46 Lessons in Early Geometry, part 5/10 / provisional version in freestyle English / a corrected version will follow in March, April or May (hopefully) / Franz Gnaedinger / February 2003 / www.seshat.ch

 

early geometry 1 / early geometry 2 / early geometry 3 / early geometry 4 / early geometry 5 / early geometry 6 / early geometry 7 / early geometry 8 / early geometry 9 / early geometry 10

 

 

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 / polygon 1a / polygon 1b / polygon 1c / polygon 1d

 

 

Lesson 20

 

Let a square measure 10 by 10 royal cubits or 70 by 70 palms or 280 by 280 fingers. The diagonals measure practically 99 palms or 396 fingers.

 

The radius of the inscribed circle measures 5 royal cubits or 35 palms or 140 fingers.

 

Combine the square with a grid of 10x10 small squares measuring 1 by 1 royal cubit or 7 by 7 palms or 28 by 28 fingers each:

 

 

. . . . . d . . . . .

. . e . . . . . c . .

. f . . . . . . . b .

. . . . . . . . . . .

. . . . . . . . . . .

g . . . . + . . . . a

. . . . . . . . . . .

. . . . . . . . . . .

. h . . . . . . . l .

. . i . . . . . k . .

. . . . . j . . . . .

 

 

Twelve points of the grid mark the circumference of the circle: a b c d e f g h i j k l. Four points define the axes: a-g and d-j, while the distances of the eight remaining points b c e f h i k l from the axes and from the center measure 3-4-5 royal cubits or 21-28-35 palms or 84-112-140 fingers each.

 

The twelve points define four short arcs: b-c e-f h-i k-l, and eight longer ones: a-b c-d d-e f-g g-h i-j j-k. The short arcs measure nearly 40 fingers each (exact number 39.731...), the longer arcs measure practically 90 fingers each (exact number 90.090...), yielding a circumference of 880 fingers or 220 palms:

 

4 x 40 fingers + 8 x 90 fingers = 880 fingers = 220 palms

 

Divide the circumference by the diameter 70 palms and you obtain a fine approximate value for pi:

 

220 p / 70 p = 22/7 = 3 1/7 = '7 of 22 = 3 '7

 

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 / polygon 1a / polygon 1b / polygon 1c / polygon 1d

 

 

Lesson 21

 

 

. . . . . d . . . . .

. . e . . . . . c . .

. f . . . . . . . b .

. . . . . . . . . . .

. . . . . . . . . . .

g . . . . + . . . . a

. . . . . . . . . . .

. . . . . . . . . . .

. h . . . . . . . l .

. . i . . . . . k . .

. . . . . j . . . . .

 

 

If you connect the 12 points a-b-c-d-e-f-g-h-i-j-k-l-a with straight lines you obtain a polygon with four short and eight long sides.

 

The sides are diagonals to the square 1x1 and rectangle 1x3:

 

1x1 + 1x1 = 2

1x1 + 3x3 = 10 = 2 x 5

 

The short side is given by the square root of 2. The long side is given by the product of the square roots of 2 and 5. These roots can be approximated by means of my number columns:

 

1 1 2 1 1 5

2 3 4 2 6 10

5 7 10 1 3 5

.. .. .. 4 8 20

2 4 10

1 2 5

3 7 15

10 22 50

5 11 25

16 36 80

8 18 40

4 9 ..

 

The arcs of the circle are slightly longer than the sides of the polygons. We may hope to counterbalance this by choosing values for the square roots of 2 and 5 that are slightly bigger than the actual values:

 

10/7 x 10/7 = 100/49 = 2 plus 2/49

9/4 x 9/4 = 81/16 = 5 plus 1/16

 

Using the values 10/7 and 9/4 we obtain the following periphery or circumference:

 

4 x 10/7 plus 8 x 10/7 x 9/4 equals 220/7

 

Divide this number by the diameter 10 and you find again

 

22/7 or 3 1/7 or '7 of 22 or 3 '7

 

The same value can be obtained by means of a number sequence. Write 4 above 1, and add repeatedly 3 above 1:

 

4 (plus 3) 7 10 13 16 19 -22- 25 28

1 (plus 1) 2 3 4 5 6 7 8 9

 

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 / polygon 1a / polygon 1b / polygon 1c / polygon 1d

 

 

Lesson 22

 

Imagine a finer grid of 50 by 50 smaller squares. The radius of the inscribed circle measures 25 units. The circumference will pass 20 points of the grid, four of which mark the axes, while the distances of the remaining points from the axes and from the center are given by the triples 15-20-25 and 7-24-25.

 

The resulting polygon has 12 shorter and 8 longer sides.

 

The short sides are diagonals to the rectangle 1x7 or to the square 5x5. The long sides are diagonals to the rectangle 4x8:

 

1x1 + 7x7 = 50 = 5 x 5 x 2

5x5 + 5x5 = 50 = 5 x 5 x 2

 

4x4 + 8x8 = 80 = 4 x 4 x 5

 

The side lengths are the square roots of 5x5x2 and 4x4x5:

 

sr 5x5x2 = 5 sr2 sr 4x4x5 = 4 sr5

 

The polygon has the following periphery:

 

12 sr2 plus 8 sr5

 

The arcs of the circle are again slightly longer than the sides of the polygon. Let us compensate this by choosing the values 17/12 and 9/4 for the square roots of 2 and 5:

 

17/12 x 17/12 = 289/144 = 2 plus 1/144

9/4 x 9/4 = 81/16 = 2 plus 1/16

 

The periphery or circumference measures then

 

12 x 17/12 + 8 x 9/4 = 157 units

 

Divide this result by the diameter 50 units:

 

157 / 50 = 3.14 = 3 1/10 1/25 = 3 '10 '25

 

The first polygon yielded 22/7, now we obtain the slightly smaller ratio 157/50. These two values are linked by another number sequence. Write 3 above 1 and add repeatedly 22 above 7:

 

3 (plus 22) 25 47 69 91 113 135 -157- 179 201

1 (plus 7) 8 15 22 29 36 43 50 57 64

 

223 245 267 289 311 333 355 377

71 78 85 92 99 106 113 120

 

 

25/8 = 3 '8 --- 113/36 = 3 '9 '36 --- 157/50 = 3 '10 '25

 

--- 201/64 = 3 '8 '64 --- 311/99 = 3 '9 '33 ---

 

377/120 = 3 '8 '60 = 3 '10 '24 --- --- --- 22/7 = 3 '7

 

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 / polygon 1a / polygon 1b / polygon 1c / polygon 1d

 

 

Lesson 23

 

What if we proceed to ever finer grids?

 

Consider the grid 250x250. The radius of the inscribed circle measures 125 units, while the circumference will be marked by 4+8+8+8 = 28 points of the grid, namely by the 4 ending points of the axes, and by 8+8+8 points provided by the triples

 

75-100-125 35-120-125 44-117-125

 

Using the still finer grid 1250x1250 you will obtain a circle of radius 625. The circumference will be marked by 4+8+8+8+8 = 36 points of the grid, namely by the four ending points of the axes, and by 8+8+8+8 points provided by the triples

 

375-500-625 175-600-625 220-585-625 336-527-625

 

The resulting polygons have sides of two or three different lengths, which are again integer multiples either of the square root of 2, or of the square root of 5, or of their product.

You can go on refining the grid by a factor of 5 as long as you like and will always obtain polygons with the same properties.

 

The ever rounder polygons are built on a sequence of 1, 2, 3, 4, 5, 6 ... triples, beginning with the Sacred Triangle 3-4-5:

 

3-4-5 15-20-25 75-100-125 375-500-625 ...

7-24-25 35-120-125 175-600-625 ...

44-117-125 220-585-625 ...

336-527-625 ...

...

 

If you know a triple a-b-c and wish to find the next one you may calculate the following terms:

 

+- 4a +- 3b +- 3a +-4b 5c

 

The long terms yield four values each. Choose the positive one that ends on 1, 2, 3, 4, 6, 7, 8 or 9 (neither on 5 nor on 0):

 

fourth triple 336-527-625

 

+- 4x336 +- 3x527 = +- 2925 or +- 237 --- 237

+- 3x336 +- 4x527 = +- 3116 or +- 1100 --- 3116

 

5 x 625 = 3125 fifth triple 237-3116-3125

 

+- 4x237 +- 3x3116 = +- 10296 or +- 8400 --- 10296

+- 3x237 +- 4x3116 = +- 11753 or +- 13175 --- 11753

 

5 x 3125 = 15625 sixth triple 10296-11753-15625

 

 

Every triple can be given by a generator. No number is allowed to be a multiple of 5; cc means: add the first numbers of the lines crosswise in order to find the first numbers of the next generator; sa means: subtract the first numbers of the upper line and add the first numbers of the lower line:

 

c 1 1 2

c 2 3 4 1x3=3, 2x2=4, 1x1+2x2=5

 

cc 1+2=3, 1+3=4

 

s 3 1 6

a 4 7 8 1x7=7, 4x6=24, 3x3+4x4=25

 

sa 3-1=2, 4+7=11

 

s 2 9 4

a 11 13 22 9x13=117, 11x4=44, 2x2+11x11=125

 

sa 9-2=7, 11+13=24

 

c 7 17 14

c 24 31 48 17x31=527, 24x14=336, 7x7+24x24=625

 

cc 7+31=38, 17+24=41

 

c 38 3 76

c 41 79 82 3x79=237, 41x76=3116, 3125

 

cc 3+41=44, 38+79=117

 

s 44 73 88

a 117 161 234 73x161=11753, 117x88=10296, 15625

 

cc 73-44=29, 117+161=238

 

and so on

 

 

key figure 1 / key figure 2 / polygon 1 / polygon 2 / polygon 3 / polygon 4 / polygon 5 / polygon 1a / polygon 1b / polygon 1c / polygon 1d

 

 

Lesson 24

 

There is also a geometric way of how to find my polygons:

Imagine a Sacred Triangle 3-4-5

 

let it rotate around the right angle, thus you obtain a cross

 

draw a circle around the center of the cross, thus you obtain the first four corners

 

now let the Sacred Triangle rotate around the smaller angles, starting from both axes

and moving in both directions: thus you obtain the rays of the 8+8+8+8+ ... further corners

 

The lines of the cross and the further rays divide the slowly rounding polygon into 4+8+8+8+8+... triangles whose heights approximate the radius of the circle, while the bases converge to the circumference.

 

The area of a single triangle is found as follows:

 

'2 x height x base

 

Calculating the sum of equally high triangles:

 

'2 x height x sum of the bases

 

In the case of the ever rounder polygon made up of ever more slender triangles we obtain the limit

 

maximal area = '2 x radius x circumference

 

The first and second polygon provided the values 22/7 and 157/50 for pi. Using them we find the following circumference and area of the circle:

 

circumference = diameter x 22/7 = 2 x radius x 22/7

 

area = '2 x radius x 2 x radius x 22/7

 

= radius x radius x 22/7 = diameter x diameter x 11/14

 

 

circumference = diameter x 157/50 = 2 x radius x 22/7

 

area = '2 x radius x 2 x radius x 157/70

 

= radius x radius x 157/50 = diameter x diameter x 157/200

 

 

The side length of the unit square measures 1 unit, and so does the diameter of the inscribed circle. The periphery of the unit square measures 4 units, while the circumference of the inscribed circle measures 22/7 or 157/50 units, yielding the ratios

 

22/7 divided by 4 equals 11/14 or 0.7857142...

 

157/50 divided by 4 equals 157/200 or 0.785

 

The area of the unit square measures 1 square unit, while the area of the inscribed circle measures

 

1 x 1 x 11/14 = 11/14 or 0.7857142... square units

 

1 x 1 x 157/200 = 157/200 or 0.785 square units

 

 

The area of polygon 1 in the grid 10x10 = 100 squares measures 74 square units. Dividing 74 by 100 we obtain 0.74. The area of polygon 2 in the grid 50x50 = 2500 smaller squares measures 1930 smaller square units. Dividing 1930 by 2500 we obtain 0.772, what is already a good deal closer to 157/200 or 11/14 (pi/4).

 

 

PS The above triples can also be found by means of a number column. Begin with 1 and 1, use a factor of minus 4, and consider every second line:

:

1 1 -4

2 -3 -8

-1 -11 4

-12 -7 48

-19 41 76

22 117 -88

139 29 -556

168 -527 -672

...................

 

 

x = -3 y = -8/2 = -4 r = 5 233.130 degrees

x = -7 y = 48/2 = 24 r = 25 106.260 degrees

x = 117 y = -88/2 = -44 r = 125 339.390 degrees

x = -527 y = -672/2 = -336 r = 625 212.520 degrees

............................................................

 

(The ratios provided by each line of the above number column can bee seen as the intersecting points of a rotating cross with the x-axis of the reals. The center of the cross lies on the y-axis of the imaginary numbers, marking 2i, while the cross rotates counter-clockwise, by an angle of arctan 1/2 = 26.5605 degrees.)

 

 

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