Pi OF THE CIRCLE Vol. II

Transcription

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2 Pi OF THE CIRCLE Vol. II (Published papers in the International Journals) By R.D. Sarva Jagannadha Reddy Retired Zoology Lecturer 19--S7-37, S.T.V. Nagar, Tirupati , India. August, 01 (For copies, send please, to the author:

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4 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.

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6 Preface has been used as a value for the last 000 years. This number actually represents polygon of Exhaustion Method of Archimedes (0 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, 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, 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 The World of Mathematics has recognized very recently, that Madavan is the first to invent infinite series for the derivation of 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 (188), Von K. Weirstrass and David Hilbert have called as a transcendental number. The basis for their proof was Euler s formula e i (Leonhard Euler, Swiss Mathematician, ). With their proofs, squaring of circle has become, without any doubt, an unsolved geometrical problem. Thus, the present thinking on is the number. is, value which is an approximation and squaring of circle is impossible with the At this juncture, the true and an exact value equal to was discovered by the grace of God in March, 1998 after a struggle of 6 years (from 197) 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 i

7 not discarded, by this author. He continued and confirmed 1 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 or 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 1 years and titled Pi of the Circle in 010, and is available in the website The second responsibility before this author was also, to search for any flaw in the derivation of the present value of equal to 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, 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 and not. Why? It has been shown in earlier paragraph that Euler s formula is the basis in calling as a transcendental number. In the formula e i + 1 0, refers to radians equal to and not constant 3.1. constant has no place for it in the above formula. When 3.1 is involved in the Euler s formula, the formula becomes wrong. Is it acceptable then to call constant as transcendental number even though this has no right of its participation in the above formula? However, it is acceptable still, if one agrees that radians constant 3.1 or radians is identical to constant 3.1 Mathematics may not accept this howler. Thus, on two counts i.e., is a polygon number and calling it as a transcendental number, based on radians 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 is also done, by IT s grace. value. It is exact and is an algebraic number. Squaring of circle ii

8 In this book there are two unpublished papers also, one is to show the first method the base of this work after 6 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. International Journal of Mathematics and Statistics Invention 3. International Journal of Engineering Inventions. International Journal of Latest Trends in Engineering and Technology 5. IOSR Journal of Engineering (IOSRJEN) While writing Pi of the Circle, Mr. A. Narayanaswamy Naidu, and while writing these published papers of this book, Mr. M. Poornachandra Reddy, have helped this author in simplification of formulas. The Editors of above Journals have published this author s work after refining the papers, keeping in mind the standards expected in the original research. This author could complete his University Education ( ) because of his mother only. He led a happy life for years with his wife. Now he is leading a peaceful life because of his second daughter R. Sarada out of three children by staying at her house, after the death of this author s wife two and half years ago. Mr. Suryanarayana of M/s. Vinay Graphics, Balaji Colony, Tirupati, has done DTP work perfectly well. This author, therefore, is greatly indebted to these well-wishers and prays to the God to bless them with good health. This author requests the readers to send their comments and they will be gratefully received and acknowledged. To end, the quantum of contribution of this author in this work is equal to, square root of less than one, in the square of trillions of 1 trillions, i.e. trillions of trillions. Author iii

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10 How did this Zoology Lecturer get encircled himself in 197 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 197, while reading an encyclopedia he saw square, triangle, trapezium and so on and found constant for circle alone in r and 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 6 years. Being a government college teacher, he was transferred in the mean time, from Piler ( ) to Kadapa, to Nagari, to Anantapur and finally to Chittoor (in December 1995). No answer to his question of 197. He was 6 when question came, waited another 6 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 iv

11 area and circumference of circle without equal to /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 /7 and it was 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 /7, 3.116, 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 as value. This author thanks the reader for this attentive reading. v

12 CONTENTS Page No. 1. Preface i-iii. How did this Zoology Lecturer get encircled himself in 197 in Mathematics? iv-v 3. Gayatri Method (unpublished) 1. times of area of the circle is equal to area of the triangle Arthanaareeswara method (unpublished) 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) 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) 7. Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan Kaasi Visweswar method) 8. Aberystwyth University Method for derivation of the exact π value New Method of Computing value (Siva Method) Jesus Method to Compute the Circumference of A Circle and Exact Value 11. Supporting Evidences To the Exact Pi Value from the Works Of Hippocrates Of Chios, Alfred S. Posamentier And Ingmar Lehmann 7 9

13 1. New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ In A Square 13. Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square 1. To Judge the Correct-Ness of the New Pi Value of Circle By Deriving The Exact Diagonal Length Of The Inscribed Square 15. The Natural Selection Mode To Choose The Real Pi Value Based On The Resurrection Of The Decimal Part Over And Above 3 Of Pi (St. John s Medical College Method) 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) 17. Hippocratean Squaring Of Lunes, Semicircle and Circle Durga Method of Squaring A Circle The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the actual length of the circumference of the circle and Pi Home page of the Author 73

14 Gayatri Method ABCD Square AB Side a 1 JG diameter a 1 OF OG radius a 0.5 FG hypotenuse a DE EF GH CH EH a a FG a a 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 + a a 1a a d a 1a a 1 This is the second method which came to life after this author s struggle for 6 years i.e. in March This author saw 1 method. He just presumed this number might be the failures he confirmed this 1 as the true for the first time by this value. After many value with Siva method (Page No. 5) where the total area of the square and the total area of the inscribed circle in square were calculated for the first time.

15 times of area of the circle is equal to area of the triangle (Arthanaareeswara method) D 1 C Square ABCD Side AB Diagonal AC A 1 B Take a paper and construct a square whose side is 1 (10 cm) and diagonal. 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 the folded paper finally looks like a paper crown. Let us find out the area of the circle. Now Circumference d; d ; Area d 1 Area of the triangle 1 ; x Area of circle 1 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 ; d ; x Area of circle 1 1 The interrelationship between two areas of circle and triangle by, shows that is not a special number called transcendental number.

16 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: x. Volume 10, Issue Ver. I (Jul-Aug. 01), PP -8 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 is derived from the Exhaustion method of Archimedes (0 BC) of Syracuse, Greece. It is the only one geometrical method available even now. The second method to compute 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 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 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, a and a 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. 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. Diagonals AC BD a 3. O Centre, EF diameter side a. 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/ 6. Triangle GOF. GF hypotenuse OG a a GF * This author studied B.Sc., (Zoology as Major) and M.Sc., (Zoology) during the years in the Sri Venkateswara University College, Tirupati, Chittoor district, Andhra Pradesh, India. And hence this author as a mark of his gratitude to the Alma Mater, this method is named after University s Honour. Page 3

17 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 a a 1 9. JC JB CB CJ Side hypotenuse a JC a a a, CB side a a 10. Bisect JB twice of CB side of Fig- JB JL + LB JK + KL + LM + MB a a 8 a 16 IJ GF 11. Similarly, bisect IA twice, of AD side of Fig- IA IP + PA IQ + QP + PN + NA 1. 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. 1. Area of DIJC rectangle DI x IJ a a a 15. The lower four rectangles are of same area. For example one rectangle IQKJ IQ x QK 16. Area of rectangles a a 16 a 16 5 Page

18 Pi treatment for the constituent rectangles of the superscribed square in the study of exact area of.. IQKJ + QPLK + PNML + NABM 17. Area of the square ABCD DIJC + bottomed rectangles a a a a 16 a 16 a Part-II 18. Let us repeat that Area of the ABCD square a d a Area of the inscribed circle ; where diameter side a 19. When side diameter a 1 Area of the ABCD square a 1 x 1 1 d a 11 Area of the inscribed circle 0. Corner area in the square (of Figs 1,, and 3) Square area circle area 1 1. It is true that any bottomed rectangles, is equal to the corner area of the square of Figs 1, and 3. Thus, bottomed rectangle corner area a Let us prove it i.e. S. No. 1 Part-III 3. The inscribed circle is equal to the sum of the areas of upper most rectangle DIJC S.No. 1 and next lower 3 rectangles IQJK, QPLK and PNML, and each is equal to 1 a a 3 a 16 a 16 a. Area of the inscribed circle Area of the corner region (S.No. 0) Area of the inscribed circle + corner area square area + 5. The sum of the areas of bottomed rectangles Square area Uppermost rectangle DIJC where a 1 1 a 16 a of S.No. 15 of 6 Page 5

19 Pi treatment for the constituent rectangles of the superscribed square in the study of exact area of.. a a a and S. No. 1 this is equal to S.No As the area of the corner region is equal to any one of the bottomed rectangles, then it is a (S.No. 0 & 1) 7. Then the sum of the areas of bottomed rectangles a a 8. Finally, Area of the uppermost rectangle DIJC Square area bottomed rectangles 9. CJ length a a 3 a 3a Side AB IJ a 30. Area of the upper most rectangle DIJC 3 a a 3a CJ x IJ 31. Thus, the areas of five rectangles which are interpreted in terms of above, are Uppermost rectangle DIJC 3a bottomed rectangles a Area of the ABCD square Uppermost rectangle + bottomed rectangles 3 a a a Area of the inscribed circle Uppermost rectangle DIJC + 3 bottomed rectangles 3 a 3 a a This is the end of the process of proof. 3. As the corner area is equal to 1. Arithmetically a 16 and. in terms of then S.No. 0 S.No. 1 where a 1 III. Conclusion It is well known, that a is the formula to find out area of a square or a rectangle. In this paper besides a, formulae, in terms of, of the inscribed circle in a square, are obtained, and equated to the classical arithmetical values of a. One has to admire the Nature, that, a circle s area can also be represented exactly equal, by the areas of rectangles, thus, the arithmetical values of these rectangles, are equated to that of a circle, 1 which thus give rise to new value This author stands and bow down and 7 Page 6

20 Pi treatment for the constituent rectangles of the superscribed square in the study of exact area of.. dedicates this work to the Nature. The Nature is the visible speck of the infinite Cosmos. The Creator exists in the invisible Energy form of this infinite Cosmos. We call this Creator as GOD and this author offers himself, surrenders himself totally and prays to THE LORD of the Cosmos of His/ Hers/ It s infinite goodness, as an infinitesimally, a small living moving body, as a mark of humble gratitude to THE LORD. References [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN []. Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Page. 5 prometheus Books, New York [3]. RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. []. RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [5]. RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [6]. RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [7]. RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [8]. RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [9]. RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p- ISSN: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [10]. RD Sarva Jagannada Reddy (01), 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: , p- ISSN: , Volume 3, Issue 11 (June 01) PP: [11]. RD Sarva Jagannadha Reddy (01), Pi of the Circle, at 8 Page 7

21 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: X. Volume 10, Issue Ver. III (Jul-Aug. 01), PP 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, 1 in this unification process, the real value is identified, and is, 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 , while the reciprocal b b 5 1 a Notice the relationship between the decimals. It suggests that 1 1. (A.S. Posamentier and I. Lehmann, 00, : A Biography of the World s Most Mysterious Number, Page 16). Archimedes (0 BC) of Syracuse, Greece, called the area arbelos that is inside the larger semi circle, but outside the two smaller semi circles of different diameters. By its shape it is also called as a shoemaker s knife. The Golden Ratio and the arbelos of Archimedes are different concepts. But in this paper by the grace of God, it has become possible to see that these two concepts too have an interesting and unexpected inter relationship between each other (one). Further, this relationship has an extended relationship also with the circle (two). It is a well known fact that there exists simple and understable relationship between circle and square (three). As circle, square are related, their combined interrelationship has been extended to value also (four). There is, thus, a divine chain of bond (of four interconnecting relations) exists, among square, circle, Golden Ratio, arbelos and value. (Here value means a true/ real/ exact/ line-segment based value. The stress here, on the adjectives to, has become necessary, because of Polygon is attributed or thrust on circle. In other words, this number to circle is a borrowed number from polygon and its 1 existence thus can not be seen in the radius of the circle, naturally. However, the new value (unlike with official value ) is inseparable with radius and is, here, humbly submitted to the World of Mathematics: Area of the circle 7r r r r r r Circumference of the circle 6r r 1 In support of the above formulae, this paper also chooses and confirms that the real value is Page 8

22 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, equal to half of the side AB of the square, whose length is 1. AB Side EN diameter 1 AC BD Diagonal. E is the mid point of AD AE 1, AB 1 Triangle EAB, EB hypotenuse EH 1 ; HB EB EH Golden Ratio HB EN Diameter 1 EJ HB JN EN EJ 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 51 and 3 5 the diameter of the JN semicircle 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) Page 9

23 A study that shows the existence of a simple relationship among square, circle, Golden Ratio.. 8. Draw a full circle with diameter KJ. It has already been established that this area of the full circle is equal to the area of the shaded region called arbelos. 9. To calculate the area of the arbelos we have the following formulas. q d q h and where h perpendicular line KJ 5 diameter of the circle LKM h radius of LKM circle. 10. Now, let us see the first formula q d q 3 5 q JN d EN diameter The conventional formula is r. KJ diameter h 5 Radius diameter h x x x Part-II 1. value is known and hence, it is possible to find out the area of the arbelos either from h As there are two values now and to find a way to decide which number actually represents value. 1. The following formula helps in deciding the real value. The formula is Side a diameter d 1 Diagonal a Perimeter of thesquare 1 Half of 7 times of side of square th of diagonal a 16 7a a q d q or, the time has come, 35 Page 10

24 A study that shows the existence of a simple relationship among square, circle, Golden Ratio.. (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 (S. No. 10) x x (S. No. 11) 16. With Gayatri value (S.No. 10) x (S.No. 11) 17. Finally, we obtain two different values representing same area of the arbelos of Archimedes. Official value gives: and Gayatri value gives: 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 Let us divide area of the arbelos of S.No. 17 with the Cube of Golden Ratio and 16 multiply it with of the formula, derived in the S.No.1, which finally gives the area of the square 1 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 Area of the arbelos obtained with official value (S.No. 17) Let us use simple calculator for the value of cube of Golden Ratio, which gives Area of the arbelos obtained with Gayatri value (S.No. 17). Let us repeat steps of S.No. 0 here: Page 11

25 A study that shows the existence of a simple relationship among square, circle, Golden Ratio.. As the exact area of the superscirbed square is obtained, it is clear, therefore, that, the real value is 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 References [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN []. Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Prometheus Books, New York [3]. RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. []. RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [5]. RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [6]. RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [7]. RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [8]. RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [9]. RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p- ISSN: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [10]. RD Sarva Jagannada Reddy (01), 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: , p- ISSN: , Volume 3, Issue 11 (June 01) PP: [11]. RD Sarva Jagannadha Reddy (01), Pi of the Circle, at [1]. R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: X. Volume 10, Issue Ver. I (Jul-Aug. 01), PP Page 1

26 IOSR Journal of Engineering (IOSRJEN) ISSN (e): , ISSN (p): Vol. 0, Issue 07 (July. 01), V3 PP 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 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 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 a and a, 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 r and r, 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 In March 1998, it was discovered the exact value from Gayatri method. This new value is In 188, C.L.F. Lindemann and subsequently, Vow. K. Weirstrass and David Hilbert (1893) said that 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 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 0 miles long. S. Ramanujan International organization of Scientific Research 63 P a g e 13

27 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan With the discovery of squaring of circle has become very easy and is done here. Archiemedes (0 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. Semicircle on EF EF diameter d side a 1 Semicircle on EG EG diameter a 5 5 Semicircle on GF EF EG GF diameter a Arbelos is the shaded region. International organization of Scientific Research 6 P a g e 1

28 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan 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 Draw a circle with diameter GH 5 d d Area of the G.H. circle Area of the G.H. circle Area of the arbelos 1 1 So, area of the arbelos Part II: Squaring of circle present in the ABCD square 6. Diameter EF d a 1 d Area of the circle To square the circle we have to obtain a length equal to. It has been well established by many methods more than one hundred different geometrical constructions that value is 1. Let us find out a length equal to. 8. Triangle KOL OK OL radius d a 1 KL hypotenuse DJ JK LM MC Side a a 1 So, DJ d a hypotenuse a 9. JA DA DJ a a Bisect JA twice JA JN + NA NP + PA International organization of Scientific Research 8 So, PA DP DA side AP a So, JA 65 P a g e 15

29 11. Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan 1 DP (As per S.No. 7) Draw a semicircle on AD diameter 1 AP, DP Draw a perpendicular line on AD at P, which meets semicircle at Q. Apply Altitude theorem to obtain PQ length PQ AP DP 1. Join QD Now we have a triangle QPD PQ , Apply Pythagorean theorem to obtain QD length 15. QD PQ PD 1 1, where, Side 1 PD is the length of the side of a square whose area is equal to the area of the inscribed circle 1 a Area of the square a 1 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 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. of S.No. Area of the arbelos Area of the circle with diameter GH 5 So, , where To square the arbelos, we have to obtain a length of the side of the square whose area is equal to area 1 of the arbelos EG diameter 5.. I is the mid of EG. International organization of Scientific Research 66 P a g e 16

30 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan EI + IG EG So EI Small square STBR Side RB EI 5 Inscribe a circle with diameter 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. 0. Triangle K ZL ZK ZL radius 1 5 K L hypotenuse RB 5 1. L U Side hypotenuse. So, L U 10 BU BT Side of the square 5 UT BT BU So, UT Bisect UT twice UT UV + VT VX + XT So, XT 0. BT 5 ; XT 0 BX BT XT 5 0 BX Draw a semi circle on BT with 5 as its diameter. 6. Draw a perpendicular line on BT at X which meets semicircle at Y. International organization of Scientific Research 67 P a g e 17

31 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan XY length can be obtained by applying Altitude theorem 1 XY BX XT XY 7. Triangle BXY BX 1 0, XY BY can be obtained by applying Pythagorean Theorem BY BX XY BY is the required side of the square whose area is equal to the area of the arbelos of Archimedes. Side 1 10 a 1 Area of the square on BY a 1 of S.No 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. 8. New value 1 value gives the area of the arbelos as gives area of the arbelos as d x d x d x Whereas the official d GH 5 of S.No Thus, the following are the two different values for the same area of the arbelos. Official value gives New value gives Diameter of the arbelos circle GH d 5 Square of the diameter d Reciprocal of the square of the diameter d Area of arbelos, if multiplied with 5 Area of the circle d we get the area of the inscribed circle in the ABCD square International organization of Scientific Research 68 P a g e 18

32 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan d a 1, Area of the arbelos reciprocal of the square of the arbelos circle s diameter Area of the inscribed circle in ABCD square S. No. 16 S.No.9 S.No Let us derive the following formula from the dimensions of square ABCD ABCD square, AB side a 1 AC BD diagonal a, Perimeter of of ABCD square a Perimeter of ABCDsquare 1 Half of 7 times of ABside of square th of diagonal a 16 7a a In this step, above steps (S.No. 9 and 3) are brought in Arbelos area x Area of the ABCD square, equal to 1. 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 and 1 Arbelos area of new value This process is done by understanding the actual and exact interrelationship among, 1. area of the ABCD square,. area of the inscribed circle in ABCD square and, 3. area of the arbelos of Archimedes. 3. 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 1 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 as value. The new value 1 has done it. The arbelos of Archimedes has also chosen the real value in association with the inscribed circle and the ABCD superscribed square. International organization of Scientific Research 69 P a g e 19

33 Squaring of circle and arbelos and the judgment of arbelos in choosing the real Pi value (Bhagavan REFERENCES [1] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN [] Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Prometheus Books, New York [3] David Blatner, The Joy of Pi (Walker/Bloomsbury, 1997). [] RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. [5] RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [6] RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [7] RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [8] RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p- ISSN: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [9] RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [10] RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [11] RD Sarva Jagannada Reddy (01), 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: , p-issn: , Volume 3, Issue 11 (June 01) PP: [1] R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: X. Volume 10, Issue Ver. I (Jul-Aug. 01), PP -8. [13] RD Sarva Jagannada Reddy (01), 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: P-ISSN: , Volume Issue 7, July. 01, PP [1] RD Sarva Jagannadha Reddy (01) 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: , p-issn: Volume, Issue 1 (July 01) PP: 3-37 [15] R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: X. Volume 10, Issue Ver. III (Jul-Aug. 01), PP [16] RD Sarva Jagannadha Reddy (01) Aberystwyth University Method for derivation of the exact value. International Journal of Latest Trends in Engineering and Technology (IJLTET) Vol. Issue July 01, ISSN: 78-61X, PP: [17] R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: X. Volume 10, Issue Ver. III (Jul-Aug. 01), PP [18] RD Sarva Jagannadha Reddy (01), Pi of the Circle, at International organization of Scientific Research 70 P a g e 0

34 International Journal of Latest Trends in Engineering and Technology (IJLTET) Aberystwyth University Method for derivation of the exact π value R.D. Sarva Jagannadha Reddy Abstract - Polygon s value of Exhaustion method has been in vogue as π of the circle for the last 000 years. An attempt is made in this paper to replace polygon s approximate value with the exact π value of circle with the help of Prof. C.R. Fletcher s geometrical construction. Keywords: Circle, diagonal, diameter, Fletcher, π, polygon, radius, side, square I. INTRODUCTION The official π value is 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. a, a, ½ab etc are the formulas of square and triangle which are derived based on their respective line-segments. Similarly, radius is a line-segment and a need is there to have a formula with radius alone and without π. The following formulas are discovered (March, 1998) from Gayatri method and Siva method. 7r r 1. Area of Circle r πr r r. Circumference of Circle 6r + π r ; where r radius and d 1 1 πd d d d where d diameter side of the superscribed square In the Fletcher s geometrical construction there are two line-segments. They are radius and corner 1 length. To find out the area of the shaded region in which corner length is present Professor has given 1 π. Fig-1: Professor s Diagram (by courtesy) II. CONSTRUCTION PROCEDURE OF SIVA METHOD Vol. Issue July ISSN: 78-61X 1

35 International Journal of Latest Trends in Engineering and Technology (IJLTET) Fig-: Siva Method Draw a square ABCD. Draw two diagonals. O is the centre. Inscribe a circle with centre O and radius ½. Side of the square is 1. E, F, G and H are the midpoints of four sides. Join EG, FH, EF, FG, GH and HE. Draw four arcs with centers A, B, C and D and with radius ½. Now the circle-square composite system is divided into 3 segments of two different dimensions, called S 1 segments and S segments. Number them from 1 to 3. There are 16 S 1 and 16 S segments in the square and 16S 1 and 8S segments in the circle. Square: ABCD, AB Side 1, AC Diagonal ; Circle : EFGH, Diagonal diameter AC JK 1 JK Diameter 1 Side; Corner length ; OL OK radius 1 1 ; LK OK OL 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 S segments 19 and 0 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-. AC and BD, diagonals ( ) ; Vol. Issue July ISSN: 78-61X

36 International Journal of Latest Trends in Engineering and Technology (IJLTET) diagonal diameter AC JK 1 KC corner length Area of the square (a 1 x 1 1) and 3 constituent segments of the square. Their relation are represented here in a formula and is equated to a 1 Professor s formula 3 1 π (of Fig.1, where radius 1) 1 1 where, 1 π has been derived with radius equal to 1, and naturally, the diameter side of the square. With this, the above formula becomes π 1 1 π The accepted value for π is With this π, the area of shaded region is equal to And, with the new π value derived above, the area of the shaded region is equal to So, this method creates a dispute now ? Which π value is right i.e. is or The study of this method is extended further to decide which π value is the real π value? To decide which π is real, a simple verification test is followed here. What is that? We have a line segment LK LK is part of the diagonal along with the corner length KC. So, in the Second step, an attempt is made to obtain the LK length, from the area of the shaded region. How? Let us take the reciprocal of the area of the shaded region; 1 1 Area of theshaded region of official π and with new π Vol. Issue July ISSN: 78-61X 3

37 International Journal of Latest Trends in Engineering and Technology (IJLTET) Area of theshaded region Then, this value when divided by 3, we surprisingly get KL length. It may be questioned why one should divided that value. The answer is not simple. Certain aspects have to be believed, without raising questions like what, why and how at times. Official π New π of new π value is in total agreement with LK of Fig.. i.e and differs however with of official π from 3 rd decimal onwards. If this argument is accepted, the present π value is not approximate value from its last decimal place, but it is an approximate value from the 3 rd decimal. IV. CONCLUSION From the beginning to the end of this method, various line-segments are involved. Professor Fletcher s construction is analyzed arithmetically with the line-segments of the Siva method. This arithmetical interpretation has resulted in the derivation of a new π value, equal to 1. The new value is exact, algebraic number. REFERENCES [1] C.R. Fletcher (1971) The Mathematical Gazettee, December, Page, London, UK. [] R.D. Sarva Jagannadha Reddy (01), Pi of the Circle, at Vol. Issue July ISSN: 78-61X

38 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9 New Method of Computing value (Siva Method) RD Sarva Jagannada Reddy I. Introduction equal to is an approximation. It has ruled the world for 0 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/. Diameter is a. Draw 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/ and with centres A, B, C and D. Now the circle square composite system is divided into 3 segments and number them 1 to 3. 1 to 16 are of one dimension called S 1 segments and 17 to 3 are of different dimension called S segments. III. Calculations: ABCD Square; Side a, EFGH Circle, diameter a, radius a/ 6 Area of the S 1 segment a ; Area of the S 18 segment a ; 18 Area of the square 16 S S Area of the inscribed circle 16S 1 + 8S General formula for the area of the circle 6 16 a 16 a a a 8 a a d a 1 a 16 1 ; where a d side diameter IV. How two formulae for S 1 and S segments are derived? 16 S S a area of the Square Eq. (1) 8 Page 5

39 a 16 S S area of the Circle Eq. ().. a a a a a (1) () 8S a 3 3 a ()x 3 S S Eq. (3) 16 S S a Eq. (1) a a a (3) (1) 16S 1 a S New Method of Computing value (Siva Method) S V. Both the values appear correct when involved in the two formulae a) Official value b) Proposed value 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 ( - ) in S 1 segment s formula and second line-segment equal to the value of ( - ) in S segment s formula are searched in the above construction. a) Official : Proposed : The following calculation gives a line-segment for 6 and no line-segment for IM and LR two parallel lines to DC and CB; OK OJ Radius a ; JOK triangle a JK Hypotenuse Third square LKMC; KM CM Side? KM IM JK a 1 a a; Side of first square DC a DC + CM 6 a a a b) Official Proposed - No line-segment for in this diagram. MB line-segment is equal to. How? Side of the first square CB a MB CB CM a a a VII. Conclusion: This diagram not only gives two formulae for the areas of S 1 & S segments andalso shows two linesegments for ( - ) and ( - ). Line-segment is the soul of Geometry. 9 Page 6

40 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP 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 and this value has been made official. From 15 th century (Madhava (150) of South India) onwards infinite series has been used for more decimals to compute 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 (178), S. Ramanujan (191), 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: i i0 16 8i 1 8i 8i 5 8i 6 Using above formula Yasumasa Kanada of Tokyo University, Japan, calculated trillions of decimals to 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 violates the definition of. This is about the value of. Next, about the nature of. C.L.F. Lindemann (188) has said is a transcendental number based i on Euler s formula e 1 0. In Mathematical Cranks, Underwood Dudley has said s only position in mathematics is its relation to infinite services (and) that has no relation to the circle. Lindemann proclaimed the squaring of the circle impossible, but Lindemann s proof is misleading for he uses numbers (which are approximate in themselves) in his proof. Hence, pre-infinite series days of geometrical method is approached again to find out exact value and squaring of circle. This author has struggled for 6 years (197 to 1998) and calculated the exact value of in March, 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 a, Diameter is also a d. 1) Straighten the square. Perimeter a Perimeter Sum of the lengths of two diagonals a a esp end segment of the perimeter of the square. esp 58 Page 7

41 Jesus Method To Compute The Circumference Of A Circle And Exact Value ) 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/. So, the radius is 1/8 th of the perimeter of the square. ) 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. a Thus as radius of the inscribed circle is to the perimeter of the square (a), 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 esp 3 a esc x a a 8 8 5) Circumference of the circle d a (where a d diameter) 1a a a II. end segment of the perimeter of the square 8 1a a x 1 Conclusion value, derived from the Jesus proof is algebraic, being a root of differs from the usually accepted value in the third decimal place, being x 56x 97 0 but also that it 59 Page 8

42 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 Supporting Evidences To the Exact Value from the Works Of Hippocrates Of Chios, Alfred S. Posamentier And Ingmar Lehmann R.D. Sarva Jagannadha Reddy /D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati , A.P., India Abstract: Till very recently we believed was the final value of.and no body thought exact value would be seen in future. One drawback with is, that it is not derived from any line-segment of the circle. In fact, 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 that exact value differs from from its 3 rd decimal onwards, being Two geometrical constructions of Hippocrates of Chios, Greece (50 B.C.) and Prof. Alfred S. Posamentier of New York, USA, and Prof. Ingmar Lehmann of Berlin, Germany, are the supporting evidences of the new value. They are detailed below. Keywords: value, lune, triangle, area of curved regions I. Introduction In the days of Hippocrates, value 3 of the Holy Bible was followed in mathematical calculations. He did not evince interest in knowing the correct value of. He wrote a book on Geometry. This was the first book on Geometry. This book became later, a guiding subject for Euclid s Elements. He is very famous for his squaring of lunes. Prof. Alfred S. Posamentier and Prof. Ingmar Lehmann wrote a very fine collaborative book on. They have chosen two regions and have proved both the regions, though appear very different in their shapes, still both of them are same in their areas. These areas are represented by a formula r 1. The symbol r is radius., here must be, the universally accepted 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 r does not agree with the value of line-segments of two regions. However, the new value 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 r 1 with the linesegments, 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 9 Page 9

43 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/; F mid point of AC N mid point of arc AC d d d d d d NF ; DM ; MC 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. d 1 1. Area of ANC Area of CPB 1 3. Area of AMB Areas of ANC + CPB (Hippocrates) 3. Area of ACM Area of BCM 8 d Area of ACB triangle 1 d d 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) d d 1 1 d i.e d 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 d 7. Area of the semicircle 8 Areas of ANC + CPB + ACM + BCM + AMB 8 Area of thesemicircle 8. d d d 1 16 d d Alfred S. Posamentier s similarity of the two areas and decimal similarity between an area and its line-segments Prof. A.S. Posamentier has established that areas of A and B regions are 10 Page 30

44 Supporting Evidences To The Exact Value From The Works Of Hippocrates Of Chios, Alfred S. equal. His formula is r 1 for the above regions. This author is grateful to the professor of New York for 1 the reason through his idea this author tries to show that his new value equal to 1 is exactly right. 1. Arc BCA; O Centre; OB OA OC Radius r. r Semicircles : BFO AFO; E and D Centres; ODDA BE OE radius OF r r ; FC OC OF r r r r r r r r 3. Petal OKFH; EK ; ED ; EJ ; JK EK EJ r r ; JK JH, HK JH + JK r r. So, FC of region A HK of region B 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 ) r Formula for A and B is r 1 Here r radius 1 From March 1998, there are two values. The official value is and the new value is and which value is exact and true? Let us substitute both the values in r, then r r Official value (It is universally accepted that is approximate at its last decimalplace however astronomical it is in its magnitude.) r New value r 6. FC HK (HJ + JK) line segments r r 7. FC HK r r 1 Half of HC and HK are same r r Area of A/B region equal to is similar in decimal value of half of FC/HK line segment i.e Formulae a, a of square and ½ab of triangle are based on side of the square and altitude, base of triangle, respectively. In this construction, FC and HK are the line segments of A and B regions, respectively. 11 Page 31

45 Supporting Evidences To The Exact Value From The Works Of Hippocrates Of Chios, Alfred S. As the value which is half of FC or HK is in agreement with the area value of A/B region 1 equal to in decimal part, it is argued that new value equal to is exactly correct. The decimals of the official value does not tally beyond 3 rd decimal with the half the lengths of HK and FC, whose value is , 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 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 invariably appears. Hence, 1 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. []. P. Dedron and J. Itard (1973). Mathematics and Mathematicians, Vol., translated from French by J.V. Field, The Open University Press, England. [3]. Alfred S. Posamentier&Ingmar Lehmann (00). A Biography of the World s Most Mysterious Number. Prometheus Books, New York, Pages 178 to 181. []. RD Sarva Jagannadha Reddy (01), Pi of the Circle, a Canto on-line edition, in the free website: 1 Page 3

46 International Journal of Mathematics and Statistics Invention (IJMSI) E-ISSN: P-ISSN: Volume Issue 5 May. 01 PP New Value: Its Derivation and Demarcation of an Area of Circle Equal to In A Square R.D. Sarva Jagannadha Reddy ABSTRACT: value is 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 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 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 (0 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 1 length of the circumference of a circle over and above its 3 diameters as equal to equal to 1. 1/7 of Archimedes of Gayatri method and its tail-end of the 1 when the diameter is 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 1 Archimedes concept of less than 1/7 by giving. The denominator part of the fraction, is actually, less than 7 of 1/7. He is a great mathematician. This fraction 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 0 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/. Draw four equidistant tangents on the circumference. They intersect at four prints called A, B, C 33 P a g e 33

47 New Value: Its Derivation and Demarcation of and D, creating a square ABCD. The diameter of the circle EF is also equal to the AB side of the square. Draw two diagonals AC and BD. Their values are a. Perimeter of the square is equal to a. In this circle-square composite system there are now 3 types of straight lines and one circumference which is a curvature. The straight lines are 1. Perimeter of the square,. Two diagonals and 3. Diameter of the circle. The values of these straight lines are known and exact. The length of the circumference is unknown and hence this method is to find out its exact length with the help of known lengths of three types of straight lines. Let us repeat here again the perimeter (a) of the square and the sum of the lengths of two diagonals a are the outcome of the tangents on the circumference. It is clear therefore that the curved circumference reflects its true length in all the straight lines of square and circle. We know very well that there are three diameters and some length called tail-end in the circumference. Circumference 3 diameters + tail end length Tail-end length is unknown and hence it is called x. 3d + x circumference Diameter is equal to side of the square i.e. d a Let us rewrite 3d + x as 3a + x To find out the length of the x We take the help of all the straight lines. The reciprocal of the two diagonals plus the perimeter of the square and when this product is multiplied by the square of the diameter ( side), will give x value. 1 1 a a x a a 1 1 Circumference 3 diameters (3a) + tail-end length (x) 3a a a Circumference of the circle d a So, 1 a a P a g e 3

48 New Value: Its Derivation and Demarcation of The length of the circumference is obtained by superscribing a square. The correct understanding of the relationship among perimeter, diagonals of the square and the diameter of the circle ( side of the square) results in knowing the exact length of the tail-end of the circumference (x), what the great mathematician 1 1 Archimedes has said is less than 1/7 is proved now to be The denominator of Gayatri method is less than 7 of 1/7 of 3 1/7 of Archimedes. METHOD-: Computation of Segmental Areas (An Elementary Approach) 1. BD AB OB ; CD. Area of ABC Area of a + b ;. Area of OAC 1 8 ; OAB 1 1 ; Sector OAD Area of segment a OAD OAB Area of segment b OAC OAD a is larger than b. So, a is greater than half of ABC b is smaller than a. So, b is lesser than half of ABC Half of ABC 3 y 8. Let a y s, y b t; s t 9. Let us assume BDAC s ; CDAB t P a g e 35

49 10. a y s; a s + y y b t; b y t a + b ABC 13. Inscribed circle consists of 16a and 8b 6 d so, Where d 1 New Value: Its Derivation and Demarcation of METHOD-3: DEMARCATION OF AREA OF CIRCLE WITH ITS STRAIGHT-LINED BOUNDARY From Gayatri method, the world came to know in March 1998, for the first time, the length of the circumference of the inscribed circle, demarcated in the perimeter of the square. The demarcated length of the 1 circumference of the circle is BA + AD + DC + CH a + a + a + a a The area of the circle is now bounded by a curvature called the circumference. In this method this area can have a straight-lined boundary. In this process the value of plays important role. The official value is With this value let us locate the area: 1. Square ABCD, Side AB a; AC BD diagonals a ; O Centre. Inscribed a circle with centre O and radius a ; side diameter a 3. OF OG Radius; FOG triangle; OF a ; FG hypotenuse a ;. EH Side Parallel to CD a EH FG a 1 5. DE EF GH CH a a 1 6. BA + AD + DC + CH a + a + a + a a circumference of the inscribed circle (from Gayatri method). 7. Let us try to demarcate the extent of area of the inscribed circle with the help of official value a ; Let us suppose a 1 9. S.No. 7 S.No The area equal to cannot be located here. It has become impossible. 11. As an alternate, let us try to locate, the area with the guidance of the Circumference of the Gayatri 1 method i.e. a, and with this, the Gayatri value is P a g e 36

50 1. Let us repeat S.No. 7, 8, 9 with the Gayatri value ; 1. 3a 3 ; where a 1; S.No. 13 S.No The area equal to have to be located now. Let us see how 16 AB a 1; K mid point of AB New Value: Its Derivation and Demarcation of 17. AB AK + KB a a Bisect KB into KL and LB a AL AK + KL a a 3a DM AL 3a 18. So DM 3a ; Join ML LB MC a NH MC 19. There are two rectangles ALMD and MNHC 3a 3 0. Area of ALMD rectangle AD x AL a a a 1. Area of MNHC rectangle MN x NH a a P a g e 37

51 New Value: Its Derivation and Demarcation of. In serial No. 10 we have said, getting an area equal to is impossible: With the Gayatri value 1, getting an area equal to of S.No. 15 is thus possible Area of the inscribed circle 3 1 Areas of rectangles ALMD+MNHC a a a (S.No. 0) (S.No. 1). Circle and square both can be inscribed and/ or circumscribed with each other. It means, both must be finite entities, having finite magnitudes, and to be represented by finite numbers. 5. Official value cannot demarcate circle s area. Whereas, Gayatri value demarcates. So is or the real value? 6. It is another way of squaring a circle. III. CONCLUSION New value is exact and it is an algebraic number and squaring of circle has be done by demarcating, the area of a circle in a square. REFERENCES [1]. Lennart Berggren, Peter Borwein, Jonathan Borwein, Pi: A Source Book (Springer, 1996). []. David Blatner, The Joy of Pi (Walker/Bloomsbury, 1997). [3]. Alfred S. Posamantier, Pi: A mysterious number (Prometheus Books, 00). []. RD Sarva Jagannadha Reddy, Pi of the Circle ( 01) P a g e 38

52 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP 17-0 Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square R. D. Sarva Jagannadha Reddy /D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati , A.P., India Abstract: It is universally accepted that as the value of. It is thought an approximation, at its last decimal place. It is a transcendental number and squaring a circle is an unsolved problem with this number. A new, exact, algebraic number is derived and verified with a proof that is followed for Pythagorean theorem. It is proved here squaring of circle and rectification of circumference of a circle are possible too. Keywords: Pythagorean thorem, circle, square,. I. Introduction Official value is It is obtained from the Exhaustion method, which is a geometrical method. This method involves the inscription of a polygon in a circle and increased the sides of the polygon, until the inscribed polygon touches the circle, leaving no gap between them. The value is actually the length of the perimeter of the inscribed polygon. And it is not the value of circle. There was no method till yesterday to measure the circumference of a circle, directly or indirectly has four characteristics: 1) It represents polygon, ) It is an approximation, 3) It is a transcendental number and ) It says squaring a circle is impossible. And such a number is attributed and followed as of the circle based on limitation principle, because of the impossibility of calculating, the length of the circumference of circle and in such a situation this field of mathematics has been thriving for the last 000 years. From 150 Madhavan of South India and a galaxy of later generations of mathematicians have discarded geometrical construction and have introduced newly, the concept of infinite series. In this paper geometrical constructions are approached again, for the derivation of value. New value has been derived. It is It is an exact value, an algebraic number and makes squaring of circle possible and done too. II. Procedure Siva method for the area of the circle of 1 st square ABCD Construction procedure Draw a square ABCD. Draw two diagonals. O is the centre. Inscribe a circle with centre O and radius ½. E, F, H and J are the mid points of four sides. Join EH, FJ, FH, HJ, JE and EF. Draw four arcs taking A, B, C and D as centres and radius ½. Now the circle square nexus is divided into 3 segments. Number them 1 to 3. 1 to 16 segments are called S 1 segments. 17 to 3 segments are called S segments. 17 to, S segments are outside the circle. 5 to 3, S segments are inside the circle. Draw KP, a parallel line to the side DC which intersects diagonals at M and N. Square ABCD Side AB 1 EH diameter Areas of S 1 and S segments S 1 6 ; S S S Area of square Page 39

53 Pythagorean way of Proof for the segmental areas of one square with that of rectangles of.. 16S 1 + 8S Area of circle Square has 3 constituent parts Fig-1: Segmental areas calculated; Fig-: Areas of Rectangles are calculated Both values are same This method is taken from the book Pi of the Circle of this author (available at In this method there are two squares of same sides. First square has an inscribed circle divided into 3 segments of two dimensions called S 1 and S segments, each category of 16 in number. And areas of these segments are calculated using the following two formulas a a S S 1 and 3 3 which are obtained by solving two equations (in Square 1) 16 S S a Area of the square (Eq.1) 16 S S a / Area of the inscribed circle (Eq.) This method is called as Siva method. In the present method: Siva Kesava method, second square is joined to the 1 st square. One side CB is common to both the squares. The second square is similarly divided, as in the case of 1 st square, into 3 rectangles. Rectangles are also of two dimensions each category of 16 numbers. The areas of each type of rectangle is equal to S 1 and S segments of the 1 st square. These rectangles are formed, based on the division of common side of the both the squares. The areas of rectangles agree cent percent with the above two formulas of Siva method, where value is 1. Thus, the division of 1 st square is exactly duplicated in the second square, except for the difference, in the 1 st square, 3 segments are curvy linear, and in the nd square, 3 segments are rectangles, naturally, of straight lines. Now let us see how the common side CB is divided. 1. Squares 1 ABCD, BZTC. Side diameter of the inscribed circle 1 3. KP Parallel side to the side DC. OM ON radius ½ 5. MON triangle; MN hypotenuse KP MN 6. DK KM NP PC So, CP, PB CB CP 1 18 Page 0

54 Pythagorean way of Proof for the segmental areas of one square with that of rectangles of.. 8. Bisect PB. PB PQ + QB QR + RB QB 8 6, CB 1, CQ CB QB CQ Bisect CQ CS + SQ We have started with side 1, divided next, into CP PB is bisected into PQ 8 and QB 8 6 and SQ After divisions, finally we have CB Side divided into parts, and PB. In the third step CQ 6 8. In the second step is bisected into CS CS 6, SQ 6, QR and RB nd Square BZTC is divided horizontally into four parts: CS, SQ, QR and RB. 1. Now BZ side of nd square is divided into 8 parts. So, each length is 1/ Finally, the nd square is divided into 16 rectangles of one dimension equal in area to S 1 segments of 1 st square and 16 rectangles of nd dimension, equal in area to S segments of 1 st square. 16. Square BZTC consists of first two rows are of S 1 and 3 rd & th rows are of S segments. 17. Area of each rectangle S 1 segments of 1 st square Sides of rectangles of 1 st & nd rowstwwx 6 16 and other side Area of each rectangle S segment of 1 st square Sides of rectangles of 3 rd & th rowsxyyz and other side The areas of 16S 1 and 16S segments of 1 st square a a Area of the 1 st square; where a The area of all the 3 rectangles Area of the nd square 1. The area of the inscribed circle in the 1 st square 16S 1 segments + 8S segments Area of the circle in the 1 st square 16. The areas of the 1 st, nd and 3 rd rows of rectangles of nd square d 3. Thus area of the circle from 1 st and nd squares is 16 1 where side diameter d 1 19 Page 1

55 Pythagorean way of Proof for the segmental areas of one square with that of rectangles of... When is equal to 1, the length of the inscribed circle in the 1 st square is d a where side diameter 1 5. Perimeter of the rectangle QXWTCS is equal to the circumference of the inscribed circle in the 1 st square. QX 1 TC; XW WT CS SQ QX + XW + WT + TC + CS + SQ In the first square we have seen that the length of the circumference of the inscribed circle is the outer edge of the 16 S 1 segments. In the nd square also the outer edges of the 1 st and nd rows of 16 rectangles are equal to Thus, Siva Kesava Method supports the value 1 obtained by earlier Gayatri, Siva, Jesus methods. 7. And also, the curvy linear 16S 1 and 16S segments of 1 st square are all squared in the nd square. III. Conclusion Two squares of same sides are drawn with one common side. Circle is inscribed in one square. Areas of square and its inscribed circle are calculated from their constituent curvy linear segments. The correctness of areas of constituent segments are verified with that of the areas of rectangles of the adjoining square. All the values thus are proved correct. 0 Page

56 International Journal of Mathematics and Statistics Invention (IJMSI) E-ISSN: P-ISSN: Volume Issue 7 July. 01 PP-01-0 To Judge the Correct-Ness of the New Pi Value of Circle By Deriving The Exact Diagonal Length Of The Inscribed Square R.D. Sarva Jaganndha Reddy ABSTRACT : Circle and square are soul and body of the subject of Geometry. All celestial bodies in the Cosmos are spherical in shape. Four equidistant tangents on a circle will give rise to a square. A circle can be inscribed in a square too. Thus, circle and square are two inseperable geometrical entities. is a fundamental mathematical constant. The world believes as value for the last 000 years. Yet it is an approximate value. Continuous search for its exact value is going on, even now. God has been kind. The exact 1 value is not a myth. It has become real with the discovery of. It is a very tough job to convince the world that the new finding is the real value. In this paper, the exact length of the diagonal of the inscribed square form the length of an arc of the superscribed circle is obtained as a proof. KEYWORDS: Circle, circumference, diameter, diagonal, perimeter, square I. INTRODUCTION The geometrical constant, called, is as old as human civilization. In the ancient days, contributions on was very admirable from the Eastern parts of the World. The Founding Father of Mathematics: Hippocrates of Chios (50 BC) has not touched the value of. But his work on the nature of circular entities such as squaring of lunes, semicircle and full circle is unparalled in the History of Mathematics.The present value, could not understood Hippocrates work and its greatness. What is the reason? This number is not the of the circle. It is the value of the polygon. It was derived using Pythagorean theorem. Pythagorean theorem gives exact length to a hypotenuse which is a straight line. Circumference of a circle is not a straight line. It is a curvature. Hence, of polygon has failed to understand the greatness of Hippocrates. In other words, is not a number at all. After a long waiting of 000 years by trillions of scholars NATURE has revealed its true length of a 1 circumference of circle and its value. The value is derived from Gayatri method. It was discovered in March This worker is the first and fortunate humble man to see this fundamental truth, and at the same time, made this author responsible to reveal to the whole world, its total personality. He was cautioned by the Nature, through Inner Voice, further, not to shirk his responsibility till the end, till the world welcomes it, and continue to search speck by speck naturally, respecting the Cosmic mind of the Nature, with indomitable determination, and fight single handedly against the conservative attitude and die like a warrior defeated, if situation of reluctance to acceptance still persistsfrom the remote past to the present, the period of study of is divided into two periods: the period of geometrical dependence in the derivation of value and the nd period from 1660 AD onwards till today, the period of dependence on infinite series, dissociating totally geometrical analysis in the derivation of value. In the present and second period, the number is considered as a number only and nothing to do with the definition of the ratio of circumference and diameter of a circle. The paradox is, geometry is forgotten in the derivation of value but searched for the same, in squaring of a circle. Finally, this number has gained four characteristics: 1. Though it is a polygon number, established itself as a number of circle.. Though an approximate number it ruled the world for many centuries as a final value. 3. Though it doesn t belong to a circle has commented squaring of circle as an unsolved geometrical problem.. Though it has gained a status of transcendental number with the definition of the fact that cannot be calculated by a combination of the operations of addition, subtraction, multiplication, division, and square root extraction (Ref.1) is derived actually in Exhaustion method, applying the Pythagorean theorem, and invariably with the involvement of 3. Further, the moment this number discared the association of Geometry, it has started a new relationship in 1 P a g e 3

57 To Judge The Correct-Ness Of The New Pi Value Euler s formula e i +1 0 with unrelated numbers which are themselves approximate numbers. C.L.F. Lindeman (188) has called number as a transcendental number based on Euler s formula. Here again, there is a discrepancy in choosing, in the Euler s formula. In the formula, means radians and not, constant 3.1 Are they both, radians and constant are identical? constant is divine, whereas radians equal to is, human creation and convenience. In the Euler s formula radians is involved and the resultant status of transcendence has been applied to constant. Let us rewright e i as e 1 0. Is it right and acceptable?it may not be wrong in calling that is a transcendental number but it is definitely wrong when number is called a transcendental number and with a consequential immediate statement squaring of a circle is impossible. The major objection is, the very definition of is the ratio of circumference and diameter of a circle. So, this discrepancy led to the conclusion that is not, infact, a number.in its support, we have the work of Hipporcrates. Hippocrates had squared a circle even before any text book on Mathematics was born; His text book is the basis of Euclid s Elements too. As Hippocrates did square the circle it implied an algebraic number. Further, Lindemann is also right in calling as transcendental number and not right, if he calls number as transcendental. Here, there is a clarity of opinion thus is a polygon number (and attributed to circle). In this paper the diagonal length is obtained from the actual length of the circumference of the circle. II. PROCEDURE Draw a square ABCD. Draw two diagonals AC and BD. O is the centre. AB Side a; AC BD diameter a Draw a circle with centre O and with radius d Diameter AC a d Perimeter of ABCD square x a a diameter d d a 1/ th of the circumference CB where d diameter diagonal a of the ABCD square. [1] Let us find out 1/ th of circumference of circle CB, with the present value, [] d a a a [3] where diameter of the circle is a [] Let us use the following formula to get the length of the diagonal (known of course i.e., a ) from the above value. [5] Perimeter of thesquare 1 Half of 7 times of side of square th of diagonal a P a g e

58 [6] a 7a a 7 1 To Judge The Correct-Ness Of The New Pi Value 16 1 [7] In the 3 rd step, let us multiply the above value with the 1/ th of the circumference of the circle, to give the diagonal AC of the square ABCD. [8] [9] a a ( )a [10] The value is [11] It is clear therefore, that the value does not give exact length of the diagonal AC of ABCD square whose value is a [1] 1 Now, let us repeat the above steps with the new value [13] d a a a 16 [1] 1/ th of Circumference of circle CB [15] a. Perimeter of thesquare 1 Half of 7 times of side of square th of diagonal a 16 7a a 1 [16] 3 rd Step: Multiplication of values of & 5 S. Nos 1 16 [17] a a 16 1 [18] As the exact length of the diagonal AC of square ABCD equal to a is obtained with the new value, 1 so, is the real value. III. CONCLUSION In circle, there are circumference, radius and diameter. In square, there are perimeter, diagonal and side. When a circle is superscribed with the square, circumference, side, diagonal and perimeter of square coexist in an interesting relationship. In this paper, this relationship is studied and the diagonal length is obtained from the arc of the circumference. This is possible only when the exact length of the circumference is known. A wrong length of circumference obtained using a wrong value gives only wrong length of diagonal. REFERENCES [1] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN [] Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Prometheus Books, New York [3] RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. [] RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP P a g e 5

59 To Judge The Correct-Ness Of The New Pi Value [5] RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [6] RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [7] RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [8] RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [9] RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [10] RD Sarva Jagannada Reddy (01), 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: , p-issn: , Volume 3, Issue 11 (June 01) PP: [11] RD Sarva Jagannadha Reddy (01), Pi of the Circle, at [1] R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: x. Volume 10, Issue Ver. I (Jul-Aug. 01), PP P a g e 6

60 International Journal of Engineering Inventions e-issn: , p-issn: Volume, Issue 1 (July 01) PP: 3-37 The Natural Selection Mode To Choose The Real Pi Value Based On The Resurrection Of The Decimal Part Over And Above 3 Of Pi (St. John s Medical College Method) R. D. Sarva Jagannadha Reddy This author though a non-medical graduate (Zoology) was offered a Medical post, Tutor in Physiology in St. John s Medical College, Bangalore, India Abstract:, 3.1, 3.116, are being used as values at school level calculations and at 7 the research-level calculations. Many more numbers are found in the literature for. A method, therefore, is necessary to decide which number is, the real value. The following method chooses as the real value. Keywords: Circle, corner area, diameter, side, square. I. Introduction Circle and square are basic geometrical constructions. To find out perimeter and area of a square there are present two formulae a and a, where a is the side of the square. Similarly, to calculate the circumference and the area of a circle, there are two formulae, r and r, where r is the radius and is a constant. The concept represents, the ratio of the circumference and the diameter of its circle. Thus, the constant is a natural and divine concept. We have radians equal to 180 0, which is a human creation and convenience. For the last 000 years, has been ruling the mathematical world as the value. In March 1998, a new value was discovered by Gayatri method and supported by more than one hundred different geometrical proofs in the last 16 years. The time now has come to decide, which value, is present value or is new value, real? Here is a simple procedure. The nature has created a square and a circle. All the celestial bodies in the Cosmos are spherical in shape. It shows the basic architectural design of the physical world from the Cosmic mind. We see on paper that it has not only created an exact relationship between the circumference and the diameter of a circle and also, the Nature has established an interesting relationship between square and its inscribed circle too. This relationship is taken as the guiding principle to decide the real value from many numbers. Hence, this method is called the Natural Selection. Page 3 7

61 The Natural Selection Mode To Choose The Real Pi Value Based On The II. Procedure Draw a square. Inscribe a circle. Square ABCD, Side AB a, AC BD diagonal, O centre. EF diameter d a Area of the square a Area of the circle d a where a d Square area Circle area Corner area Square area Corner area a x a a a a Divide x by 3. In Siva method, it is found that when the circle square composite geometrical construction is divided symmetrically, the number of segments are x Thus, we obtain finally, the formula which is equal to the value over and above 3. As 3 8 of the diagonal of a square, so also is for the circumference of the inscribed circle, and it is an established fact by this work. And further, the number 3 represents a common associating factor of the inscribed circle and the square. The procedure followed here is in steps. Step 1, Calculates the areas of square and circle Step, Obtains corner area by deducting circle area from the square area Step 3, Squarearea Corner area and Square area 1 Step, Corner area 3 At the 1 st step while calculating the area of the circle, known value is used. In this paper two values are chosen: the official value and the new value 1 Any value enters at the 1 st step, and it s decimal part reappears at the th step. Thus, the resurrection of the decimal part of value is observed at the th step. And this happens only when the real value is taken in the 1 st step. Any other number, if used, does not reappear fully, at the th step. Side diameter a 1 Area of the square : a 1 x 1 1 I. With official value Area of the circle d x 1 x 1 x Square area Circle area Corner area Square area Corner area Square area 1 Corner area Page 35 8

62 The Natural Selection Mode To Choose The Real Pi Value Based On The The decimal part of the official value is Only first two decimals 0.1 reappeared in the th step, instead of all the decimals. II. Let us repeat the above process with the new value d Area of the circle x 1 x 1 x Square area Circle area Corner area Square area Corner area Square area Corner area 3 3 The decimal part of the new value is All the decimals have now reappeared in the th step. There are some more numbers if one looks at the Internet. Prominent numbers that are attributed to, besides of Archimedes, are (Laxman S. Gogawale), 3.15 (Mohammadreza Mehdinia), (from PHI) etc. All these values too have failed when processed in the above steps, to resurrect at the th step. S. No. Proposed/accepted numbers to Resurrected decimal part over and above 3 of value (Official value) Archimedes is much nearer to the real value than though it has been 7 considered as final value to is farthest low value to. number of Golden Ratio is the next closest to the real value attributed to. Out of all the numbers attributed to value detailed in the above Table, only one number of 1 has resurrected itself at the end, with its all the decimals, over and above 3. Hence 1 is the true value. Other numbers have succeeded in coming back with one or two first decimals only beyond 3. Hence, these numbers have, failed in the race to qualify themselves, in the selection, by the natural geometrical construction as a true value. III. Conclusion There are many values in the literature. Two values and are very popular. 7 Third value equal to is added now to the existing values saying that the new Page 36 9

63 The Natural Selection Mode To Choose The Real Pi Value Based On The value is the only true value. In this paper a simple method is found, to choose, the real value. This method chooses 1 as the real value, which is the exact and an algebraic number. REFERENCES [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN []. Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Prometheus Books, New York [3]. RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. []. RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [5]. RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [6]. RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [7]. RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [8]. RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [9]. RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p- ISSN: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [10]. RD Sarva Jagannada Reddy (01), 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: , p-issn: , Volume 3, Issue 11 (June 01) PP: [11]. RD Sarva Jagannadha Reddy (01), Pi of the Circle, at [1]. R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: x. Volume 10, Issue Ver. I (Jul-Aug. 01), PP Page 37 50

64 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: X. Volume 10, Issue Ver. III (Jul-Aug. 01), PP 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 * ) R.D. Sarva Jagannadha Reddy Abstract: Circle, square and triangle are basic geometrical constructions. constant is associated with the circle. In this paper, circle-triangle interlationship chooses the real value of and calculating the area of the triangle involving of the inscribed circle. The alternate formula to find the area of the triangle is 3 3 d. This formula has a geometrical backing. 1 Keywords: Altitude, base, circle, diameter, perimeter, triangle I. Introduction The official value is It is an approximation, inspite of having trillions of its decimals. A new value to was discovered in March The value is Both the values have their own supporting arguments. Triangle is another geometrical entity. Its area is calculated using the formula ½ ab, where a altitude and b base. In this paper, a different formula has been derived to find out the area of the equilateral triangle based on the of its inscribed circle. The formula ½ ab gives the area of the triangle. No other formula is necessary for the area of triangle. The main purpose of derivation of new area formula of triangle is, to test the correctness of value involved in the new formula. One advantage in using the new formula for area of the triangle is, the resulting value tally s exactly with the value of ½ ab only, when the chosen value is correct one. If the wrong/ approximate value is involved in the new formula, it does not give exact value of the triangle. In other words, out of the two numbers and only one number, gives the exact area of the triangle. The number which fails in giving exact value to the area of the triangle is decided as the wrong value. Procedure Draw a circle with centre O and radius d. Draw three equidistant tangents on the circle. The tangents intersect at A, B and C, creating an equilateral triangle ABC. DE is the hypotenuse of DOE triangle or the chord of the circle. Calculations: Centre O Radius OD OE d Diameter DF d Triangle ABC Side AB BC AC a 3 d * This author was awarded Merit Scholarship by Atomic Energy Commission, Trombay, Bombay for his M.Sc., (Zoology) study in S.V. University College, Tirupati, Chittoor Dt. A.P.,India, during This author is highly indebted to the AEC and hence this paper is named in the AECs Honour. 13 Page 51

65 An Alternate Formula in terms of Pi to find the Area of a Triangle and a Test to decide the.. Altitude DC 3d Radius OE OD d Hypotenuse Chord DE d d Area of the triangle by Conventional formula 1 ab 1 DC AB 1 3d 3 3 3d d New formula to find the area of ABC triangle Perimeter of the ABC triangle 3 x AB 3 3d 3 3d times of perimeter of the ABC triangle 1 times of diameter DFof circle times of chord DE 3 3d d 1d To find out the area of the ABC triangle, multiply the area of the circle with Area of the circle d Area of the ABC triangle Area of the circle x 3 3 Thus, d 1 d d where d diameter of the inscribed circle is the new formula to find out the area of the superscribed triangle about a circle. Where, may be or Area of the ABC equilateral triangle besides from the ½ ab formula is where a 3d 1 3d 3d 3 3 d 3 3d 3 a 3 3 d 3 3 So, the value d as the area of the ABC triangle, should also be obtained, using the above new formula, derived in terms of d should be equal to 1 d 1 Page 5

66 An Alternate Formula in terms of Pi to find the Area of a Triangle and a Test to decide the.. The unknown one in the above is. Let us write formula with two different values d d OR should be d , we have obtained this value instead of where d 1 So, official value has failed to give 3 3 otherhand, new value 1 value 1 is the real value. Equating conventional formulas ½ ab naked truth of latter s correctness. has given 3 3 as the value of the area of ABC triangle. On the as the area of ABC triangle. This shows, that the new 3 a 3 3 to new formula d 1 is itself a justification and a II. Conclusion 3 3 ½ ab is the formula to find out the area of a triangle. In this paper, a new formula d is 1 derived. This formula by giving the exact area of ABC triangle shows, that circle and the equilateral triangle are clearly interrelated and are not very different as we have been believing. The benefit we derive by using this formula is, this formula chooses the real value, discarding other approximate values attributed to. This method alone, for the first time in the history of mathematics, acts as a testing method of and boldly says 1 Archiemedes upper limit of 3 or is a lower value compared to the real number. 7 7 Post script As it is proved in this paper that the real value is 1 based on the area of the triangle ABC, it has made possible to demarcate the length of the circumference of the inscribed circle in the straight-lined perimeter of the ABC triangle. How? Let us know first the length of the circumference of the inscribed circle with the known value 1 Diameter d Circumference d 15 Page 53

67 1 An Alternate Formula in terms of Pi to find the Area of a Triangle and a Test to decide the.. When the diameter is equal to 1, the circumference value Let us search for the line-segments equal to 1 Diameter DF d Altitude of ABC triangle DC 3d. Radius OH d DE hypotenuse Chord G is the mid point of DE So, DG GE OG OG d d d GH Radius OG d d d CI DC 3d IJ GH d 1 Circumference of the circle d d, So, the demarcated length DCJ in the perimeter of the ABC triangle is equal to the circumference of the inscribed circle. DC + CI + IJ 3d 3d 1 d d Thus the straight lined length equal to length of the curvature of the circumference of the inscribed circle is called rectification of circumference of a circle. It was done never before. People tried earlier in the perimeter of the square but never in the perimeter of the triangle. References [1]. Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN []. Alfred S. Posamentier & Ingmar Lehmann (00),, A Biography of the World s Most Mysterious Number, Prometheus Books, New York [3]. RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. []. RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [5]. RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [6]. RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [7]. RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP Page 5

68 An Alternate Formula in terms of Pi to find the Area of a Triangle and a Test to decide the.. [8]. RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [9]. RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p- ISSN: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [10]. RD Sarva Jagannada Reddy (01), 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: , p- ISSN: , Volume 3, Issue 11 (June 01) PP: [11]. RD Sarva Jagannadha Reddy (01), Pi of the Circle, at [1]. R.D. Sarva Jagannadha Reddy (01). 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: , p-issn: x. Volume 10, Issue Ver. I (Jul-Aug. 01), PP Page 55

69 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 Hippocratean Squaring Of Lunes, Semicircle and Circle R. D. Sarva Jagannadha Reddy /D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati , A.P., India Abstract: Hippocrates has squared lunes, circle and a semicircle. He is the first man and a last man. S. Ramanujan is the second mathematician who has squared a circle upto a few decimals of equal to The squaring of curvature entities implies that lune, circle are finite entities having a finite magnitude to be represented by a finite number. Keywords: Squaring, lune, circle, Hippocrates, S. Ramanujan,, algebraic number. I. Introduction Hippocrates of Chios was a Greek mathematician, geometer and astronomer, who lived from 70 until 10 BC. He wrote a systematically organized geometry text book Stoicheia Elements. It is the first book. And hence he is called the Founding Father of Mathematics. This book was the basis for Euclid s Elements. In his days the value was 3 of the Holy Bible. He is famous for squaring of lunes. The lunes are called Hippocratic lunes, or the lune of Hippocrates, which was part of a research project on the calculation of the area of a circle, referred to as the quadrature of the circle. What is a lune? It is the area present between two intersecting circles. It is based on the theorem that the areas of two circles have the same ratio as the squares of their radii. His work is written by Eudemus of Rhodes (335 BC) with elaborate proofs and has been preserved by Simplicius. Some believe he has not squared a circle. This view has become very strong with the number a polygon s value attributed to circle, arrived at, from the Exhaustion method (EM) prevailing before Archimedes (0 BC) of Syracuse, Greece, and refined it by him, hence the EM is also known as Archimedean method. This number has become much stronger as value, and has been dissociated from circle-polygon composite construction, with the introduction of infinite series of Madhavan (150) of South India, and independently by later mathematicians John Wallis (1660) of England, James Gregory (1660) of Scotland. With the progressive gaining of the importance of as value from infinite series, the work of squaring of circle of Hippocrates has gone into oblivion. When the prevailing situation is so, in the mean time, a great mathematician Leonhard Euler ( ) of Switzerland has come. His record-setting output is about 530 books and articles during his lifetime, and many more manuscripts are left to posterity. He had created an interesting formula e i +1 0 and based on his formula, Carl Louis Ferdinand Lindemann ( ) of Germany proved in 188 that was a type of nonrational number called a transcendental number. (It means, it is one that is not the root of a polynomial equation with rational coefficients. Another way of saying this is that it is a number that cannot be expressed as a combination of the four basic arithmetic operations and root extraction. In other words, it is a number that cannot be expressed algebraically). Interestingly, the term transcendental number is introduced by Euler. When all these happened, naturally, the work on the Squaring of circle by Hippocrates was almost buried permanently. This author with his discovery in March 1998 of a number from Gayatri method, and its confirmation as value, from Siva method, Jesus proof etc. later, has made the revival of the work of Hippocrates. Hence, this submission of this paper and restoring the golden throne of greatness to Hippocrates has become all the more a bounden duty of this author and the mathematics community. II. Procedure I. Squaring of Lunes-(1) Hippocrates has squared many types of lunes. In this paper four types of lunes are studied. Consider a semi-circle ACB with diameter AB. Let us inscribe in this semi-circle an isosceles triangle ACB, and then draw the circular are AMB which touches the lines CA and CB at A and B respectively. The segments ANC, CPB and AMB are similar. Their areas are therefore proportional to the squares of AC, CB and 39 Page 56

70 Hippocratean Squaring Of Lunes, Semicircle and Circle AB respectively, and from Pythagora s theorem the greater segment is equivalent to the sum of the other two. Therefore the lune ACBMA is equivalent to the triangle ACB. It can therefore be squared. The circular arc AMB which touches the lines CA and CB at A and B respectively can be drawn by taking E as the centre and radius equal to EA or EB. 1. AB diameter d. DE DC radius d/ 3. F midpoint of AC. N midpoint of arc AC. 5. NF d d d d d d, DM, MC d Area of ANC Area of CPB d 8 7. Area of AMB Areas of ANC + CPB d Area of ACM Area of BCM 9. 1 d d Area of ACB triangle d 10. According to Hippocrates the area of the lune ACBMA is equal to the area of the triangle ACB. Area of Lune ACBMA Area of triangle ACB (ANC + CPB + ACM + BCM) d d d Squaring of Lunes-() 11. Let ABC be an isosceles right angled triangle inscribed in the semicircle ABOC, whose centre is O. AB and AC as diameters described semicircles as in figure. Then, since by Ecu. I, 7, Sq. on BC Sq. on AC + Sq on AB. Therefore, by Euc. XII,, Area semicircle on BC Area semicircle on AC + Area semicircle on AB. Take away the common parts Area triangle ABC Sum of areas of lunes AECD and AFBG. Hence the area of the lune AECD is equal to half that of the triangle ABC. 1. BC diameter d, 13. OB OC radius d/ 1. AB AC d diameter of the semicircle ABF ACD d d 15. GI EJ 16. Sq. on BC Sq. on AC + Sq. on AB 17. d Area of the larger semicircle BAC Area of the smaller semicircle ABF ACD Diameter d On the 0 Page 57

71 Hippocratean Squaring Of Lunes, Semicircle and Circle Area d d d a) Areas of two smaller semicircles d d Area of the triangle ABC 1 base altitude Base BC d, Altitude OA d 1 d d Area d 0. Segment AIBG Segment AJCE d d d Lune AGBF lune AECD. Area of the lune (AGBF or AECD) Semicircles (ABF & ACD) Segments (AIBG & AJCE) d d d 8 8 Areas of AIBG + AJCE Squaring of lunes-(3) There are also some famous moonshaped figures. The best known of these are the crescents (or lunulae) of Hippocrates. By the theorem of Thales the triangle ABC in the first figure is right angled: Thus p m + n. The semicircle on AB p has the area A AB p /8; the sum of the areas of the semicircles on AC and BC is A AC +A BC (n + m )/8 and is thus equal to A AB. From this it follows that: The sum of the areas of the two crescents is the area of the triangle. 3. AB diameter d. BC radius d/ 3d 5. AC 6. DF d/ 7. DE d 3d 3d d 8. EF 9. GH d/ 30. GJ 3d 3d d 3d d 31. HJ GJ GH 3. Area of the semicircle BDCF d d 1 d Area of the semicircle AGCJ 3d 3d 1 3 d 8 3 where BC diameter d where AC diameter 3d 1 Page 58

72 3. Area of the triangle ABC 1 AC BC 1 3d d 3 d Area of the curvature entity BDCE d Area of the curvature entity AGCH Circle AGCH 1 3 d d 16 d Area of the triangle d, where side AC 16 d 37. Area of the lune BECF 3 3 Semicircle BDCF BDCE segment d d 3 8 (S.No. 3) (S.No. 35) 38. Area of the lune AHCJ Semicircle AGCJ AGCH segment d d 3 8 (S.No. 33) (S.No. 36) 39. Sum of the areas of two lunes area of the triangle (S.No. 37) + (S.No. 38) (S.No. 3) d d d 8 Squaring of lunes () The sum of the areas of the lunes is eqal to the area of the square. 0. AB side d 1. DE EC d/. AO OC 3. EF. FG d d d d d 5. d Area of the circle Where diameter d d 1 d d d 6. Area of the semicircle DECG d Where DC diameter d d Area of the curvature entity DECF d 8 8. Area of the lune DFCG d Semicircle DECG Curvature entity DECF d d The sum of the areas of lunes the area of the square Hippocratean Squaring Of Lunes, Semicircle and Circle 6 3 d d 96 Page 59

73 Hippocratean Squaring Of Lunes, Semicircle and Circle d d d 8 8 III. Squaring of a semicircle Hippocrates next inscribed half a regular hexagon ABCD in a semicircle whose centre was O, and on OA, AB, BC, and CD as diameters described semicircles. The AD is double any of the lines OA, AB, BC and CD, Sq. on AD Sum of sqs. On OA, AB, BC and CD, Area semicircle ABCD sum of areas of semicircles on OA, AB, BC and CD. Take away the common parts. Area trapezium ABCD 3 lune AEFB + Semicircle on OA. 50. DA diameter d d 51. Area of the semicircle DABC d 8 8 DA 5. radius of larger semi circle d AB 53. AB d diameter of smaller semi circle ABE d d d 1 d Areas of semicircle on OA, AB, BC and CD d Area of sector OAFB d 6 3 d Area of the triangle OAB d Area of the segment AFB Sector Triangle d d d Area of lune AEBF Semicircle on AB AFB segment d d d x Area of one lune x Area of 3 lunes 3 d d Area of 3 lunes + semicircle on OA d d d d d Area of trapezium 3 x OA triangle (S.No. 56) d d Area of 3 lunes + semicircle on OA Area of trapezium d d (S.No. 60) 3 Page 60

74 Hippocratean Squaring Of Lunes, Semicircle and Circle IV. Squaring of circle Consider two concentric circles with common centre O and radii such that the square of the radius of the larger circle is six times the square of the radius of the smaller one. Let us inscribe in the smaller circle the regular hexagon ABCDEF. Let OA cut the larger circle in G, the line OB in H and the line OC in I. On the line GI we construct a circular segment GNI similar to the segment GH. Hippocrates shows that the lune GHIN plus the smaller circle is equivalent to the triangle GHI plus t he hexagon. 63. OA radius of the smaller circle d 6. OH radius of the larger circle 65. Third circle: GI radius GK + KI OH OI d 67. OK OH 6 d 68. KI 3 OI OK d 69. Radius of the third circle 3 GI x KI d d 6 6 d 70. Area of the GHI triangle 1 GI HK OH 6 HK d 1 3 d d d Area of the AOB triangle OA AB d OA d ; AP PB AB AP 3 d ; Area 1 1 d 3 3 OA PB d d Area of the hexagon Area of the triangle AOB x d 6 d d d Area of the smaller circle d 7. Area of the segment GH Segment HI 75. d Area of the larger circle Where d d 6d 6d 6d d Area of the larger circle is divided into 6 sectors d d 6 Page

75 Hippocratean Squaring Of Lunes, Semicircle and Circle Area of the triangle OGH OHI GHI d 8 (S.No. 70) 78. Area of the GH segment HI segment 3 3 Sector Triangle (OGH) 3 3 d d 8 d 8 (S.No. 76) (S.No. 77) There are two segments GH and HI d d Similarly, GNIK is also another segment which is the part of the sector GNIQ. It consists of the triangle GIQ and GNIK segment. 80. To find out the area of the sector GNIQ, let us first find out the area of the circle whose diameter is equal to that of the third circle. 3 Diameter of the third circle riadus x d 3 d (S.No. 69) d Area 3 d3 d 18 d d Then let us find out the area of the sector th d d d Now let us find out the area of the triangle GIQ 1 GI KQ Where KI GI 3 d 1 3 d GI QI GQ Radius of the third circle. 3 3 KQ QI KI d d 3 6 d Area 1 GI KQ d d d Area of the segment GNIK Sector Triangle (S.No. 81) (S.No. 8) d d d Now it has become possible to calculate the area of GHIN segment Triangle GHI Segment GNIK (S.No. 70) (S.No. 83) d d d Area d 85. Area of the lune GHIN Segments + Segments + Circle GH & HI GHIN S.No. 78 S.No. 8 S.No d d d d 86. Area of the triangle GHI + Area of the hexagon ABCDEF (S.No. 70) (S.No. 7) 5 Page 6

76 Hippocratean Squaring Of Lunes, Semicircle and Circle d d d Area of lune + Circle d Area of triangle+ hexagon d (S.No. 85) (S.No. 86) So, the sum of the areas of lune and circle is equal to the sum of the areas of triangle and hexagon. V. Post Script The following are the points on which some thinking is necessary: is accepted as value is a transcendental number. 3. As this polygon s value is accepted as of the circle, circle and its value have become transcendental entities.. The concept of transcendental number vehemently opposes squaring of circle. Latest developments is the new value is the exact value. 7. This number is an algebraic number, being the root of x 56x Squaring of circle is done with this number. Conclusion 9. Hippocrates did square the circle is a transcendental number it is correct can not square a circle, - is also correct. Final verdict 1. As Hippocrates did the squaring a circle, it amounts to confirming that circle and its value are algebraic entities. It implies that as is a borrowed number from polygon and attributed to circle, called a transcendental number, said squaring a circle an unsolved geometrical problem, the final verdict is, all are correct, except one, i.e. attributing of polygon to circle. Hence is not a value at all. References [1]. T. Dantzig (1955), The Bequest of the Greeks, George Allen & Unwin Ltd., London. []. P. Dedron and J. Itard (1973) Mathematics and Mathematicians, Vol., translated from French, by J.V. Field, The Open University Press, England. [3]. W.W. Rouse Ball (1960), A short Account of the History of Mathematics, Dover Publications, New York. []. W.G.H. Kustner and M.H.H. Kastner (1975). The VNR Concise Encyclopedia of Mathematics, Van Nostrand Rusinhold Company. [5]. R.D. Sarva Jagannadha Reddy (01). Pi of the Circle at 6 Page 63

77 IOSR Journal of Mathematics (IOSR-JM) e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 Durga Method of Squaring A Circle RD Sarva Jagannadha Reddy Abstract: Squaring of circle is an unsolved problem with the official value with the new value 1/ (1- ) it is done in this paper. Keywords: Exact Pi value 1/ (1- ), Squaring of circle, Hippocrates squaring of lunes. I. Introduction Squaring a circle is defined as constructing a square having an area equal to that of a given circle. It is also called as quadrature of the circle. This concept has been there from the days of Rhind Papyrus (1800 B.C) written by a scribe named Ahmes. Hippocrates of Chios (50 B.C) has squared lunes, full circle and semicircle along with lunes. He fore saw the algebraic nature of the value. value has failed to find a place for it in the squaring of lunes. Though the World of Mathematics has accepted his squaring of lunes, they became silent for why is a misfit in his constructions. Further, there is a false opinion that Hippocrates could not square a circle. However, Hippocrates did square a full circle and a semicircle along with a lune. In both the cases squaring a lune, squaring a circle along with a lune the new value, 1 has explained perfectly well the constructions of Hippocrates. Thus the propositions of Hippocrates which remained theoretical all these 00 years, have become practical constructions with the discovery of 1. It is clear therefore, we have misunderstood Hippocrates because, we believed as the value of. I therefore apologize to Hippocrates on behalf of mathematics community for the past mistake done by us. And to atone the academic sin committed by us, I bow my head and dedicate the explained parts (for details: Pi of the Circle, last chapter: Latest work, Pages from 73 to 81) to Hippocrates, in James Gregory (1660) has said squaring of circle is impossible. His view has been confirmed by C.L.F. Lindemann (188) based on Euler s formula e i Von K. Weiertrass ( ) and David Hilbert (1893) have supported the proof of Lindemann by their proofs. S. Ramanujan (1913) has squared a circle upto some decimals of Prof. Underwood Dudley doesn t accept Lindemann s proof because this is based on numbers which are approximate in themselves. Now, the exact value is discovered. It is 1 1. It is an algebraic number. The following is the procedure how to square a circle. whose value is II. Procedure We have to obtain a side of the square ; when 1, then a a CD a, OK OF radius, FK, JK FG GC, GC JG KF 1 a 1 a 1 Page 6

78 a ; GB BC GC a a a ; Bisect GB. GH HB a 8. Bisect HB. a 16 HI BI 1 1 CI BC BI a a 16 a 16 ; Area of the circle a 16 CB diameter a; Durga Method Of Squaring A Circle Draw a semicircle on CB, with radius a and center O ; CO O B 1 where a 1 Draw a perpendicular line on CB at I, which meets semicircle at Y. Apply altitude theorem to obtain IY length IY CI IB Connect YC which is the side of the square CYUT whose area is equal to that of the inscribed circle in the square ABCD. Apply Pythagorean theorem to get CY from the triangle CIY. Side of the square CY Area of the square CYUT CI IY area of the inscribed circle in the square ABCD. III. Conclusion 1 is the exact value of circle. Hence, squaring of circle is done now. The misnomer Circle squarer will sink into oblivion. Hippocrates will now gets his deserving throne of greatness though delayed unfortunately for 00 years. 15 Page 65

79 International Journal of Engineering Inventions e-issn: , p-issn: Volume 3, Issue 11 (June 01) PP: 9-35 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the actual length of the circumference of the circle and Pi R. D. Sarva Jagannadha Reddy Abstract: There is only one geometrical method called Exhaustion Method to find out the length of the circumference of a circle. In this method, a regular polygon of known number and value of sides is inscribed, doubled many times until the inscribed polygon exhausts the space between the polygon and circle as limit. In the present paper, it is made clear, that the value for circumference, i.e of polygon, attributing to circle is a lower value than the real value, and the real value is adopting error-free method. Key words: Circle, corner length, diameter, diagonal, polygon, side, square. I. Introduction The Holy Bible has said value is 3. The formula for circumference of a circle is d, where is a constant and d is the diameter. When the diameter is 1 the circumference is equal to value. When, a hexagon is inscribed in a circle of ½ as its radius, the perimeter of hexagon is equal to 3. It means, the circumference is greater than 3, because hexagon is an inscribed entity in the circle. With the unit diameter of circle, circumference and/or value is exchangeable, because, both are represented by a single number. There were many values for (Chung Hing, 50 AD), (Wang Fau, 50 AD), (Liu Hui, 63 AD), (Aryabhata, 99 AD). From Francois Viete (1579) onwards the value has become an accepted value. It is a well established fact that the value is However, two things which are associated with this number have not satisfied some scholars of mathematics. They are is a borrowed number of polygon, attributed to circle, believing in the logic of limitation principle and being an approximate number, made super computers too have failed to find the exact value and. Assertively, this number has said, squaring a circle impossible, being it is a transcendental umber. High school students, now and then ask, when mathematics is an exact science, how is it possible we have many values are being used, for example, 3.1, 3.116, Nature has been kind. Exact value has been found at last. A few papers in support of the exact value have been published (in Reference). Surprisingly the new value is struggling very hard still for its approval. All the time, the work done on has been cited, saying, the past generation or the present workers could not be wrong. Thus, the new value (through was discovered 16 years ago, in March 1998 is yet to be accepted. From March 1998, with the discovery of , this worker has been struggling to find evidences in support of new value, and also struggling much more in search of an error in the derivation of There are some similarities and some differences between present and new values. The similarity is, in Exhaustion method a polygon is inscribed in a circle and in the new method a circle is inscribed in a square. The differences are many: 1. Present value represents the perimeter/ area of the polygon and attributed to circle; the new value represents the area of the inscribed circle only in a square (called Siva method in Reference). Here, square helps but does not lend its value to circle.. Present value is derived applying Pythagorean theorem meant for straight lines. New value is derived adopting entirely a new approach for which no previous proofs/ statements are available. 3. Present value is a transcendental number and new value is an algebraic number.. Present value says squaring of a circle is an impossible idea and Page

80 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the whereas the new value squared a circle. It also, circled a square (i.e., constructing a circle whose circumference and/or area equal to the perimeter and/ or area of a square respectively. For example, if the side of a square is 1, the perimeter would be and its area would be 1. Circling a square means we have to find out a radius geometrically of circle whose length of the circumference is and/ or area of the circle is 1. In this paper, present value-to some decimal places is obtained using the new value 1, showing, circle, polygon and square are not different geometrical entities and their interrelationship, if understood correctly, may help in the derivation of present value from the new value also. If this idea is accepted, an algebraic number like 1 is also, capable of giving rise to, a transcendental number like It leads to the creation of a new doubt, is is really a transcendental number? There are 3 examples cited below, linking present and new values and To drive to the point an elaborate explanation is chosen and here and there repetitions too appear. II. PROCEDURE Example-1 The following formula gives the present value upto 5 decimal places. When my work on deriving the new value of equal to was submitted, I was advised by many to think how to derive the present value cautioning me as the new one was wrong. In obedience, I hereby submitted the following formula giving the number exact upto five decimal places where 00 To get the same value ( ) from the Classical method of Archimedes, we have to travel a long way using the following general formula to calculate the length of the perimeter of the inscribed polygon of 1536 sides starting from 6 sides, in a circle Side sn r r r sn A geometrical line segment for Let us draw a circle with diameter 9 and radius 9/. Cut the circumference at one point and straighten it and further divide it into 00 equal segments. Similarly, divided the diameter into 3 equal segments. 1/3 rd diameter + one segment length of circumference Page 30 67

81 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the Example- Archimedes inscribed a polygon of 96 sides with the circle and gave perimeter of the polygon as equal to 3 10/ When the diameter of the superscribed circle is 1 unit, the perimeter value. Prof. Constantine Karapappopulos of University of Patra, Patra, Greece has suggested a formula for the 1 3 Archimedian minimum as equal to 3 1 in the range of The proposed formula for x to gives the circumference of the circle as equal to Perimeter of Polygon x Circle connecting link Circumference of Circle is obtained by the calculation method. One need not set aside it on the reason that the above factor is based on the calculation method. This factor or circle connecting link can be viewed as a clue or guiding factor for further analysis. On close observation one finds the denominator and the numerator of the link are symmetrical and give us a right relationship between the inscribed polygon and the superscribed circle. Example-3 Classical method involves the inscription of polygon in the circle. Starting with the known perimeter of a regular polygon (here we start with a regular hexagon) of n sides inscribed in a circle, the perimeter of the inscribed regular polygon of n sides can be calculated by the application of Pythagorean theorem. Let C be a circle with centre O and radius r, and let PQ s be a side of a regular inscribed polygon of n sides having a known perimeter. Then the apothem, OM u is given by r s side RQ w of he inscribed polygon of n sides is found from polygon is known. hence the sagetta, MR v r u is known. Then the w v s Page 31 hence the perimeter of this 68

82 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the S. No. n First Side PQ s ns Perimeter Circumferen ce s/ t Apothem r t u Sagetta r u v Succeeding side v t w Let us analyse the above Table. The calculation is started with a regular polygon of 6 sides having the perimeter equal to 3. The sides of polygon are doubled for 9 times and now the inscribed polygon has 307 sides and its perimeter circumference of circle value is equal to Thus, the number of sides are increased 51 times finally The length of the perimeter has increased from to The correct value is From the above Table, it is possible to find the exact length only upto i.e. 6 decimal places only with 307 sides of the polygon. ABCD Square; EFGR Circle; AB Side NQ diameter 1; AC Diagonal ; AN QC Corner length; Diagonal diameter AC NQ 1; Hexagon STGUVE In the Classical method of calculating perimeter/ circumference/ in the above table, number of sides of the inscribed polygon plays an important role. The derivation is started with 6 sides and ended with 307 sides. The perimeter of the hexagon is 3 and the perimeter of the polygon of 307 sides has increased to Page 3 69

83 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the Starting no. of sides of the polygon 6 Ending no. of sides of the polygon 307 Starting perimeter 3 Ending perimeter Present value New value In the following formula a relation is shown between two values: present value and new value as in the above two examples. Squareof thestarting no.of sides Ending no.of sides Present value (correct upto 6 decimals) New value Diagonal diameter The lengthy process involved in the above Exhaustion method is represented in a single formula In the Exhaustion method Pythagorean theorem is applied invariably appears in every aspect of calculation. The result obtained from the above formula using is much more accurate at 7 th decimal place than what it is ( ) obtained in the Exhaustion method in the table. In the Exhaustion method RQ is the hypotenuse and is the side of the inscribed polygon. And in the above formula diagonal diameter corner lengths on either side of the diameter also play an important role. Hence, it has become possible that the new value has given the present value (to some decimals). Let us understand, in much more clear terms, the above formula. There are four factors in the Exhaustion method. They are No. of sides of the hexagon 6 Perimeter of the hexagon ½ x 6 3 No. of sides of the final polygon 307 Perimeter of the final polygon Let us divide final sides by the sides of the beginning polygon i.e. hexagon Let us divide corner length AN + QC into 51 parts. AB Side NQ Diameter 1 AC diagonal Corner length AC NQ diagonal diameter 1 Let us divide above length into 51 parts Multiply the above value with the no. of sides of the beginning polygon (hexagon) Deduct the above value from the new value value (to some decimals) So, the corner length 1 is divided into 51 parts which gives the present Page

84 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the 6. In this process corner length 1 is taken for consideration. 7. In the Exhaustion method the hypotenuse RQ in Fig. is taken for consideration and proceeded successively for many times. 8. AC is the hypotenuse of the triangle DAC of Fig. and RQ is the hypotenuse of the triangle RMQ. 9. How does the length over and above 3 diameters of the circumference of the circle is arrived in deriving the new value? The answer is very simple. Let us divide the corner length QC only of Fig- by Side AB diameter NQ 1 Corner length QC Diagonal diameter AC NQ 1 QC Divide QC by Now, the length of circumference 3diameters and add to the 3 diameters. 10. When we compare two ways of arriving the exact length of circumference of a circle, it is clear in Exhaustion method the perimeter of the inscribed polygon is increased slowly by doubling the number of sides of the previous polygon. Thus, the number of sides have been increased from 6 to 307, it means it has been increased 51 times. In other words, we have divided corner length into 51 parts. In the second approach in the arrival of length of the circumference, the corner length is Corner length QC divided at one stroke with So, 3 sides + QC Thus there are two values for the length of the circumference of the circle of Exhaustion method and of Gayatri method 1. We can thus visualize a diagram of, containing a hexagon whose perimeter STGUVE is 3, next to hexagon, of inscribed polygon of 307 sides and further next and the circle EFGR, whose circumference is equal to with radius ½. III. Conclusion 1 There are now two values and In the Exhaustion method, perimeter of the polygon is attributed to the circumference of the circle. As the inscribed polygon is smaller one, the value must be lesser than the exact length of the circumference of circle. This paper clearly shows the calculation error involved in the arrival of the actual length of the circumference ( ) of the circle. Hence, Exhaustion method is not suitable in arriving the of the circle. However, this study shows that both the values have a common-ness in their nature. In Exhaustion method while doing calculations involving squaring, square root etc. and only a few decimals have been taken. All the numbers are infinite numbers. So, the prolongation of round-off-error is universal throughout the calculations. And this has resulted in a lower value instead of the actual value. However, the new work has tried to overcome the error supposed to be in the Exhaustion method by adopting Gayatri method, Siva method, Jesus Page 3 71

85 The unsuitability of the application of Pythagorean Theorem of Exhaustion Method, in finding the method etc. and found the actual length of the circumference and real value, i.e REFERENCES [1.] Lennart Berggren, Jonathan Borwein, Peter Borwein (1997), Pi: A source Book, nd edition, Springer-Verlag Ney York Berlin Heidelberg SPIN [.] RD Sarva Jagannada Reddy (01), New Method of Computing Pi value (Siva Method). IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 8-9. [3.] RD Sarva Jagannada Reddy (01), Jesus Method to Compute the Circumference of A Circle and Exact Pi Value. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. I. (Jan. 01), PP [.] RD Sarva Jagannada Reddy (01), 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: , p-issn: Volume 10, Issue Ver. II (Mar-Apr. 01), PP 09-1 [5.] RD Sarva Jagannada Reddy (01), New Pi Value: Its Derivation and Demarcation of an Area of Circle Equal to Pi/ in A Square. International Journal of Mathematics and Statistics Invention, E-ISSN: P-ISSN: Volume Issue 5, May. 01, PP [6.] RD Sarva Jagannada Reddy (01), Pythagorean way of Proof for the segmental areas of one square with that of rectangles of adjoining square. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 3 Ver. III (May-Jun. 01), PP [7.] RD Sarva Jagannada Reddy (01), Hippocratean Squaring Of Lunes, Semicircle and Circle. IOSR Journal of Mathematics, e- ISSN: , p-issn: Volume 10, Issue 3 Ver. II (May-Jun. 01), PP 39-6 [8.] RD Sarva Jagannada Reddy (01), Durga Method of Squaring A Circle. IOSR Journal of Mathematics, e-issn: , p-issn: Volume 10, Issue 1 Ver. IV. (Feb. 01), PP 1-15 [9.] R.D. Sarva Jagannadha Reddy (01), Pi of the Circle, at Page 35 7

86 R.D. SARVA JAGANNADHA REDDY HOME PAGE 1. Mother : Dhanalakshmi. Father : Venkata Reddy 3. Date of Birth : Native Place : Bandarupalli Village Yerpedu Mandal, Chittoor District, AP, India. 5. Phone No. : rsjreddy13196@gmail.com 7. Present Address (Temporary) : /D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati , INDIA 8. Education : B.Sc., Zoology (Major), Botany, Chemistry (minors) M.Sc., Zoology at S.V. University College, Tirupati. 9. Books : 1. Origin of Matter. Origin of the Universe 3. Organic Bloom (on Animal Evolution). Pi of the Circle Telugu Books 5. Sarvam Pavithram 6. Pavana Prapancham 7. Mahabhagavatham Maanavaavirbhavam 8. Abhinandana 9. Mattipella 10. Janthu Pravarthana (Animal behavior) for B.Sc., 11. Kachhapi 10. Wife : Late Savithri Children : Shyam Sundar Reddy, Gowri Devi, Sarada 11. Profession : Lecturer in Zoology, Retired on

87 1. Donation : As a mark of Gurudakshina to my Alma Mater Sri Venkateswara University, Tirupati a granite stonesphere of 6 feet diameter and another granite stonesphere of 3 feet diameter to Govt. Junior College, Piler, Chittoor district (where, this author got the idea in 197, while working as Junior Lecturer in Zoology, one can get formulae for the computation of area and circumference of a circle without using constant /7 as in r and r) have been humbly donated. Stone at S.V.U. Mathematics Department, Tirupati, A.P., India. Stone at Govt. Jr. College, Piler, Chittoor District, A.P., India. 7

88 Donated 11 Feet high Sivalingam to S.V. Higher Secondary School, Prakasam Road, Tirupati, and consecrated at TTD s, S.V. Dhyanaramam, Opposite to Regional Science Centre, Alipiri, Tirupati. Donated 6 Feet high Sivalingam to Beriveedhi Elementary School, Tirupati and consecrated at Rayalacheruvu Katta, 15 km away from Tirupati. 75

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