Deep Secrets

In the same way that the constancy of the Pi ratio is an unexpected property of circles, so too is the fact that the length of the radius of any circle, when inscribed around the inside of its circle, will fit precisely six times. Diagram 6 shows this done, and the inscribed hexagon that results. Though perhaps a bit of an extra step, this relationship between a circle's radius and its inscribed hexagon is clearly one that is easily discoverable. By drawing radii from the circle's center to each spot where the sides of the hexagon meet the circle, six equal triangles can be formed.

These triangles each have all three of their sides of equal length, and such triangles are referred to as being 'equilateral'. Another property of equilateral triangles is that all three of their angles are also equivalent. Furthermore, each of these equal angles contains 60, or one sixth of a full rotation. (We measure a full circle as comprising 360, and therefore one sixth of a complete rotation is denoted as having 60.) We will leave aside for the moment a discussion of how angular measure may have been dealt with in ancient Egypt, and focus merely on the comparative aspects of the various line segments involved in these triangles.

To facilitate further discussion, an enlargement of one of the above equilateral triangles is presented in Diagram 7. A new line (AM) has been drawn from the center of the circle to a side of the hexagon, to meet it at a right angle. As can be satisfactorily proven through empirical measurement, the addition of this line divides the equilateral triangle directly in half, creating two equal right triangles. Since this means that the line MB equals line MC, each must be half of a unit in length (BC being originally equal in length to one radius). This then means that we now know the length of two of the three sides of each of these two right triangles (AB and AC = 1, and MC and MB = 1/2).

The Pythagorean Theorem then allows the derivation of the length of the third side (AM in the diagram) to be easily achieved.¹⁴

AM² + (1/2)² = (1)²

This leads to: AM² = 1 - 1/4 = 3/4

And so, AM = 3/2 = .8660254...

An ancient Egyptian researcher could now have understood that a right triangle whose height and base were in the ratio of 3/2 to 1/2 would have a base angle of 60, or 1/6th of a full circle. If the reverse were true, and the height and base were in a ratio of 1/2 to 3/2, then he would know that the base angle was 30, or 1/12th of a full circle.

Furthermore, from a study of the manner in which a square's diagonal divides a square in half (see Diagram 2 in previous section), he would note that in the two resulting right triangles made by a square's diagonal the ratio of the height to the base in each triangle is 1 to 1, and that the base angles accompanying such a ratio will then each be 45 (i.e., half of a right angle), or 1/8th of a full circle rotation. Surely with these observations having been made the question would be asked: "Are there yet other right triangles besides the 30 and 45 right triangles that can be diagrammatically derived which would also permit one to know the relative lengths of the triangle's sides?"

Diagram 8 shows that with a little extrapolation there is still more that can be straightforwardly gleaned from the inscribed hexagon.

Line AM has been extended to touch the circle (at D). Lines BD and CD are drawn in, and a new perpendicular is dropped from point A to meet BD at point E. As before, we have a situation where two equivalent right triangles are formed, AEB and AED. We know that AB equals AD (since both are radii of the circle), and both triangles share AE as a side. Therefore (via the Pythagorean equality), their third sides must also be equal. Thus we find that EB must equal ED, and that AE divides the 30 angle at A in half to create two 15 angles. (Triangles such as DAB, which have two sides of equal length, are known as "isosceles" triangles).

Is there now a way to find the comparative lengths of the line segments AE and ED? Indeed there is. Triangle BMD is a right triangle, and we already know that MB equals 1/2 the length of the radius. Also, it is clear that MD equals AD minus AM. Therefore:

MD = 1 - 3/2 = 1 - .866025 = .133975

The Pythagorean Theorem now can be used to solve for BD:

BD² = MB² + MD² = (1/2)² + (.133975)² = .267949, and so BD = .517638

ED is then half of this, and with this knowledge, the length of AE can be found, again by using the Pythagorean Theorem. (It turns out to be .96593. Bear in mind that all of these results would have been computed, and expressed, utilizing the Egyptian unit fraction method. This is a task within the computational capability of the ancient Egyptian scribes.)

It is at this point that the Egyptian priests may well have felt that they were very much on to something, considering that they would have now computed right triangle ratios for angles of 15, 30, 45, 60 and 75. Using the techniques here described, Diagram 8 can even be further developed to supply ratios for 7 1/2 and 82 1/2. What we see taking shape from these diagrams is nothing other than the beginning of a trigonometric table. All of these correlations are naturally occurring constants. Regardless of how big or how small the initial circle is made, the ratio of the sides of the resultant right triangles remains the same. The same constancy holds true here as was seen with the 2 phenomenon, where we saw that the v2 relationship holds true regardless of a square's size. Would the priests not have thought that a profound secret of creation had thus far been revealed, and that more of the hidden secrets of the measureable world were simply waiting to be learned - waiting only for the ingenuity to be mustered to find them?

Certainly more work would have been done to see what else could be extracted from diagrams which stem from the inscribed hexagon.

If the scribe had set his dividers to subtend the length of the equilateral triangle's perpendicular (line AM in Diagrams 7 and 8, and blue line in Diagram 9), he would have found that it appears to mark off a perfect 7 equal-sided figure (heptagon) around the inside of the circle. AM is, in fact, only .001715 smaller than the side of a truly accurate equal-sided inscribed heptagon, but this deficiency is not readily detectable empirically.

The inscribed heptagon would have allowed for the nearly exact computation of the sides of right angle triangles associated with 1/14th, 1/28th, and 1/56th, etc., of a full rotation. (Due to the fact that AM - whose relative length is .8660254 - is a close approximation to the exact length of a side of a true inscribed heptagon, computations based on the .8660254 length would necessarily be incorrect by a correspondingly small amount).

Furthermore, as explained in the Appendix, the inscribed hexagon can also provide a means (though a slightly more complex one) for deriving a very nearly accurate inscribed eleven equal-sided figure (an "endecagon"), with associated computable right triangles.¹⁵

It is my opinion that the heptagon and endecagon would not have initially played a central role in either the derivation of trigonometry nor in the design process of the Great Pyramid.¹⁶ I say this largely because there was a much more elegant means at hand for expanding trigonometric capabilities, and also because this latter method arises from a numerical constant whose existence in nature may possibly have been seen as being even more mysterious than that of the Pi phenomenon.

Next Section: Pentagon and Trignometry

for this diagram.
16. Note that the slope of the Great Pyramid has been measured to have been 51 52' plus or minus 2'. (See W.M. Flinders Petrie, The Pyramids and Temples of Giza, p. 13). The angle that a side of an inscribed equal-sided heptagon subtends from the center of its circle is 51 25.7'. I believe it quite likely that these matters played an increased role in the Egyptian pyramid design concerns of a much later date - but this is a subject which will have to wait for a future discussion.