Pyramids: Newsletter Challenge (08/21/07)

Hi Randall

That is unusual for you in that I couldn't follow it. The following is my misunderstanding of what you wrote:

I assume that when you write "equilateral top triangles" you mean "congruent....". But the only way I can see to create two tetrahedra from a square-ended pyramid is if we cut the pyramid along a diagonal of the rectangular face - and the two halves only have equal volumes if the apex is perpendicularly disposed to the centre of the rectangular face; as I understand you have already defined the apex to be perpendicularly disposed to one corner of the rectangular face, that is not possible.

An additional difficulty will be proving that each of these halves has the same volume as the cheese-pared tetrahedron that never formed a part of the pyramid.

Fyz

You're right I meant congruent not equilateral.

Have you realised that I'm only dealing with a quarter of the design: the same arguments need to be applied to all four corners. If the apex had been above the centre of the base then the four corners would have been identical: just thought I'd try to kill two birds with one stone.

The apex is vertically above the "inner" corner of the cuboid I have chosen, so the slicing of the two parts of the pyramid is along the diagonal from this inner corner to the near right corner of the pyramid, and, of course the sloping edge/"corner" of the pyramid runs within this slicing plane.

Yes, I realise it's one section (only a quarter if the pyramid is symmetrical) - hence the apex being vertically above one of the corners. You can cut the pyramid using either diagonal of the corner. I've assumed the cut goes through the orthogonal edge. Only the triangles in the base and on the shared cut are necessarily congruent - the other sloping sides are not. So these are not identical tetrahedra - they only have identical bases and heights. To know the volumes are equal you either have to assume that sliding the top around at the same height leaves the volume unchanged (this relies on a concept of adding zero-thickness slices - primitive intuitive calculus), or to rely on Euclid's calculus-based proof.

Have I missed something here?

Fyz

The whole point is that although the two tetrahedrons are very different shapes, any horizontal slice through them both reveals two congruent triangles: A & B. This is a projection from above.

Add all the congruent triangles up and the two solids must be the same volume. I know this is an infinite sum (calculus like), but, I think it would make sense to: say a pyramid designer.

I was thinking that all my drawing and disecting of pyramids might be wrong way to go about this.

Firstly I thought make a pyramid ( scale model ofcourse ) drop it in to a full container of liquid what spills out is the volume of the pyramid ( you could make several to check accuracy )

Then I was thinking why not just take a block of dimensions Length x Bridth x Height pick a spot on top for an apex and cut away the four slopes.

Weigh the block before and after cutting ( again several may be done to check accuracy )

Then use the proportion to calculate the full scale pyramid.

See # 172. Once you've accepted the scaling assumption, dissection and simple calculations are all you need to develop the general case.

Fyz

The technique I used relies on scaling: i.e. the relationship between the volumes of similar* shapes. I'm pretty sure it would have been unproved at the time, except for specific shapes such as parallelepipeds and prisms. But the question said "calculate", and I believe the idea would have been available for that purpose.

Fyz

*= shapes between which all equivalent linear dimensions are scaled by the same multiple, and all equivalent angles are the same

N.B. that my objection to the methods that used a scaled model and measured the liquid displacement was a combination of it being primarily measurement, of applying only to the specific shape of pyramid for which the model was made. (I wasn't too happy that the reference period was about 1500 years before Archimedes either). Had I been worried about the scaling I would have mentioned that too.

I already suggested we use scaling of "similar" figures, where all corresponding angles between the similar figures are equal and all corresponding lengths are in a fixed proportion. Also that you only need this to work with a factor of two (or any single fixed non-unity factor, for that matter).

Now, about that frustum?

How about proof by analogy?

We know rcapper's solution is correct for rectangular pyramids, in which the apex is above the center.

So, we take a vertical plane, and use it to cut through one sloping wall of the pyramid, so that the edge of the cut is a trapezoid. Now we skew the pyramid to the right and note that the trapezoid has skewed, but not changed in area: it still has the same a major and minor sides and same height. (We also know that the volume of the skewed box that would contain this pyramid remains the same after skewing.)

So, just as we know it is sufficient to describe the infinite points along a line with just the end points, we also know that an infinite number of such sections could be cut into the prism. We know that, at each one, the same effect will be observed, namely that skewing these trapezoids does not change their area. Therefore, just as the same formula applies to a cube or a rhomboid, the same formula applies to a pyramid or a skewed pyramid.

I think it is reasonable to assume that the Egyptians knew there were an infinite number of points along a line, and similarly an infinite number of planes across a solid.

I think the intended solution is simpler. See post #172 for a starting point?

Fyz (and you have to catch me before you can burn me)

By George, I think I've got it. No fooling!

The solution requires taking ANY rectangular based pyramid with base of L and W and height of H, put it in a box of L, W, and H, then construct a mirror image box and pyramid on top, apex to apex. Now you can take the ends of the new double box rectangular prism and write a general formula for pyramids, V_p= n x L x W x H, where n is the unknown factor (we all know it is 1/3 of course). Now if you add the 6 volumes they must add up to the double box, V_b = 2 x L x W x H.

I did this and solving for n gave me 1/3.

I will have to send the rest of the proof later because I have to get home!

Hey STL,

I think you have it, but don't you have to use complementary pyramids? I like the wirror plane part. Clever

Tom

Here's my high-tech attempt to replicate a low-tech method for finding the volume of a pyramid.

Let's say the Egyptians wanted to know the volume of a pyramid before they built one to full scale and they decided to create a formula using a cubical block and experimenting. Now let's take a cubic block that is 10 units x 10 units x 10 units. The block will look like this:

We know the cube has a volume of 1000 cubic units.

Now draw a line across the top surface from the midpoint of the front surface to the midpoint of the back surface. Draw lines from this bisector line to the bottom corners of the front and back surfaces where thy meet the right profile surface. Remove material from the block in the area outlined by these lines. The block now looks like this:

Now, what is the volume of the block?

Note that the top surface has lost half of its area, but the bottom retains its full area. So the lost volume must be somewhere less than half than original.

Also note that the front and back surfaces lost area. The edges of the slanted surface are diagonals that bisect rectangles formed by the front and back surfaces and the bisector line on the top. Because a diagonal divides a rectangle into halves, we see that the front and back surfaces lost one fourth of their areas. If a rectangular block is split along a diagonal, it loses half of its volume. We can think of a cube as two rectangular blocks joined at their largest faces. So, this block now represents one and a half rectangular blocks, or three fourths the volume of a cube. We tell our trusty Egyptian scribe to record the formula in the current form as V = LWH – LWH/4.

Now we draw diagonal lines from the first bisector to the bottom corners of the front and back surfaces at the left profile, and again remove the material within the outline. The result is a prism:

Because we have removed the same amount of material as we did in the first step, we tell the scribe to adjust the formula thusly: V = LWH – LWH/4 – LWH/4 = LWH – LWH/2 = LWH/2.

Now we draw diagonal lines from the midpoint of the top edge to the bottom corners of the back surface, and again remove material within the outline to this shape:

We have a similar situation to the first time we removed material. We have removed part of one half of the prism, but it is not one half of that one half. What is the relationship we need to find the volume of this item?

What I believe is that volume is directly proportional to surface area. I have not confirmed this yet, as I need to run some numbers. I will post my findings later.

And yes, I did finish my pyramid. I just haven't finished my math yet.

Using the hint, it looks easy if we can make the leap that V=λabh where λ is some constant number. This is what makes STL's solution (and others, of course, earlier on) so attractive in that it uses boxes whose volume are known to be of the form abh. Is there another way around this? I know we can can take corners of a cube and show the abh/3 form there, but how do I say it applies to all pyramids?

I know we can can take corners of a cube and show the abh/3 form there, but how do I say it applies to all pyramids?

I will show you how my method will apply to ANY rectangular based pyramid, because I will do it without using a cube! I will use a rectangular prism, so the solution is more general.

However, I think I will start with a clean sheet! Look for my next post with the proof.

I will show you how my method will apply to ANY rectangular based pyramid, because I will do it without using a cube! I will use a rectangular prism, so the solution is more general.

But only somewhat more general. Fyz's proof will also pertain to pyramids in which the apex is not over the center of the base.

Also, not to oversell Fyz's proof in any way... but it is said that his proof will turn atheists into believers and believers into atheists.

But only somewhat more general. Fyz's proof will also pertain to pyramids in which the apex is not over the center of the base.

Uhhh.....Ken? Have you looked at my solution in post #205? After all, your quote above is from your post #210. My solution IS totally general. It does cover the general case where the apex may be anywhere over the center of the base.

Geez, I thought I had done it and found a solution that followed all the rules, but I am being ignored, or worse, trivialized! And people are still submitting "solutions" based on the use of a cube instead of a general rectangular prism or worse, fluid or semi-fluid, or even weight analysis. How can fluid or weight measurements be considered a "calculation" solution? Even if calculations are used subsequently to "scale up", it is still based on a measurement other than the Length, Width, and Height of the pyramid!

How about it Fyz? Do you see anything wrong with my solution? It is totally calculation oriented and uses no calculus, only basic algebra and geometry. Heck, I did not even have to use trig! And has anyone presented this complete solution with proof before me? If so, I will be the first to congratulate them, but I have not seen it.

Surely my solution and proof is not hard to follow (unlike some others I have seen), since it uses only basic algebra!

Do I REALLY need to draw you guys a picture?

But only somewhat more general. Fyz's proof will also pertain to pyramids in which the apex is not over the center of the base.

Uhhh.....Ken? Have you looked at my solution in post #205? After all, your quote above is from your post #210. My solution IS totally general. It does cover the general case where the apex may be anywhere over the center of the base.

Geez, I thought I had done it and found a solution that followed all the rules, but I am being ignored, or worse, trivialized! And people are still submitting "solutions" based on the use of a cube instead of a general rectangular prism or worse, fluid or semi-fluid, or even weight analysis.

STL, you wretched moronic nitwit!! I'm kidding, of course, and understand that when you wrote this it was late at night. I think you did not absorb what I wrote, namely: But only somewhat more general. Fyz's proof will also pertain to pyramids in which the apex is not not not not over the center of the base. I hope that you can see that your response to that, namely "My solution IS totally general. It does cover the general case where the apex may be anywhere >>> over the center <<< of the base," strikes me as nonsensical . The more general case is where the apex is, to repeat, NOT over the center of the base.

For example:

As I hope you can see, a rectangular prism which has the same base as the pyramid does not fully contain such a pyramid. So your proof fails from nearly the first statement. I don't think anyone has trivialized your "proof" ... perhaps they are politely ignoring it (letting you feel smug in your knowledge that you have proven something) given that long ago Fyz stated that his proof would cover both right pyramids and oblique pyramids, as well as pyramids with non-regular bases.

Rcapper's proof pertaining to pyramids which fit into a cube clearly, simply, and elegantly proves the formula for the special case of pyramids that fit into a cube in the stated way. Your proof adds little to that: your are still dealing only with right pyramids, but your proof takes up far more space.

As far as others posting possible proofs that rely on measurements, etc... Isn't it unreasonable to expect people to thoroughly read all 200 plus posts here? It is also reasonable to assume that in 1800 BC proof methods might be quite different than they are now. It is very likely that Egyptians simply arrived at the formula empirically (a view which some historians hold) and that formal proofs were of no use to them. They would no more "prove" (in the Euclidean sense) that their formula is correct than we would "prove" (in the Euclidean sense) that F=MA. So any post that proposes a method the Egyptians might have used seems like a valid contribution. We can only gain from such posts -- one can certainly choose to ignore them, but there may be some interesting insight in them. I especially liked the post from the writer of the upcoming book: It can make one wonder about the difference between weighing in the conventional sense and the balancing of equal weights. The latter intuitively proves scalability, but the former does not.

First of all, let's not change the rules in the middle of the game. Fyz stated that it must include the case where the apex does not lie directly over the center of the base. Fine. My solution does that. Now you want to ADD that it must include cases where the apex does not lie over the base AT ALL! Now who is doing things the Egyptians would never have considered? ("Hey Pharaoh, that new pyramid is looking a bit lopsided, don't you think? Aren't you afraid it might fall over?")

Also, as pointed out, using the cube to find the answer requires you to ASSUME that the factor is universal and extends beyond the simple cube example, whereas my solution does not use the cube, but actually includes positioning the apex ANYWHERE over the base. And it PROVES that the solution is universal, not ASSUMES it. If you want to use the term "oblique pyramid", fine. I believe you will find the definition does NOT put the apex outside of the base, just that it cannot be over the CENTER of the base, which is exactly how Fyz stated it in #22, so why are you pushing this "outside the base requirement"?

Finally, when I read the challenge, the question of Egyptians was to state that calculus was not needed, as the Egyptians did not have calculus and the question was simply stated as:

But the volume of a rectangular-based pyramid can readily be calculated without calculus. How?

So once again, I ask, how does my solution NOT meet the challenge as stated in the original question, even including Fyz additional requirement, or clarification, in #22? It may not be what Fyz has in mind, but it certainly solves the challenge.

Well, I finished my math, and did not find any direct porportionality between area and volume, at least to say that if a solid loses a given amount of surface area it will lose the same porportion of volume. That relationship may be true when the loss does not create a surface with a different angle than before. I still have more ruminating to do on this.

IF you mean what I think you mean, you won't find one. For example: if you remove material from all surfaces in a way that allows the solid to remain the same shape, the relationship is V = k.A^1.5
(Where k is a proportionality constant that depends on the specific shape.)

Fyz

Ok, Tom (Guest #194). I don't know if this includes the "complementary pyramids" you mentioned, but, as I promised in #192 earlier here goes:

First construct a rectangular pyramid with a base of length L and width W and a height of H. Then construct a box (rectangular prism) which encloses the pyramid, with the same dimensions, L, H, and W.

Next construct a mirror image box and pyramid above and touching (at the mirror plane) the first two, corners to corners (of the boxes) and apex to apex (of the pyramids). Now you have a new, double box of dimensions L, W, and 2H containing both pyramids. You will notice that the faces of the new double box are now each a base for new rectangular pyramids, all of which meet at the same apex on the mirror plane. We will describe the volume of each pyramid by a general formula, V_P = n H L W, leaving out the multiplication signs for brevity. The universal constant, n, is to be found.

If the apex is located at any point on the mirror plane such that it divides the length, L, and width, W, into two shorter lines, L₁, L₂ and W₁, W₂, such that L= L₁ + L₂ and W = W₁ + W₂, these shorter lines will be the heights of the other four pyramids.

Now we can easily see that the volume of the double box, V_B = 2H W L, as each single box has a volume of V = H W L. The double box exactly contains all six pyramids, so that:

V_B = V₁ + V₂ + V₃ + V₄ + V₅ + V₆ and the pyramids' volumes are:

V₁ = V₂ = n H W L

V₃ = n 2H W L₁

V₄ = n 2H W L₂

V₅ = n 2H L W₁

V₆ = n 2H L W₂

Substituting these volume formulas in the summation equation above we get:

2H W L = (n H W L + n H W L) + n 2H W L₁ + n 2H W L₂ + n 2H L W₁ + n 2H L W₂

Simplifying:

2H W L = 2Hn (WL + WL₁ + WL₂ + LW₁ + LW₂)

Then, WL = n (WL + WL₁ + WL₂ + LW₁ + LW₂) and

1/n = WL/WL + (WL₁ + WL₂)/WL + (LW₁ + LW₂)/WL and

1/n = 1 + (L₁ + L₂)/L + (W₁ + W₂)/W = 1 + L/L + W/W = 1+ 1 + 1 = 3

So, if 1/n = 3 then n = 1/3 and the universal formula is V_P = 1/3 H L W.

Q.E.D. !!!!!

Is this what you had in mind, Fyz?

I had to read that a couple of times, but it looks good. Is the assumption of a 'universal constant' valid though ? I think it probably is, but I have a nagging feeling that Fyz has sometjing very elegant lined up, that can probably be expressed in just a few lines. I can visualize what you mean in my head, but it would be nice (if you have time) to see one of your cool drawings just to re-inforce the description. In essence, I think the idea was mentioned before though I'd have to trawl through this thread to find where.

"I can visualize what you mean in my head, but it would be nice (if you have time) to see one of your cool drawings just to re-inforce the description."

Well, I don't have any CAD at home, so here's the best I could do with MS Paint.

Numbers 1 through 6 show the enclosed pyramids and Box 1 and Box 2 are the stacked rectangular prisms that enclose them. L, W, and H are Length, Width, and Height.

Oh, although some of the pyramids may look like square based pyramids, as H+H looks very close to W, it isn't necessarily so. Imagine H as very large or very small and it makes no difference.

If I understood how to "Fill the voids between adjacent pyramids using 8 of those 1/4 pyramids", I'd be right with you. I really can't see how they fit. That doesn't necessarily mean it can't be done - just that I can't see it.

Fyz

Not what I had in mind. Once you assume a universal constant, you only require a single case. So you only need the six pyramids in the perfect cube that was proved very early on. If you think about it, the universal constant assumption already implies both morphing and linear effects of linear stretching....

The proposed method only requires scaling of identical shapes (an admittedly unjustified assumption, but it's a calculation, not a proof, and this is the smallest I know to make), plus cutting-and-reassembly, knowing the volumes of prisms (itself easily proved by cutting-and-eassembly), and arithmetic.

"Not what I had in mind. Once you assume a universal constant, you only require a single case."

Granted, it may not be what you had in mind, but doesn't it correctly address all the particulars in your challenge? And, I did not ASSUME a universal constant, I proposed it, then proved it by creating formulae that covered the general case for a rectangular pyramid, and solved for the constant. How does that NOT solve the challenge?

If you assume a universal constant, you have proved nothing. Solving for a specific case using the cube only proves that it works for that example, not for all. I don't see how solving for algebraic equations that are the parameters of the general case can be considered "morphing" or "linear stretching". Maybe I just don't understand how you are using those terms.

I don't see how solving for algebraic equations that are the parameters of the general case ...

... if only it were the general case. The general case is an oblique pyramid. Your "coffee creamer" is a tetrahedron, which is also an example of an oblique triangular pyramid.

Back in post 22, Fyz wrote:

The challenge is to find a way that works for any rectangular-based pyramid, regardless of relative height or whether the apex is over the centre of the base.

... if only it were the general case. The general case is an oblique pyramid. Your "coffee creamer" is a tetrahedron, which is also an example of an oblique triangular pyramid.

Whoa, there! You are confusing my earlier attempt with my final solution, which has nothing to do with "creamers" and everything to do with the apex being anywhere, not just over the center of the rectangle. Don't you see that? My solution does meet the requirements he posted in #22.

And this is a good place to point out that I am the numbskull who screws up terminology. Now that I understand what an oblique pyramid is, what do you call one where the apex is not above the base at all? That's the one I was busy worrying about.

Tom

Tom,

As Blink correctly pointed out, that case where the apex is not above the base at all is also an oblique pyramid, but it is usually not considered. Such a structure would be cantilevered and not necessarily self-supporting if it were constructed with uncemented blocks. At some point, and I am sure it depends on many factors, including block size, weight, perhaps even friction (as the blocks may have to slip somewhat before falling off) if the apex were far enough outside of the base the upper portion of the pyramid would topple over.

I remain unconvinced that the challenge, as stated by Fyz, must include such hyper-oblique pyramids.

Actually I wasn't jumping back to your earlier attempt to suggest that your current attempt relied on coffee creamers. I was just using your catchy terminology to elucidate what I meant by an "oblique pyramid".

Your current attempt (post 205) seems problematic in the following ways:

It starts by saying to construct a rectangular pyramid. So far so good. Here's my pyramid:

Next, you say: "Then construct a box (rectangular prism) which encloses the pyramid, with the same dimensions, L, H, and W." At this point I'm stuck, because I can find no rectangular prism that both encloses my pyramid, and has the same dimensions, L, H, and W. Can you?

If I could get past that, then I could move on to where you claim there is a universal constant that pertains to all the pyramids in your construction (and presumably to all pyramids in the real world). You appear to be starting out with that which is to be proved: namely that there is a universal constant. The math would work out the same, if there were actually six different constants, n,o,p,q,r, and s with their average being 1/3. How do you counter the argument of someone who says "A highly oblique pyramid looks very skinny; therefore it must have a different constant, maybe .25. In your construction, the fatter looking pyramids have a larger constant... maybe .4. Cones, being round, must therefore have another constant." I can't see that the math proves that n (and only n) applies to each of your 6 pyramids -- you seem to be assuming that.

I'm probably being obtuse.

Look, we could argue all day about whether or not your extremely oblique (apex not over base at all) pyramid should be included in the challenge. I will admit that it would not work for those. My point is that this does solve the challenge for all ordinary rectangular pyramids that might normally be constructed. In all the examples which I have researched on web that attempt to teach the mathematics and geometry of pyramids, you almost NEVER see your example given.

But to say that my methodology is wrong because I am starting out with that which is to be proved, then finding the proof by solving for a single number mathematically is incorrect,...well I just don't understand your thinking. When I went to school we were taught not just, "here is the theorem, learn it and use it", but also, "here is the proof of that theorem". So I approached the problem in that way.

The hypothesis is that there could be a single universal constant that fits all cases of the normal (non-extreme) rectangular pyramid. Formulas were developed for the general case where six pyramids just fit inside a box, using the same three variables. This allows us to also add the formula for the sum of those values and set it equal to the volume of the box. If these equations are solvable for the constant, then any other pyramid that uses those same variables can be solved by the same equation, because none of them are a distinct and limited case, like the regular pyramid would be. These equations in fact conform to the parameters where the apex may lie anywhere over the base.

I then proceeded to mathematically prove that my constant, n, was equal to a single real rational number. Where did I throw out any possibilities that n could be any other number? I did not assume that. It worked out by itself. If there were any other numbers possible, then the answer would have been some polynomial or other formula and not a single constant.

I am still waiting for Fyz to comment on the viability of my solution. It may not be the same as his, after all, there can be many ways to skin a cat (ooh, sorry, Del! ), but that does not mean it is wrong.

I am still looking forward to see what Fyz had in mind tomorrow!

Hi STL - all that is irrelevant. An overhanging pyramid can be created by subtraction of suitably formed pyramids with a vertical right corner if need be.

The real problem is that, so far as I can tell, your "proof" would only work for the original square-base half-height pyramid (six per cube). The reason is that otherwise you have (at best) three different pyramids duplicated in reflected pairs, and the only way to equate their volumes is to assume that it doesn't matter which dimensions make the base and which is the height - which is effectively assuming the universal formula. And, if the original pyramid does not have its apex above the centre of the rectangle, the volumes of at least one of the opposing pairs will in general* be different - see my post #260.

If I've misunderstood what you have done, please provide more explanation.

Fyz

*There are some special cases where they are the same, but they are not particularly interesting for this purpose.

"The reason is that otherwise you have (at best) three different pyramids duplicated in reflected pairs"

NO,NO,NO. If in my diagram the apex were very near the left side, for example, or even on the edge, the opposing pyramids would not be reflected pairs. They would hardly be equal in size, one could even be zero in volume! Only the top and bottom would be equal in size. The other four pyramids do NOT have to be half-height for this to work. The mirror box was constructed simply to make the identification of the other general pyramids simple, and to simplify the math needed to solve for n. Since I can find the same n for non-half-height pyramids as well as for half-height pyramids, it must be a general case. The cube solution does not do that.

Maybe you are confused because I used L1 or L2 and W1 and W2 for the height of those other four pyramids, but I could just have easily used other variables, z, y, x, w, v, u, t, s, or r! (since a through q are normally used for constants). I thought it would simplify the thought processes needed to visualize the geometry, but maybe I was wrong.

Hi STL

I must be misunderstanding what you have done.

What I see is a pyramid whose apex is off-centre reflected about a plane through the apex that is parallel with the base. This creates a second pyramid and a frame for a rectangular parallelepiped. Within that frame you have six pyramids. But the volumes of these pyramids aren't equal; for example the apex could be directly above one of the edges, in which case (at least) one of the pyramids has zero volume.

Please clarify.

Fyz

Yes, Fyz, exactly right. That pyramid would have zero volume, since its height (above its base) would be zero, but the other five pyramids would still add up to the volume of the "double box" which exactly contains them. Put the apex at a corner of the base rectangle, with two pyramids of zero volume. It still works. That is the beauty of solving this problem algebraically using variables, all cases are covered by the general formulas. That is why I split the Length, L, into L1 and L2, and the width, W, into W1 and W2, which would represent the choice of any point over the base of the first pyramid. The formulas (formulae?) for all the pyramids were then put in terms of these variables.

Now, do you see any error in my logic or algebra?

I had assumed I had communicated what I take to be the fallacy, which was why I was assuming the drawing was being misinterpreted. Either I hadn't communicated the fallacy, or you disagree, or I'm still missing something.

What I am saying is that you are assuming that a single formula exists that connects the base area and the heights of the pyramids in your drawing. In fact, you are assuming more than that, - you are assuming it uses the product of the base area and the height to the apex in a direction orthogonal* to the base and a fixed constant will give the correct volume. Without assuming such a formula, I don't see how you are calculating the relationships between the volumes that make up the rectangular parallelepiped.

If you do assume such a formula, I can't see why you can't simply apply it to any pyramid to find the constant. In which case so you could base it on the pyramid-in-a-cube (sounds like a formulaic restaurant chain) without going to the trouble of making this construction.

What am I missing?

Fyz

*To put a rather ridiculous counter-example, how do we know in advance, for example, that the formula should use the base area times the orthogonal height, rather than the average area of two opposing faces multiplied by their internal orthogonal separation nearest the base? Or that the constant is not a shape-dependent variable, or...

"To put a rather ridiculous counter-example, how do we know in advance, for example, that the formula should use the base area times the orthogonal height, rather than the average area of two opposing faces multiplied by their internal orthogonal separation nearest the base? Or that the constant is not a shape-dependent variable, or..."

Because this is a theorem, a proposition. A theorem, like a theory, does not need to be known for certainty in advance. It can be educated guesswork. But then you must prove the theory or theorem, by using either formulas which represent a general case, or by a multitude of examples such that the preponderance of evidence leads to the conclusion that there is a high degree of probability that the theory is correct, or a combination of these. I believe that I showed mathematically by giving a general case as well as specific cases as formulas which allowed me to solve for the constant, finding not a polynomial or multiple solutions, but a single, rational number, 1/3.

Any more than that I cannot say without a risk of repeating myself, over and over again for the umpteenth time, just as you are, so one of us, or both is missing something! IF I saw how I made a mistake, I would admit it, but I just don't see it!

Perhaps I was being a bit hard. As I understand it, what you have done is to show that the proposition does not lead to inconsistencies for this type of assemblage*. So it lends support to the proposed solution. Coupled with measurements on volumes of various pyramids, this would indeed be enough to support the use of this formula as a physical model - just as we used "Newton's laws of motion" in their original form until problems were observed at very high velocities. But that is not the same as proof. (Nor, to be fair, is mine - unless you accept the additional assumption I stated. But the assumption is much less stringent than requiring that a formula exists for the generalised shape)

Hopefully, we can agree on that?

Fyz

*I'm not certain that the Egyptians would have had the algebraic tools to do what you've done - but that's fair enough, as it was not an explicit requirement of the question.

Here is a simple tabletop demonstration to "prove" that the volume of a pyramid is 1/3 time the base times the height.

Equipment and materials:

two 5lb blocks of cheese (usually square and long like bar stock)

a long knife

and a beam balance scale.

Nice flat table top

Proceedure:

Lay a block of cheese down on the table with the square end facing you and the long side running away.

Make one cut perpendicular to the table, corner to corner the long way of the cheese. Now turn each piece a quarter of a turn to the left or right. This should lay each half "on its back". Now make another perpendicular cut on each half.

This leaves you with two pieces that are quarters of the pyramid and two pieces of "waste". If you do the same to the other block of cheese you will have four quarters that make a full pyramid and four pieces of "waste".

Now for the magic: Two pieces of waste weigh the same as one quarter of the pyramid. You can check this with your balance. Thus 10 pounds of cheese produced the equivalant of 6 equal parts (or each 5 lb block produced 3 equal parts.)

Thus the pyramid itself weighs four sixths (two thirds) of 10 pounds or 6.6666 lbs. And the volume equal to the base times the height would weight 20 pounds. Thus the volume of the pyramid is one third the base times the height. QED

It is likely that brick mud was the first material used for this sort of demonstration. Bricks were the primary building material of Mesopotamia prior to Egypts use of limestone. It is likely that the origin of this "formula" as implied in a problem in the Rhind Papyrus (see Gillings) circa 1800BC was Mesopotamia. There are artifacts that consist of turned alibaster vases found in the palace at Ebla that date to the early dynastic period of Egypt. This is evidence of trade and contact in the 3000BC to 2500 BC time frame, before the pyramid builders began on the Giza Plateau.

Contrary to popular belief, the Egyptians were very exacting mathematicians. The so called "Recto Table" of the Rhind Papyrus is exact when breaking fraction of numerator two into the sum of unit fractions (numerator one).

Look for my forthcoming book for other facinating stories / discoveries in "Mathematics Before the Greeks"

It would be interesting to first cut off the ends of the two pieces of cheese obliquely to form two parallelepipeds, and start the demonstration from there. Would the pyramid quarters formed then actually fit together? Would it be possible to generalize the demo from right pyramid to oblique pyramid?

When you publish, do let us know -- I'm sure you'll find some readers here.

Ken,

If you can show the right pyramids to be OK (and that clearly depends on that universal constant), then i think it is easy to generalize to obliques.

In the Rhind papyrus, they use subtraction, so that is understood at least in part. They have no zero, so this may be a little shaky.

In the Moscow papyrus, they clearly add volumes.

So, with a small/tiny/miniscule leap of faith, we might assert that they knew you could subtract volumes. And, to some extent, problem 14 in the Moscow papyrus does lend itself to a subtraction of volumes.

If you know you can subtract volumes, then any right pyramid can be turned into an oblique pyramid by subtracting another, smaller, right pyramid, or even two if you want to make it doubly oblique (BTW, Egyptians had no need of oblique pyramids since stone work does not lend itself well to such structures). If the universal constant is OK, both right pyramids can be well described in terms of one of the aforementioned boxes, either of which has a well-defined volume.

As always, the applicability of 1/3 to ALL pyramids is the possible weak point and I suspect Fyz has an answer based on proportions. The figure in problem 14 suggests an awareness of proportion and I think they were aware of Pytagorean techniques.

We'll see. Hopefully, I didn't get the papyri mixed up.

Tom

Going back to the square base pyramid with central apex being half the height above the base as the sides are long.

Knowing that six of these fit into a cube the base being each surface. Which proves by simply observation the formula for this pyramid as being L x B x 2H / 6 or

L x B x H / 3.

Take two of these pyramids.

Divide one by 4 through the apex horizontally & vertically you are left with 4 pieces a quarter of the base size and volume. (4 sided base)

Take the second pyramid and divide this one diagonally through the apex leaving 4 pieces a quarter of the base size and volume. (3 sided base)

All eight of these pieces are still covered by the above equation but now two things have happened.

Firstly the apex is no longer centred and secondly the shape of the base does not matter.

This could be further explain by a front or/and elevation of the pyramid when positioning the of apex along a line parallel to the base the area seen does not change on the front, end or plan views & if the areas haven't change surely the volume remains the same?

"Going back to the square base pyramid with central apex being half the height above the base as the sides are long.

Knowing that six of these fit into a cube the base being each surface.."

Enough with the cubes and square based prisms, already! Please take a look at my solution and proof in post #205 and tell me if that doesn't meet the challenge to find a solution by calculation for any rectangular based pyramid! I think it does!

Yes STL you are right your formula does work well done.

I was just pointing out that it is possible to reason that the formula work for all 'right' pyramids e.g. by halfing asquare one you will be left with rectangular.

Not variable proportions - just linear scaling* ("similar" figures in the strongest sense), volumes of right prisms (that already needed transfer, which I see as subtraction and re-addition??), cuts and re-assembly, addition, and integer subtraction (not via zero, however).

*An extra clue? Only a factor two scaling is needed

I hadn't read translations of the papyri, until just now. Great stuff! I particulary like the fractional doubles. They seem so strange, from our perspective. It would be great to be able to go back in time, to actually live with this system for a while.

This gives an approximate result for the shape you have chosen. You can make a number of measurements on a number of different shapes and draw the conclusion that is is likely that the 1/3*base-area*height applies to all rectangular pyramids.

But you can do that without resorting to (essentially imprecise) weighings, and using only the much less tight assumption that the multiplication constant is the same under linear scaling (which also means that you don't need to know which dimensions to multiply before you start). That is the intended challenge, and the basic knowledge was certainly available to the ancient Egyptians.

Hi Danmoody

It's an elegant approximate demonstration for any particular case. But applying it to a large pyramid means assuming that scaling actually works between the convenient mud prototype and the large pyramid. It is possible to produce a demonstration that requires only that linear scaling by a factor gives volume scaling by a factor of eight. All the other techniques and background information were known - i.e. cutting, reassembly, addition, volume of a prism. You'll see that my solution is expressed using subtraction, which fits better with current notation - but that is not really needed.

The technique works for tetrahedra as well. The reason that I chose rectangular based pyramids is partly that these were of interest to the Egyptians, and partly because it allows the simplest method of cutting. However, the application to tetrahedra is actually quite straightforward, and it is easy to develop the volumes for overhanging tetrahedra from tetrahedra with a right-angled corner - though this does require basic subtraction.

A question on the use of unity numerators: - I can imagine that the notation for fractions could have advantages in avoiding ambiguities if the standard communication medium was in some sense linear. However, in that case it seems a bit perverse not to use 1/2+1/6 for 2/3. To my mind, that makes this looks more like communications jargon to keep a mathematical elite secure than an intrinsic belief that this is a good way to express fractions. Do you have a view on that?

This is a solution I've only checked for when the top of the pyramid is in the center and the sides L and W not necessary the same, rectangular.

If the distance from any side to the centre is the base of a triangle and the height is the other side and the pyramid surface is the hypotenuse, then if you think of how to add up numbers 1 to 9 = 9+1 and 2+8 and 7+3 etc and apply this to the triangles working to the corner, then:

half the length of the pyramid side * half the distance to the centre * the height.

do this for each side and add them up, this then equals the volume?

Don't understand some of the others post, so hope this is not a repeat?

regards JD.

Sorry, but I, in turn, don't understand what you are doing.

Fyz

Yes done it on the run, I dont know what I'm doing either.Adding areas not units so it does not work.

Sorry Fyz and everybody. Regards JD

I read as many of the previous replies as my early morning eyes would allow, but if someone already stated this I did not catch it.

The pyramids are a series of rectangles stacked atop one another. calculate the volume of each rectangular section from bottom to top and add results.

or..

calculate the volume of each rectangular section from top to bottom and add results.

do I win? ooh, I hope I did. what's the prize? is it coffee? I really would like some coffee!

cr3

Too late - loads of pyramids built from rectangles already. Anyway, this requires one of the following:

that you already know the volume of a frustum (a rectangular plate with sloping sides);

that you look for improving approximations as you increase the number of layers towards infinity (= calculus); or

that you are explicit about how you calculate the volume of the frustum relative to the volume of the pyramid.

Frustum?

I thought the blocks for construction were rectangular? Simply rectangular.

Now where is my coffee?!

You'll have to get your own coffee The question was about pyramids, not about stepped structures roughly in the shape of pyramids, and the "avoiding calculus" reference was to proposals that the ancient Egyptians "must have had" calculus, because apparently they left a record that included an exact factor of 1/3. The figure of 1/3 is not accurate for a structure built entirely from rectangular blocks - which obviously isn't a true pyramid anyway.

Anyway, this requires one of the following:

that you already know the volume of a frustum...

OK , so my construction arrives at a frustrum that is 7/8 of a full pyramid (truncated at half full height). That's got to be worth at least a biscuit, even if the art-work was slightly lacking.

Ha! The evil deed is done. Decomposition of a rectangular parallelpiped was truly a wild duck (oops, I mean goose) chase. Can someone please tell me how to insert drawings (sketches) into a message.

Tom

*It would be surprising if I was really, because the only reason I came up with this was to satisfy myself that the claim that "the ancient Egyptians must have had a primitive form of calculus to come up with the formula" could be countered without suggesting they resorted to guesswork based on limited measurements (i.e. didn't adhere to their own standards).

It all depends upon their definition of volume. It is obvious, but not so easy to prove, that as all cross-sections meeting any plane parallel to their base of two tetrahedra with congruent bases and the same height are a pair of congruent triangles, the two tetrahedra have equal volume.

Euclid, on the other hand, insists on only using congruent triangles or solids and when they are not available uses reductio ad absurdum to prove his point. If the Egyptians had a more practical and less rigorous definition my "proof" would be adequate (the revised version might be valid even by Euclid's standards but as I cannot draw and cannot even cut triangles of cheese to fit together - I did try - it's difficult to show that the two left-over tetrahedra are congruent and not just equal)

sometimes I don't feel very smart.

That still does not help with KLMN, which does not usually have a face that is congruent to a face of either of the others. If I'm wrong about this, please show me.

Fyz

No, it does not deal with KLMN (I don't see how that can be congruent to the other two in normal circumstances, having tried several times)

I was just replying to your comment about whether the Egyptians adhered to their own standards.

Just fill a bucket with liquid, then dunk any irregularly shaped object in it and measure the volume of liquid that spills over the side. That's how to calculate the volume with out calculus or any math.

that simple

(height)*(area of the base)/3

How the ***** did you find that out without someone having resorted to calculus?

It is simple.

Consider a cuboid with rectangular base A x B and height 2H. Volume of such cuboid is Vc = A x B x 2H, what was, I hope, obvious fot the ancient people.

Now take the central point inside the cuboid, H away both from the bottom and top of the cuboid and imagine the pyramides which are made by the cuboid's walls and the central point. There are six (smooth side) pyramides:

two pyramids A x B base and H height (top and bottom pyramide),

two pyramids A x 2H base and 1/2 B height,

two pyramids B x 2H base and 1/2 A height.

This pyramids fully fill the cuboid's volume.

Of course the volume of a pyramide Vp, eg. A x B base and H height, is not simply Vp = A x B x H but is smaller and is:

Vp = epsilon x A x B x H , where epsilon is a constant factor for pyramid.

Now you can write:

Vc = A x B x 2H = sum of the volume Vp of the six pyramids =

= epsilon x (6 x A x B x H)

From this you can easily see that : epsilon = 1/3 ,

so the volume of pyramid A x B base and H height is 1/3 x A x B x H.

Ready!

Now, Ramzes, 12 amphors filled with wine please!

Piotr Grabowski

Piotr,

You have solved for only one case, where the apex of your 6 pyramids lies directly over the center of their bases. These are known as right pyramids. This is far from the generalized case which I presented in #205, and illustrated in #218, but you got the idea. At least you did not limit yourself to a square-based cube. Your "cuboid" would be more correctly called a rectangular prism, which is the common box shape. It is also known as a right (or 90°) parallelepiped, a six-sided polyhedron where all of the faces are parallelograms AND all angles are right angles, or in other words, all the faces are rectangles.

Unfortunately, there are some who say even my general solution does not solve the challenge, but I guess we agree to disagree. My solution includes oblique (non-centered apexes) pyramids, but only if the apex is over the base. Extremely oblique pyramids, where the apex lies outside (not directly over) the base are not proven by this solution. However, in light of the reference to the Egyptians and pyramid building (did they ever even build non-square pyramids?), it is hardly fair to include these extreme cases.

I hope you will check my solution and see how it is more general then what you have proposed.

Request denied.

Even if this works, it only solves the problem for right pyramids (with the vertex above the centre of the base). If you can justify your statement that "you can write Vp = epsilon x A x B x H, where epsilon is a constant factor ..." you can regard the job as done for these pyramids. Although that is unnecessarily restricted it would be good enough for practical purposes, as that is the only kind of pyramid we intend to build. Therefore, if you can do that without making use of the "official" solution, 3 amphora filled with wine will await you by the foot of the pyramid currently under construction (at 1800 BC). Otherwise, in view both of your having repeated suggestions made by others and rejected for precisely the above reason, and of your temerity in proposing your own reward, you will head the list of slaves to be tortured to death and subsequently interred immediately prior to the next sealing.

Ramesses

Otherwise, in view both of your having repeated suggestions made by others and rejected for precisely the above reason, and of your temerity in proposing your own reward, you will head the list of slaves to be tortured to death and subsequently interred immediately prior to the next sealing.

Yikes! If that's how Fyzaoh "Ramesses" treats his servants, me and the other Israelites are outta here! Red Sea, shmed sea. We'll cross that river when we come to it. 40 years wandering in the wilderness? Heck, after this gig we can do that standing on our heads! Who needs Pyramids anyway, I got some plans for this really nice Temple...

- Mosses

We are a civilised society - we don't torture faithful and obedient slaves here.

But did you hear about this years harvest? Not too much straw. After those comments, guess which slaves are not going to have first access for brickmaking (favourite adopted son or not).

Did I hear "time and a half for overtime"? Sounds about right - off you go to the slammer.

(I reckon you always were a basket-case)

Fair O

Unfair O,

Well, that's the last straw! (he, he)

Sounds like a bunch of Bull-rushes to me!

Hey, do you like frogs? Bloody rivers? Locusts? Dead babies? I hopes so, cause "It's go time!"

Go find yourself some Nubian scabs, cause the UPW (United Pyramid Workers) is shutting you down, O Mighty One! Oh, and good luck getting your wimpy son Mernephtah to ramrod those scab workers and finish your great projects. He couldn't cut his way out of a papyrus bag! His idea of a scarab amulet is a real dung beetle on a string, including the dung. I'm telling you, that boy's steps don't go to the top of his Pyramid! He's got the hieroglyphics of an "L" on his forehead. He is NOT the sharpest spear in your armory, to say the least. When his Greek tutor asked him to define "Geometry", he said, "Well, I got a trunk, some branches, twigs, and a lot of leaves. Gee, Ah'm a tree!"

<chanting>Let my people, GO! Let my people GO! Let my people, GO! Let my people GO!

To complete the picture as regards direct solution for pyramids with the apex overhanging, plus tetrahedra and polygon-based pyramids.:

Take a reference pyramid, and extend its sloping edges a factor two to make a double-height pyramid of eight times the volume. Cut the reference pyramid away from the top of the the extended pyramid, to leave a frustum. Now make two full-size cuts through the frustum so that they separate a pyramid that is identical to the reference pyramid from one of the corners (N.B. that, for an overhanging apex, there may only be one corner where this is possible). This will simultaneously separate three prisms - one with a rectangular base, and two with triangular bases. The total volume of these three prisms is 2.w.l.h (you can cut and reassemble each of the three prisms to make "right prisms" to confirm this if you think it is necessary). This (2.w.l.h) is equal to the total volume of the double height pyramid (= 8 reference pyramids) minus the volume of the two reference pyramids we have removed; that is 6 reference pyramid volumes. So a single reference pyramid has volume 2.w.l.h/6 = w.l.h/3.

This method will work for any pyramid whose base is a parallelogram, but not with arbitrary shapes such as tetrahedra.
The basic method (using vertical cuts to create right prisms) works with non-overhanging tetrahedra, where we need up to 10 cuts:
. The first cut horizontal to separate the original tetrahedron
. Three further vertical cuts from the top edges of the frustum, just sufficiently long to separate a vertical right triangular prism
. Two outward-going vertical cuts orthogonal to each of the three immediately previous cuts (six in all). These will define three right prisms resting their sides, and three quadrilateral pyramids.
.The three quadrilateral pyramids can be assembled to make one reference tetrahedron.
. By apportioning the area of the base triangle (4 x the area of the base of the reference prism), we can see that the sum of the prisms is 2 x base-area x height (the dimensions again being those of the reference prism), and the sums continue as for the rectangular pyramid above.

We can then extend the situation to any overhanging tetrahedron by filling empty space to create a non-overhanging tetrahedron, and then removing at most two other non-overhanging tetrahedra to leave the original tetrahedron whose volume was to be calculated. We can then return to Euclid's methods to extend the tetrahedral result to cover any polygonal based prism.

So I have this mental picture of Edward G. Robinson staring down at you in a mud pit, saying "So, Moses. How do you like your proof now?"

Tom

I believe Mr. Robinson to be history, since Moses cast his staff on the ground at his feet, and it bit him. Then Moses climbed out (and by all accounts, lived to 120).

Regarding my 'proof', if you accept the assumptions it seems fine to me. I.e. unless you know different, I think it "does what is says on the tin.". If you have reservations, spit 'em out. Otherwise, it would be good if someone who is handy with a drawing package would illustrate the processes (it's always hard to follow a verbal description of a 3-dimensional process).

Fyz

And you thought you had trouble following MY solution! My graphics are a piece of cake compared to what it would take to illustrate what you just described!

ROFLMAO at Fyz's "simple solution".

Re the official solution:

What an elegant, simple solution, clearly and concisely explained. Wow.

Even without drawing it, it's easy to visualize. STL has pretty good drawing skills, and I bet he'll volunteer to draw it for those who need help with the visualisation.

For cheese lovers, it would be a simple matter of starting with a largish pyramid, cutting off the top half ("half" as measured by height) and setting that aside to serve as a reference for the original single-sized pyramid (of volume = "Volume"). Then slice off the 4 pieces that, once removed, reveal the rectangular prism over which the smaller prism was perched (before the knife came out). Then, just cut off the four corner pieces, which clearly can be assembled into a perfect replica of the single sized pyramid.

How cool.

With your description and STL's drawing (the one I think he'll volunteer to do), all nicely colored to show how the four triangular prisms equal the rectangular prism, etc., this really should be published on one of the math sites so students everywhere have access to it.

Well done proof, and a great stimulating challenge question.

"What an elegant, simple solution, clearly and concisely explained. Wow.

...this really should be published on one of the math sites so students everywhere have access to it.

Well done proof, and a great stimulating challenge question."

Fyz! Don't sit down too quickly. You just might just squash his head and kill poor Ken!

Sorry, can't oblige on the drawing, even if a dumb idiot like me could understand this "elegant, simple solution", I am just too busy eating my sour grapes, sulking in my corner, and doing my Rodney Dangerfield impression.

I don't get the "formulas" that Fyz used. Maybe if he used proper notation it would all become much more clear. Surprisingly inerudite for a Physicist. "bits-of-pyramid", indeed! And he had to use 8 pyramids while I only used 6. I would say my proof was much more efficient! <grin>