Dave L Wrote:
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> It's certainly possible that residual information
> of value is contained in the problem that is not
> part of the main exercise.


Hi Dave:
Actually this problem reminds me of Da Vinci
The answers are the beginning of the problem...to be read in reverse.

You must start with the 7 '2 '4 '8 (7.875), consider it as the side of a square and multiply it by four. Your answer puts you onto the right track.

The second portion of the math uses the number 4'2 (4-1/2) and the author selects the 1 and 1/4 amounts to total 5'2'8 (5-5/8).
But you must realize that 4-1/2 does not exist on the illustration...so where did it come from?

Answer:
He took the 2'4 (2-1/4) and doubled it then took the number 5 and divided it by 4 (1-1/4) to get the half area of a right-angle triangle.

Area of a (2-1/4 x 5) rectangle = area of a (4-1/2 x 2-1/2) rectangle. The "half" area = (4-1/2 x 1-1/4) = 5-5/8...as the author states.

He proves this by multiplying the 4-1/2 by 1, 2, 1/2, and 1/4 but neither of the 2 or 1/2 values are required...at least that is what we believe...but they are an integral part of the puzzle to be reckoned with later.

Back to the 4-1/2 and 1-1/4 selected by the author.
Multiplying them together is definitely an area calculation, but these numbers can also be written as 2-1/4 x 2-1/2 and now you may begin to see what is going to transpire. This is the area of a "triangle" that is produced from a 2-1/4 x 5 based triangle (area = [h x b]/2 = 2-1/4 x 5/2)...the two measures of 5 and 2-1/4 in the illustration are now included in the calculations...Chace couldn't see it.

But you must constantly question why these two numbers were selected...they didn't fall out of the sky...there is a deeper meaning to the values.

If you draw a right-angle triangle of sides 5 and 2-1/4 then the hypotenuse is equal to the square root of 30 actually [30.06] and that equals 5.483.
The "perimeter" of this triangle is (5.483 + 5 + 2.25) = 12.733....!
Don't know if you recognize the number but 4/pi = 1.2732.

That's the first section of the puzzle and it gets better as you apply the answers to the illustration.

By the way.
7'2'4'8 x 4 as first stated is equal to 31.5 (10 x pi).

>
> The problem is operating within a cultural
> technical system, where repetition and tradition
> will also have had an impact.

Correct...Chase and all couldn't escape from theirs.


> so it is possible to
> consider that the factor 7 or 14 for the radius of
> the triange is of interest.

7 is indicated at the point of the triangle but not part of its dimensions.

> there is not detailed explanation
> for this problem in Gillings, and I don't follow
> Chace's transcription

There are no directions and that's the beauty of the whole problem.
You are given sufficient to solve, but you must first resolve. A perfect example of a problem given to a student to work on until all answers are discovered...a duplicate of the Giza dilemma to the "t".

> on the pages that I have
> detailed in the photograph's of the original
> publication below, which I obtained earlier in the
> year.

Dave:
Any possibility of e-forwarding these copies to me...most appreciated if you can...need book references also.

> In particular, I don't understand why the
> scribe has written 6 beside two transverse lines
> that are clearly not equal in length.

'ah...now you are thinking...the two sixes are for a reason.

> I don't even understand how to do the problem in modern math to
> be honest,

Modern mathematicians are shooting in the dark...if they follow the math as set out on the papyrus then they can "only guess"

> At least you have confirmed to me that the problem
> is about areas, setats, and not lengths!

Actually it involves linear, area and volume measures of a sphere.

Best.
Clive