15 PUBLISHED PAPERS ON PI.pdf

THE GAYATRI Pi VALUE (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])

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

Eulers 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 Eulers 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 Eulers 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 ITs grace.

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 authors 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

authors 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 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. Johns 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

1

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 Universitys 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 2 a4

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

= IQKJ = IQ x QK = 2 2

a a16

=

22 2 a16

16. Area of 4 rectangles

4



= IQKJ + QPLK + PNML + NABM = 22 24 a

16

= 22 2 a

4

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

= 2 2 22 2 2 2a 4 a a

4 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 a

16

= 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 a

4

of

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

16

of S.No. 15

2 22 2 2 2a 3 a4 16

=

2214 2 aa

16 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 2a a

4

= 22 2 a

4

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 a

4

(S.No. 20 & 21)

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

= 244 a

4

= 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 24

3 a 3 a a4 4

This is the end of the process of proof.

32. As the corner area is equal to

1. Arithmetically = 22 2 a

16

= 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 circles 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/ Its 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, 2nd edition, Springer-Verlag Ney York Berlin

Heidelberg SPIN 10746250.

[2]. Alfred S. Posamentier & Ingmar Lehmann (2004), , A Biography of the Worlds 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


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 Worlds 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 shoemakers 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 2rr r

2 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 q4

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 q4

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

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 Worlds 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.

[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.

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 KZL

ZK = ZL = radius = 1

5

KL = hypotenuse = 1

25 =

2

5

RB = 2

5

21. LU = Side hypotenuse

2

=

2 2 1

5 5 2

=

2 2

10

22. So, LU = 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

Area of the circle =

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 circles 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 Worlds 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 value

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

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

I. INTRODUCTION

The official 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, ab 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 . The following formulas are discovered (March, 1998) from Gayatri method and Siva method.

1. Area of Circle = 27r 2r

r r2 4

=

and

2. Circumference of Circle = 2r 2r6r 2 r

2

+ = ; where r = radius

2d 1 1 dd d d2 2 4 4

=

where d = diameter = side of the superscribed square In the Fletchers 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 114

.

Fig-1: Professors 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 . 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 . 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 = 12

; LK = OK OL = 1 2 2 22 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 12

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

Professors formula

2a132 142 12 2

2

=

(of Fig.1, where radius = 1)

where, 114

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 12 2

2

=

14 24

=

The accepted value for is 3.14159265358 With this , the area of shaded region is equal to 11 3.14159265358 0.21460183661...4

=

And, with the new 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 value is right i.e. is 3.14159265358 or

14 2 3.14644660942...?4

=

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

To decide which 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 1Area of theshaded region 0.21460183661of official

=

= 4.65979236616

and with new



23

1 1 16 4.68629150101...Area of theshaded region 2 2 2 2

16

= = =

+ +

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 = 4.65979236616 0.14561851144...

32=

New = 4.68629150101 0.1464466094...

32=

0.14644660942 of new 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 from 3rd decimal

onwards. If this argument is accepted, the present 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 Fletchers construction is analyzed arithmetically with the line-segments of the Siva method. This arithmetical

interpretation has resulted in the derivation of a new 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 2 216 16

128 128a a a

Area of the inscribed circle = 16S1 + 8S2 = 2 2 26 2 2 2 14 216 8

128 128 16a 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 segments formula and second line-segment equal to the

value of (4 - ) in S2 segments formula are searched in the above construction.

a) Official : - 2 = 3.1415926 - 2 = 1.1415926. Proposed : - 2 = 14 2

24

= 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 24 4

a 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 15

th 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 Eulers 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 Lindemanns 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 1/8

th 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 3.1415926is, 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 Euclids 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. Posamentiers 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 1 14 24

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 22

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 22

r r

7. Half of HC and HK are same 2 2 12 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 ab 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 1 14 24

= 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 Worlds 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

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