Pyramids: Newsletter Challenge (08/21/07)

"I generally try to fudge the formula so the two legs are equal and multiply by 1.5--close enough to the square root of two and yet easier to juggle in one's head."

Hey, be careful! Someone might accuse you of being an engineer!

The above diagram* illustrates how a prism can be split into 3 parts of equal volume. The volume of each will thus be 1/3 of it's triangular base x height (ie 1/3 of the prism volume). If this leap of faith is accepted, consider the middle of those 3 parts ; 4 identical such pieces could be used to construct a square based pyramid, the volume of which would be 1/3 of it's base area x height. Clearly the height of the original prism can be adjusted so as to construct any height of square pyramid. I'm not so sure about the 'leap of faith' but the extrapolation of idea was mine**.

*Not my own artwork of course. Credit to AnnaGreta Nordvall

** Inspired by toy pyramid that once fell from a Christmas cracker - a small puzzle to assemble 2 identical pieces into a tetrahedron.

If this leap of faith is accepted,

"WHAT?? WHAT?? Are you trying to pull a fast one, or what?" says ken, as he draws his knife. How would I know each of those three pyramids is of equal volume?

It will take more than pretty pictures to win this challenge.

Well, bearing mind that 2 are the same, you have to slice them in your mind and juggle the shapes a little, then figure out the 3rd !!

Ah ha! An excellent link. The conical pyramid within has inspired me.

Re your link: be scared, be really really really .... scared
_________________
This quote intentionally left empty

Yeah, now imagine everyone of those kids, in about 5-10 years, carrying either an automatic rifle or an RPG. I hope they're on our side, not coming at us!

From the film "Red Dawn":

Col. Andy Tanner: ...The Russians need to take us in one piece, and that's why they're here. That's why they won't use nukes anymore; and we won't either, not on our own soil. The whole damn thing's pretty conventional now. Who knows? Maybe next week will be swords.
Darryl Bates: What started it?
Col. Andy Tanner: I don't know. Two toughest kids on the block, I guess. Sooner or later, they're gonna fight.
Jed Eckert: That simple, is it?
Col. Andy Tanner: Or maybe somebody just forget what it was like.
Jed Eckert: ...Well, who is on our side?
Col. Andy Tanner: Six hundred million screaming Chinamen.
Darryl Bates: Last I heard, there were a billion screaming Chinamen.
Col. Andy Tanner: There were.
[he throws whiskey on the fire; it ignites violently, suggesting a nuclear explosion]

Kris,

That is, I think, a correct answer. I have done much the same, except starting with a right rectangular solid, but couldn't think of a way to "sell" it as proven.

Tom

Yes. I have a problem with what Kris describes as the "leap of faith". Euclid apparently thought it needed proving, and used calculus...

Hi Tom. I think the drawing approach is the way to go. I had started to experiment with a slab of cheese this morning, but I was dragged elsewhere and then lunchtime approached.....! With a bit of dimensional tuning, and some additional slicing, I think the problem can be solved. Proving the volumetric trisection of a prism as shown, is possibly easier than doing so directly with a pyramid.

Ach du lieber! You had it with the movement of one diagonal and then all three pieces would be identical. I should have read this closer but I was intent on my rectangular parallelpipeds.

Tom

What calculus?

A volume of a pyramid is simply the base area X the height divided by 3. Where do you see the calculus? is a bucket full of water means calculus?

Wangito.

Brilliant! How did they find that out in the first place? 19C BC copy of the rubber bible, perhaps?

"...And Moses came down from Mt. Sinai, saying "God gave me these fifteen....<CRASH!>, no, ten, TEN COMMANDMENTS"

From the banned Book of Moses: And Moses thus spake to Aaron, "God wrote that the volume of a Pyramid shall be one third of the area of its base multiplied by it height. That's how we knew when the Egyptians were working us beyond our contract and we had to go out on strike, because we measured all those gigantic rocks they made us haul up their silly Pyramid. Hey, nobody's gonna cheat the Israelites on my watch!"

Nice - and that's some trade's unionist, citing the law before it's been written.

Always got to pick those nits, eh, Fyz? Even in a joke? Since it's my joke, I can decree that God gave Moses the "10 Rules of Pyramid Building Labor Negotiation" before they left Egypt! But then if you have to go to that detail, you take the fun out of it.

Nice!

Okay, I cheated*..., not having time to think about this, and being 200 miles away from my senior school notes to see if we actually covered the proof, or just learnt the formula. Maybe I'll play over the weekend.

The formula given in the link doesn't use limits, but it does use "adding up rows" (I'm sure there's a fancy term for that, but I've forgotten it) so I'm not sure if it meets the Challenge. Adjudication Fyz?

*Don't look if you want to keep on playing!

Like Euclid's proof, it depends on the use of limits, and in a way that is merely an explicit presentation of the limit-to-zero-size-step that you would find in integral calculus (if you were to make the approximations for this particular case).
It's a good proof, but the object of the challenge is to solve this problem without using a method of limits.

BTW, I still haven't managed to create the congruent tetrahedra claimed by John77 when the three dimensions are different. If that can be done, it's a flawless proof.

Otherwise, we are down to my original method, which does require one assumption that I believe was commonplace at the time; I'll reveal that on Friday evening (Greenwich meridian based timing). It could prove a bit embarrassing, as I suspect that many will regard that assumption as being one too far.

Fyz

I know what Fyz's reaction will be to this: namely that it is a basic form of calculus.

I proposed a similar solution in post 10, but rather than approaching a limit, said that the area is half way between a ten layer wedding cake just outside the pyramid, and a nine layer wedding cake (same layer thickness) just inside the pyramid. I think that if you do the math, you come up with .335 for the constant in the formula. The Egyptians, having a preference for ratios of small integers, would call that 1/3. (If 22/7 was OK, then why not 3141596/1000000?)

I might argue that slicing up oddly shaped things into rectangles or rectangular prisms (and then adding up areas or volumes of the pieces) is a basis for calculus, but it is not "calculus". Calculus, as we know it, was formalized by Newton*. I suppose an analogy would be that while an engine might be the basis for a car, you don't need to have a car to make a lawnmower. Simple arithmetic is a basis for calculus, but it is not "calculus".

I think the idea of approximating the volume of a pyramid via the wedding cake technique would have been intuitively obvious to Egyptians, just as approximating the area of an oddly shaped figure by counting graph paper squares is obvious to a fourth grader. The fourth grader sees, quickly, that using graph paper with smaller squares provides a more accurate answer. Does that mean the fourth grader is "using calculus"? I'd say no.

No one knows how the Egyptians arrived at their formula for calculating the volume of a pyramid. I'd guess that they did so empirically and intuitively, rather than by "calculus". Play with some wedding cake layers, and you can see that the volume of a pyramid with a given base does not change as the slices are skewed one way or another. Slice the steps off at the midpoint of each step, and you can see that the pieces you've sliced off can be flipped over to fill gaps, making a side smooth (except, of course at the corners). As Egyptians built larger and larger pyramids, and as they went from stepped to smooth, I'd bet they could see that 1. the wedding cake approach got them pretty close 2. that using more layers got them closer, and that 3. the constant in the formula certainly appeared to be approaching 1/3. I think a bright fourth grader could arrive at the same conclusion in a day's time spent cutting up cakes.

Did Egyptians do formal proofs in 1800 BC?

* I'm writing that only because you are from the UK. If I were writing to Andy Germany, I'd write "Leibniz".

I'm writing that only because you are from the UK. If I were writing to Andy Germany, I'd write "Leibniz".

OMG, Newton vs. Leibniz! Please don't open that can of worms!

Calculus* in the sense I was using** is the use of infinitesimal differences for the purposes of exact calculation. More to the point, the question was posed with reference to the Wikipedia article, which would presumably have implied Euclid's proof using infinitesimals, or possibly something cruder.

Whatever, the challenge is intended to be: Prove the factor 1/3 without using infinitesimals or morphing (for this purpose, I say we take between 33332/99998 and 33334/100000 etc to be using infinitesimals)

Did the Egyptians use formal proofs: how should I know? - so much of their papyrus for things that did not require permanent record has been composted over the years. Clearly, accounts which showed who owned what would be kept much safer than mathematical archives that could be recreated.

*Regardless of whether you prefer Newton or Leibniz - I'd generally go for Leibniz because his notation was clearer for the general case; perhaps also because Newton was at Trinity

**Of course, the real meaning of "using calculus" is "getting stoned", but I don't see how that helps here.

I'm with you on the wedding cake approach - my first thoughts were along those lines; as you say, it is an intuitive approach. As posts had already been made on that and ruled "out" by Fyz, I went looking for something else. What I found is not different from the wedding cake, it just puts some numbers/algebra to it and comes up with a series and formula, hence giving a proof. I was pretty sure Fyz would disallow it.

I have come up with three more solutions to the problem (but not the challenge).

Guess what? It always ends up with a prism and three pyramids and i have to proove that one triangular pyramid is the half of the rest. Duhhh.

And a question: we have a right angle rectangular-based pyramid and we split it in half along the diagonal that corresponds to the right angle. Now we have two right angle triangular pyramids wich have the same base dimensions but they have the right angle on a different corner. Can we say that those two pyramids have the same volume because of exact same base dimensions, exact height plus the fact that they are right angled? I mean they are the same volume but i don'tknow how much anti-calculus that might be.

If yes, that leads to a solution to the challenge

We have two lines vertical to a plane. Can we proove that these two lines are parallel without extending them to infinity(Ohhh no, that seems like kkk-kk-calculus? Aaaaaa)

In no particular order:

Parallel lines were used at the time, though the definition was perhaps dubious.

Congruence is OK - if the pyramids have identical rectangular bases and the apex is the same height orthogonally above a corner they are either identical or mirror images - therefore congruent and equal volume. (I haven't found a way to get congruent groups of three, however).

N.B. For convenience, I'm trying to use the term "tetrahedron" whenever I have a "triangular based pyramid".

Fyz

Length X width X Height / 2

????

Have you tried this to see if it works?

Not only does this miss the point ("how do you create the formula"), but this also looks to be the formula for the volume of a prism?

"...this also looks to be the formula for the volume of a prism?"

Only for a triangular-based prism. A rectangular prism is Length x Width x Height, but NOT divided by two!

Well, just by looking at the rectangular-based pyramid, you will notice that its volume is 1/3 the volume of the rectangular solid that the base would create

"Well, just by looking at the rectangular-based pyramid, you will notice that its volume is 1/3 the volume of the rectangular solid that the base would create"

Oh, you can tell that, "just by looking"? Wow, you must be some kind of geometric idiot savant, you can juggle and rearrange 3-D spatial volumes in your head? Unless you can show how you can add two more identical pyramids to form "the rectangular solid that the base would create" I don't see how you can tell "just by looking" that the volume is 1/3.

Please explain, Rain Man! (I give you this name since you only appear as "Guest".)

What do you say to this, Fyz?

I had assumed it was intended as a joke.

If not, I'd say of your comment:
"Je d'irai meme plus*. On ne peut pas le voir.

But of course I can tell just by looking, same as I can see that the stars are stuck to the sky and the moon shrinks as it rises in the sky.

In any case, I wonder what tolerance you would put on "just by looking"?

(But I try to avoid the term "idiot" except when referring to my brother [matter of principle] and where my own personal mistakes are concerned)

*Dupont - or was that Dupond?

Fyz

We used to say of someone who had the capability to visually estimate lengths and volumes, that they had "calibrated eyeballs", and there is some truth to that, to a degree, but not so that one can easily see a pyramid is one third of a rectangular prism with the same base and height, methinks!

This looks like a good place to insert my solution since I think I'll get a few potshots.

The original manuscript uses h/3 without explanation and that suggests eiher calculus or simple borrowing from somebody else. As I read through 8 or 9 problems, I think these Egyptians were applied mathematicians and engineers rather than pure math types. So, here's my version of an applied proof, and I'm even gonna try to duplicate their language a little:

{Since I don't know how to insert a drawing, use your imagination. There are 3 cylindrical vessels, a, b, and c in order. b is connected to both a and c by a tube at the bottom which has a valve and a check valve so that you can get flow from b to a but not from a to b (and the same for c). Both sets of tubes, valves, and check valves are identical.b has Nile water in it.}

Lo! Observe the quantity. Divide the quantity (open the valves) so that a and b are to be the same. Divide the quantity so that c and b are to be the same. Let all division to be now all at the same time. See! The quantity is now in 3 vessels. Every vessel has an equal amount. You have found one-third. Let it be written! Praise Pharoah! Let it known this works for any n. So spread this word throughout the land.

Now, what you have found is that division of a quantity into 2 equal parts simultaneous with a second division of like manner produces thirds without the need for communication between the new parts. After weeks of trying this with sand, water, bad wine, etc, everyone agrees this seems to work and it is entered into the Egyptian Everyday Math Papyrus, using hieroglyphics that include a bird that looks like a duck.

So, you then apply this to a rectangular parallelpiped (or whatever the locals called this back then) where you do the division with diagonals on two perpendicular faces. You know the volume of the rectangular parallelpiped to be abh. So now you clearly have abh/3. Four of these remainder thingys will make any right pyramid. To get frustri, simply subtract a little one from a big one.

Fire away.

Tom

You could easily be right that that's analogous to what the Egyptians actually did.

But it's measurement followed by calculation; measurement was not mentioned in the challenge, and neither was it intended (or indeed needed, though you may regard the basic assumption as pushing things a little).

Fyz

I don't think the actual measurement is central. The idea is the definition of a fraction and the application to the volume and those can be seen, I believe, in the ideal. But, I take your point about this not being specifically an answer to the challenge. I'm certain you can do this exact same thing in Boolean, but I'm not up to that.

Good question, though. If not the best, close to it. Lots and lots of sidetracks, research, possibilities. Plenty of thinking here.

You are clearly distrusting simplicity, so I applaud your internal consistency. The proof, however, appears to take a logical leap or two that even Superman could not manage.

If your divided solid is a parallelepiped, then where do the "perpendicular faces" come from? And are the "diagonals" lines or planes?

I think 3Doug is using the terms correctly: he did write "rectangular parallelepiped".

Fyz

Are we all going loony? When you wrote "3Doug" did you mean "TVP45"?

I didn't see (or at least read) the "rectangular". I must have automatically jumped to the word with lots of syllables, like a crow to shiny objects.

Take a model of the pyramid whos base area is 1 by 1+n units, place it in a full tank of water saving the displacement, put the displaced water into a empty tank whos base is 1 by 1+n units and measure it's depth.

The volume of the pyramid = (area of the base) * hieght * 2 * (depth of displaced water).

This is a variation of the displacement method described in post #44 (mine). Using your method the depth of the displaced water will be 1/3 the hieght of the pyramid. The formula would be correct except for the factor of 2. Also for a rectangular pyramid the area of the base is just L x W so the resulting formula is

Volume of the pyramid = (L x W x H) / 3

Each time you do this, you can provide an approximate result for a single shape of pyramid. We can certainly provide a more confident approximation (without invoking infinitesimals) by using the wedding-cake method - we just need to define the required accuracy and use sufficient layers.

So far, we have three specific pyramidal shapes for which we can assess the volume exactly - by dissecting a cube.

"So far, we have three specific pyramidal shapes for which we can assess the volume exactly - by dissecting a cube."

Fyz,

Are you saying this is the answer, to develop an assessment of each particular pyramidal shape by dissecting a cube, or do you have a more generic, fits-all-shapes approach?

By the way, you never responded to my earlier partial solution (#35) where I reassembled the four quarters of the pyramid into a hollow rectangular (square) prism, with a piece missing inside that looked like this:

This piece, which I call the "creamer" from its resemblance to those coffee shop cream packets (sour cream also, I think) would be a special case of an irregular tetrahedron based on an isosceles triangle made from side of the pyramid. The two short ends are equal to the diagonals of the base of my rectangular prism and the four long edges match the edges of my pyramidal sections.

Now if only I had an easy way to calculate the volume of this!

Aha! I found the "creamer". It is a package called, what else, "Tetra Classic", from the company "Tetra Pak".

Hi STL

What I'm saying is that the challenge has (so far) demonstrably produced three shapes for which the challenge has been solved. These are RCapper's dissection of a cube in post #13 which produces a pyramid with a square base and apex that is above the centre of the base; the other two shapes are the one-half and one-quarter sections of that pyramid mentioned in later posts.

My solution is completely general for a rectangular-based pyramid. Actually, it also works for a tetrahedron, so it can be extended from there to any polygonal pyramid. The reason for choosing a rectangular base was merely that it is easier to visualise.

However, the tendered solution does rely on an assumption that I believe was fully accepted at the time - probably without proof. So you may regard it as half-way to the 'morphing' methods that I believe were not accepted at the time.

I didn't respond to the partial solution because I too couldn't see a way forward - but that doesn't mean that none exists, so I wouldn't want to be discouraging.

Fyz

STL,

I'm being serious here. No puns. In my sectioning of rectangular parallelpipeds, when I goofed up on where I put my diagonals, I sometimes got volumes like that. I couldn't figure out anything to do with them so I just threw them away. If I recall correctly, their volume was about 1/24 of abh. The next time I get to work on this, I'll try to remember to keep one and examine it.

Tom

6 identical square based pyramids + 4 'creamers' = 1 large square based pyramid.

1 square based pyramid = 2 creamers.

Anybody got a scalpel ?

< Kris rapidly scampers off >

That might lead somewhere if you can create a rectangular block using identical creamers and pyramids.

Fyz

I got to 1 creamer = 2 neolithic axe-heads, and my Play-Do got too squidgy ! I was hoping for another pyramid.. Then I.....well, I think you get the drift fyz.

Yes the creamer as you call it! I pattented this in early 2000. Your right it is the center space of a pyramid and also a diamond! but the volume of this well i call it a "nook & cranny" or "perpetual block" can exist in any geometrical shape! Now the important thing to note is that the nook and cranny inside the pyramid was designed for a reason but not concluded by slicing a wedding cake so to speak as you remarked. This is the ideal chamber for zero resistance. This is a chamber designed to harnes an energy source of great magnitude. Possibly even Virtual reality without the headset. Its ideal for sound containing no right angles! What is fascinating about sound is it travels the same way as way in the same manor and same fashion except light can pass through a vacume. You see this particular volume is the key to pi this is the structure of a containment field to transport matter through time space continuum and or create an atmosphere in wich a vessel can be sheilded during interstellar travel. The points or vwerticies are where the nodes would emmit from and recieve with. This is the only way to regulate a constant energy source without spastic fluctuations and the only way to accuratley monitor the acitivty within the chamber or containment field. I saw you mentioned its extremely difficult to measure the volume of such an entity. Also as you mentioned inner layers and exterior layers ...ok think of an atom and shells of electrons each electron has a reciporacle and an orbit. The verticies here are those electrons and they fix the trajectory from one shell with electrons orbiting to another. please contact me I am looking for people to help produce this product wich i have tons of ideas and pattents for! This product will revolutionize every market known to man. it will even advance technology in such a way that could be very dangerous for our safety. but there are many harmless products for everyday use as well!

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oh and to answere how this fits into a payramid its simple the pyramids actually represent a hemispher so you would need two pyramids to complete what you only see half of on the surface. each pyramid must be stacked on top of one another with the bease of each pyarmid connected. I believe this volume is actually 1/32 which would make sense pertaining to a surveior usind mils on a scale of 64 wich half is 32 and can be converted into degress! So if 32 of these were massed each would represent a bandwidth of degrees making 32 degrees or 32 bandwidths of 11.25 degrees. wich leaves a margin for era of aproximatley 1.25 degress hence half of 100 = 50 and half of 50 = 25 and half of 25 = 12.5. 12.5-11.25 = 1.25. 1.25*2=2.5. 25-2.5= 22.5 = 11.25+11.25. the notation is ment to express the increase in algorythmic exponention with each increase in capacity leaving the containment field to harness thenergy within parameter of each range in incremtes of 1.25 starting at 11.25. You can take these numbers and use the same formula to equate the distribution and integrity of energy contained within the containment feild. This would predicate the neccessary impulses weather electrical sound light and possible all of them to be re introduced to the chamber as required for constant masking and phasing so the energy field in containment does not expand or collapse to the point of fussion or fission overloading and breaking containment or withering into nothing.

did not see the earlier post.

regards JD.

Find the area of one of the triangular walls and multiply the area by the base of one of the triangular walls.

How do I decide which wall to choose - except for a symmetrical pyramid with a square base, they give different (and, except for coincidences, wrong) results?

The question as it appears in the 08/21 edition of Specs & Techs from GlobalSpec:

We have records that the volume of a pyramid was known by about 1800 BC. Wikipedia speculates that some form of early calculus-type of system was used. But the volume of a rectangular-based pyramid can readily be calculated without calculus. How?

The answer to this question will be posted on the morning of August 28th, right here on CR4.

A pyramid of equal sides and half that in height will fit six times in to a cube of side length cubed therefore L x B x 2H x 6 or L x B x H / 3 .

A pyramid of equal sides and half that in height will fit six times in to a cube of side length cubed therefore L x B x 2H / 6 or L x B x H / 3 .

SJAM,

You are only proving the equation for a special case of square-based pyramid with a height that is half of one side. The challenge calls for a general solution to ANY rectangular based pyramid.

Besides, several posters have present identical solutions to the one you have, so presenting it as your own solution, even if incomplete, without acknowledging those that beat you to it, is quite lame.

At least you did not sign in as a Guest!

Fyz might just have us stumped on this one.

STL Eng.,

I realise that a Square base is a special case however it is an easily made shape and it may have been noticed that all six would fit together furthermore if you double the lenght of one side , four of these fit together with two of doubled height.

instead of 6 pyramids of L x B x H e.g. where: L=12 , B=12 , H=6

you can have 4 pyramids L=12 , B=24 , H=6 . With 2 pyramids L=12 , B=12 , H=12

AND BY DOUBLING A DIMENSION IN EACH PYRAMID THIS SHOULD GIVE TWICE THE VOLUME IN THE ASSEMBLED CUBOID.

Once you have 6 unequal pyramids inside a block you lose the concept of the proof for the 6 equal pyramids inside a cube; that is the volume of each pyramid equals 1/6 the volume of the cube. I believe rcapper in post #13 was the first to present this proof.

You have the volume of a block that equals the volume of the 4 equal pyramids plus the volume of the 2 equal pyramids. You need to know the relationship between two different pyramids but you cannot use (L x W x H/2) / 3 because that has only been established for a pyramid with a square base and a hieght of 1/2 of lenght of the base.

Fyz,

You are fiendish! Diabolical! Clever and deceptive! A perfect role model.

I've been worrying about how the Egyptians did it instead of how we might do it.

OK, here goes (without proofs):

1. It can be shown that the corner pyramid of a cube = 1/6 the volume of the cube. From the easy to calculate geometry of a cube, it can be shown that the area of that base times the height of that corner must be multiplied by 1/3 to get the volume.

2. It can be shown, pre-Cavalieri, that this corner has the same volume as a rectangular base pyramid with the same base area and same height. It is trivial to show that there are an infinite number of such pyramids all with the same ratio of base area and height. One of those is a right pyramid with a base that has an aspect of 2:1. That pyramid can be bisected into 2 equal pyramids of half the volume, now with a new ratio of base area to height.

3. By the same method, it may be shown that there are an infinite number of these pyramids, including one that may be easily bisected, leading to 1/4 the volume, ad infinitum.

4. Although this shows only the original volume, 1/2, 1/4, 1/8, 1/16,... I appeal to exhaustion.

I think this problem can also be solved by Heron's formula, but, after all the wise quacks about the duck, I'm not introducing anymore birds.

In closing, I think I should rename you after the James Bond villian: Dr. Know (double entendre intended). I've lost more sleep on this one than I did on Mr. Truman Brain's vacuum sphere.

By the way, do you have access to Gillings?

Tom

what / who is / are Gillings ?

Hi Tom

Do you mean Dr. Knowall?

I'm unclear how your method generalises to include (for example) any pyramids with rectangular bases other than square or 2x1 - have I missed a trick (or maybe several)?

Unless you mean the bakery (there are none near where I live), the reference to Gillings is way over my head.

Fyz

Hello,

The equivalent approach, as in Cavalieri, works for any shape so long as it has the same area..

Sorry about the reference. Didn't mean to be arcane. Gillings is the primary reference in the WIKI article and apparently explains how the Egyptians might have done this. Thus my question. I don't have easy access to good books.

No, I meant Dr. Know. Every time I get in a discussion with either you or Jorrie, I feel like Watson talking to Holmes. On the other hand, both of you are patient and explain stuff. It's a great education for an OF like me.

Tom

I can get hold of RJ Gillings book when I visit the nearest university where i have reading rights - I just checked on-line, and it does have a copy available to borrow. Unfortunately, it's not the easiest place to visit from where I live, so I only go when I really need to. If you are desperate enough to spend a few dollars, Amazon claim to have several copies available in the US. However, the real problem here is likely to be lack of evidence; I imagine that at least one of the Wikipedia writers will have trawled the book before compiling the article. By the way, Giddings's* other published work is "The Mathematics of Denis Diderot", which sounds to be similarly speculative (though I may be doing Giddings an injustice here).

Hiu Liu would take you a few years before Cavalieri, but still nothing like far enough. And I don't think the "equal area at all heights" fully solves the problem of the pyramids - you need either the height proportionality extension or the area proportionality extension as well.

*Apostrophe construction especially for English Rose

Yes, I'm not sure of time lines and have run out of time to follow up for now. So, I must await your clue tomorrow.

Recognising your pun: I imagine your fudge was more tasty than most of the other fudges in this thread.

So why not try following the clue (no-one else appears to have done so)?

Create a pyramid whose edges are scaled a factor of two up from an "Original Pyramid". Now cut an Original Pyramid from the top of the double-pyramid, and see what happens to the frustum when you cut out some prisms. If the apex of the pyramid is vertically over some point on the base, and you keep these cuts vertical and parallel to the edges of the base, you may not even need butter to visualise what's going on.

Fyz

.

Hi their

this is my first post

building a stair piramid with bloks

first layer four bloks and one on top would mean following to the volume

1² +2² = 1+4 is 5 bloks (volume piramid)

2³ =8 the volume of cube

volume of the piramid would be 5/8 of the cube 0.625

next layer of 3 by 3 bloks under it would give

1²+2²+3²= 1+4+9=14 bloks (volume piramid)

3³=27 bloks volume of the cube

volume of the piramid would be 14/27 of the cube 0.518518518

The higer the piramid gets the closer it wil come to 0,33333333333333 so 1/3 of the

cube surounding the piramid but as long it is staired it wil never come to exactly 1/3

regards

satyoda

A standard method taught in many schools is to select the volume of the blocks so that with N layers the blocks are all inside the pyramid, and with N+1 layers no part of the pyramid is outside the blocks. Then the difference between the Volume of the blocks and the expected volume of the pyramid is calculated. It is then shown that the differences between the volume of the blocks and the "expected" volume of the pyramid tends to zero as the thickness of the blocks reduces. That is a way of doing integration for this special case without being familiar with the standard techniques of calculus. But it is calculus, nonetheless.

Apologies if this has already been suggested and rejected on the grounds that it comes too close to using calculus.

Pick any base and apex.......................And complete the pyramid

......

Choose any corner.......................And discard the other three.

...........

Re-draw chosen corner.....And complete the cuboid on the base

.........

Slice the cuboid into two congruent cheeses...........................................

........

Chop off the waste to get back to the corner of the pyramid..

Now use a vertical plane on the diagonal to cut the corner of the pyramid into two different irregular tetrahedron, but, notice that any horizontal slice cuts the two solids with equilateral top triangles.

Add up all the equilateral triangles: and the two solids must be the same volume. This seems a bit dodgy from the point of view of the challenge, but, look at it from the point of view of the irregular pyramid builder: he must need the same amount of stone to make each half of the corner.

Exactly the same method applies to the waste section and the front half of the corner.

Just for completeness the the corner is now 2/3 of the cheese and the cheese is half the cuboid.