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Replies: 86   Last Post: Jan 28, 2013 5:19 AM

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 Franz Gnaedinger Posts: 330 Registered: 4/30/07
Posted: Dec 15, 2011 2:42 AM

I don't know what problem number 31 of the Rhind
Mathematical Papyrus has to do with Mayan astronomy,
but if I must I can also shed light on this problem.
In my opinion, the Rhind Mathematical Papyrus offers
problems that can be solved on several level. On the
first level, beginners learn how to handle unit fraction
more demanding problems, and on the highest level they
are being told about theoretical insights. RMP 31 on the
a fine example of Egyptian wit, plus a theoretical insight:

RMP 31 - a granary on a ring

33 divided by 1 "3 '2 '7 equals 14 '4 '56 '194 '388 '679
'776

Imagine a regular hexagon whose side measures 66 fingers.
Inscribe and circumscribe a circle. The two circles
form a ring. The radius of the outer circle measures 66
fingers. How long is the radius of the inscribed circle
in palms?

Consult the number column for the approximation of the
square root of 3 whose first lines are 1 1 3 / 2 4 6 /
1 2 3 / 3 5 9 / 3 5 9 / 8 14 24 / 4 7 12 / 11 19 33 /
30 52 90 / 15 26 45 / 41 71 123 / 112 194 336 /
56 97 168 ...

The side of the hexagon measures 66 fingers. Multiply 66
fingers by 168/97 and you obtain the diameter of the
inscribed circle in fingers. Multiply 33 fingers by
168/97 and you obtain the radius in fingers. Divide 33
fingers by 97/42 = 1 "3 '2 '7 and you obtain the radius
of the inscribed circle in palms:

33 fingers divided by 1 "3 '2 '7 equal 14 '4 '56 '97 '194
'388 '679 '776 palms

The area of the ring is given by the difference

area circumscribed circle minus area inscribed circle

The area of the circumscribed circle is found as follows:

Use the pi sequence

3/1 (plus 22/7) 25/8 47/15 ... 311/99

and the value '99 of 311 and you'll obtain

66 fingers x 66 fingers x '99 x 311 = 13,684 square
fingers

Now for the area of the inscribed circle. It measures

14 '4 '56 '194 '388 '679 '776 palms times 14 '4 '56 '194
'388 '679 '776 palms times pi

Is anyone prepared to carry out that multiplication ???

Ahmes would smile and offer a much simpler solution based
on a fine theorem:

Imagine a regular polygon of 3, 4, 5, 6, 7 ... equal
sides. The circumscribed circle and the inscribed circle
form a ring. Draw a circle around a side of the polygon.
Its area equals the area of the ring.

The side of the regular hexagon measures 66 fingers, the
radius of the circle around a side measures 33 fingers,
and the area of the ring measures

33 fingers x 33 fingers x '99 x 311 = 3,421 square fingers

The area of the outer circle measures 13,684 square
fingers, the area of the ring measures 3,421 square
fingers, and the area of the inner circle measures
13,684 - 3,421 = 10,263 square fingers. Comparing
these areas reveals the following proportions:

area inner circle / area ring / area outer circle
= 3 / 1 / 4

Build a granary on the ring. If the height measures
5 '2 royal cubits or 154 fingers, the volume of the
wall measures practically 24 cubic cubits, and the
capacity 72 cubic cubits or 108 khar or 540 quadruple-
hekat or 2160 hekat.

The secrets promised in the opening lines of the Rhind
Mathematical Papyrus can only be found by means of the
experimental approach to early mathematics.

Date Subject Author
11/17/11 Franz Gnaedinger
11/17/11 Milo Gardner
11/18/11 Franz Gnaedinger
11/18/11 Milo Gardner
11/19/11 Franz Gnaedinger
11/19/11 Milo Gardner
11/20/11 Franz Gnaedinger
11/20/11 Milo Gardner
11/20/11 Milo Gardner
11/21/11 Franz Gnaedinger
11/22/11 Franz Gnaedinger
11/22/11 Milo Gardner
11/23/11 Franz Gnaedinger
11/24/11 Franz Gnaedinger
11/24/11 Franz Gnaedinger
11/24/11 Franz Gnaedinger
11/24/11 Milo Gardner
11/25/11 Franz Gnaedinger
11/26/11 Franz Gnaedinger
12/2/11 Franz Gnaedinger
12/2/11 Milo Gardner
12/3/11 Franz Gnaedinger
12/4/11 Franz Gnaedinger
12/4/11 Milo Gardner
12/5/11 Franz Gnaedinger
12/5/11 Milo Gardner
12/7/11 Franz Gnaedinger
12/8/11 Milo Gardner
12/10/11 Franz Gnaedinger
12/12/11 Franz Gnaedinger
12/12/11 Milo Gardner
12/13/11 Franz Gnaedinger
12/13/11 Milo Gardner
12/15/11 Franz Gnaedinger
12/15/11 Milo Gardner
12/15/11 Milo Gardner
12/16/11 Franz Gnaedinger
12/16/11 Milo Gardner
12/18/11 Franz Gnaedinger
12/18/11 Milo Gardner
12/19/11 Franz Gnaedinger
12/20/11 Franz Gnaedinger
12/20/11 Milo Gardner
12/21/11 Franz Gnaedinger
12/22/11 Franz Gnaedinger
12/23/11 Franz Gnaedinger
12/24/11 Franz Gnaedinger
12/29/11 Franz Gnaedinger
1/2/12 Franz Gnaedinger
1/3/12 Milo Gardner
1/4/12 Franz Gnaedinger
11/28/11 Velev, Petyr
1/6/12 Franz Gnaedinger
1/6/12 Milo Gardner
1/9/12 Franz Gnaedinger
1/17/12 Franz Gnaedinger
1/19/12 Franz Gnaedinger
1/19/12 Milo Gardner
1/27/12 Franz Gnaedinger
2/10/12 Franz Gnaedinger
2/28/12 Franz Gnaedinger
3/2/12 Franz Gnaedinger
3/23/12 Franz Gnaedinger
3/24/12 Milo Gardner
4/9/12 Franz Gnaedinger
4/10/12 Franz Gnaedinger
4/13/12 Franz Gnaedinger
4/17/12 Franz Gnaedinger
4/18/12 Franz Gnaedinger
4/18/12 Franz Gnaedinger
5/5/12 Franz Gnaedinger
5/7/12 Franz Gnaedinger
5/7/12 Milo Gardner
5/8/12 Franz Gnaedinger
5/8/12 Milo Gardner
5/8/12 Franz Gnaedinger
5/8/12 Franz Gnaedinger
5/9/12 Franz Gnaedinger
5/10/12 Franz Gnaedinger
8/14/12 Franz Gnaedinger
1/13/13 Franz Gnaedinger
1/19/13 Franz Gnaedinger
1/23/13 Franz Gnaedinger
1/23/13 Franz Gnaedinger
1/24/13 Franz Gnaedinger
1/26/13 Franz Gnaedinger
1/28/13 Franz Gnaedinger