The Solution of the Pyramid Problem Part 11

I shall set forth the geometric significance of this star, as far as my general subject warrants me, and show that it is the _geometric emblem of extreme and mean ratio_, and the _symbol of the Egyptian Pyramid Cheops_.

A plane geometric star, or a solid geometric pyramid, may be likened to the corolla of a flower, each separate side representing a petal. With its petals opened and exposed to view, the flower appears in all its glorious beauty; but when closed, many of its beauties are hidden. The botanist seeks to view it flat or open in its geometric symmetry, and also closed, as a bud, or in repose:--yet judges and appreciates the one state from the other. In the same manner must we deal with the five pointed star, and also with the Pyramid Cheops.

In dealing with so quaint a subject, I may be excused, in pa.s.sing, for the quaint conceit of likening the interior galleries and chambers of this pyramid to the interior whorl of a flower, stamens and pistil, mysterious and incomprehensible.

Figure 67 (page 101), is the five pointed star, formed by the unlapping of the five slant sides of a pyramid with a pentagonal base.

Figure 70 (page 106), is a star formed by the unlapping of the four slant sides of the pyramid Cheops.

The pentagon GFRHQ, (_Fig._ 67) is the base of the pyramid "_Pentalpha_"

and the triangles EGF, BFR, ROH, HNQ and QAG, represent the five sides, so that supposing the lines GF, FR, RH, HQ and QG, to be hinges connecting these sides with the base, then by lifting the sides, and closing them in, the points A, E, B, O, and N, would meet over the centre C.

Thus do we close the geometric flower Pentalpha, and convert it into a pyramid.

In the same manner must we lift the four slant sides of the pyramid Cheops from its star development, (_Fig._ 70) and close them in, the four points meeting over the centre of the base, forming the solid pyramid. Such transitions point to the indissoluble connection between plane and solid geometry.

As the _geometric emblem of extreme and mean ratio_, the pentangle appears as an a.s.semblage of lines divided the one by the others _in extreme and mean ratio_.

To explain to readers not versed in geometry, what extreme and mean ratio signifies, I refer to Figure 65:--

Fig. 65.

Let AB be the given line to be divided in extreme and mean ratio, _i.e._, so that the whole line may be to the greater part, as the greater is to the less part.

Draw BC perpendicular to AB, and equal to half AB. Join AC; and with BC as a radius from C as a centre, describe the arc DB; then with centre A, and radius AD, describe the arc DE; so shall AB be divided in E, in extreme and mean ratio, or so that AB: AE:: AE: EB. (Note that AE is equal to the side of a decagon inscribed in a circle with radius AB.)

Let it be noted that since the division of a line in mean and extreme ratio is effected by means of the 2, 1 triangle, ABC, therefore, as the exponent of this ratio, another reason presents itself why it should be so important a feature in the Gzeh pyramids in addition to its connection with the primary triangle 3, 4, 5.

Fig. 66.

To complete the explanation offered with figure 65, I must refer to Fig.

66, where in constructing a pentagon, the 2, 1 triangle ABC, is again made use of.

The line AB is a side of the pentagon. The line BC is a perpendicular to it, and half its length. The line AC is produced to F, CF being made equal to CB; then with B as a centre, and radius BF, the arc at E is described; and with A as a centre, and the same radius, the arc at E is intersected, their intersection being the centre of the circle circ.u.mscribing the pentagon, and upon which the remaining sides are laid off.

We will now refer to figure 67, in which the pentangle appears as the symbolic exponent of the division of lines in extreme and mean ratio.

Thus: MC : MH :: MH : HC AF : AG :: AG : GF AB : AF :: AF : FB

while MN, MH or XC: CD:: 2: 1--being the geometric template of the work.

Thus every line in this beautiful symbol by its intersections with the other lines, manifests the problem.

Note also that

GH = GA AE = AF DH = DE

I append a table showing the comparative measures of the lines in Fig.

67, taking radius of the circle as a million units.

Fig 67.

Table Showing the Comparative Measures of Lines.

(_Fig. 67._)

ME = 2000000 = diameter.

AB = 1902113 = AD DB

MB = 1618034 = MC + MH = MP + PB

AS = 15388415

EP = 1453086 = AG + FB

AF = 1175570 = AE = GB

MC = 1000000 = radius = CD + DX = CH + CX

AD = 9510565 = DB = DS

PB = 854102

QS = 8122985

MP = 763932 = CH 2 = base of Cheops.

AG = 726543 = GH = XH = HN = PF = FB = Slant edge of Cheops. = slant edge of Pent. Pyr.

DE = 690983 = DH = XD = apothem of Pentagonal Pyramid.

{apothem of Cheops.

MH = 618034 = MN = XC = {alt.i.tude of Pentagonal Pyramid.

{side of decagon inscr'd in circle.

MS = 500000

{mean proportional between MH and HC 485868 = { {alt.i.tude of Cheops.

OP = 449027 = GF = GD + DF

HC = 381966 = half base of Cheops.

SO = 3632715 = HS

CD = 309017 = half MH