-
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.
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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.
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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
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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
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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*)
R.D. Sarva Jagannadha Reddy
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
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Pi treatment for the constituent rectangles of the superscribed
square in the study of exact area of ..
www.iosrjournals.org 45 | Page
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
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Pi treatment for the constituent rectangles of the superscribed
square in the study of exact area of ..
www.iosrjournals.org 46 | Page
= 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
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Pi treatment for the constituent rectangles of the superscribed
square in the study of exact area of ..
www.iosrjournals.org 47 | Page
= 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
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Pi treatment for the constituent rectangles of the superscribed
square in the study of exact area of ..
www.iosrjournals.org 48 | Page
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
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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 www.iosrjournals.org
www.iosrjournals.org 33 | Page
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
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A study that shows the existence of a simple relationship among
square, circle, Golden Ratio ..
www.iosrjournals.org 34 | Page
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
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A study that shows the existence of a simple relationship among
square, circle, Golden Ratio ..
www.iosrjournals.org 35 | Page
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
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A study that shows the existence of a simple relationship among
square, circle, Golden Ratio ..
www.iosrjournals.org 36 | Page
(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
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A study that shows the existence of a simple relationship among
square, circle, Golden Ratio ..
www.iosrjournals.org 37 | Page
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)
R.D. Sarva Jagannadha Reddy
Abstract: - value 3.14159265358 is an approximate number. It is
a transcendental number. This number
says firmly, that the squaring of a circle is impossible. New
value was discovered in March 1998, and it is
14 2
4
= 3.14644660942.. It is an algebraic number. Squaring of a
circle is done in this paper. With
this new value, exact area of the arbelos is calculated and
squaring of arbelos is also done. Arbelos of
Archimedes chooses the real value.
Keywords: - Arbelos, area, circle, diameter, squaring, side
I. INTRODUCTION Circle and square are two important geometrical
entities. Square is straight lined entity, and circle is a
curvature. Perimeter and area of a square can be calculated
easily with a2 and 4a, where a is the side of the square. A circle
can be inscribed in a square. The diameter d of the inscribed
circle is equal to the side a of
the superscribed square. To find out the area and circumference
of the circle, there are two formulae r2 and
2r, where r is radius and is a constant. constant is defined as
the ratio of circumference and diameter of
its circle. So, to obtain the value for , one must necessarily
know the exact length of the circumference of the circle. As the
circumference of the circle is a curvature it has become a very
tough job to know the exact value
of circumference. Hence, a regular polygon is inscribed in a
circle. The sides of the inscribed polygon doubled
many times, until, the inscribed polygon reaches, such that, no
gap can be seen between the perimeter of the
polygon and the circumference of the circle. The value of
polygon is taken as the value of circumference of
the circle. This value is 3.14159265358
In March 1998, it was discovered the exact value from Gayatri
method. This new value is 14 2
4
=
3.14644660942.
In 1882, C.L.F. Lindemann and subsequently, Vow. K. Weirstrass
and David Hilbert (1893) said that
3.14159265358 was a transcendental number. A transcendental
number cannot square a circle. What is squaring of a circle ? One
has to find a side of the square, geometrically, whose area is
equal to the area of a
circle. Even then, mathematicians have been trying, for many
centuries, for the squaring of circle. No body could succeed except
S. Ramanjan of India. He did it for some decimals of 3.14159265358
His diagram is shown below.
Then the square on BX is very nearly equal to the area of the
circle, the error being less than a tenth of an inch when the
diameter is 40 miles long.
S. Ramanujan
13
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 64 | P a g
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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
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 65 | P a g
e
Draw a perpendicular line at G on EF diameter, which meets
circumference at H. Apply Altitude theorem to
obtain the length of GH.
GH = EG GF = 4 1
5 5 =
2
5
4. Draw a circle with diameter GH = 2
5 = d
Area of the G.H. circle =
2d
4
=
2 2
4 5 5 25
5. Area of the G.H. circle = Area of the arbelos
So, area of the arbelos = 14 2 1
25 4 25
= 14 2
100
Part II: Squaring of circle present in the ABCD square 6.
Diameter = EF = d = a = 1
Area of the circle =
2d
4
= 1 1
4 4
7. To square the circle we have to obtain a length equal to
4
. It has been well established by many
methods more than one hundred different geometrical
constructions that value is 14 2
4
. Let
us find out a length equal to 4
.
8. Triangle KOL
OK = OL = radius = d
2 =
a
2 =
1
2
KL = hypotenuse = 2d
2 =
2a
2 =
2
2
DJ = JK = LM = MC = Side hypotenuse
2
= 2a 1
a2 2
=
2 2a
4
= 2 2
4
So, DJ = 2 2
4
9. JA = DA DJ = 2 2
a a4
= 2 2
a4
. So, JA = 2 2
4
Bisect JA twice
JA JN + NA NP + PA
= 2 2
4
2 2
8
2 2
16
So, PA = 2 2
16
10. DP = DA side AP = 2 2
116
= 14 2
16
15
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 66 | P a g
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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
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 67 | P a g
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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
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 68 | P a g
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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
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Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 69 | P a g
e
d = a = 1, 14 2
4
= 14 2 1
1 14 4
=
14 2
16
31. Area of the arbelos reciprocal of the square of the arbelos
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
-
Squaring of circle and arbelos and the judgment of arbelos in
choosing the real Pi value (Bhagavan
International organization of Scientific Research 70 | P a g
e
REFERENCES
[1] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997),
Pi: A source Book, 2nd edition, Springer-Verlag Ney York Berlin
Heidelberg SPIN 10746250.
[2] Alfred S. Posamentier & Ingmar Lehmann (2004), , A
Biography of the 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
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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
International Journal of Latest Trends in Engineering and
Technology (IJLTET)
Vol. 4 Issue 2 July 2014 134 ISSN: 2278-621X
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
International Journal of Latest Trends in Engineering and
Technology (IJLTET)
Vol. 4 Issue 2 July 2014 135 ISSN: 2278-621X
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
International Journal of Latest Trends in Engineering and
Technology (IJLTET)
Vol. 4 Issue 2 July 2014 136 ISSN: 2278-621X
24
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IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV.
(Feb. 2014), PP 48-49
www.iosrjournals.org
www.iosrjournals.org 48 | Page
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)
www.iosrjournals.org 49 | Page
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
-
IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. I.
(Jan. 2014), PP 58-59
www.iosrjournals.org
www.iosrjournals.org 58 | Page
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
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Jesus Method To Compute The Circumference Of A Circle And Exact
Value
www.iosrjournals.org 59 | Page
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
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IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. II
(Mar-Apr. 2014), PP 09-12
www.iosrjournals.org
www.iosrjournals.org 9 | Page
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
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Supporting Evidences To The Exact Value From The Works Of
Hippocrates Of Chios, Alfred S.
www.iosrjournals.org 10 | Page
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
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Supporting Evidences To The Exact Value From The Works Of
Hippocrates Of Chios, Alfred S.
www.iosrjournals.org 11 | Page
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
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Supporting Evidences To The Exact Value From The Works Of
Hippocrates Of Chios, Alfred S.
www.iosrjournals.org 12 | Page
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
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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
R.D. Sarva Jagannadha Reddy
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