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
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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
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
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
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
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
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
Lesson 24
There is also a geometric way of how to find my
polygons:
Imagine
a Sacred Triangle
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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