100% Exact area of circle = (17 - 8√3) r2

Algebraic & Geometric proofs

Laxman Gogawale PUNE : 411043

Mob. : 919890646560

Email id:

Research paper published in two international journals.

We know that, where C /D = A/R2 = approximate value of pi = 3.1415926535897…which is endless value.

C = circumference of circle, D= diameter of circle, A = area of circle, R= radius of circle,

If we calculate C / D we cannot measure end point of circumference. So it will give approximate results.

I started research to find exact area of circle using A /R2 method.

There are different proofs to find the old value of pi but no one get 100% exact answer.

Like using number series, Trigonometry, Dividing circle into infinite parts, as practically we can’t measure endpoint etc. In order to find 100% exact area of circleI found the new method.

I have made number of proofs but here I am giving simple proof out of it. If you are interested in studding complete thesis then please contact me.

Reason why I am 100% sure about my research is that I estimated pi value by number of different algebraic methods. Why should we discuss pi is transcendentaloralgebraic? This is not important. But we are more concentrated on calculating 100% exact answer which is most important.

References,

Exact area of equilateral triangle formula = (√3 ÷ 4) × side2

Complete thesis of my research titled as ‘‘Exact value of pi ’’ is being published in following journals:

IOSR(international journal of scientific research) journal of mathematics in May-June 2012.

IJERA(international journal of Engineering research and applications) in July-August 2013.

Soft copy of my thesis is now also available on internet and one can get it by making search with following key words:

“LaxmanGogawale.

“Pi value Gogawale.

=

Fig.1 fig. 2

Note: let a, b, c & d each part shows area

Area of square = (12a + 12b + 12c + 4d) = (16a + 16b) = 4r2 (from fig. 1 & 2)

(12a + 12b + 12c + 4d) – (16a + 16b) = 0

= (- 4a - 4b + 12c + 4d)= 0

From fig. no. 2

16a = [2r × 2(√3/2) r] = (2√3) r2 a= (2√3) r2 /16 = (0.125√3) r2

16b = [2r – 2(√3/2) r] × 2r = (4 - 2√3) r2 b=(4 - 2√3) r2/ 16 = (0.25 – 0.125√3) r2

Problems faced during the research of pi that:

How to estimate the exact values of part c& part d?

I found that (a + b – 3c – d = 0) (14b – 2a – 3c = 0) (3a – 13b – d = 0)

All above equations are equal & have value equal to zero.Which helps in determining value of 3cd in terms of ab

i.e.3c = (14b – 2a) = [14(0.25 – 0.125√3) r2 – 2(0.125√3) r2] = (3.5 – 2√3) r2

d = (3a – 13b) = [3(0.125√3) r2 - 13(0.25 – 0.125√3) r2] = (2√3 – 3.25) r2

a = (0.125√3) r2 b = (0.25 – 0.125√3) r2 3c = (3.5 - 2√3) r2 d= (2√3 – 3.25) r2

In order to verify this addition or subtraction of any multiple of previous equation to main equation of area of square will results into area of square itself.

By using above method I have prepared following equations. (Note:x, y, z is any number)

Area of square =(12a + 12b + 12c + 4d)+ [x (a + b – 3c – d)]= area of square

=(12a + 12b + 12c + 4d)+ [y (14b – 2a – 3c)] = area of square

=(12a + 12b + 12c + 4d)+ [z (3a – 13b – d)] = area of square

Area of square = (12a + 12b + 12c + 4d) + [x (a + b – 3c – d)] + [y (14b – 2a – 3c)] = area of square

= (12a + 12b + 12c + 4d) + [x (a + b – 3c – d)] + [z (3a – 13b - d)] = area of square

= (12a + 12b + 12c + 4d) + [y (14b – 2a – 3c)] + [z (3a – 13b – d)] = area of square

= (12a + 12b + 12c + 4d) + [x (a + b – 3c – d)] + [y (14b – 2a – 3c)] + [z (3a – 13b – d)] = area of square

As well as if we put value of 3cd as per below equations in new equations (obtained after addition or subtraction) it will give answer equal to (17 - 8√3) r2

Area of circle = (12a + 12b + 12c) = 3r2 + 12c

Area of inscribed dodecagon = (12a + 12b + 12c) – 12c = 3r2 = area of circle – 12c

Area of square = (12a + 12b + 12c) + 4d = 4r2 = area of circle + 4d

For example (x =7, y = 9, z = - 11)

Area of square = (12a + 12b + 12c + 4d) + 7(a + b – 3c – d)

= (12a + 12b + 12c + 4d) + (7a + 7b – 21c – 7d)

= 19a + 19b – 9c - 3d equation no. 1

= (12a + 12b + 12c + 4d) + 9(14b – 2a – 3c)

= (12a + 12b + 12c + 4d) + (126b – 18a – 27c)

= - 6a + 138b – 15c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 11(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (33a – 143b – 11d)

= - 21a + 155b + 12c + 15d equation no. 3

Total of 3 equations = area of 3 square

= (19a + 19b – 9c –3d) + (- 6a + 138b – 15c + 4d) + (- 21a + 155b + 12c + 15d)

= (- 8a + 312b – 12c + 16d)

= (- 8a + 312b) + (area of 1 inscribed dodecagon – area of 1 circle = - 12c)

+ (Area of 4 square – area of 4 circles = 16d)

= (- 8a + 312b) + (3r2) + 4(4r2) – area of (1 + 4) circle

= (- 8a + 312b) + 3r2 + 16r2 – area of 5 circle

Area of 3 square + area of 5 circle = - 8a + 312b + 19r2

Area of 5 circle = - 8a + 312b + 19r2 – area of 3 square

= - 8a + 312b + 19r2 – 3(4r2)

= - 8a + 312b + 7r2

Use value of a & b = - 8(0.125√3) r2 + 312(0.25 – 0.125√3) r2 + 7r2

= - (√3) r2 + (78 – 39√3) r2 + 7r2

= (85 – 40√3) r2

Area of circle = (85 – 40√3) r2/ 5 = (17 - 8√3) r2

For example X = 4, y = 8, z = 6

Area of square = (12a + 12b + 12c + 4d) + 4 (a + b – 3c – d) + 8 (14b – 2a – 3c)

= (12a + 12b + 12c + 4d) + (4a + 4b – 12c – 4d) + (112b – 16a – 24c)

= 128b – 24c equation no. 1

Area of square = (12a + 12b + 12c + 4d) + 4 (a + b – 3c – d) + 6 (3a – 13b - d)

= (12a + 12b + 12c + 4d) + (4a + 4b – 12c – 4d) + (18a – 78b - 6d)

= 34a – 62b – 6d equation no. 2

Area of square = (12a + 12b + 12c + 4d) + 8 (14b – 2a – 3c) + 6 (3a – 13b – d)

= (12a + 12b + 12c + 4d) + (112b – 16a – 24c) + (18a – 78b – 6d)

= 14a + 46b – 12c – 2d equation no. 3

Area of square = (12a + 12b + 12c + 4d) + 4 (a + b – 3c – d) + 8 (14b – 2a – 3c) + 6 (3a – 13b – d)

= (12a + 12b + 12c + 4d) + (4a + 4b – 12c – 4d) + (112b – 16a – 24c) + (18a – 78b – 6d)

= 18a + 50b – 24c – 6d equation no. 4

Total of 4 equations = area of 4 square

= (128b – 24c) + (34a – 62b – 6d) + (14a + 46b – 12c – 2d) + (18a + 50b – 24c – 6d)

= 66a + 162b – 60c – 14d

= 66a + 162b + (area of 5 inscribed dodecagon – area of 5 circle = - 60c)

+ (Area of 3.5 circles – area of 3.5 square = - 14d)

= 66a + 162b + 5(3r2) – 3.5(4r2) + area of (- 5 + 3.5) circle

= 66a + 162b + 15r2 – 14r2 – area of 1.5 circle

= 66a + 162b + r2 – area of 1.5 circle

Area of 4 square + area of 1.5 circle = 66a + 162b + r2

Area of 1.5 circle = 66a + 162b + r2 – area of 4 square

= 66a + 162b + r2 – 4(4r2)

= 66a + 162b – 15r2

Use of a & b value = 66(0.125√3) r2 + 162(0.25 – 0.125√3) r2 – 15r2

= (8.25√3) r2 + (40.5 – 20.25√3) r2 – 15r2

= (25.5 - 12√3) r2

Area of circle = (25.5 - 12√3) r2 /1.5

= (17 - 8√3) r2

Total Area of(3 + 4) square = (- 8a + 312b – 12c + 16d) + (66a + 162b – 60c – 14d)

= 58a + 474b – 72c + 2d

= 58a + 474b + (area of 6 inscribed dodecagon - area of 6 circle = - 72c)

+ (Area of 0.5 square – area of 0.5circles = 2d)

= 58a + 474b + 6(3r2) + 0.5(4r2) - area of (6 + 0.5) circle

= 58a + 474b + 20r2 – area of 6.5 circle

Area of 7 square + area of 6.5 circle = 58a + 474b + 20r2

Area of 6.5 circle = 58a + 474b + 20 r2 – area of 7square

= 58a + 474b + 20r2 – 7(4r2)

= 58a + 474b – 8r2

Use of a & b value = 58(0.125√3) r2 + 474(0.25 – 0.125√3) r2 – 8r2

= (7.25√3) r2 + (118.5 – 59.25√3) r2 – 8r2

= (110.5 - 52√3) r2

Area of circle = (110.5 - 52√3) r2 /6.5

= (17 - 8√3) r2

12c = Area of circle – area of inscribed dodecagon

= (17 - 8√3) r2 – 3r2 = (14 - 8√3) r2 3c = (14 - 8√3) r2 /4 = (3.5 - 2√3) r2 = 14b – 2a = 3c

4d = area of square – area of circle

= 4r2 - (17 - 8√3) r2 = (8√3 - 13) r2 d = (8√3 - 13) r2 /4 = (2√3 – 3.25) r2 = 3a – 13b = d

a = (0.125√3) r2 b = (0.25 – 0.125√3) r2 3c = (3.5 - 2√3) r2 d= (2√3 – 3.25) r2

Area of square = (12a + 12b + 12c + 4d) + 54(a + b – 3c – d) more examples

= (12a + 12b + 12c + 4d) + (54a + 54b - 162c - 54d)

= 66a + 66b - 150c - 50d equation no. 1

= (12a + 12b + 12c + 4d) + 78(14b - 2a - 3c)

= (12a + 12b + 12c + 4d) + (1092b - 156a - 234c)

= - 144a + 1104b - 222c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 86(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (258a – 1118b – 86d)

= - 246a + 1130b + 12c + 90d equation no. 3

Area of 3 square = (66a + 66b - 150c - 50d) + (- 144a + 1104b - 222c + 4d) + (- 246a + 1130b + 12c + 90d)

= - 324a + 2300b - 360c + 44d

= - 324a + 2300b + (area of 30 inscribed dodecagon – area of 30 circle = - 360c)

+(Area of 11 square -area of 11 circles = 44d)

= - 324a + 2300b + 30(3r2) + 11(4r2) - area of (30 + 11) circle

= - 324a + 2300b + 90r2 + 44r2 - area of 41 circle

= - 324a + 2300b + 134r2 - area of 41 circle

Area of 3 square + Area of 41 circle = - 324a + 2300b + 134r2

Area of 41 circle = - 324a + 2300b + 134r2 – area of 3 square

= - 324a + 2300b + 134r2 – 3(4r2)

= - 324a + 2300b + 122r2

= - 324(0.125√3) r2 + 2300(0.25 – 0.125√3) r2 + 122r2

= (- 40.5√3) r2 + (575 – 287.5√3) r2 + 122r2

= (697 - 328√3) r2

Area of circle = (697 + 328√3) r2/41

= (17 - 8√3) r2

Area of square = (12a + 12b + 12c + 4d) - 142(a + b – 3c – d)

= (12a + 12b + 12c + 4d) - (142a + 142b - 426c - 142d)

= - 130a - 130b + 438c + 146d equation no. 1

= (12a + 12b + 12c + 4d) + 274(14b - 2a - 3c)

= (12a + 12b + 12c + 4d) + (3836b - 548a - 822c)

= - 536a + 3848b - 810c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 396(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (1188a – 5148b – 396d)

= - 1176a + 5160b + 12c + 400d equation no. 3

Area of 3 square = (- 130a - 130b + 438c + 146d) + (- 536a + 3848b - 810c + 4d) + (- 1176a + 5160b + 12c + 400d)

= - 1842a + 8878b - 360c + 550d

= - 1842a + 8878b + (area of 30 inscribed dodecagon – area of 30 circle = - 360c)

+(Area of 137.5 square -area of 137.5 circles = 550d)

= - 1842a + 8878b + 30(3r2) + 137.5(4r2) - area of (30 + 137.5) circle

= - 1842a + 8878b + 90r2 + 550r2 - area of 167.5 circle

= - 1842a + 8878b + 640r2 - area of 167.5 circle

Area of 3 square + Area of 167.5 circle = - 1842a + 8878b + 640r2

Area of 167.5 circle = - 1842a + 8878b + 640r2 – area of 3 square

= - 1842a + 8878b + 640r2 – 3(4r2)

= - 1842a + 8878b + 628r2

= - 1842(0.125√3) r2 + 8878(0.25 – 0.125√3) r2 + 628r2

= (- 230.25√3) r2 + (2219.5 – 1109.75√3) r2 + 628r2

= (2847.5 - 1340√3) r2

Area of circle = (2847.5 + 1340√3) r2/167.5

= (17 - 8√3) r2

Area of square = (12a + 12b + 12c + 4d) + 423(a + b – 3c – d)

= (12a + 12b + 12c + 4d) + (423a + 423b - 1269c - 423d)

= 435a + 435b - 1257c - 419d equation no. 1

= (12a + 12b + 12c + 4d) + 741(14b - 2a - 3c)

= (12a + 12b + 12c + 4d) + (10374b - 1482a - 2223c)

= - 1470a + 10386b - 2211c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 985(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (2955a – 12805b – 985d)

= - 2943a + 12817b + 12c + 989d equation no. 3

Area of 3 square = (435a + 435b - 1257c - 419d) + (- 1470a + 10386b - 2211c + 4d) + (- 2943a + 12817b + 12c + 989d)

= - 3978a + 23638b - 3456c + 574d

= - 3978a + 23638b + (area of 288inscribed dodecagon – area of 288 circle = - 3456c)

+(Area of 143.5 square -area of 143.5 circles = 574d)

= - 3978a + 23638b + 288(3r2) + 143.5(4r2) - area of (288 + 143.5) circle

= - 3978a + 23638b + 864r2 + 574r2 - area of431.5 circle

= - 3978a + 23638b + 1438r2 - area of 431.5 circle

Area of 3 square + Area of 431.5 circle = - 3978a + 23638b + 1438r2

Area of 431.5 circle = - 3978a + 23638b + 1438r2 – area of 3 square

= - 3978a + 23638b + 1438r2 – 3(4r2)

= - 3978a + 23638b + 1426r2

= - 3978(0.125√3) r2 + 23638(0.25 – 0.125√3) r2 + 1426r2

= (- 497.25√3) r2 + (5909.5 – 2954.75√3) r2 + 1426r2

= (7335.5 - 3452√3) r2

Area of circle = (7335.5 + 3452√3) r2/431.5

= (17 - 8√3) r2

Area of square = (12a + 12b + 12c + 4d) + 285693(a + b – 3c – d)

= (12a + 12b + 12c + 4d) + (285693a + 285693b - 857079c - 285693d)

= 285705a + 285705b - 857067c - 285689d equation no. 1

= (12a + 12b + 12c + 4d) + 572867(14b - 2a - 3c)

= (12a + 12b + 12c + 4d) + (8020138b - 1145734a - 1718601c)

= - 1145722a + 8020150b - 1718589c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 985246(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (2955738a – 12808198b – 985246d)

= - 2955726a + 12808210b + 12c + 985250d equation no. 3

Area of 3 square = (285705a + 285705b - 857067c - 285689d) + (- 1145722a + 8020150b - 1718589c + 4d) + (- 2955726a + 12808210b + 12c + 985250d)

= - 3815743a + 21114065b - 2575644c + 699565d

= - 3815743a + 21114065b + (area of 214637inscribed dodecagon – area of 214637 circle = - 2575644c)

+(Area of 174891.25 square -area of 174891.25 circles = 699565d)

= - 3815743a + 21114065b + 214637(3r2) + 174891.25(4r2) - area of (214637 + 174891.25) circle

= - 3815743a + 21114065b + 643911r2 + 699565r2 - area of389528.25 circle

= - 3815743a + 21114065b + 1343476r2 - area of 389528.25 circle

Area of 3 square + Area of 389528.25 circle = - 3815743a + 21114065b + 1343476r2

Area of 389528.25 circle = - 3815743a + 21114065b + 1343476r2 – area of 3 square

= - 3815743a + 21114065b + 1343476r2 – 3(4r2)

= - 3815743a + 21114065b + 1343464r2

= - 3815743(0.125√3) r2 + 21114065(0.25 – 0.125√3) r2 + 1343464r2

= (- 476967.875√3) r2 + (5278516.25 – 2639258.125√3) r2 + 1343464r2

= (6621980.25 - 3116226√3) r2

Area of circle = (6621980.25 - 3116226√3) r2/389528.25

= (17 - 8√3) r2

Area of square = (12a + 12b + 12c + 4d) + 6(a + b – 3c – d)

= (12a + 12b + 12c + 4d) + (6a + 6b – 18c – 6d)

= 18a + 18b – 6c - 2d equation no. 1

= (12a + 12b + 12c + 4d) - 18(14b – 2a – 3c)

= (12a + 12b + 12c + 4d) + (- 252b + 36a + 54c)

= 48a - 240b + 66c + 4d equation no. 2

= (12a + 12b + 12c + 4d) - 18(3a – 13b – d)

= (12a + 12b + 12c + 4d) - (54a – 234b – 18d)

= - 42a + 246b + 12c + 22d equation no. 3

Area of 3 square = (18a + 18b – 6c - 2d) + (48a - 240b + 66c + 4d) + (- 42a + 246b + 12c + 22d)

= 24a + 24b + 72c + 24d

= 24a + 24b + (area of 6 circle – area of 6 inscribed dodecagon = 72c)

+ (Area of 6 square – area of 6 circles = 24d)

= 24a + 24b – 6(3r2) + 6(4r2) + area of (6 – 6) circle

= 24a + 24b – 18r2 + 24r2 – area of 0circle

= 24a + 24b + 6r2

= 24(0.125√3) r2 + 24(0.25 – 0.125√3) r2 + 6r2

= (3√3) r2 + (6 – 3√3) r2 + 6r2

= 12r2

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