As a minor problem, I have been looking in the net for some
proof of the volume if the truncated square pyramid but all
of them use a more general concept, the volume of pyramid itself!
V = (h/3) * (B + sqrt (Bb) + b).
You can find the derivation of this theorem in any textbook of solid
geometry. In order to understand it, you need to know that if a
pyramidal surface is intersected by a plane parallel to its base, the
ratio of the area of the intersection to the area of the base equals
the square of the ratio of the distances of those sections from the
vertex. The algebra involved in the proof is a bit messy, but
certainly within the capability of a high school student. You can
prove this theorem in another way, using the prismoidal theorem [V =
(h/6)(B + 4M + b)], and you can also find this derivation in most
solid geometry textbooks.
Hope this helps.
Let me to explain further. It was not a question of level of the student, but
of level of the mathematics. I try to avoid any proof requiring integration
or other kind of infinite sumation or limiting process, so no exhaustion,
no devil' staircases, etc... Even the "previous" theorem you quote, I am not
sure if it requires the use of exhaustion principle.
On other hand, there are proportions where you can calculate the volume
just by cutting and pasting pieces of the pyramid. For
instance, the trivial case B=b, for every h. But
also Sqr(B)= Sqr(b)+ 2 h, where the corners paste into a small cube.
So I wonder if there are more proportions solving the problem
In principle there is a theoreme telling that a object can be cutted and
pasted into a cube only if its "Dehl Invariant" is zero; but I am having
a bad time doing this calculation.
8 x ARCCOS(1/A) + 4 A x ARCCOS(-1/A*A)
with A=SQRT[[SQRT(5)+3]/2], and tensor product, x, over the naturals.
The only way to cancel the invariant should be to have both ACOS
to be rational multiples of PI. Are they? I believe not.
pana...@otenet.gr (Eur Ing Panagiotis Stefanides) wrote in message news:<knaio88f16o1@legacy>...