"why was the height of the KC walls increased to 80 palms?
The usual and most obvious possible answer to this is because it then makes the perimeter of the Chamber's side (north and south) walls equal* to the length of the walls multiplied by 3 1/7 (height = length 20rc x 3.143 รท 2 - length 20rc = 11.429rc)."
Well, I think you have answered your own question. I would add this. By making each course 16 palms high, the designers made sure that the dimensions of the wall should be read in palms. So, we have three dimensions for the chamber walls as 80p x 70p x 140p. As you have implied, the perimeter of the end walls then becomes 80 + 80 + 140 + 140 = 440p. This is the same number of palms as the cubits in the base. The ratio 440:140 reduces to 22:7, which we recognize as "pi". This was Petrie's original discovery, as I remember, but I think it is suggestive rather than persuasive.
"Interestingly, if the walls of the Chamber were designed to be 11.429rc and the vertical distance between the surface of the raised floor and the Chamber ceiling was intended to be half the diagonal length of the Chamber's floor plan (i.e. 11.18rc), then the appearance of Phi and the 3:4:5 triangle (or ratios very, very close to them) in this Chamber could be fairly argued to be purely unintentional."
I think the appearance of phi is unintentional. In fact, I can't really see phi in these dimensions. You would have to add 5 cubits to 11.18 to get 10 x phi, and there is no dimension of 5 cubits in the chamber (though there is in the antechamber!).
Taken in isolation, it may appear that the 3-4-5 triangle is unintentional - and I would like to see that argument - but taken in concert with the dimensions of the mortuary temple of 75c x 100c, which is exactly 5 times the dimensions of the 3 x 4 rectangle contained in the King's Chamber, it has to be admitted that the architect was familiar with the 3-4-5 triangle.
Actually, I have another problem with the dimensions of the chamber which I have not been able to reconcile. The height of the first course above the floor is 16p - 1.75p = 57 fingers, or 1 finger greater than 2 cubits. As I understand it, the top of the shafts is supposed to be 2 cubits above the floor, and flush with the top of the first course of the wall. There is a discrepancy here of 1 finger, and I can't account for it.