The Verification of Archimedes' Pi - cakravala.in.coocan.jpcakravala.in.coocan.jp/archimedespi.pdf · 1 The Verification of Archimedes' Pi July 31, 2000 Ryoji Tatsukawa 1. Introduction

1

The Verification of Archimedes' PiJuly 31, 2000

Ryoji Tatsukawa

1. Introduction

The famous genius Archimedes played a very active role in science in about 200B.C. He calculated pi, which was correct to two places of decimals, 3.14.

It was so carefully calculated that a more accurate one could not devised for 1,400years after his death.

I verified his calculations in the interval of my study and admired his talent ofcalculation, accuracy, cautiousness and patience.

Without calculators, he had completed such complicated works. Thanks to a

computer, I have rather easily done this work. I affixed the reference marks to theupper right of quotations and illustrations to trace their original sources.

2. Inequality:

According to some books, this inequality is used in ancient Greek timesto get the value of after A is modified as follows:

Proof: Setting

we get

a ± 2a ± 1b

< a2 ± b < a ± 2ab

…………(*)

A

A = a2 ± b ( b > 0 )

1), 2), 3)

a2 ± b = a ± c （a：positive integer c：positive decimal）a2 ± b = a2 ± 2ac + c2 > a2 ± 2ac∴ ±b > ±2ac

As a > 0, then ± 2ab

> ±c

n)

The Verification of Archimedes' Pi

2

that is,

Therefore,

Note: They say that the ancient Babylonians approximated as follows:

Proof:

But the equality holds if and only if

3. Approximating square roots by ArchimedesThere is no descriptions how Archimedes approximated square roots except for

its results. However, some historians speculate that he might use the abovementioned inequality * . Let us try the calculation with this inequality.( )

∴ a2 ± b = a ± c < a ± 2ab

…………………………①

a2 ± b = a ± c 2 = a2 ± 2ac + c2

As 0 < c < 1, then c2 < c.

a2 ± b < a2 ± 2ac + c = a2 ± c 2a ± 1

∴ ±b < ±c 2a ± 1

As a ≧ 1, then 2a ± 1 > 0

∴ ± 2a ± 1b

< ±c

a ± 2a ± 1b

< a2 ± b < a ± 2ab

1)a2 + b ≒ a +

2ab

A

a2 + b = a ×a2 + b

a (geometric mean)

≦

a +a2 + b

a2

(arithmetic mean)

= a +2ab

b = 0.

by using ① and ②

a ± 2a ± 1b

< a ± c = a2 ± b …………………………②


3

1 Approximating( )

( )a

We set 2 for approximate value of because is between 1 andBy using inequality * , let so that( )

( )b

By using a , we get that is,( )

So, we are able to set 5 for approximate value ofBy using , let so that we get(＊)

( ) …………………………c

By using b , we get that is,( )

So, we are able to set 26 for approximate value ofBy using c , let so that we get( )

3

3 = 22－1

2 － 2×2－11

< 22－1 < 2 － 2×21

5 < 27 <214

.

5 +2×5 + 1

2< 52 + 2 < 5 +

2×52

leading to5711

< 3 3 <265

∴1911

< 3 <2615

265153

< 3 <1351780

(a)

675 .

675 = 262－1

26 － 2×26－11

< 262 －1 < 26 － 2×261

leading to132551

< 15 3 <135152

27.

∴53

< 3 <74

53

< 3 <74

27 = 52 + 2

3 1),3)

3

1911

< 3 <2615

28511

< 675 < 26.

4 .

5×33

< 3 3 <7×3

4;

19×1511

< 15 3 <26×15

15;


4

Note: 1 Approximation is made to both sides alternately.)

2 is correct to 4 places of decimals;)

( ) …………………………2

Note: He selected rather than

( ) …………………………3

∴265153

< 3 <1351780

1.7320261 < 3 < 1.732052.

349452 > 59118

(b)

349452 = 5912 + 169

> 591 +2×591 + 1

169

= 591 +1183169

= 59117

> 59118

17

.18

13739433364

> 117218 (c)

13739433364

=18

87932385

=18

93772 + 4256

>18

9377 +2×9377 + 1

4256

=18

9376 +2301118755

> 117218

3


5

( )4

…………………………

Note: He selected

5472132161

> 233914 (d)

5472132161

=14

87554113

=14

93572 + 664

>14

9357 +2×9357 + 1

664

=14

9356 +1937918715

> 233914

(5) 9082321 < 301334 …………………………(e)

9082321 = 30142－1875

< 3014－ 2×30141875

= 3014－14

= 301334

3380929 = 18392－992

< 1839－ 2×1839992

= 183826863678

< 1838119

= 3014－18756028

(6) 3380929 < 1838119

…………………………(f)

14

.


6

Note: He selected rather than

( ) …………………………7

( ) …………………………8

1018405 < 100916

(g)

1018405 = 10092 + 324

< 1009 +2×1009

324

< 100916

4069284361

< 201714 (h)

4069284361

=16

146494225

=16

121022 + 35821

<16

12102 +2×12102

35821

=16

12102 +3582124204

<16

12102 +32

= 201714

119 3

4.


7

4. Measuring pi

There are fifteen writings which Archimedes wrote. "The Measurementof the Circle" is one of them. Referring to circumference of circle, it says as follows."It is shorter than three times of its diameter and an excess, one-seventh of itsdiameter" and "It is longer than three times of its diameter and an excess, tenseventy-firsts of its diameter."

This means that is,We realize that he calculated pi correctly to two places of decimals.

1 An outline of the calculation( )

To make a start, we draw a circumscribed regular hexagon to a given circle O.

Doubling its sides of circumscribed regular polygons, from hexagon, dodecahedron,24, 48 to a 96 sided-polygon, we find that each perimeter is decreasingly gettingcloser and closer to of the diameter.

Similarly, we draw a inscribed regular hexagon to a circle O. Doubling its sides

of inscribed regular polygon to a 96 sided regular polygon, we find that eachperimeter is, increasingly, getting closer and closer to of the diameter.

We conclude that

Let us verify his calculations.

2 Calculating perimeter( )

i Circumscribed 96 sided-polygon( )

The next figure shows a circle with a center O and a radius OB; and the parts

31071

< p < 317

,

317

∴ 31071×

(Diameter of circle O）< p×(Diameter of circle O）< 317×

(Diameter of circle O）

31071

< p < 317

.

3.1408 < p < 3.1429.

1),2),3),4)

31071


8

of five regular polygons circumscribed out of circle at a point B.

Because △A1OB is a half part of regular triangle, by using (a), we get

A1BBO

= 3 >265153

,A1OA1B

= 2 …………………………①

Because OA1 is a bisector of ∠A1OB, we get

A

A

A

A

A

1

2

3

4

5

B

B

B

B

B

…

…

…

…

…

a

a

a

a

a

half

half

half

half

half

side

side

side

side

side

of

of

of

of

of

regular

regular

regular

regular

regular

circumscribed

circumscribed

circumscribed

circumscribed

circumscribed

hexagon

dodecahedron

24

48

96

sided

sided

sided

out

polygon

polygon

polygon

of

out

the

out

out

out

of

circle

the

of

of

of

the

the

the

circle

circle

circle

circle


9

The following results hold in general.

∴ A2BBO

=A1O + BO

A1B

=A1OA1B

+A1BBO

(：)A2OA2B

2

=

=

A2B2 + BO2

A2B2

1 +A2BBO 2

(Pythagorean theorem) Note:

cos h1

Let ∠A22

=

=

1

h,

+

then

tan2 h

(・)AnBBO

=An~1OAn~1B

+An~1BBO

(：)AnOAnB

2

= 1 +AnBBO 2

(・)A2BBO

=A1OA1B

+A1BBO

by using ①

>265153

+ 2

=571153

(：)A2OA2B

2

= 1 +A2BBO 2

> 1 +571153

2

=349450

1532

∴A2OA2B

>349450153

by using (b)

(・)A1OBO

=A1A2

A2B∴

A1O + BOBO

=A1A2 + A2B

A2B(componendo)


10

>591

18

153

∴A2BBO

>571153 ,

A2OA2B

>591

18

153 …………………………②

(：)A3OA3B

2

= 1 +A3BBO 2

(・)A3BBO

=A2OA2B

+A2BBO

by using ②

>591

18

153+

571153

=1162

18

153

> 1 +1162

18

153

2

=137394

3364

1532

∴A3OA3B

>137394

3364

153by using (c)

>1172

18

153

∴A3BBO

>1162

18

153 ,A3OA3B

>1172

18

153 …………………………③


11

(・)A4BBO

=A3OA3B

+A3BBO

by using ③

>1162

18

153+

117218

153

=2334

18

153

(：)A4OA4B

2

= 1 +A4BBO 2

> 1 +2334

14

153

2

=5471632

161

1532

∴A4OA4B

>

5471632161

153by using (d)

>2339

14

153

∴A4BBO

>2334

14

153 ,A4OA4B

>2339

14

153 …………………………④

(・)A5BBO

=A4OA4B

+A4BBO

by using ④

>

2339 14

153+

233414

153

=

4673 12

153


12

Setting L as the perimeter of 96-sided circumscribed polygon, then we get thefollowing results.

As the perimeter of circle is

ii Inscribed 96-sided regular polygon( )

The next figure shows a circle with a center O and a radius OB, and parts offive regular polygons inscribed in the circle at a point B.

∴ A5B <4673

12

153BO …………………………⑤

L = 2×A5B×96 by using ⑤

< 2×4673

12

153OB×96

= 313359347 ×

(2BO)

< 317×

(2BO)

p×(2BO),

p×(2OB) < 317×

(2OB)

∴ p < 317 …………………………(＊＊)


13

① ②By using and

The following results hold in general.

△A2BC∽△A1DC∽△A2BD

∴A2CA2B

=A1DA1C

=A2DA2B

…………………………①

Because DC is a bisector of ∠C in △A1BC,

A1CBC

=A1DBD

∴ A1DA1C

=BDBC

…………………………②

A2CA2B

=A1DA1C

=BDBC

…………………………③

(・)A2CA2B

=A1D + BDA1C + BC

（componendo）

=A1C + BC

A1B

(：)A2BBC 2

=A2B2 + A2C2

A2B2 = 1 +A2CA2B

2

Note:

then

Let

cos h1

∠A2 = h,

= 1 + tan2 h

(・)AnCAnB

=An-1CAn-1B

+An-1BBC

(：)AnBBC 2

= 1 +AnCAnB

2

Because △A1BO is a regular triangle,A1CA1B

= 3 <13517801

,A1BBC

= 2 …………………………④

A

A

A

A

A

1

2

3

4

5

B…

B…

B…

B…

B…

a

a

a

a

a

side

side

side

side

side

of

of

of

of

of

regular

regular

regular

regular

regular

inscribed

inscribed

inscribed

inscribed

inscribed

hexagon

dodecahedron

24-sided

96-sided

48-sided

in

polygon

polygon

polygon

the

in

circle

the

in

in

in

the

the

the

circle

circle

circle

circle


14

(・)A2CA2B

=A1CA1B

+A1BBC

<1351780

+ 2

=2911780

(：)A2BBC 2

= 1 +A2CA2B

2

< 1 +2911780

2

=9082321608400

∴A2BBC

=9082321

780by using (e)

<3013

34

780

A2CA2B

<2911780 , A2B

BC<

301334

780 …………………………⑤

(・)A3CA3B

=A2CA2B

+A2BBC

<2911780

+3013

34

780

=1823240

(：)A3BBC 2

= 1 +A3CA3B

2

< 1 +1823240

2


15

=3380929

2402

∴A3BBC

<3380929

240by using (f)

<

1838119

240

A3CA3B

<1823240 , A3B

BC<

1838119

240 …………………………⑥

(・)A4CA4B

=A3CA3B

+A3BBC

<1823240

+

1838119

240

=100766

=3661

119

240

(：)A4BBC 2

= 1 +A4CA4B

2

< 1 +100766

2

∴A4BBC

<1018405

66by using (g)

<1009

16

66

=1018405

662


16

Because the perimeter of 96-sided regular inscribed polygon is we have

=2016

16

66

A4CA4B

<100766 , A4B

BC<

1009 16

66 …………………………⑦

(・)A5CA5B

=A4CA4B

+A4BBC

<100766

+1009

16

66

(：)A5BBC 2

= 1 +A5CA5B

2

< 1 +2016

16

66

2

=4069284

361

662

∴A5BBC

<4069284

361

66by using (h)

<2017

14

66

∴66×BC

201714

= A5B

p×BC,


17

By using and ,(＊＊) (＊＊＊)

Reference Books) アルキメデスを読む上垣渉著１９９９日本評論社1) 円周率の歴史平山諦著１９９０大阪教育図書2) ギリシャ数学史Ⅰ Ⅱ 平田寛訳１９５９共立出版3 ,

Thomas L. Heath, A MANUAL OF GREEK MATHEMATICS, OXFORD, 1931

) ギリシャの数学彌永昌吉著１９７９共立出版4

31071

< p < 317

p×BC = A5B×96

>2017

14

6336BC

=2017

14

6336BC

= 311378069

BC

> 31071

BC

∴ p > 31071 …………………………(＊＊＊)


Documents

The Verification of Archimedes' Pi - cakravala.in.coocan.jpcakravala.in.coocan.jp/archimedespi.pdf · 1 The Verification of Archimedes' Pi July 31, 2000 Ryoji Tatsukawa 1. Introduction