You're all looking at the wrong pyramid. Khafre is where the action is. Its slope angle is 53.13 degrees, the apex angle of a triangle in a square, like this, and this is where to find Phi, and the decimal system, and a few wee other bits and pieces as well. This particular puzzle is unpicked by the use of the Definer circle geometry method.
This is how it unfolds.
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Normally, the Definer circle is determined by the shortest distance to the corner or apex, depending on the particular triangle being analysed. This time I am using the longer distance from I, the triangle leg -baseline circle intersection point. The reason will become clear later on, but the end result is unaffected by this choice.
As can be seen, the circle IC is too large to cut the leg BC, so the leg will have to be extended to meet it at J, as below.
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This is the result, and the smaller Definer circle JB fills the gap.
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This smaller Definer turns out to be a perfect fit for leg BC, dividing it five times.
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On leg AC I have used the usual shorter radius IA Definer method, and as you can see it matches the top Definer off leg BC, so it really doesn't matter which method is used, but there may be a good reason to choose one or the other on occasion, like now.
Also visible here are three different size Definers, all valid for this construction. I have decided to name them the Primary, for the largest taken from I(IC here); secondary, for the smaller taken from I(I'A here); and tertiary etc for any other Definers that follow. Some triangles have several reductions before a match is found for the leg length.
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All the tertiary Definers are added to both legs here, and their centres labelled. They will mostly be removed for clarity next.
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Lines are drawn from A to I and A to J, forming the red triangle here. Its legs are populated with the same tertiary Definers, and they divide the legs up in the proportions 3:4:5.
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These are the angles produced, with two 3-4-5 triangles and one 5-6-5 triangle. At the apex C can be seen a vesica piscis formed along the line M'M. This makes this line 1 unit long.
![]()
This is how it unfolds.
Normally, the Definer circle is determined by the shortest distance to the corner or apex, depending on the particular triangle being analysed. This time I am using the longer distance from I, the triangle leg -baseline circle intersection point. The reason will become clear later on, but the end result is unaffected by this choice.
As can be seen, the circle IC is too large to cut the leg BC, so the leg will have to be extended to meet it at J, as below.
This is the result, and the smaller Definer circle JB fills the gap.
This smaller Definer turns out to be a perfect fit for leg BC, dividing it five times.
On leg AC I have used the usual shorter radius IA Definer method, and as you can see it matches the top Definer off leg BC, so it really doesn't matter which method is used, but there may be a good reason to choose one or the other on occasion, like now.
Also visible here are three different size Definers, all valid for this construction. I have decided to name them the Primary, for the largest taken from I(IC here); secondary, for the smaller taken from I(I'A here); and tertiary etc for any other Definers that follow. Some triangles have several reductions before a match is found for the leg length.
All the tertiary Definers are added to both legs here, and their centres labelled. They will mostly be removed for clarity next.
Lines are drawn from A to I and A to J, forming the red triangle here. Its legs are populated with the same tertiary Definers, and they divide the legs up in the proportions 3:4:5.
These are the angles produced, with two 3-4-5 triangles and one 5-6-5 triangle. At the apex C can be seen a vesica piscis formed along the line M'M. This makes this line 1 unit long.
