
CHAPTER 11 Properties of the Great Pyramid’s Geometrical Configuration It is common to find in general literature about the Great Pyramid that a measurement in one section of the structure is equivalent, related or proportional to measurements other sides of the structure. There are no coincidences in these measurements. This is the effect of the geometrical configuration of its design, as it will observe in the following examples. Note in figure 168, that the position of point X divides the vertical diameter in two sections (QX and XK). These sections are in a ratio, which is equivalent to f, that is, (594.13 / 367.19) = f. Additionally, it will be shown that the location of the point identified as X, appears to be a control point in other pyramid designs. There are many interesting geometric relations in this configuration. In figure 155, which shows the Great Pyramid the measurements, apparently do not show any relation among them. For example, the elevation of the King Chamber’s floor over the base line (140.26') does not show anything extraordinary. But if the vertical distance from the base line to point X (113.47') is added, the elevation of the floor’s chamber over point X will be 253.73'. Now, this measurement, indeed, becomes important in the design, and establishes a direct relation with the others. Note that the inclined distance from point X to the base line level (point S) is also equivalent to 253.73'. It can be inferred, that the elevation of the King Chamber’s floor was set by projecting to the vertical axis, a distance equivalent to that from point X to point S. It will be noted that the horizontal distance between the intersection of the descending and ascending passages (point I), to the pyramid’s axis, is also equivalent to 253.73'. But that is not all with the 253.73' measurement. If this number is divided by f, the result is 156.81'. Now, 156.81' represents the inclined distance between the north and the south walls of the Grand Gallery. If, instead, 156.81' is multiplied by f, the product is 410.54', which represents the vertical distance of the Queen Chamber’s floor underneath the projection of the top (apex) of the Great Pyramid. Note that the difference between 410.54' and 253.73 is also 156.81'. The product of 410.54 and 156.81 is equivalent to (253.73)². With the relation of these numbers, it can be said that the inclined length between the Grand Gallery’s north and south walls (156.81') multiplied by f , represents the elevation of the King Chamber’s floor over point X, that is, 253.73'. All of these mathematical relations happen as a result of the unique characteristics of the f value present in the right triangle having its sides in the ratio 1:2 (see Chapter 9).
Figure 168 When the value of the diameter of the circle is divided by f, and then, the result divided again by f, the results of the measurements, are values easy identifiable inside the pyramid’s structure. For example: R = radius of the circle = the pyramid’s height D = (2)(R) = Circle’s diameter = 961.33' f = Golden Number = (1.61803...) 961.33 / f = 594.13' = Distance from point Q to point X. 594.13 / f = 367.19' = Distance from point X to point K. 367.19 / f = 226.94' = Twice the distance from point X to the center O, of the circle. 226.94 / f = 140.26' = Elevation of King Chamber’s floor over the base line. If the same procedure is done with the pyramid’s height, a new set of identifiable measurements becomes noticeable. 480.66 / f = 297.07' = Distance from point Q to point X. 297.07 / f = 183.60' = Elevation of the Queen Chamber’s floor over point X. 183.60 / f = 113.47' = Vertical distance from point X to point O, center of the base. 113.47 / f = 70.13' = Elevation of the Queen Chamber’s floor over point O. These numbers show that it exists a numerical and geometrical relation between the different sections in the pyramid, not attributable to incidental occurrences, but to the factor of f. This is the reason why the dimensions of the Great Pyramid are so related. The location of point X in this special type of configuration, as the division point of the vertical diameter in the f proportion, appears to be the most important location in the configuration, and could represent a control point for the pyramid designs. Figure 169 show the basic geometric configuration of the pyramid, without its internal structural elements. There are other interesting properties on this configuration. In the examples, the radius (AO) of the circle was set equal to one (1); consequently, the diameter is equal to two (2). The length of the sides of the square base (I, II, III, IV), is equivalent to (b, where (b) = (D / Öf). It can be noticed that if both sides are squared, b² = D² / f. Furthermore, the equation can be transformed to read f = D² / b². From this equation, it can be seen that the area of the square that inscribes the circle (between points 1, 2, 3, and 4), divided by the area of the base of the pyramid, is equal to f. Additionally, calculating the square root in both sides of the formula, the result is Öf = D / b. I have already established that the ratio of D / b represents the value of the function of the tangent of the angle of the faces of the pyramid. This value is Öf = 1.27201965. The angle with this function is 51.82729237º (= 51º 49' 38.25'’). It means that the angle of elevation of the faces of the pyramid is 51º 49' 38.25'’.
Figure 169 I developed the following formulas, which are useful to calculate the proportional measurements in any pyramid, with the same configuration as the Great Pyramid, based on its height (R). Any value of R, with the specified unit of measurement, can be used in the formulas. The calculation will give the corresponding proportional measurement, to the Great Pyramid. R = Radius = Pyramid’s height = R Chords: (ar) = (bf) = (cd) = (eh) = (Ö5 1)(R) QX = Vertical distance from point Q to point X = (Ö5 1)(R) b = Sides of the square ( I, II, III, IV) = (2 / Öf)(R) PQ = QN = b = Sides of the square = (2 / Öf)(R) KP = KN = Inclined distance KP and KN = (Ö5 1)(R) HQ = QT = Apothem (inclined distance from the center of the base to the top) = (Öf)(R) OX = Distance from point O to point X = (1 / f³)(R) XK = Distance from point X to point K = (3  Ö5)(R) PN = Base (Side of triangle PQN) = (4 / Öf³)(R) As an example, consider the calculation of the measurements corresponding to the Great Pyramid. Assuming R = 480.66', the following results are obtained for each of the indicated section. R = Radius = Pyramid’s height = 480.66' Chords: (ar) = (bf) = (cd) = (eh) = 594.13' D = Circle’s diameter = 961.33' b = Sides of the square = 755.75' PQ = QN = b = Sides of the square = 755.75' QX = Vertical distance from point Q to point X = 594.13' KP = KN = Inclined distance KP and KN = 594.13' HQ = QT = Apothem (inclined distance from the center of the base to the top) = 611.41' OX = Distance from point O to point X = 113.47' XK = Distance from point X to point K = 367.19' PN = Base (Side of triangle PQN) = 934.16' * The property of the «almost equal» mathematical relation between the circumference of the circle and the perimeter of the base, as found in the Great Pyramid configuration, although minimal, is constant for all pyramid heights. The circumference of a circle (C) is equal to (D)(p). Using a unitary radius (R = 1), the value of C = 2 p. The perimeter (P) of the base is equal to 4 (b), that is, P = 4 (2 / Öf) = 8 / Öf. Therefore, the difference is (2 p  8 / Öf) / 2 p = 0.000950022, figure that represents less than one tenth of one percent (0.095 %). * The slope angle of the faces is determined, as I explained before, by the formula D / b. Then, D = 2 and b = 2 / Öf (figure 156). The calculated angle will be 51° 49' 38.3". * The sine function for the angle 51° 49' 38.3" is equivalent to 1 / Öf, while the cosine function is 1/ f. Consequently, the tangent is Öf. Since the radius is equal to 1, and (b / 2) is 1 / Öf, using the Pythagorean triangle: (HQ)² = 1² + (1 / Öf)² = f. This means that the apothem, represented by HQ, is equal to Öf. * Another important property of the configuration is that the surface area of each one of the faces is equivalent to the area of the square having the height of the pyramid as its side length. Some authors attribute this same property as coming from Herodotus, as one of the Great Pyramid’s characteristics. In figure 170, the triangular area of the inclined face between points Q, II and III, can be calculated as half the base length multiply by the inclined distance from the center of the base to the pyramid’s top (apothem). The length (b) of the base is equivalent to D / Ö f, and the apothem (h’) is R Ö f. Therefore, the surface area of the face of the pyramid is (1 / 2) (D / Öf) (R Öf), that is, (D) (R) / 2 = (R) (R) = R². Figure 171 shows the square area established with the pyramid’s height as one of its sides. The area of the square is equal to (R)(R) = R², which corresponds with the area shown for the surface of the face.
Figure 170
Figure 171
As an example, with sides 755.7488" and apothem (Qt = h’) = 611.4136', the surface area (assuming leveled) of the face would be equal to (1/2) (755.7488)(611.4136), that is, 231,037.55 squared feet. The area of the square formed (figure 158) having its side’s equal to the height of the pyramid is (480.6637)², which is also equal to 231,037 squared feet. * Another property of this special geometrical configuration is that the ratio of the surface area of the four faces, to the area of the base, is equal to f. If A1 is the total area of the faces, and A2 the area of the base, then, A1 = (4)(R²), and A2 = [D/Öf]² = D²/ f= (4 R²) / f. The ratio would be A1 / A2 = (4 R²) / (4 R² / f) = f. Consequently, the surface area of the faces divided by the area of the base is equivalent to f. * Interesting as well, is the property shown in figure 172. The length of the lines PQ and QN are equivalent to the length of the base (b) of the pyramid. Therefore, it can be stated that the squares constructed over these lines are equivalent to the square of the base of the pyramid.
Figure 172 The length of the line KM (in figure 173), used to delineate the geometric configuration, is equivalent to Ö51. Plato, the Greek mathematician and philosopher, indicated in his writings that the Creator of the Universe utilized the value of Ö5 1 among his formulas to create the Cosmos. As it will be noticed, this same formula is the origin of the geometric configuration I presented, as the solution to the Great Pyramid’s geometry. To prove that PQ and QN are equal to HT (the length of the sides of the square), first it must be established that PQ is equal to PN. In figure 173, KM and KP is equal to Ö5  1 by construction. The triangle KPQ is inscribed in the semicircle KAQ, with center at O. This triangle is defined, as a right triangle, based on the theorem, which states that a triangle inscribed in a semicircle, with the diameter as its hypotenuse, is a right triangle. The triangle QNK is also inscribed in the semicircle QBK, with center O. By means of the same theorem, the triangle QNK is also a right triangle. Since KP = KN, the radius is common for both semicircles, and the right angles of the triangles KPN and QNP are equal, they are congruent triangles. Therefore, line PQ = QN.
Figure 173 Using the Pythagorean theorem: (P Q ) ² + ( P K ) ² = (2 R)² = ( 2 ) ² (P Q ) ² = 4  ( Ö5  1 ) ² = 4  ( 5  2 Ö5 +1 ) (P Q ) ² = 2 Ö5  2 = 2 ( Ö5 1 ) = 4 ( Ö5  1 ) / 2 Since = ( Ö5  1 ) / 2 = 1 / f Then (P Q ) ² = 4 / f Therefore: PQ = QN = 2 / Öf In reference to triangle QHO, since QO is equivalent to R = 1 and HO to b/2 (half the square’s side). The hypotenuse HQ is equal to Öf. In accordance with the same theorem: (1)² + (b/2)² = (Öf) (Öf)² (b/2)² = (1)² f  b²/4 = 1 b²/4 = f  1 Substituting f  1 = 1 / f, b = 2 / Öf, Therefore b = PQ = QN * In figure 174, the chord (ar) is a segment of the side IIV of the square and its length is equivalent to (Ö5  1). To verify the length of this chord:
Figure 174 The distance HO is equal to b / 2 = 1 / Öf. Using the rectangular triangle HaO, and the Pythagorean theorem, the following equation can be established: (aH)² + (HO)² = (Oa)². (aH)² + (1 / Öf)² = (1)² (aH)² = 1  (1 / f) = (f  1) / f, Since (f  1) = 1 / f
(aH)² = 1 / f² (aH) = 1 / f (ar) = 2 times (aH), that is, 2 (1 / f) = (2 / f ) = Ö5  1 Therefore, the length of chord (ar) is equivalent to (Ö5  1). * Figure 175 shows some geometric relations in the configuration expressed in terms of f.
Figure 175 * Observe that the distance between points V and U at the base of the drawing is equal to 2b = 4 / Öf = 3.14460. This value represents twice the value of the base (b), that is, 2 (1.572303) = 3.14461. Note that the ratio 4 / Öf is an approximate value for p. * As shown in figure 176, the area of the base of the pyramid is equivalent to the rectangular area between point’s Q, Q’ M’ and K. Since the radius of the circle is assumed as one (1), then, the area of the rectangle is equivalent to 2 multiplied by (Ö5 1). The area of the base of the pyramid is equivalent to b² = 2 (Ö5 1), therefore, both areas are equal. * It can be also observed in figure 176, that the rectangle between the points Q, X, U, and U’, creates an area which is also equivalent to that of the base of the pyramid, since QX = (Ö5 1), and XU = 2, the area would be equal to 2 (Ö5 1) = b².
Figure 176 * Figure 177 shows the vertical sectional plane through the diagonal plane of the pyramid (triangle QUU’). The circle with radius OU inscribes the projection of the square base of the pyramid. The radius of this circle is equivalent to the horizontal distance from the center O, to the corners of the pyramid, as seen through the diagonal cross sectional plane. This distance (OU) is 534.395'. That is, 53.73' more than the radius OA from the interior circles that passes through the pyramid’s top. The length of the inclined line (UQ), from the top of the pyramid to the corner of the sides is 718.76', and has a slope angle of 41.9699° (41° 58' 11") in relation to the horizontal base. The interior angle in the top of the pyramid, in the diagonal view, is 96.0602° (96° 3' 7"). The diagonal measurement between the opposite corners of the base is 1,068.79'. * The area of the square, which circumscribes the outer circle, that is, between points 1, 2, 3 and 4 (figure 178), is equal to twice the area of the base of the pyramid (between points I, II, III, and IV). This means that (1,068.79)² = (2)(755.75)², which is correct.
Figure 177
Figure 178 * The following information about the Great Pyramid’s geometry seems to be very interesting. Figure 179 shows a pyramid with a height of 480.66'. The radius (OQ’) corresponds to the circle traced tangent to the pyramid’s faces, and is equal to 153 (p) / f = 297.06'. Notice that the vertical distance from point Z (the elevation of the platform at the south end of the Grand Gallery) to point Q’ (the intersection of the circumference of the circle tangent to the faces and the vertical axis), is equivalent to (297.07'  140.26') = 156.81'. The elevation of 140.26' corresponds to point Z over the base line. Notice that the vertical distance of 156.81' is exactly equal to the longitudinal length from the north to the south wall of the Grand Gallery. This means that if the distance ZQ’ is projected from point Z to the inclined floor of the Grand Gallery, the length of the Grand Gallery will be established as it was originally design, and before it was displaced to the south of the central pyramid’s axis (see Chapter 2  Grand Gallery).
Figure 179 * The formula to find the vertical distance from point O, at the center of the base line, to point X, as stated before in this chapter, is (R) (1 / f³). If the height of the pyramid is 480.66', that is, (R = 153 p), the value of OX would be 153 p / f³ = 113.47'. Variations in the Geometrical Configuration * Another method to create the triangle that symbolizes the cross sectional view of the pyramid as seen through the center of its sides, is by using one of the methods to divide the square in the Golden Section. Figure 180 shows a square between points I, II, III and IV. Trace the horizontal line AB and the vertical QK joining the midpoints of its sides. Trace a line from point I to point B. With center B, and with a radius equivalent to the distance from point B to point II, trace a semicircle from point II to point III. Mark point D at its intersection with line IB. From point I, using radius ID, trace an arc to intercept the side (IIV) of the square. Mark point C at its intersection. From C, trace a horizontal line to intercept side IIIII of the square, mark point S at its intersection. Mark point X at the intersection of this line with the vertical diameter. The ratio between QX and XK, IC and CIV, IIS and SIII, is equal to f =(1.618033...). This is one of the different methods known to define the Golden Section in a square.
Figure 180 As is show in figure 181, inscribe a circle inside the square shown between points I, II, III and IV. From Q trace inclined lines to the intersection points of the circumference of the circle and the horizontal line CS. Mark these points as P and N. The lines QP and QN will establish the triangle PQN. Set points H and T at the intersections of line PQ and QN, respectively, with the horizontal diameter. The triangle QHT symbolizes the cross sectional view of the pyramid, as seen through the center of its faces. The position of point X divides the vertical diameter in two segments, QX and XK. Their ratio is equivalent to the Golden Number (QX / XK ) = f.
Enlarged Section
Figure 181 However, the case with the horizontal diameter is different. Since both the horizontal and the vertical diameters are equal, their relation with f can be calculated in the horizontal direction. The horizontal diameter is also equivalent to the horizontal distance between points C and S. The length of CS represents the sum of the segments CP, PN and NS. In accordance with the calculations, the length PN = 2 (b) / f. Consequently, (CP+NS) = D  2 (b) / f, or , R [1  (2R / Öf³)]. These formulas can be applied to any circle inscribed in a square having the same configuration of the Great Pyramid. For the Great Pyramid, with R = 480.66 feet (height), the length of CP = NS would be 480.66 [1  (2)(480.66) / Öf³)] = 13.58'. Therefore, the horizontal diameter would be D = 2(13.58) + 934.16 = 961.33'. These formulas and analysis show that the relation of f is different for the square and for the circle. Points C and S established the Golden Section line for the square; however, the inclined lines (faces) from the apex (point Q) to join point P and point N, in the circle’s circumference, establish the f configuration in reference to the figure of the circle. I have always emphasized the importance that could have the location of point X in the different geometrical configuration designs. It should be interesting to investigate what the location of this point signified for the ancient Egyptian designers. The fact that most of the inclined descending passages, after reaching certain location, change the direction to horizontal and end up at a chamber, seems to indicates that the descending passage was originally aligned to certain location in the vertical axis, and then, this alignment was changed to horizontal, apparently to deceive as to the location of that point. In special, in the Great Pyramid’s model, presented in Chapter 2, where this location coincides with that of the point X. Other pyramids, like the Bent and Red Pyramid, apparently also follow this arrangement. It seems that the descending passage is aligned from the entrance to the pyramid, to a point established already in the design, possibly point X. The location of point X varies according to the particular geometrical configuration used for the pyramid. For those interested in the location of point X for any particular pyramid, I will explain my formula to determine this value for any pyramid designed using the figure of the circle and having a square base. The vertical measurement between point O (at the center of the base) and the underground location of point X, as shown in figure 181, will be defined by the formula that I developed and which shows that the vertical distance OX = (4  k²) (R) / (4 + k²), where R is equivalent to the pyramid’s height, (b) represents the length of the base, and k = (b / R). For example, for the Great Pyramid, with a height R = 480.66' and sides b = 755.75', then: OX = (4  k²) (R) / (4 + k²) OX = [4  (755.75 / 480.66)²](480.66) /4 + (755.75/480.66)² = OX = 113.47' The formula to find the vertical distance OX, for the particular configuration of the Great Pyramid, is also equivalent to OX = R (1 / f³), that is, 480.66 / (1.618033)³ = 113.47'. The results of the calculations for the value of OX, using the abovementioned formulas, are shown for the following pyramids. Chephren : (height 471', sides 707') OX = 131.57' Micerynus: (height 214', sides 346') OX = 44.80' Bent Pyramid : (see Chapter 12) (height 336', sides 620') OX = 27.00' Bent Pyramid (lower section) : (height 438.41', sides 620') OX= 146.14' Bent Pyramid : (upper section) : (height 336', sides 711.66') OX=  19.25' Red Pyramid : (height 341', sides 722') OX=  19.41' Note: When the value of OX is negative, the side’s length of the pyramid is larger than the circle’s diameter; therefore, point X will be located over the center of the base (point O). Nevertheless, due to the symmetry, apparently, the same value was used, but under point O. The descending passages will change to horizontal near, or at the baseline level, where the chamber will be build. From another side, in my studies I found one mathematical expression that is an approximation of p, in terms of f, which can be easily traced geometrically. This is p = f / sine of 31°. As shown in figure 182 (a), determine the midpoint (g) of the radius AO. From point g, using radius gQ, trace an arc to intersect the diameter AB, mark point Y’ at the intersection. The distance from point A to point Y’ represents the value of f = 1.6180.
Figure 182 As shown in figure 182(b), from point A set an inclined line at 59º to intersect the vertical line through point Y’. Define point Y at the intersection point. The length of the line AY will be equal to 3.14158... value that is a good approximation of p (= 3.14159...). * Figure 183 illustrates that the type of geometrical configuration applied to the Great Pyramid provides an easy and fast way to divide the circumference of a circle in ten (10) equal segments. With center in point X, trace a circle using QX as the radius. From point K, trace another circle using KX as the radius and mark point (a) and (b) in the intersection of the two circumferences. The angle formed between point X and point (a), and X and point (c), measure 36º. This angle is also equal to the angle between point X and point (c), and point X and point (b). Use the chord from (a) to (c) or from point (c) to (b) to divide the circumference of the circle with center X, in ten (10) equal segments. With (b) as centers, and the distance from (b) to (c) as the radius, trace a circle to cut the circumference of the circle with center X, mark the intersection as point (d). Point (d) will be the next center and the radius the distance from (d) to (b). Repeat the same procedure around the circumference of the circle center X. The circumference will be equally divided in ten segments, by the ten chords.
Figure 183 * Another interesting figure traced with the same geometric configuration is shown in figure 184. It shows a harmonious figure of the triangle PQN, where its form is maintained, but the lateral, vertical and horizontal dimensions are reduced successively in proportion and harmony with the function of f. The geometrical arrangements of the sides of the triangles corresponds to the process of dividing, consecutively each side, by the value of f. Also, each following radius, is equally divided by f. This process, produces the following circle tangent to the inclined sides of the triangle previously created. I used the Great Pyramid’s height for the initial circle, in order that the dimensions of the radiuses and triangles in the figure could be easily recognized.
Figure 184 It should be understandable that if the Great Pyramid was designed with the geometric configuration here exposed and explained, all its dimensions and general characteristics should be in agreement with my designed pyramid’s model. What it would be really surprisingly, is that both designs having equal angles, measurements and characteristics, that they do not have any relation at all, and everything is attributable to just coincidences.

