HC = 381966 = half base of Cheops. SO = 363271·5 = HS CD = 309017 = half MH PR = 277516 GD = 224513·5 SP = 263932 The triangle DXH represents a vertical section of the pentagonal pyramid; the edge HX is equal to HN, and the apothem DX is equal to DE. Let DH be a hinge attaching the plane DXH to the base, now lift the plane DXH until the point X is vertical above the centre C. Then the points A, E, B, O, N of the five slant slides, when closed up, will all meet at the point X over the centre C. We have now built a pyramid out of the pentangle, whose slope is 2 to 1, altitude CX being to CD as 2 to 1. Apothem DX = DE Altitude CX = HM or MN Altitude CX + CH = CM radius. Apothem DX + CD = CM radius. Edge HX = HN or PF Note also that MP -- = CH 2 OP = HR Let us now consider the _Pentangle as the symbol of the Great Pyramid Cheops_. The line MP = the base of Cheops. The line CH = half base of Cheops. The line HM = apothem of Cheops. The line HN = slant edge of Cheops. Thus: Apothem of Cheops = side of decagon. Apothem of Cheops = altitude of pentagonal pyramid. Slant edge of Cheops = slant edge of pentagonal pyramid. Now since apothem of Cheops = MH and half base of Cheops = HC then do apothem and half base represent, when taken together, extreme and mean ratio, and altitude is a mean proportional between them: it having already been stated, which also is proved by the figures in the table, that MC : MH :: MH : HC and apoth: alt :: alt : half base. Thus is the four pointed star _Cheops_ evolved from the five pointed star _Pentalpha_. This is shown clearly by Fig. 68, thus:-- Fig. 68. Within a circle describe a pentangle, around the interior pentagon of the star describe a circle, around the circle describe a square; then will the square represent the base of Cheops. Draw two diameters of the outer circle passing through the centre square at right angles to each other, and each diameter parallel to sides of the square; then will the parts of these diameters between the square and the outer circle represent the four apothems of the four slant sides of the pyramid. Connect the angles of the square with the circumference of the outer circle by lines at the four points indicated by the diameters, and the star of the pyramid is formed, which, when closed as a solid, will be a correct model of Cheops. Calling apothem of Cheops, MH = 34 and half base, HC = 21 as per Figure 6. Then-- MH + MC = 55 and 55 : 34 :: 34 : 21·018, being only in error a few inches in the pyramid itself, if carried into actual measures. The ratio, therefore, of apothem to half-base, 34 to 21, which I ascribe to Cheops, is as near as stone and mortar can be got to illustrate the above proportions. Correctly stated arithmetically let MH = 2. Then HC = √5 - 1 MC = √5 + 1