A systematic search for other diagrammatic means by which the circle could be uniformly partitioned eventually would have led to the construction we see in Diagram 10.

In the course of an ancient scribe's trial and error probing, a square is drawn using the circle's radius as the length of its side. Next, this square is cut in half (line TA) creating two equivalent rectangles having side lengths of 1 and 1/2. No new information here so far, but something curious does occur when a diagonal is drawn in one of these rectangles (e.g., line CA).

With dividers set to the length CA, one finds that four of these lengths will apparently inscribe within the circle precisely 3/4ths of a rotation (clockwise from C to G, for instance). Although the length of CA can be computed via the Pythagorean Theorem (it equals 5/2 ), these results, though interesting, are not of immediate benefit in that they do not lead to a uniform partitioning of the full circle. They are, however, a sign that further investigation is warranted.

With a search underway for line lengths that could inscribe equal-sided polygons, Diagram 11 is the breakthrough construction that would sooner or later have been arrived at. In it, line CA has been swung down from A, to extend along line OH to a point B such that AB equals AC. Since AB and AC are the same length, the length of OB is simply AC minus AO. (As shown above, OB therefore equals 5/2 -1/2, which in decimal notation equals .618034.) ¹⁶

With the dividers set to the length of CB, the scribe would then have hit paydirt with the finding that this length (CB) appeared to inscribe exactly five chords (a 'chord' is a line segment touching the circle at two points) around the inside of the circle! Double-checking, it will be found that the more care taken in drawing this diagram, the more one sees that this result is in fact precisely accurate. In addition, if next the dividers are set to the length OB, one finds the perhaps even more amazing result that the length of OB will inscribe exactly ten times around the inside of the circle!! ¹⁷

As it so happens, on any given length of line there is one and only one point such that the length of the whole line divided by the larger segment equals the length of this larger segment divided by the shorter segment. Point B is just such a division of the line that is the radius OH. (In the diagram above, this means that OH÷OB = OB÷BH)¹⁸ The ratio of 1 to .618034 (i.e., OH ÷ OB) is today popularly known as "the golden ratio", or "the divine proportion". Let us now look at some of the doors that this unique relationship can open.

In Diagram 12, the inscribed pentagon has been entered after first diagrammatically deriving the length of CB (as was shown in Diagram 11), and then using CB to delineate the pentagon's sides. Radii to each of the pentagon's five corners have also been drawn in. Line OM is dropped at a right angle to LN thus creating two equivalent right triangles. Since LN equals CB (with CB equal to 1.17557 - see footenote 17), and since ML and MN are each half of LN, the Pythagorean Theorem can be used to solve for OM, thereby giving the right triangle ratio (i.e., MN ÷ OM) for an angle that is 1/10th of a full rotation (= 36 ).

A further look at the diagram reveals that by drawing in NQ perpendicular to OH, a rectangle can be formed in which QN = OM and OQ = MN. This arrangement graphically displays a number things. The first is that angle NOQ must equal 1/4th minus 1/10th of a full rotation (i.e., 5/20ths - 2/20ths = 3/20ths of a full rotation, or 90- 36 = 54 in our system). Second, that the ratio of the sides of the right triangle for the 54 angle (i.e., QN ÷ OQ) is the reverse of the ratio that is found for its complementary angle of 36 .

Since the angle that subtends each face of the pentagon is 1/5th of a full rotation (= 72), angle POH must equal 1/5th minus 3/20ths (i.e., 1/20th) of a full rotation, or 18. One way to find the ratio for 18 is by extending OM to meet the circle and putting into use the same procedure that was used to find the ratio for 15 in the Hexagon section (where the angle of 30 was halved). However, a short cut is at hand. We already know that the length of the side of an equal-sided decagon (ten sides) is .618034 of the radius (refer to Diagram 11 and footnote 16). Half of .618034 will therefore be the length of the shorter side of the associated right triangle for 18 (see PR in Diagram 12). With OP = 1, the Pythagorean Theorem then finds the length of OR.

Using these techniques, the inscribed pentagon allows the ratios for the following angles to be fairly straightforwardly derived: 9, 18, 27, 36, 54, 63, 72, and 81. To this must be added the ratios for 15, 30, 45, 60, and 75 that we have derived earlier. (Note that values for 4 1/2, 7 1/2, 22 1/2, etc. can then also be found.)

With the addition of the discovery of the golden ratio, the ability to inscribe equal-sided polygons of 4, 5, 6, 7, 8 (8, from further division of the square) and 10 sides could then have been known. Finding a way to inscribe an equal-sided nonagon (nine-sided) figure would even add a few more angles to the list (20, 10, 5, 70, 80 and 85. See Appendix for this derivation.)¹⁹ However, what would really be most desirable at this point is a mechanism that could be used to fill in any and all gaps in the right triangle ratio tabulations, for otherwise, gaps there would always be. I think it possible that the master scribes could have reached this juncture in their researches, and I believe that if they had made it this far, then they certainly would have had the ingenuity and perseverance to devise a diagrammatic means to proceed.

One way in which they may have moved forward would have been to recognize that if they could devise a method for determining the right triangle ratio associated with any angle A - B where the ratios for both angles A and B were known, then the list of ratios could be tabulated to one and a half degree increments (as from 9 - 7 1/2 = 1 1/2 = increments of 1/240th of a rotation). Diagram 13 shows how this question can be handled graphically.

As with some of the previous constructions, this diagram may at first appear to some readers as forbiddingly complex. With a little attention to detail, I think it will be found that it is in fact not overly difficult at all.

Triangle OZT is the right triangle resulting from angle A. Triangle OWK is the right triangle resulting from angle B. In consequence, triangle OET is the right triangle defined by the angle (A - B). The issue is: if we know the lengths of the sides of the right triangles on angles A and B, is there a way to compute the sides of the right triangle on angle (A - B)?

To resolve this question, we need to be able to find the lengths of the sides OE and TE in terms of quantities we already know. So, what do we already know? We know that OT and OK are both radii, and so both equal 1. Also, we are assuming that we are in a situation where the lengths of the sides: OZ, TZ, OW, KW are also known. For convenience and clarity, the following representations can then be used: OZ = C; OW = D; KW = B; TE = H; TJ = L; JZ = Y; TZ = (L+Y) = side A; OJ = P; and JE = Q.

A close look at the diagram tells us that it includes three right triangles which have the same three angles as each other. These are triangles OZJ, OWK, and TEJ. As we saw in the hexagon discussion, the ratio of the sides of a right triangle remains the same for all right triangles having the same angles, regardless of their size.

Could the Egyptians of the Old Kingdom have come up with this understanding? As stated before, it is true that we are not aware of their use of this style of representational formula. However, they clearly did use what has been termed "non-symbolic arguments" which are "quite rigorous" in their mathematics.²¹ They also had the mathematical ability to handle very involved computations, and there is little question that they had the ingenuity to have devised a method on the order of what is here described.

It is additionally possible that they could well have contrived a way to refine a trigonometric table to increments even smaller than 1 1/2 degrees. Reference can be made to the Appendix for this discussion.

You might ask, "so what is the big deal about the possibility of the ancient Egyptians having developed trigonometry?". Other than the obvious benefit trigonometry lends to such disciplines as architecture and land surveying, it also makes possible the ability to accurately determine the size of the Earth.
Next Section: Measuring the Earth

16.  As explained in the Square Root Derivation section in the Appendix, I believe the scribe would have found the 5 to be 2+1/8+1/9. Hence, in his notation he would find 5/2 -1/2 to be equal to 1/2+1/16+1/18. 
17.  CB is in fact the correct length for the side of an equal-sided inscribed pentagon, and OB is the correct length for an equal-sided inscribed decagon. Using the Pythagorean Theorem, CB computes to be 1.17557 the size of the radius. The Egyptians could have expressed this value as 1+1/9+1/18+1/112. See Euclid's derivation of the 36 angle (and hence of the inscribed equal-sided pentagon) in the Appendix.
18.  Numerically, this statement means that: 1 ÷.618034 = .618034 ÷ (1-.618034) = 1.618034.
19.  Via trial and error persistence, a near perfect nonagon can be derived, again using the pentagon. See Appendix.
20.  This is usually written: Sine(A-B) = CosB x SineA - SineB x CosA, and is known as the Sine Subtraction Formula.

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