816339

4 4

8. 3a 30.75

4 4 ; Let us suppose ‘a’ = 1

9. S.No. 7 – S.No. 8 = 0.78539816339 – 0.75 = 0.03539816339

10. The area equal to 0.03539816339 cannot be located here. It has become impossible.

11. As an alternate, let us try to locate, the area with the guidance of the ‘Circumference of the Gayatri

method i.e. 14 2

a4

, and with this,

the Gayatri value is 14 2

4

36



12. Let us repeat S.No. 7, 8, 9 with the Gayatri value 14 2

4

13. 14 2

4

;

14 2 1 14 2

4 4 4 16

14. 3a 3

4 4 ; where a = 1;

15. S.No. 13 – S.No. 14 = 14 2 3 2 2

16 4 16

16. The area equal to 2 2

16

have to be located now. Let us see how

AB = a = 1; K = mid point of AB

17. AB = AK + KB = a a

2 2

Bisect KB into KL and LB = a

4

AL = AK + KL = a a 3a

2 4 4

DM = AL = 3a

4

18. So DM = 3a

4; Join ML

LB = MC = a

4 = NH = MC

19. There are two rectangles ALMD and MNHC

20. Area of ALMD rectangle = AD x AL = 23a 3

a a4 4

21. Area of MNHC rectangle = MN x NH = 22 2 a 2 2

a a4 4 16

37



22. In serial No. 10 we have said, getting an area equal to 0.03539816339 is impossible: With the Gayatri

value 14 2

4

, getting an area equal to

2 2

16

of S.No. 15 is thus possible.

23. Area of the inscribed circle

Areas of rectangles ALMD+MNHC = 2 2 23 2 2 14 2

a a a4 16 16

(S.No. 20) (S.No. 21)

24. Circle and square both can be inscribed and/ or circumscribed with each other. It means, both must be

finite entities, having finite magnitudes, and to be represented by finite numbers.

25. Official value cannot demarcate circle’s area. Whereas, Gayatri value demarcates. So is

3.14159265358… or 14 2

4

= 3.14644660942… the real value ?

26. It is another way of squaring a circle.

III. CONCLUSION New value is exact and it is an algebraic number and squaring of circle has be done by demarcating,

the area of a circle in a square.

REFERENCES [1]. Lennart Berggren, Peter Borwein, Jonathan Borwein, Pi: A Source Book (Springer, 1996).

[2]. David Blatner, The Joy of Pi (Walker/Bloomsbury, 1997).

[3]. Alfred S. Posamantier, Pi: A mysterious number (Prometheus Books, 2004).

[4]. RD Sarva Jagannadha Reddy, Pi of the Circle (www.rsjreddy.webnode.com, 2014).

38



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

39

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

40




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

41



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.

42

International Journal of Mathematics and Statistics Invention (IJMSI)

E-ISSN: 2321 – 4767 P-ISSN: 2321 - 4759

www.ijmsi.org Volume 2 Issue 7 || July. 2014 || PP-01-04


To Judge the Correct-Ness of the New Pi Value of Circle By

Deriving The Exact Diagonal Length Of The Inscribed Square

R.D. Sarva Jaganndha Reddy

ABSTRACT : Circle and square are soul and body of the subject of Geometry. All celestial bodies in the

Cosmos are spherical in shape. Four equidistant tangents on a circle will give rise to a square. A circle can be

inscribed in a square too. Thus, circle and square are two inseperable geometrical entities. is a fundamental

mathematical constant. The world believes 3.14159265358… as value for the last 2000 years. Yet it is an

approximate value. Continuous search for its exact value is going on, even now. God has been kind. The exact

value is not a myth. It has become real with the discovery of 14 2

4

. It is a very tough job to convince the

world that the new finding is the real value. In this paper, the exact length of the diagonal of the inscribed

square form the length of an arc of the superscribed circle is obtained as a proof.

KEYWORDS: Circle, circumference, diameter, diagonal, perimeter, square

I. INTRODUCTION The geometrical constant, called , is as old as human civilization. In the ancient days, contributions

on was very admirable from the Eastern parts of the World. The Founding Father of Mathematics:

Hippocrates of Chios (450 BC) has not touched the value of . But his work on the nature of circular entities

such as squaring of lunes, semicircle and full circle is unparalled in the History of Mathematics.The present

value, 3.14159265358… could not understood Hippocrates’ work and its greatness. What is the reason ? This

number is not the of the circle. It is the value of the polygon. It was derived using Pythagorean theorem.

Pythagorean theorem gives exact length to a hypotenuse which is a straight line. Circumference of a circle is

not a straight line. It is a curvature. Hence, 3.14159265358… of polygon has failed to understand the greatness

of Hippocrates. In other words, 3.14159265358… is not a number at all.

After a long waiting of 2000 years by trillions of scholars NATURE has revealed its true length of a

circumference of circle and its value. The value is 14 2

4

= 3.14644660942… derived from Gayatri

method. It was discovered in March 1998. This worker is the first and fortunate humble man to see this

fundamental truth, and at the same time, made this author responsible to reveal to the whole world, its total

personality. He was cautioned by the Nature, through Inner Voice, further, not to shirk his responsibility till the

end, till the world welcomes it, and continue to search speck by speck naturally, respecting the Cosmic mind

of the Nature, with indomitable determination, and fight single handedly against the conservative attitude and

die like a warrior defeated, if situation of “reluctance to acceptance” still persistsFrom the remote past to the

present, the period of study of is divided into two periods: the period of geometrical dependence in the

derivation of value and the 2nd

period from 1660 AD onwards till today, the period of dependence on infinite

series,

dissociating totally geometrical analysis in the derivation of value. In the present and second period, the

number 3.14159265358… is considered as a number only and nothing to do with the definition of “the ratio of

circumference and diameter of a circle”. The paradox is, geometry is forgotten in the derivation of value but

searched for the same, in squaring of a circle. Finally, this number 3.14159265358… has gained four

characteristics: 1. Though it is a polygon number, established itself as a number of circle. 2. Though an

approximate number it ruled the world for many centuries as a final value. 3. Though it doesn‟t belong to a

circle has commented „squaring of circle‟ as an unsolved geometrical problem. 4. Though it has gained a status

of transcendental number with the definition of “…the fact that cannot be calculated by a combination of the

operations of addition, subtraction, multiplication, division, and square root extraction…” (Ref.1) is derived

actually in Exhaustion method, applying the Pythagorean theorem, and invariably with the involvement of 3 .

Further, the moment this number discared the association of Geometry, it has started a new relationship in

43

To Judge The Correct-Ness Of The New Pi Value…


Euler’s formula e i

+1 = 0 with unrelated numbers which are themselves approximate numbers. C.L.F.

Lindeman (1882) has called number as a transcendental number based on Euler‟s formula. Here again, there

is a discrepancy in choosing , in the Euler‟s formula. In the formula, means radians 1800 and not,

constant 3.14… Are they both, radians and constant are identical ? constant is divine, whereas radians

equal to 1800 is, human creation and convenience. In the Euler‟s formula radians 180

0 is involved and the

resultant status of transcendence has been applied to constant. Let us rewright ei

+ 1 = 0 as

1 3.14e 1 0 . Is it right and acceptable ?It may not be wrong in calling that 3.14159265358… is a

transcendental number but it is definitely wrong when number is called a transcendental number and with a

consequential immediate statement “squaring of a circle” is impossible. The major objection is, the very

definition of is “the ratio of circumference and diameter of a circle”. So, this discrepancy led to the conclusion

that 3.14159265358… is not, infact, a number.In its support, we have the work of Hipporcrates. Hippocrates

had squared a circle even before any text book on Mathematics was born; His text book is the basis of Euclid’s

Elements too. As Hippocrates did square the circle it implied an algebraic number. Further, Lindemann is

also right in calling 3.14159265358… as transcendental number and not right, if he calls number as

transcendental. Here, there is a clarity of opinion thus. 3.14159265358… is a polygon number (and attributed

to circle).

In this paper the diagonal length is obtained from the actual length of the circumference of the circle.

II. PROCEDURE

Draw a square ABCD. Draw two diagonals AC and BD. „O‟ is the centre.

AB = Side = a; AC = BD = diameter = 2a

Draw a circle with centre „O‟ and with radius d

2=

diameter d

2 =

2a

2

Diameter = AC = 2a = d

Perimeter of ABCD square = 4 x a = 4a

1/4th

of the circumference CB = d

4

=

2a

4

where d = diameter = diagonal = 2a of the ABCD square.

[1] Let us find out 1/4th

of circumference of circle CB, with the present value, 3.14159265358…

[2] d

4

=

2a

4

=

3.14159265358 2a

4

=

1.11072073453a

4

[3] where diameter of the circle is 2a

[4] Let us use the following formula to get the length of the diagonal (known of course i.e., 2a ) from the

above value.

[5] Perimeter of thesquare


4

44



[6] 4a

7a 2a

2 4

4

7 2

2 4

4

14 2

4

16

14 2

[7] In the 3rd

step, let us multiply the above value with the 1/4th

of the circumference of the circle, to give the

diagonal AC of the square ABCD.

[8] 3.14159265358 2a 16

4 14 2

[9] 1.11072073453 a

1.271275345344

= 1.41203188536… = (1.41203188536)a

[10] The 2 value is 1.41421356237…

[11] It is clear therefore, that the value 3.14159265358… does not give exact length of the diagonal AC of

ABCD square whose value is 2a = 1.41421356237.

[12] Now, let us repeat the above steps with the new value = 14 2

4

[13] d

4

=

2a

4

14 2 12a

4 4

=

14 2 2a

16

[14] = 1/4th

of Circumference of circle CB

[15] Perimeter of thesquare


4

a. 4a

7a 2a

2 4

16

14 2

[16] 3rd

Step: Multiplication of values of 4 & 5 S. Nos

[17] = 14 2 2 16

a 2a16 14 2

[18] As the exact length of the diagonal AC of square ABCD equal to 2a is obtained with the new value,

so, 14 2

4

is the real value.

III. CONCLUSION In circle, there are circumference, radius and diameter. In square, there are perimeter, diagonal and

side. When a circle is superscribed with the square, circumference, side, diagonal and perimeter of square co-

exist in an interesting relationship. In this paper, this relationship is studied and the diagonal length is obtained

from the arc of the circumference. This is possible only when the exact length of the circumference is known.

A wrong length of circumference obtained using a wrong value gives only wrong length of diagonal.

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] 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.

[4] 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.

45



[5] 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

[6] 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.

[7] 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.

[8] 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 [9] 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

[10] 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.

[11] RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com. [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.

46


International Journal of Engineering Inventions

e-ISSN: 2278-7461, p-ISSN: 2319-6491

Volume 4, Issue 1 (July 2014) PP: 34-37

www.ijeijournal.com Page | 34

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)

R. D. Sarva Jagannadha Reddy This author though a non-medical graduate (Zoology) was offered a Medical post, Tutor in

Physiology in St. John’s Medical College, Bangalore, India

Abstract: 22

7, 3.14, 3.1416, 3.14159265358… are being used as values at school–level–calculations and at

the research-level calculations. Many more numbers are found in the literature for . A method, therefore, is

necessary to decide which number is, the real value. The following method chooses 14 2

4

=

3.14644660942… as the real value.

Keywords: Circle, corner area, diameter, side, square.

I. Introduction Circle and square are basic geometrical constructions. To find out perimeter and area of a square there

are present two formulae 4a and a2, where ‘a’ is the side of the square. Similarly, to calculate the circumference

and the area of a circle, there are two formulae, 2r and r2, where ‘r’ is the radius and is a constant. The

concept represents, the ratio of the circumference and the diameter of its circle. Thus, the constant is a

natural and divine concept. We have radians equal to 1800, which is a human creation and convenience.

For the last 2000 years, 3.14159265358 …. has been ruling the mathematical world as the value. In

March 1998, a new value 14 2

4

=3.14644660942… was discovered by Gayatri method and supported by

more than one hundred different geometrical proofs in the last 16 years. The time now has come to decide,

which value, is present value or is new value, real ? Here is a simple procedure.

The nature has created a square and a circle. All the celestial bodies in the Cosmos are spherical in

shape. It shows the basic architectural design of the physical world from the Cosmic mind. We see on paper

that it has not only created an exact relationship between the circumference and the diameter of a circle and also,

the Nature has established an interesting relationship between square and its inscribed circle too. This

relationship is taken as the guiding principle to decide the real value from many numbers. Hence, this method is called the Natural Selection.

47

The Natural Selection Mode To Choose The Real Pi Value Based On The


II. Procedure Draw a square. Inscribe a circle.

Square = ABCD, Side = AB = a, AC = BD = diagonal, ‘O’ centre.

EF = diameter = d = a

Area of the square = a2


2 2d a

4 4

where a = d

Square area – Circle area = Corner area =

22 a

a4

=

24a

4

Squarearea

Corner area =

2

2

a 4x

4 4a

4

Divide x by 32. In Siva method, it is found that when the circle – square composite geometrical

construction is divided symmetrically, the number of segments are 16 + 16 = 32

x

32=

4 1

4 32

=

4

32 4

=

1

8 4

= 1

32 8

= - 3

Thus, we obtain finally, the formula 1

32 8 which is equal to the value over and above 3. As 2

of the diagonal of a square, so also 2 is for the circumference of the inscribed circle, and it is an established

fact by this work. And further, the number 32 represents a common associating factor of the inscribed circle

and the square.

The procedure followed here is in 4 steps.

Step 1, Calculates the areas of square and circle

Step 2, Obtains corner area by deducting circle area from the square area

Step 3, = Squarearea

Corner area and

Step 4, = Squarearea 1

Corner area 32

At the 1st step while calculating the area of the circle, known value is used. In this paper two values are chosen:

3.14159265358… the official value and

3.14644660942… the new value = 14 2

4

Any value enters at the 1st step, and it’s decimal part reappears at the 4th step. Thus, the resurrection of the

decimal part of value is observed at the 4th step. And this happens only when the real value is taken in the 1st step. Any other number, if used, does not reappear fully, at the 4th step.

Side = diameter = a = 1

Area of the square : a2 = 1 x 1 = 1

I. With official value 3.14159265358…


2d

4

= 3.14159265358 x 1 x 1 x

1

4 = 0.78539816339

Square area – Circle area = Corner area = 1 – 0.78539816339

= 0.21460183661

Squarearea

Corner area =

1

0.21460183661 = 4.65979236616

Squarearea 1

Corner area 32

= 4.65979236616

32 = 0.14561851144

48



The decimal part of the official value is 0.14159265358… Only first two decimals 0.14 reappeared in the 4th step, instead of all the decimals.

II. Let us repeat the above process with the new value 3.14644660942…


2d

4

= 3.14644660942 x 1 x 1 x

1

4

= 0.78661165235

Square area – Circle area = Corner area = 1 – 0.78661165235 = 0.21338834765

Squarearea 1

Corner area 0.21338834765 = 4.68629150097

Squarearea 1

Corner area 32

= 4.68629150097

32 = 0.1464466094

The decimal part of the new value is 0.14644660942 All the decimals have now reappeared in the 4th step.

There are some more numbers if one looks at the Internet. Prominent numbers that are attributed to , besides

22

7 of Archimedes, are 17 – 8 3 (Laxman S. Gogawale), 3.125 (Mohammadreza Mehdinia), 3.144605511

(from PHI) etc. All these values too have failed when processed in the above steps, to resurrect at the 4th step.

S. No. Proposed/accepted numbers to Resurrected decimal part over and

above 3 of value

1. 22

7 = 3.142857142857

0.1458333333

2. 3.14159265358 (Official value) 0.14561851144

3. 17 – 8 3 = 3.1435935396

0.14595873078

4. 3.125 0.14285714285

5. 3.144605511029 0.14613140674

6. 3.2 0.15625

7. 14 2

4

=3.14644660942

0.14644660942

Archimedes’ 22

7 is much nearer to the real value than 3.14159265358… though it has been

considered as final value to . 3.125 is farthest low value to . number of Golden Ratio is the next closest to

the real value attributed to . Out of all the numbers attributed to value detailed in the above Table, only

one number 3.14644660942… of 14 2

4

has resurrected itself at the end, with its all the decimals, over and

above 3. Hence 14 2

4

is the true value. Other numbers have succeeded in coming back with one or two

first decimals only beyond 3. Hence, these numbers have, failed in the race to qualify themselves, in the

selection, by the natural geometrical construction as a true value.

III. Conclusion

There are many values in the literature. Two values 22

7 and 3.14159265358 … are very popular.

Third value equal to 14 2

4

= 3.14644660942… is added now to the existing values saying that the new

49



value is the only true value. In this paper a simple method is found, to choose, the real value. This method

chooses 14 2

4

as the real value, which is the exact and an algebraic number.

REFERENCES [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, 2



[2]. Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the World’s Most Mysterious Number, Prometheus Books,

New York 14228-2197.



[4]. 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.

[5]. 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

[6]. 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.



2014), PP 17-20.





[10]. 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.

[11]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com.



5728, p-ISSN:2319-765X. Volume 10, Issue 4 Ver. I (Jul-Aug. 2014), PP 44-48.

50



e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 4 Ver. III (Jul-Aug. 2014), PP 13-17 www.iosrjournals.org


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*)


Abstract: Circle, square and triangle are basic geometrical constructions. constant is associated with the

circle. In this paper, circle-triangle interlationship chooses the real value of and calculating the area of the

triangle involving of the inscribed circle. The alternate formula to find the area of the triangle is

23 3d

14 2

. This formula has a geometrical backing.

Keywords: Altitude, base, circle, diameter, perimeter, triangle

I. Introduction

The official value is 3.14159265358… It is an approximation, inspite of having trillions of its

decimals. A new value to was discovered in March 1998. The value is 14 2

4

= 3.14644660942… Both

the values have their own supporting arguments. Triangle is another geometrical entity. Its area is calculated using the formula ½ ab, where a = altitude and b = base. In this paper, a different formula has been derived to

find out the area of the equilateral triangle based on the of its inscribed circle. The formula ½ ab gives the area of the triangle. No other formula is necessary for the area of triangle. The main purpose of derivation of

new area formula of triangle is, to test the correctness of value involved in the new formula. One advantage in using the new formula for area of the triangle is, the resulting value tally’s exactly

with the value of ½ ab only, when the chosen value is correct one. If the wrong/ approximate value is involved in the new formula, it does not give exact value of the triangle. In other words, out of the two numbers

3.14159265358… and 14 2

4

= 3.14644660942… only one number, gives the exact area of the triangle.

The number which fails in giving exact value to the area of the triangle is decided as the wrong value.

Procedure

Draw a circle with centre ‘O’ and radius d

2. Draw three equidistant tangents on the circle. The tangents

intersect at A, B and C, creating an equilateral triangle ABC. DE is the hypotenuse of DOE triangle or the

chord of the circle.

Calculations:

Centre = O

Radius = OD = OE = d

2

Diameter = DF = d

Triangle = ABC

Side = AB = BC = AC = a = 3 d

* This author was awarded Merit Scholarship by Atomic Energy Commission, Trombay, Bombay for his

M.Sc., (Zoology) study in S.V. University College, Tirupati, Chittoor Dt. A.P.,India, during 1966-68. This

author is highly indebted to the AEC and hence this paper is named in the AECs’ Honour.

51

An Alternate Formula in terms of Pi to find the Area of a Triangle and a Test to decide the …..


Altitude = DC = 3d

2

Radius = OE = OD = d

2

Hypotenuse = Chord = DE = d

22 =

2d

2

Area of the triangle by Conventional formula

1ab

2 =

1DC AB

2 =

1 3d3d

2 2 =

23 3d

4

New formula to find the area of ABC triangle

Perimeter of the ABC triangle 3 x AB = 3 3d = 3 3d

4 timesof perimeter of theABCtriangle

14 timesof diameter DFof circle 2 timesof chord DE

= 4 3 3d

2d14d 2

2

= 12 3

14 2

To find out the area of the ABC triangle, multiply the area of the circle with 12 3

14 2


2d

4

Area of the ABC triangle = Area of the circle x 12 3

14 2

=

2d 12 3

4 14 2

=

23 3d

14 2

where d = diameter of the inscribed circle

Thus, 23 3

d14 2

is the new formula to find out the area of the superscribed triangle about a circle.

Where , may be 3.14159265358 or 3.14644660942 = 14 2

4

Area of the ABC equilateral triangle besides from the ½ ab formula is = 23

a4

where a = 3d

= 1 3d

3d2 2 =

23 3d

4

= 23

3d4

= 23 3

d4

So, the value 23 3

d4

as the area of the ABC triangle, should also be obtained, using the above new

formula, derived in terms of

23 3d

14 2

should be equal to 23 3

d4

52



The unknown one in the above is . Let us write formula with two different values.

2

2

2

3 3 3.14159265358d

14 23 3

d4

14 23 3

4 d14 2

OR should be

3 3 3.141592653581 1

14 2

where d = 1

= 1.29703410738, we have obtained this value

instead of 3 3

4 = 1.29903810567

So, official value 3.14159265358 has failed to give 3 3

4 as the value of the area of ABC triangle. On the

otherhand, new value 14 2

4

has given

3 3

4 as the area of ABC triangle. This shows, that the new

value 14 2

4

is the real value.

Equating conventional formulas ½ ab = 23

a4

to new formula 23 3

d14 2

is itself a justification and a

naked truth of latter’s correctness.

II. Conclusion

½ ab is the formula to find out the area of a triangle. In this paper, a new formula 23 3

d14 2

is

derived. This formula by giving the exact area of ABC triangle shows, that circle and the equilateral triangle

are clearly interrelated and are not very different as we have been believing. The benefit we derive by using

this formula is, this formula chooses the real value, discarding other approximate values attributed to . This

method alone, for the first time in the history of mathematics, acts as a testing method of and boldly says

Archiemedes’ upper limit of 1

37

or 22

7 is a lower value compared to the real number.

Post script

As it is proved in this paper that the real value is 14 2

4

based on the area of the triangle ABC, it has made

possible to demarcate the length of the circumference of the inscribed circle in the straight-lined perimeter of

the ABC triangle. How ?

Let us know first the length of the circumference of the inscribed circle with the known value 14 2

4

Diameter = d

Circumference = d

53



= 14 2

4

When the diameter is equal to 1, the circumference = value

Let us search for the line-segments equal to 14 2

4

.

Diameter = DF = d

Altitude of ABC triangle = DC = 3d

2

Radius = OH = d

2

DE = hypotenuse = Chord = 2d

2

G is the mid point of DE

So, DG = GE = OG = 2d

4

OG = 2d

4

GH = Radius – OG = d 2d

2 4 =

2 2d

4

CI = DC = 3d

2

IJ = GH = 2 2

d4

Circumference of the circle d = 14 2

d4

, So, the demarcated length DCJ in the perimeter of the ABC

triangle is equal to the circumference of the inscribed circle.

DC + CI + IJ = 3d 3d 2 2

d2 2 4

=

14 2d

4

Thus the straight lined length equal to length of the curvature of the circumference of the inscribed circle is called rectification of circumference of a circle. It was done never before. People tried earlier in the

perimeter of the square but never in the perimeter of the triangle.




[2]. Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the World’s Most Mysterious Number, Prometheus Books,

New York 14228-2197.







Ver. II (Mar-Apr. 2014), PP 09-12



2014, PP-33-38.



2014), PP 17-20.

54










[11]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com



5728, p-ISSN:2319-765X. Volume 10, Issue 4 Ver. I (Jul-Aug. 2014), PP 44-48

55



e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 3 Ver. II (May-Jun. 2014), PP 39-46



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

56



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

57



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

58



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

59



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)

60



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)

62



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

63



e-ISSN: 2278-3008, p-ISSN:2319-7676. 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

=

64


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.

65

International Journal of Engineering Inventions

e-ISSN: 2278-7461, p-ISSN: 2319-6491

Volume 3, Issue 11 (June 2014) PP: 29-35


The unsuitability of the application of Pythagorean Theorem of

Exhaustion Method, in finding the actual length of the

circumference of the circle and Pi

R. D. Sarva Jagannadha Reddy

Abstract: There is only one geometrical method called Exhaustion Method to find out the length of the

circumference of a circle. In this method, a regular polygon of known number and value of sides is inscribed,

doubled many times until the inscribed polygon exhausts the space between the polygon and circle as limit. In

the present paper, it is made clear, that the value for circumference, i.e. 3.14159265358…. of polygon,

attributing to circle is a lower value than the real value, and the real value is 14 2

4

= 3.14644660942

adopting error-free method.

Key words: Circle, corner length, diameter, diagonal, polygon, side, square.

I. Introduction

The Holy Bible has said value is 3. The formula for circumference of a circle is d, where is a

constant and d is the diameter. When the diameter is 1 the circumference is equal to value.

When, a hexagon is inscribed in a circle of ½ as its radius, the perimeter of hexagon is equal to 3. It

means, the circumference is greater than 3, because hexagon is an inscribed entity in the circle. With the unit

diameter of circle, circumference and/or value is exchangeable, because, both are represented by a single

number. There were many values for . 10 = 3.16… (Chung Hing, 250 AD), 142

45 = 3.155… (Wang Fau,

250 AD), 3.14159… (Liu Hui, 263 AD), 3.1415 (Aryabhata, 499 AD). From Francois Viete (1579) onwards

the value 3.1415926… has become an accepted value.

It is a well established fact that the value is 3.14159265358… However, two things which are

associated with this number have not satisfied some scholars of mathematics. They are 1. 3.14159265358… is a

borrowed number of polygon, attributed to circle, believing in the logic of limitation principle and being an

approximate number, made super computers too have failed to find the exact value and 2. Assertively, this

number has said, squaring a circle impossible, being it is a transcendental umber. High school students, now

and then ask, when mathematics is an exact science, how is it possible we have many values are being used, for

example 22

7, 3.14, 3.1416, 3.14159265358….

Nature has been kind. Exact value has been found at last. A few papers in support of the exact

value have been published (in Reference). Surprisingly the new value is struggling very hard still for its

approval. All the time, the work done on has been cited, saying, the past generation or the present workers

could not be wrong. Thus, the new value (through was discovered 16 years ago, in March 1998 is yet to be

accepted.

From March 1998, with the discovery of 14 2

4

= 3.14644660942, this worker has been struggling

to find evidences in support of new value, and also struggling much more in search of an error in the

derivation of 3.14159265358…

There are some similarities and some differences between present and new values.

The similarity is, in Exhaustion method a polygon is inscribed in a circle and in the new method a

circle is inscribed in a square.

The differences are many: 1. Present value 3.14159265358… represents the perimeter/ area of the

polygon and attributed to circle; the new value represents the area of the inscribed circle only in a square

(called Siva method in Reference). Here, square helps but does not lend its value to circle. 2. Present value is

derived applying Pythagorean theorem meant for straight lines. New value is derived adopting entirely a new

approach for which no previous proofs/ statements are available. 3. Present value is a transcendental number

and new value is an algebraic number. 4. Present value says squaring of a circle is an impossible idea and

66

The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the…


whereas the new value squared a circle. It also, circled a square (i.e., constructing a circle whose

circumference and/or area equal to the perimeter and/ or area of a square respectively. For example, if the side

of a square is 1, the perimeter would be 4 and its area would be 1. Circling a square means we have to find out a

radius geometrically of circle whose length of the circumference is 4 and/ or area of the circle is 1.

In this paper, present value-to some decimal places – is obtained using the new value 14 2

4

,

showing, circle, polygon and square are not different geometrical entities and their interrelationship, if

understood correctly, may help in the derivation of present value from the new value also.

If this idea is accepted, an algebraic number like 14 2

4

is also, capable of giving rise to, a

transcendental number like 3.14159265358…. It leads to the creation of a new doubt, is 3.14159265358… is

really a transcendental number ?

There are 3 examples cited below, linking present and new values 3.14159265358… and

3.14644660942… = 14 2

4

.

To drive to the point an elaborate explanation is chosen and here and there repetitions too appear.

II. PROCEDURE

Example-1

The following formula gives the present value upto 5 decimal places.

When my work on deriving the new value of equal to 14 2

4

= 3.1464466.. was submitted, I was

advised by many to think how to derive the present value 3.1415926… cautioning me as the new one was

wrong.

In obedience, I hereby submitted the following formula giving the number exact upto five decimal

places.

93 3.14159...

200

where

14 2

4

To get the same value (3.14159…) from the Classical method of Archimedes, we have to travel a long way

using the following general formula to calculate the length of the perimeter of the inscribed polygon of 1536

sides starting from 6 sides, in a circle

Side = 2 2 2

2n ns 2r r 4r s

A geometrical line segment for 3.14159…

Let us draw a circle with diameter 9 and radius 9/2. Cut the circumference at one point and straighten

it and further divide it into 200 equal segments.

Similarly, divided the diameter into 3 equal segments.

1/3rd

diameter + one segment length of circumference

9

3

+

14 29

49

200 200

= 3.14159…

67



Example-2

Archimedes inscribed a polygon of 96 sides with the circle and gave perimeter of the polygon as equal

to 3 10/71 = 3.140845… When the diameter of the superscribed circle is 1 unit, the perimeter = value. Prof.

Constantine Karapappopulos of University of Patra, Patra, Greece has suggested a formula for the

Archimedian minimum as equal to 1 3

3 13 3

in the range of 10

371

to 10

370

The proposed formula for = 14 2

4

gives the circumference of the circle as equal to

1 23 1

2 2

1 33 1

3 3

x

3 3 2 1

2 2 3 1

= 1 2 14 2

3 12 2 4

Perimeter of Polygon x Circle connecting link = Circumference of Circle

3 3 2 1

2 2 3 1

is obtained by the calculation method. One need not set aside it on the reason that the

above factor is based on the calculation method. This factor or circle connecting link can be viewed as a clue

or guiding factor for further analysis. On close observation one finds the denominator and the numerator of the

link are symmetrical and give us a right relationship between the inscribed polygon and the superscribed circle.

Example-3

Classical method involves the inscription of polygon in the circle. Starting with the known perimeter of a

regular polygon (here we start with a regular hexagon) of n sides inscribed in a circle, the

perimeter of the inscribed regular polygon of 2 n sides can be calculated by the application

of Pythagorean theorem. Let C be a circle with centre O and radius r, and let PQ = s be a

side of a regular inscribed polygon of n sides having a known perimeter. Then the apothem,

OM = u is given by

2

2 sr

2

hence the sagetta, MR = v = r – u is known. Then the

side RQ = w of he inscribed polygon of 2n sides is found from

2

2 sw v

2

hence the perimeter of this

polygon is known.

68



S.

No

.

n First Side PQ

= s

ns

Perimeter =

Circumferen

ce =

s/2 = t Apothem

2 2r t u

Sagetta

r – u = v

Succeeding

side

2 2v t w

1. 6 0.5 3.0 0.25 0.43301270

1

0.0669872

99 0.258819045

2. 12 0.258819045 3.1058828544 0.129409522 0.48296291

3

0.0170370

87 0.13052619

3. 24 0.13052619 3.132628565 0.065263095 0.49572243 0.0042775

7 0.065403128

4. 48 0.065403125 3.139350157 0.032701564 0.49892946

1

0.0010705

39 0.032719082

5. 96 0.032719082 3.141031907 0.016359541 0.49973229

3

0.0002677

07 0.016361731

6. 192 0.016361731 3.141452431 0.008180657

05

0.49993306

9

0.0000669

31

0.0081811394

95

7. 384 0.0081811394

95 3.141557566

0.004090569

748

0.49998326

7

0.0000167

33

0.0040906039

72

8. 768 0.0040906039

72 3.14158385

0.002045301

986

0.49999581

6

0.0000041

84

0.0020453062

62

9. 153

6

0.0020453062

62 3.141590424

0.001022653

133

0.49999865

4

0.0000010

4

0.0010226536

68

10. 307

2

0.0010226536

68 3.141592067 - - - -

Let us analyse the above Table. The calculation is started with a regular polygon of 6 sides having the perimeter

equal to 3. The sides of polygon are doubled for 9 times and now the inscribed polygon has 3072 sides and its

perimeter = circumference of circle = value is equal to 3.141592067… Thus, the number of sides are

increased 512 times finally 3072

5126

. The length of the perimeter has increased from 1

62

= 3.000000

to 3.141592067… The correct value is 3.14159265358… From the above Table, it is possible to find the exact

length only upto 3.141592 i.e. 6 decimal places only with 3072 sides of the polygon.

ABCD = Square; EFGR = Circle; AB = Side = NQ = diameter = 1;

AC = Diagonal = 2 ; AN = QC = Corner length;

Diagonal – diameter = AC – NQ = 2 1 ; Hexagon = STGUVE

In the Classical method of calculating perimeter/ circumference/ in the above table, number of sides of the

inscribed polygon plays an important role. The derivation is started with 6 sides and ended with 3072 sides.

The perimeter of the hexagon is 3 and the perimeter of the polygon of 3072 sides has increased to

3.141592067…

69



Starting no. of sides of the polygon = 6

Ending no. of sides of the polygon = 3072

Starting perimeter = 3

Ending perimeter = 3.141592067

Present value = 3.14159265358

New value = 14 2

4

= 3.14644660942…

In the following formula a relation is shown between two values: present value and new value as in the

above two examples.

New value – Squareof thestarting no.of sides

Diagonal diameterEnding no.of sides

= Present value (correct upto 6 decimals)

14 2 6 6 899 67 22 1

4 3072 256

= 3.14159254422…

The lengthy process involved in the above Exhaustion method is represented in a single formula 899 67 2

256

.

In the Exhaustion method Pythagorean theorem is applied. 3 invariably appears in every aspect of

calculation. The result 3.14159254422… obtained from the above formula using 2 is much more accurate at

7th

decimal place than what it is (3.141592067…) obtained in the Exhaustion method in the table.

In the Exhaustion method RQ is the hypotenuse and is the side of the inscribed polygon.

And in the above formula diagonal – diameter = corner lengths on either side of the diameter also play an

important role. Hence, it has become possible that the new value has given the present value (to some

decimals).

Let us understand, in much more clear terms, the above formula.

There are four factors in the Exhaustion method. They are

No. of sides of the hexagon = 6

Perimeter of the hexagon = ½ x 6 = 3

No. of sides of the final polygon = 3072

Perimeter of the final polygon = 3.141592067

1. Let us divide final sides by the sides of the beginning polygon i.e. hexagon 3072

5126

.

2. Let us divide corner length AN + QC into 512 parts.

AB = Side = NQ = Diameter = 1

AC = diagonal = 2

Corner length = AC – NQ = diagonal – diameter = 2 1

Let us divide above length into 512 parts = 2 1

512

= 0.00080901086

3. Multiply the above value with the no. of sides of the beginning polygon (hexagon) 2 1

6512

=

0.00485406516

4. Deduct the above value from the new value 14 2

4

= 3.14644660942 which gives the present

value (to some decimals).

14 2 6 2 6 899 67 2

4 512 256

= 3.14159254424

5. So, the corner length 2 1 is divided into 512 parts.

70



6. In this process corner length 2 1 is taken for consideration.

7. In the Exhaustion method the hypotenuse RQ in Fig.2 is taken for consideration and proceeded

successively for many times.

8. AC is the hypotenuse of the triangle DAC of Fig.2 and RQ is the hypotenuse of the triangle RMQ.

9. How does the length over and above 3 diameters of the circumference of the circle is arrived in

deriving the new value ?

The answer is very simple.

Let us divide the corner length QC only of Fig-2 by 2 and add to the 3 diameters.

Side = AB = diameter NQ = 1

Corner length QC = Diagonal diameter

2

= AC NQ 2 1

QC2 2

Divide QC by 2

= 2 1 1 2 1 2 2

2 42 2 2

Now, the length of circumference = 3diameters + 2 2

4

= 2 2 14 2

34 4

10. When we compare two ways of arriving the exact length of circumference of a circle, it is clear in

Exhaustion method the perimeter of the inscribed polygon is increased slowly by doubling the number of

sides of the previous polygon. Thus, the number of sides have been increased from 6 to 3072, it means it

has been increased 512 times. In other words, we have divided corner length into 512 parts.

In the second approach in the arrival of length of the circumference, the corner length is

divided at one stroke with 2 = Corner length QC

2

So, 3 sides + QC

2 =

2 1 13

2 2

= 14 2

4

11. Thus there are two values for the length of the circumference of the circle.

1. 3.14159265358… of Exhaustion method and

2. 3.14644660942… of Gayatri method

12. We can thus visualize a diagram of 2, containing a hexagon whose perimeter STGUVE is 3, next to

hexagon, 3.14159265358 of inscribed polygon of 3072 sides and further next and the circle EFGR, whose

circumference is equal to 3.14644660942… with radius ½.

III. Conclusion

There are now two values 3.14159265358… and 3.14644660942 … = 14 2

4

. In the Exhaustion

method, perimeter of the polygon is attributed to the circumference of the circle. As the inscribed polygon is

smaller one, the value 3.14159265358… must be lesser than the exact length of the circumference of circle.

This paper clearly shows the calculation error involved in the arrival of the actual length of the circumference

(= ) of the circle. Hence, Exhaustion method is not suitable in arriving the of the circle. However, this study

shows that both the values have a common-ness in their nature. In Exhaustion method while doing

calculations involving squaring, square root etc. and only a few decimals have been taken. All the numbers are

infinite numbers. So, the prolongation of round-off-error is universal throughout the calculations. And this

has resulted in a lower value 3.14159265358… instead of the actual value. However, the new work has tried to

overcome the error supposed to be in the Exhaustion method by adopting Gayatri method, Siva method, Jesus

71



method etc. and found the actual length of the circumference and real value, i.e. 14 2

4

=

3.14644660942…

REFERENCES [1.] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, 2nd edition, Springer-Verlag Ney York

Berlin Heidelberg SPIN 10746250.

[2.] 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.

[3.] 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. [4.] 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

[5.] 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.

[6.] 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.

[7.] 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 [8.] 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

[9.] R.D. Sarva Jagannadha Reddy (2014), Pi of the Circle, at www.rsjreddy.webnode.com

72


R.D. SARVA JAGANNADHA REDDY – HOME PAGE

1. Mother : Dhanalakshmi

2. Father : Venkata Reddy

3. Date of Birth : 13.04.1946

4. Native Place : Bandarupalli Village

Yerpedu Mandal,

Chittoor District, AP, India.

5. Phone No. : 0877-2244370

6. E-mail : [email protected]

7. Present Address (Temporary) : 19-9-73/D3,

Sri Jayalakshmi Colony,

S.T.V. Nagar,

Tirupati – 517 501, INDIA

8. Education : B.Sc., Zoology (Major),

Botany, Chemistry (minors) 1963-66

M.Sc., Zoology 1966-68

at S.V. University College, Tirupati.

9. Books : 1. Origin of Matter

2. Origin of the Universe

3. Organic Bloom (on Animal Evolution)

4. Pi of the Circle

Telugu Books

5. Sarvam Pavithram

6. Pavana Prapancham

7. Mahabhagavatham – Maanavaavirbhavam

8. Abhinandana

9. Mattipella

10. Janthu Pravarthana (Animal behavior) for B.Sc.,

11. Kachhapi

10. Wife : Late Savithri

Children : Shyam Sundar Reddy, Gowri Devi, Sarada

11. Profession : Lecturer in Zoology, Retired on 30.06.2003

73

mailto:[email protected]

12. Donation : As a mark of Gurudakshina to my Alma Mater Sri

Venkateswara University, Tirupati a granite stone-

sphere of 6 feet diameter and another granite stone-

sphere of 3 feet diameter to Govt. Junior College,

Piler, Chittoor district (where, this author got the idea

in 1972, while working as Junior Lecturer in Zoology,

one can get formulae for the computation of area and

circumference of a circle without using constant

22/7 as in r2 and 2 r) have been humbly donated.

Stone at S.V.U. Mathematics Department, Tirupati, A.P., India.

Stone at Govt. Jr. College, Piler, Chittoor District, A.P., India.

74

Donated 11 Feet high Sivalingam to S.V. Higher Secondary School, Prakasam

Road, Tirupati, and consecrated at TTD‟s, S.V. Dhyanaramam, Opposite to

Regional Science Centre, Alipiri, Tirupati.

Donated 6 Feet high Sivalingam to Beriveedhi Elementary School, Tirupati and

consecrated at Rayalacheruvu Katta, 15 km away from Tirupati.

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