Equation 6 (Area of face = height squared) can help here also because
it implies that the areas of all four faces of the Pyramid are equal (because
there is only one height). This gave me a method to find that exact spot
on the base (if equation 6 is correct) where the peak was over by using
a step and repeat procedure. The procedure goes something like this.
1. pick any place on the base and position the peak over it
2. construct lines from the four corners to the peak (defining the
faces)
3. compute the area of each face
4. if they are different, move the peak to some other point on the
base
5. keep moving the peak until all face areas are equal
I automated the procedure on a computer and plotted the result in a
3d graph.
I found that spot that produces a Pyramid with four faces all of equal
area. I then computed the lengths of the four lines that goes to the four
corner from that spot (this spot I'll refer to as Cent_eqarea). The lengths
are listed below.
NW corner to Cent_eqarea  1.11414 
NE corner to Cent_eqarea  1.11343 
SE corner to Cent_eqarea  1.11322 
SW corner to Cent_eqarea  1.11278 
I then converted the values above to the nearest unit fractions. Hence:
NW  1+1/9+1/330.171  1.1141398 
NE  1+1/9+1/432  1.1134259 
SE  1+1/9+1/474  1.1132208 
SW  1+1/9+1/600  1.1127777 
The unit fraction for the length NW to Cent_eqarea never did work out
right. I began to suspect that this length is actually a derived one. It
can be derived because the three other unit fractions help to form two
exact triangles that can be used to find the NW to Cent_eqarea length.
See figure 5.
The first thing to note in figure 5 is the two triangles on the South and East sides. Their side lengths are all integer expressions (so they can be exact parts of the model) also their placement determine the exact spot of the Cent_eqarea. These two triangles can also determine the rest of the base.
The way to do it is to consider the North base length (UFR_north) to be hinged at the NE corner and the West base length (UFR_west) to be hinged at the SW corner. These two length could then be rotated as shown (by the dashed line with arrow) until their ends met. The meeting of these length would then determine the NW, SW and NE corner angles.
This also would create the length from the NW corner to the center (dashed line with question mark) and explain why it can not be written as an integer expression (just what I found). I was happy with the above model for the base. It looked like just four more lines and the model for the full pyramid would be done.
But while rechecking the math I found a discrepancy between the apothem and the altitude of the two base triangles discussed above and shown in figure 5.
I had computed the areas of the two triangles using just the sides for the computation. They were:
South triangle's area = 0.61938
East triangle's area = 0.619744
Since for a triangle:
area= base * altitude / 2,
solving for altitude gives:
South triangle's altitude = 0.786424
East triangle's altitude = 0.787096
I then noticed that the apothem was the hypotenuse of a right triangle whose other sides were the height and this altitude. But solving for altitude gives:
South =.782375
East = .782933
The triangle' altitude does not match the altitude given by the apothem's right triangle (see figure 6). It shows that there is a choice between apothems. Equation 6 and 7 seem to lead to conflicting results.I tried converting the various decimal number (from figure 6) to unit fractions. But none of them were even close to an exact fit (they have to fit exactly because they are part of a right triangle).
The discrepancy (for me) was resolved when I took an over head view. It was then that I realized that if I use the shortest apothem (1.269689 from figure 6), it would cause the face to be slightly concave. See figure 7.
I say resolved because I had done enough research to know that the faces of the Pyramid actually are slightly concave! This was discovered around 1940 from an aerial photograph.
This gave me confidence that this model that I am pursuing is right.
I went through all the faces and found that equation 7 caused all of them
to be slightly concave. It also meant that instead of one big base triangle
associated with each face there are two smaller ones (additionally this
means that each face is split into a right and left side).
And then all of sudden I was done. At first I thought there had to be more unit fractions to complete the model. But it turned out that using just the seven unit fractions I had found, and the height squared relationship, is enough to fully specify a model (including the concave faces).
I made a 3D drawing using a CADD program, and learned the model has some subtle features. The first was that the concave faces would not show up no mater how big the drawing was.
In fact it appears that the only way that you could see the concave effect is with something like the sun's shadow ( as in the aerial photograph ).
For this reason I will use figure 10 [not included!] (the concave faces are exaggerated) to discuss some components of the model.
First a summary of the unit fractions:
Name  unit fraction  normal fraction 






























Some other relationships are:
ap_a = 2/UFR_a (equation 7)
dtc_a = SQRT(ap^2  1) (distance to center)
and of course:
height =1
The model is composed of 16 triangles ( 8 on the base & 8 on the
faces ). I went about finding lengths and angles for those. But the most
intriguing aspects of the model has to do with area.
To begin a discussion on area it will help to think of the model as actually two models, a big one and a small one. The big one is simply the same model with no concave faces and the small one has the concave faces as in figure 10.
I well discuss face area first. Equation 6 says that face area equals
height squared. This mean of course that each face area should equal one
(1^2 =1 ). I checked first the face areas of the big model (no concave
faces). The result are shown below:
North face  East face  South face  West face 

1.0019666  1.00202155  1.001971222  1.0020280353 
But I was amazed by the face areas of the small model. This is found
by adding the area of the left and right triangle for each face. The result
are:
North face  East face  South face  West face 

1.0000081955  1.0000086766  1.0000081774  1.0000086396 
This means that by shaving just a small sliver of material from each face (making them concave ) makes there area equal essentially equal one. My first question was, could the Egyptians even make such an exact area.
Each year for thousands of years before the Pyramid the Egyptians had
to redraw all the farm boundaries after the Nile flood. The main task was
to give each farmer the same area to farm as he had last season. After
a thousand years they probably got very good at it. I would say then, that
it is reasonable to suppose that when they planned a gigantic national
monument, it's various exact areas would be important to them.
With a renewed respect for area I next calculated the base triangles
area. As before, first the big model:
North base tri.  East base  South base  West base 

.620255  .619744  .619383  .619895 
And the small model (this area is the sum of two triangles):
North  East  South  West 

.617071  .616467  .616195  .616618 
Here again comparing the different values from the big and small models is instructive.Any body who has ever played around with numbers would probably recognize that the constant .6180339887.... (first mentioned by the Greeks) has something to do with the above areas. The four areas from the big model are about 0.3% larger than the constant. And the four areas from the small model are roughly 0.3% smaller than the constant.
This suggests that the Egyptians knew the constant to better than 0.15% . And considering the precision of the face area values (from above) it seems a reasonable assumption.
So that seemed to be the end of my quest. The small model with concave
faces has enough features (it's height is equal one and each face area
is also equal one) to convince me that it was the original plan for the
Great Pyramid. It sure has unity going for it!
It was after considering a simpler model I had second thoughts. The simpler model is a square pyramid with height of one and face area of one. All the base lengths equals 1.5723028 and it turns out the base triangle are exactly 0.6180339... in area (the constant from above which I will call GR [for golden ratio]).
If the Egyptian could build a complicated model with face area and height
equal one. Then they certainly could have built the simpler square pyramid
I have described. And if that is true they would know how to construct
an area exactly equal to 0.6180339... (not just to 0.15%).
Also when considering the volume of the model pyramid, the formula
started me thinking. The formula is:
volume = (area of base)*height / 3
Since the height of the model is one, if the base had slightly more area the formula could be:
Vol_Model=4*GR/3=0.8240453...
That would make the face areas, height, base area and volume all have special exact values.
THE END FOR NOW
last update 1/16/2000
P.S. Below is a sketch (not to scale) of the final model of the Pyramid I have come up with. I will explain it in another paper if there is any interest. Email comments to Terry at hawmtn@aloha.net