Pyramids: Newsletter Challenge (08/21/07)

Am I misreading the question? The volume of a pyramid is the area of the base times 1/3 the height. Same is true for a cone.

This is Euclid's proof, re a triangular prism. Given that the volume of any prism is the area of its base times its height, then this proof should apply to any prism or cone.

This proof is based on Euclid's Proposition 5, which uses limits; personally, I regard this as an early form of calculus.

This is not necessary to determine the volume of a rectangular pyramid. (N.B. Euclid is thought to be around 300 BC, so this technique may not have been available in 1800 BC).

Fyz

Well, let's see.

We can assume they were able to calculate the volume of step pyramids. So suppose, at about the time when they were transitioning from step pyramids to smooth-sided pyramids, they drew an elevation view of a smooth-sided pyramid. Then they drew a ten-step pyramid such that the bottom of each step edge coincides with the desired smooth side. Next, they drew a nine-step pyramid with the same step height, with the top edge of each step coinciding with smooth side. Thus the nine-step pyramid was completely contained within the desired pyramid, and the ten-step one had lots of little bits sticking out.

They notice that the little pieces sticking out in the ten-step pyramid equal the gaps between the nine-step pyramid and the smooth one. So they average the volumes of the two stepped pyramids, and take that to be the volume of the smooth one. If they were happy with 22/7 for pi, then they'd have to be happy with 1/3 as the relevant constant for the pyramid.

Or how about this? They take a cube and weigh it. They whack away at it to make a pyramid, and then weight the pyramid and all the shavings.

Or how about this: "Pyramids R Us" initially thought that the volume of a pyramid was equal to its base area times its height. They ordered the stone. They found they could build 3 whole pyramids of the contracted size. They sell the extra pyramids at attractive prices, and make a huge profit, labor being essentially free. They make a note to order 1/3 as much stone next time.

Or... maybe as they built pyramids with more and more steps they noticed that the amount of stone used seemed to be getting closer and closer to 1/3 of the volume of a cube with the same base.

It just occurred to me that in the above scheme you are simulating calculus. If they built a small prototype pyramid using small measurable blocks and then started filling in the angles with even smaller blocks, while keeping track of the total volume they would in effect be emulating calculus. As the blocks got smaller they could easily see that their volume was approaching a certain relationship that we take for granted today in the simple equation for a pyramid.

Also they could have used only a part of the pyramid instead of the whole. For example for the typical Egyptian four sided pyramid just build one quarter of it and use the symmetry to save time by multiplying by four.

Who knows, maybe thats where they got the idea for calculus or what ever they called it back then!

It's an interesting question.

It's hard to draw a line between what constitutes calculus and what does not, especially when imagining yourself going back in time, and when, in addition, you have in hand a formula for the volume of a pyramid that seems like simple arithmetic. Certainly, the formula for the area of a triangle involves no calculus, and never has: you don't need to slice a triangle into small parts to see that it's area is half the area of the parallelogram that would be made from two of the triangles in question.

Likewise for the volume of a triangular prism.

And so it seems it should be for a pyramid. All the edges are straight lines, all the planes are flat: how can there be any need for calculus? Someone mentioned early in this thread that calculus was required to find the position or instantaneous speed of a falling object. It's not, if the acceleration is constant. So we think calculus when we think of things are not constant... where there are curves, and especially where there are curves of changing curvature. (For example: acceleration changing by virtue of aerodynamic drag, which is in turn changing with speed.) By comparison, a pyramid seems so straightforward and linear.

You can come up with 1/3 by jumping to erroneous conclusions. As someone did earlier, you could assume that volume would be the average of the volume of a slice at the base and a slice at the top, which means the relevant constant would be 1/2. Then someone points out that that can't be possible, because that would give you the volume of a prism which could contain the pyramid with lots of extra volume at the top that would have to be cut off on either side of the apex.

So then you might start thinking about a cube and the parts to be removed to leave a pyramid. First, you cut away two prisms, leaving the prism that would contain the pyramid (the one we just mentioned that gives you the incorrect 1/2 factor). Then you turn the cube 90 degrees, and do the same cuts again. Eureka!!. There's a pyramid left, and I've removed half the material again, so the pyramid must be 1/4 of the total cube! Done... Simple.

Then someone points out that your second set of cuts didn't actually remove half of what's left, which you can see by comparing the pyramid size to the size of the cut off pieces. So you decide the constant must be half way between 1/2 and 1/4. In another mathematical leap, you pick 1/3, because 3 is halfway between 2 and 4. You then test you hypothesis with a careful experiment in which you cut a pyramid from a 10 lb block of cheese, and you weight the stuff cut away, and find that it is exactly 2/3 of 10 lbs, or 6.667 lb (which is close enough to convince you). You repeat the experiment with three more cheese blocks, and come away convinced that 1/3 is the correct factor. Furthermore, you convince yourself that your math is correct, and that you did it all with arithmetic, and you came up with exactly 1/3, not some number approaching 1/3 that you'd get with some fancy calculating system.

Then someone comes along and says "Isn't the number half way between 1/2 and 1/4 actually 3/8?" But you ignore them, because you know that 1/3 is correct, and you know that all it takes to figure it out is some arithmetic.

I agree that this is true. But is this formula not a simplification of an integral? Therefore you are using calculus to find the volume are you not?

Displacement would work also but not on the scale we are talking 1800 bc.

There has to be something else to this.

You might be on to something. for equilateral pyramids they could easily scale up the measurements.

So measure the displacemnet of a small pyramid, calculate the ratio's of heights? between the big and small pyramids and apply the same scale-up to the volume

Not sure what ratio you would use

It's easy,

They just counted the number of blocks they used to make one!

It`s not so easy,

because blocks are not equant!

Since the Egyptians invented glass, we could make a large graduated cylinder, fill it half full of water, immerse a pyramid, and measure the apparent increase in water volume.

Wait, is this a trick question?

No, it is not intended as a trick question. It is possible to produce an exact answer using "classical" solid geometry. BTW, Archimedes (who is credited with this method for measuring volume at around 250 BC) somewhat postdates 1800 BC.

Also by the way, my understanding is that the formula is given in the Egyptian document that is referenced in Wikipedia's entry "History of Calculus", but the method is not.

Fyz

Fyz,

If it isn't a trick question, why is there a duck in the formula? Sure, you can always say that's a raptor, but I see duck. And Groucho Marx taught me to be suspicious of ducks. Or, is it all a KGB plot?

Anyway, I need a clarification. Did the Egyptians use zero? If so, it's straightforward; if not, we gotta leave out the capstone.

Tom

Wait! There's a duck in the formula? The baby can't swallow that! Better say the secret "woid" and get the duck out of here! Or just duck out, I don't care!

Don McLean wrote "and while Lenin read a book on Marx, the quartet practiced in the park". I didn't know he was reading about Groucho! Or was it Lennon, not Lenin?

And isn't a KGB plot where they bury all the old Russion spies?

Tom

Either that's quackers, or it's all beyond me. But there's no need for explicit zero.

Fyz

How on earth can you fill it half full?

OK, have it your way. Fill it half empty!

ROFL

Yes!! This is a lot like the story about Edison and the light bulb. The story goes that Edison gave one of his assistants the task of determining the volume of a light bulb to know how much gas each bulb would require. The assistant promptly got his Calculus out and spent a week calculating the volume. When he presented the results to Edison, Edison quickly pointed out that all he needed was a quick estimate and using a water displacement technique was what he was expecting to see.

"But the volume of a rectangular-based pyramid can readily be calculated without calculus"

Unless the definition of the word calculated has changed, I do not think the water volume displacement method would be not a "calculation", but rather a direct measurement, would it not?

M-W says "Calculate" is "to determine by mathematical processes" in the sense it is used in the question. Where is the mathematics in measuring displacement?

Without calculus:

By proving that you can build a perfect pyramid of dimension 6x6x3 units by using 36 cube blocks of 1x1x1 units. This is analogue to proving the area of a triangle eq 1/2 base x perp. height)

The height units do not have to be the same as the base units.

place the 36 block in a 6x6 grid and number 1,1 1,2 . . .1,6 2,1 ... 2,6 ..... 6,6

take blocks 1,3 1,4 3,1 3,6 4,1 4,6 6,3 6,4 and cut in half and place on the second layer at 2,3 2,4 3,2 4,2 3,5 4,5 5,2 & 5,3

block 1,2 & 2,1 must be cut in 2 and placed in the vacant block on the second layer at 3,3, the same with 1,5 2,6 == 3,4 & 5,1 6,2 == 4,3 & 6,5 5,6 == 4,4

Block 1,1 must then be cut into pieces and placed on 1,1 2nd 2,2 and 3rd 3,3

the same must be done with the other corners.

Result = a perfect pyramid.

Therefore the volume of a pyramid = base area x 1/3 height. = 6 x 6 x 3 x 1/3

cube units.

First, assume the pyramid is a sphere. Umm. Nope, that won't work.

OK, if I take a pyramid and tessellate it in 4 dimensions, it should be trivial.

But seriously, this is a piece of cake with CAD. And that raises a good question: is geometric construction a form of calculation? Where is the bright line between measurement as in 3.17 mm and measurement as in log(13.7) laid out on a scale?

And, if you buy that line, do you suppose the Egyptians used SW-1802? And, if so, why haven't the bugs been fixed?

But seriously, this is a piece of cake with CAD.

Funny coincidence that you should mention that. I was just thinking about a 3D CAD program I had about 15 years ago which would incorrectly calculate the volume of a frustum of a pyramid. It was particularly frustrating (frustrumating?) because it would calculate many (but not all) shapes correctly. (One minute I was saying, "Wow, isn't that cool?" and the next I'd be saying "Wait... that doesn't look right.") I found that although it would have been handy to use the volume and CG calculations, I could not trust them, so I didn't use them.

Ken,

You don't have to actually calculate the volume. Any rectangular base pyramid of base area 36 and height 3 can be morphed into your solution or rcapper's. That doesn't answer the question though since I doubt the Egyptians morphed.

Another interesting way to do it is to take a rectangular solid and do a Boolean NOR of all diagonal cuts. However, I'm not smart enough to do Boolean geometry (I lay this squarely on that time down in Sonora when the pack mule carrying the nitro decided to relieve himself on an electric fence).

Tom

I think you'd find that using Boolean geometry in this particular way led to pyramids of different shapes that you would need to show had equal volumes. That would be most of the work...

just keep hitting rebuild until it works.

Here's an idea but it would only work with a pyramid with sides that were 45 degrees. If you put six together at their points you would have a cube. So area equals one sixth the cube of one side of the base.

Maybe thats how they built them? They made a big square/Cube as the first attempt. But after they all started fighting over who owned it cut it into 6 so they could all have a slice? Then all they needed to do was divide the volume of the cube (which they already new) hey presto they had tehe answer and new how much to charge per slice.

Actually it works with any angle for the pyramid, draw diagonals from opposite corners of the cuboid and you get six pyramids of equal size. If it is not a cube the pyramids are different shape but the same size axbxc/6, where h=c/2. Its easy to prove if a=b, I think that you still get to prove the equality using congruent quarter-pyramids if a does not =b but it's harder to visualise

NB Euclid was basically a recorder and compiler of known geometric proofs, so we do not know exactly when the proofs he quotes were developed, just that they were during or before his lifetime - mostly before.

You say it's "easy to prove that the pyramids are the same size". That is exactly what Euclid does for the case of a triangular prism; but he needs to use proposition5, which uses limits. If you can prove it without either using limits or a concept of linear dependence of volume on each dimension for triangular shapes (which not apparently accepted by Euclid even for calculating areas in the plane), that would be interesting.

Agreed, the work outlined by Euclid mostly pre-dates him. Maybe limits were developed 1500 years earlier, maybe not. But the challenge remains to do it without relying on limits etc. Given the conceptual requirements of most proofs, the interesting thing is that the calculations are quite straightforward, and require nothing other than construction, cuts, and arithmetic.

The first part only works for the simple case where the vertex lies vertically above the centre of the square/rectangular base.

Draw planes parallel to each face of the cuboid through the common vertex of the six pyramids. This splits each of the six pyramids into four identical right-angled tetrahedrons with triangular base a/2 by b/2 and height c/2 or base a/2 by c/2 and height b/2 or base b/2 by c/2 and height a/2.

The twenty-four tetrahedrons are congruent, so have equal volume.

If you move the vertex along a line parallel to one of the edges of the base, then you get two sets of twelve congruent tetrahedrons so the pyramid still has one-sixth of the volume of the cube (although two of the others are unequal).

If you want to go to the next stage I think you have to bisect (into two unequal parts) the pyramid at right angles to the axis of symmetry, separate the cuboid into two parts and add mirror images of the two parts of the cuboid.

The two new symmetrical pyramids are jointly one-sixth of the new cuboids, in fact each is one-sixth of its own cuboid. Move each vertex the same distance along the mirror and, as shown above each reshaped pyramid is still one-sixth of the volume of its new cuboid.

Detach the mirror images and rejoin the pyramid and conjoin the two mirror images. The two pyramids are jointly one-sixth of the two cuboids. The mirror image must be the same volume as the original, so the pyramid is one-sixth of the volume of the cuboid.

Any position for the vertex can be achieved by moving along the two axes, so this appears to prove that a pyramid on a rectangular base has one-third of the volume of the enclosing cuboid.

I don't know offhand how to extend this to an irregular quadrilateral but I expect one can construct a rectangle using opposite corners and most of its longest side, a series of rectangles based on the right-angle triangles lying between the quadrilateral and the rectangle and pyramids based on these rectangles etc.

I think that's it for stage 1, and once you've got this far you could abandon the double height. Perhaps you could also simplify the description (so that even I could follow the detail?). (I think you simply need to pass the planes through the vertex, so you end up with four groups, each group being three congruent tetrahedra)

As a proof for the vertex above the rectangle, that's much better than my original stage 1.

I think it now needs extending to cover pyramids whose vertex is not vertically above any section of the base, but that is relatively trivial.

Can you find an equivalent for tetrahedra? I haven't managed that for this method (I wish that was "yet"?)

Fyz

Whoops - I've been at the malapropisms again I wrote tetrahedra when I meant right-cornered pyramids. Apologies

Fyz

Sorry, if you want a simple clear explanation you need someone who knows what he's talking about: I just picked this one up and ran with it. I'll try to clarify but it may take some time to do so because I can see it clearly in double-height but I cannot yet see how to prove it in single height although that ought to be easier.

If we assume the previous post proves that V=axbxc/3 for a pyramid on a rectangular base with an an apex vertically above the base, then one can extend it to one with an apex above an extension of the base along one axis by simply fitting on a juxtaposed pyramid with the same apex whose base is a linear continuation of the first and which covers the point vertically below the common apex. The joint volume will be ax(b+d)xc/3 and the volume of the added pyramid will be axdxc/3and we get the answer by subtraction.

If the apex is off on a diagonal, then you need three more pyramids fitting together, one with its apex above its base, two with apex offset along one axis and the original. The total volume and, from above, that of each of the three additional pyramids is given by the formula, so the original one is also, by subtraction.

For a tetrahedron there must be (at least) one base that surrounds the perpendicular from the opposite apex. Drop a perpendicular plane from each edge onto that base and you will have a large triangle comprised of three smaller triangles supporting a large tetrahedron comprising three smaller tetrahedra each with two vertical faces, one horizontal face and one sloping face. Separate the three small tetrahedra and reflect through each vertical face in turn and you get three pyramids each of which is exactly four times the size of the small tetrahedron from which it is formed. If V is four times base times height for the pyramid then this must also be true for each small tetrahedron and hence for the original one which was the sum thereof.

Actually, I think I was wrong** in assuming that the groups of three (or six) pyramids inlcuded suitable groups of congruent triplets, except when the individual pyramidal sections have all the orthogonal sides equal.
They have the same three dimensions, but in each case a different pair of the dimensions applies to the (unique) square face - so they are not congruent. In the same way, the groups of six pyramids in the we cannot say they are equal without either resorting to "morphing" or using Euclid's proof, which I contend uses the arguments of calculus (and so should not be assumed at 1800BC.
**Not having the greatest three-dimensional vision, I checked by cutting a block of butter - but I could still have got it wrong

Unfortunately, until someone can come up with something better, we are back to needing something like my original proof - I'll post a hint at the weekend.

I should have said 48 (oops!) tetrahedra each of which is half of the twenty-four quarter-pyramids. In my original stage these are congruent but rotated. Each face is a right-angle triangle. The twenty-four quarter-pyramids are created where the introduced two planes intersect the base of a pyramid and divide it into four parts. The tetrahedra are formed by dropping a perpendicular plane onto the base through each edge Two sets have base whose sides are a/2 x b/2 and a hypotenuse and height c/2 so its vertical sides are a/2 x c/2 and hypotenuse and b/2 x c/2 and hypotenuse, the others have heights a/2 and b/2. Two right-angled tetrahedra with sides comprised of triangles whose sides are the same length are congruent.

Each tetrahedron includes one-eighth of the square or rectangular face and all share the perpendicular from the apex to the base.

It is a lot easier to visualise with regular pyramids but it still works with those on rectangular bases and different angles at the apex

Sorry, I thought they would still be congruent because they obviously were in the case of a cube, they all had mutually orthogonal sides a/2, b/2 and c/2 and a common longest side, and the ones that I tried rotated into each other. As you point out, this isn't enough to make them congruent even though all the triangles at the base of any one pyramid are congruent and given any two pyramids you can find a tetrahedron from one that rotates into a tetrahedron from the other pyramid, because you can find pairs that don't rotate into each other. The triangle on the base (external face) of one pyramid is congruent to that on an internal face of another, I end up with three sets of congruent tetrahedra. It seems obvious that they all have the same volume (axbxc/48), but my first few attempts to prove this utilised my subsequent proof (hence circular) or Euclid (disallowed for using calculus). Saying that the area of the cross section x% of the way up is (100-x)% of that at the bottom hence if the bases have equal area and the height is the same, the volume is the same uses calculus.

Still reckon this should work, but ...

After long delay, and eventually going away to do some work, I think there is a less elegant but valid way to show V=axbxh/3

Take two of the adjacent tetrahedra that are not quite congruent and jointly comprise one-quarter of the first pyramid (base right-angled triangle a/2 x b/2, height c/2. Reflect one of them through their common face (the one with sides c/2, sqrt(a^2+b^2)/2, (a^2+b^2+c^2)/2). You get two overlapping tetrahedra with the parts that are included in one and not the other being two triangular prisms. The triangles at the base of these prisms are congruent, with (if a>b) one side b/2, one (b^2)/2a and a hypotenuse (a^2 +b^2)xb^2/a^2, and those p% of the way up have all dimensions reduced by p% while remaining congruent. So the two prisms are equal. So the two tehrahedra which comprise a common part plus one or other of these prisms have equal volume without being the same shape.

Hence each of these tetrahedra is one-eighth of the volume of the first pyramid. We can rotate one or other into a tetrahedron in each of the other pyramids. If each pyramid comprises eight tetrahedra of equal volume and each of these can be rotated into a tetrahedron in another pyramid, each tetrahedron must be of equal size. So each is one forty-eighth of the total cuboid and each pyramid is one sixth or one third of base area times height.

The generalisation to pyramids with apices not over the centre of the base should still work, given equal but not uniformly congruent tetrahedra

Did everyone get that? Is that the answer, Fyz? It sounds good, but I got lost trying to construct all those figures in my head. Can anyone make some pictures of what he is tallking about?

I got lost as well. An off centre pyramid can be sliced through the apex to give 4 parts which are each 1/4 of a 'centred' pyramid. The rectangular pyramids can be used to construct a block so as to give with 2 square based pyramids, then......?

Can you scan a drawing or something in John ?

Hi John

You lost me before half-way, I'm afraid. I didn't even get any triangular prisms out of this. I could have been making the wrong cuts or reflections. I think I need pictures of some blocks of cheese (do cheese pictures work the same for genuine rodent-type rats as for Fyz's, I wonder?)

Fyz

"I think I need pictures of some blocks of cheese"

Hey guys, be careful of using cheese to model your pyramids, especially if you resort to displacement techniques. I tell you, Swiss definitely will NOT work! It tried it. Kind of makes your whole theory full of holes!

However, plain old American is Gouda enough for me! Oh, that was cheesy!

"Master, master, come quickly!"

"Igor, calm down, you will rile the Muenster!"

"But Master, you wanted to know who cut the cheese last night! I know who it was! It was Colby, Colby Jack, the Monterey Kid!"

"So he's back. I'll have to keep him away from my daughter, Brie. She really melts when she hears him singing in his soft Velveeta voice. I may just have to call in a favor from my Italian friends, Mozzarella and Provolone. They come from Gorgonzola, and are big wheels in the Cream Syndicate. They should know how to slice and dice a blockhead like him.

"But master, you should get him now, while he has spread himself thin. I heard he is hanging out with only a bunch of old crackers over at Quesadilla's Cantina. I also heard Big Bleu and Roquefort, part of his old gang, will be dressing soon."

"Very well, Igor. Bring me "The Slicer". That should take care of Colby! And I know just how to handle crumby guys like Bleu and Roquefort, you just got to fork 'em, fork 'em good!"

So my friends, that is how I cut the cheese!

I really, really cannot draw (one of the reasons why I am not an engineer). I'll try giving co-ordinates and hope someone can visualise or has CADCAM to do it for him.

Taking the vertex of the pyramid as the origin since it is the centre of the cuboid, the eight corners are +/- a/2, +/-b/2, +/-c/2. The five vertices of the first pyramid are (-a/2, -b/2, -c/2), (-a/2, b/2, -c/2), (a/2, b/2, -c/2), (a/2, -b/2, -c/2). (0,0,0). The eight tetrahedra all have two vertices in common: (0,0,0) and (0,0,-c/2); the first one also has (0,-b/2, -c/2) and (a/2, -b/2, -c/2), the second has (a/2,-b/2, -c/2) and (a/2, 0, -c/2). Each face of each tetrahedron is a right-angled triangle and the length in each orthogonal dimension is the same, namely a/2, b/2 and c/2 but they are not congruent. Each is congruent with three other tetrahedra within that pyramid. The first of the two is also congruent with the tetrahedron with co-ordinates (0,0,0) (a/2,0, 0), (a/2, b/2, 0) and (a/2, b/2, c/2) and eleven others. There are three sets of sixteen congruent tetrahedra.

We are not allowed to use integration to show that two solids on the same or equal bases with the same height have the same volume so I suppose that it isn't good enough to say that every "horizontal" cross-section (i.e the intersection of every plane parallel with a constant z co-ordinate between -c/2 and 0, of the two tetrahedra are identical but rotated through 180 degrees.

Reflecting the second tetrahedron through the common face will give you a triangle with vertices (0,0, -c/2) (a/2, -b/2, -c/2) and (-a/2xsin(B-A), -a/2xcos(B-A), -c/2) where tanA =a/b. The two non-overlapping areas are right-angled triangles with one side a/2 long and one angle (B-A) and are congruent. The two non-overlapping volumes are the tetrahedra created by these triangles and the apex of the pyramid and every cross-section is congruent. Therefore these two latter tetrahedra are equal, hence the two original tetrahedra which are the sum of a common volume plus one or other, have the same volume.

If this works, repeat for two adjacent non-congruent tetrahedra in one of the other pyramids and you get all the tetrahedra have equal volume.

There must be a more elegant solution but I just can't see it

sounds good to me

The standard formula of 1/3 of the height times the area of the base will work. The Machinery Handbook is a very handy source.

And how was this formula obtained without some form of calculus?

Perhaps I need a definitive definition of calculus. Having never taken the subject I am at a disadvantage as to what is or is not a calculus process.

This challenge (which I admit to setting) is intended to be about how you create the formula without using integration, differentiation, or zero limits. That is to say restricting the methods to those we know for certain were available to the Egyptians in 1800 BC. RCapper (post #13) has shown how it may be done for the specific case of a symmetrical pyramid with a square base and whose height is half its base dimensions. 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.

Fyz

So, we can blame this one on you, then?

Just kidding. Actually, one of the best I have seen for a while!

Good on ya', mate!

Asumming that egyptians were enlighted by Amun Ra and knew that if you split a pyramid with a rectangular base in four pieces (by two planes vertical to the base that each coincides with a diagonal of the base-i guess i made it clear-sorry if the explanation isn't good enough but my mothertongue is not English) each piece would equals a 1/4 of the pyramid volume(because of same base surface and same height).

Now imagine a solid rectangular. Draw the diagonals in order to create the six pyramids. There are three pairs of pyramids. Lets call them P1(x2), P2(x2), P3(x2). Now lets say that P1 consists of 4 A pieces, P2 consists of 4 B pieces and P3 4 C pieces. So we have a total of 8 A, 8 B and 8 C pieces. That means that the solid rectangular consists of 8 A, 8 B and 8 C pieces.

Now lets say that we are watching the solid rectangular from one side. The projected pattern that we see is a rectangular and its diagonals that split it in four triangles. Let us take the two similar triangles and extend them along the axis that is vertical to the plane of the rectangular in order to form a solid. That solid volume consists of 8 C an 4 B pieces and it's obvious that equals the 1/2 of the volume(V) of the solid rectangular. That is:

8C+4B=0.5V (1)

Let us do the same thing by watching the solid rectangular by rotating it 90 degrees around the vertical axis.

We end up with the conclusion that 8 C and 4 A pieces have a volume that equals the half of the total volume of the solid rectangular. So

8C+4A=0.5V (2)

We conclude from relations (1) and (2) that

A=B (a).

We repeat the same process by rotating the solid rectangular by 90 degrees around the horizontal axis. Guess what!

8B+4C=0.5V (3)

From relations (1) and (3) we conclude that

B=C (b).

From relations (a) and (b) we end up with the conclusion that

A=B=C.

So 8A+8B+8C=V -> 6*(4A) = 6*(4B) = 6*(4C) = V -> 6*P1 = 6*P2 = 6*P3 = V

where V=(base area)*(height) is the total volume of the solid rsctangular.

Now it is clear that the volume of a pyramid with rectangular base equals V(r.b.-pyramid) = 1/3*S(base area)*H(height).

But there might be a catch here. I used a theorem that says that the median from the right angle of an orthogonal triangle splits the triangle in two triangles that have the same surface.

I asume that this theorem might not be known to the Egyptians when they tried to calculate the volume of the pyramid. So Egyptians might indeed have calculated the volume of the pyramid with some sort of calculus.

P.S.: I'm sorry if my description of the solution didn't help very much to visualize it but i tried my best

All the work has been done by the assumptions of the first paragraph - pyramids of equal base area and equal orthogonal height have equal volume. This would be a result of a method of demonstrating the volume.

Fyz

That pretty much shows that I made no assumption.

Another simpler way to find the volume is to consider a solid rectangular and a pyramid with rectangular base ( the base of the pyramide is the bottom surface of the solid rectangular ) that its apex is the point of intersection of the diagonals of the solid rectangular's top surface.

The projection of the pyramid on a side surface of the solid rectangular looks like an M (top-down) ,that is like a W. The volume of the solid that comes up if we extend the triangle on the axis that is vertical to the surface is the half of the solid rectangular's volume. What is left if we remove the volume that is not included in that solid is a prism with a triangular base. The two parallel surfaces of the prism are placed on the sides of the solid rectangular. To form a pyramid from that prism we just need to remove two pieces. The volume of these two pyramids that we must remove in order to have the pyramid equals 1/6 of the volume of the prism according to Eucleides. But the volume of the prism equals 1/2 of the volume of the solid rectangular.

So the volume of the pyramid is:

Vpyramid = V - 1/2V - 1/6V = 1/3V

Where V is the volume of the solid rectangular.

re "That": I don't think so. Book XII, proposition 6 rests on the result for a triangular-base pyramid, which used book XII, proposition 5. As I explained earlier, this relied on an early form of calculus.

I didn't manage to follow your second explanation - perhaps someone can draw or reword it?

Fyz

You wrote about assumption which is not accurate since it's not an assumption or something i made up to solve the problem but a proposition that stands true. As for the early calculus maybe we should forget length, surface, volume, pi, fi etc. and stick with integers and points.

This is an explanatory sketch that shows what i meant. In A we have a solid rectangular. By removing the green parts we are left with B which has the half volume of the solid. Now if we remove the yellow parts according to that we get the volume of the pyramid. Again "that" is based on proposition XII.5.

The challenge says no calculus. You will of course the correct result by using XII.5, but that is not answering the challenge. (Similarly, the standard 'modern' method that uses integration doesn't answer the challenge).

Of course, the simplest way to get the result from XII.5 would be to split your pyramid into two tetrahedra...

Perhaps you might mark your post as "off topic"?

Fyz

Perhaps they just kept a record of each block, like they did with grain etc, and just added them up?

Fyz,

Well, so far, I have cut (in my mind and on 2D isometric paper diagrams) the pyramid of h, height and s, side of base, into four slices by cutting two diagonals at the base and going straight up to the peak. Then I can recombine those slices to just fit into a rectangular prism (big long cube) that is h long and has diagonals on the ends of s length and so a square base which is √(1/2) x s on each side, and has a Volume of V= s²/2 x h, but also has an empty space inside that looks like one of those cream packets that has two flattened ends (the diagonals of the big long cube's ends) rotated 90º from each other, and I can't quite figure the volume of that to subtract from the other volume to obtain the total volume of the pyramid sections.

The only other thing I know is that the flat ends of my "creamer" are s in length and it is also h in length from center of end to center of end. Obviously, that volume must be s² x h/6 , since we "know" that the volume of the pyramid is s² x h/3, but how do we calculate it without begging the question?

Am I on the right track? Any hints?

(Area base)^3 / 6

Posted by Dan

"(Area base)^3 / 6"

Dan,

I think you mean "(side of base)^3 / 6", but that only works for square base pyramids with a height equal to the side of the base. An area measurement (2 dimensional unit) which is then cubed becomes a 6 dimensional object! To get a 3-D volume unit you must either cube a single linear dimension, multiply a squared dimension unit like area by a linear unit, or multiply three linear units (for example W x L x D).

To show that "(side of base)^3 / 6", will not always result in a correct answer look at an example like this:

What if your base pyramid resembled an obelisk, like the Washington Monument, but instead of being capped by a smaller pyramid, just extended until the sides all met to form a very tall pyramid? In this case "(side of base)^3 / 6", will be only a very small fraction of the tall pyramid's volume.

I agree with the displacement method. Make a small rectangular pyramid and a cube or cuboid with the same base and height of the pyramid. Measure the displacement of both. The volume of the pyramid will be 1/3 the volume of the cuboid. The resulting formula is

(L x W x H) / 3 Which is equivalent to the formula given in a previous post.

Area of Base x 1/3 Height = L x W x 1/3 x H = (L x W x H) / 3

if you know the volume of an individual building stone, the volume of the pyramid is a simple addition problem. The volume of a pyramidal solid is the average of the top (0) and the bottom areas times the height. So whatever the area of the base is, divide it by two and multiply it by the height.

a calculus-like process could involve measuring the area of each row of building stones and multiplying each of these times the height of the row, then adding all the rows up. Calculus is just a means of estimating a smooth variation (as if the walls were chinked with alabaster or mud to make the walls smooth instead of steplike) from discrete bits of data (like the individual rows). In fact, it was a similar problem that inspired Newton to invent calculus. He knew how quickly an object would accelerate due to the force of gravity, but how could he determine at any specific point in the object's fall how fast it was going? Or if he knew how fast it was going at some point in the path, how could he determine how far it had fallen? The object was an apple, or so they say. It didn't inspire human beings to discover gravity, but it did inspire Newton to invent a new field of mathematics. The calculus had been there all along, unsuspected by mortals, but Newton found it lurking in the dark, like a hungry dog watching a picnic from the cover of a thicket, and took its measure and revealed it to the rest of us. Many people who should know better think Eve gave Adam an apple, thus earning expulsion from Eden. Did calculus expel us from an eden as well? Would there have been V-2 weapons or hydrogen bombs without calculus?

The volume of a pyramidal solid is the average of the top (0) and the bottom areas times the height.

Ooops. That doesn't seem to square well with the usual calculation, in which you'll find 1/3, not 1/2.

Your formula would work for a triangular prism, if you set it on one of its rectangular faces, and called that the base.

I think a pyramid is a triangular prism set on its rectangular side. I don't get where the 1/3 factor comes from. If it were a cube, it would be the area of the base times the height. If it were a trapezoidal solid, it would be the average of the top and base areas times the height and it wouldn't matter how irregular or oblique the sides were. A pyramid is a special case of a trapezoidal solid where the top face has an area of zero, so it should be the average of the base and top ((0+basearea)/2) times the height and it wouldn't matter if the apex were centered on the base or skewed off miles across the desert. Where does the 1/3 come from?

Bear in mind that I am a layman relying on high school solid geometry. It's possible that with the space race and generation x and the internet bubble, the average of 0 and a quantity is now derived by dividing by three.

The highest region of a triangular prism set on its rectangular side is an edge that is parallel to the base. A pyramid narrows to a point - so you can obviously place the equivalent pyramid (same base area and height) inside the prism - and with room to spare.

I think a pyramid is a triangular prism set on its rectangular side.

Guest,

No, it's not. Hopefully this will explain it for you.

Think of of the roof of a house. Two basic kinds have either two rectangular planes or four triangular planes. The type with two planes, which meet at a line along the peak, and have horizontal triangles on each end, neglecting the house below the attic is a triangular prism. It is obviously larger in volume than the other type, with four triangular sides on the same rectangular base, since you could actually slice off both ends of the two-side roof, cutting straight from the center point to the line at the bottom of the end triangles where they meet the house, to form the 4-sided pyramidal roof. I hope this helps you to visualize the difference.

Perhaps these diagrams will also help:

1 Triangular Pyramid (Tetrahedron)
2 Square based Pyramid (specific case of Rectangular based Pyramid)
3 Regular Pentagonal Pyramid
4 Regular Hexagonal Pyramid
5 Regular Octagonal Pyramid

Also, Blink, the triangular prism need not be on its rectangular base for the Volume formula to work, since you could always put two triangular prisms together on the triangular base to form a parallelogram based prism. And we know that a parallelogram, like a rectangle, has an area equal to the long side multiplied by the short side, and so if extruded to form a prism would have a volume equal to the base area times the length of that extrusion, which would be the height. Right?

Also, Blink, the triangular prism need not be on its rectangular base for the Volume formula to work, since you could always put two triangular prisms together on the triangular base to form a parallelogram based prism. And we know that a parallelogram, like a rectangle, has an area equal to the long side multiplied by the short side, and so if extruded to form a prism would have a volume equal to the base area times the length of that extrusion, which would be the height. Right?

My god, man, you gone daft! I know you know this stuff -- I think perhaps you haven't had your second cup of coffee.

the triangular prism need not be on its rectangular base for the Volume formula to work,

True for the "normal" formula. I was commenting on the guest's formula (at least I think it was a guest). If you flop a triangular prism over on a rectangular side, and call that the base, then the volume is the base area times the height/2. This is equivalent to taking the area of the "usual" base (one of the triangles) (which has the 1/2 factor in it already) and multiplying that by the entire height.

If we simplify a little, and call the triangle a right triangle with sides A, B, C (C being the hyp) and call the height (traditional, as shown in your diagram) L, Then you can come up with two formulas that are mathematically equal, but conceptually different:

area of triangular base x height = 1/2 (A*B) x L = 1/2 * A * B * L

area of rectangular side used as base * 1/2 height = (B * L) * 1/2 *A = 1/2 * A * B * L

And we know that a parallelogram, like a rectangle, has an area equal to the long side multiplied by the short side, and so if extruded to form a prism would have a volume equal to the base area times the length of that extrusion, which would be the height. Right?

Wrong. The area of a parallelogram is one side multiplied by "height" not by the other side. The only case in which the other side and "height" are equal is in the unique case of the rectangle. To prove this to yourself (although I think you already know this) draw a a 2x4 rectangle in CAD and calculate its area. Then skew it so the height remains 4 but the sides get really long, let's say around 10 units. Then calculate the area again. It's the same... whereas multiplying one side (2) by the other (10) would give you 20 square units, not the correct 8.

I know you know this stuff. I'm hoping you didn't drive to work under the influence of a coffee deficit.

<shouting at lab assistant> Igor, you impetuous fool! Get away from that keyboard and stop impersonating me! Get back in your cage! Get back, I say! <whipcrack>

So, sorry, Mr. Fry, my assistant was up to his usual tricks! Of course, you are right, the area of a parallelogram is the perpendicular height, not one of the sides, multiplied by the base line. Igor did have a point though, that triangular prism could be placed on end and the formula would still be V = 1/2 the product of the same three dimensions, since, as you noted, the 1/2 factor is used in computing the area of the base triangle.

Sorry for the disruption. Carry on.

You are absolutely right. basearea*height*(1/3). I guess I wouldn't have made it as a priest mathematician in the royal court. As regards greek geometers, egyptian surveyors used the pythagorian theorem hundreds, maybe thousands of years before pythagoras because the nile valley had to be resurveyed after each flood season from a few benchmarks left behind and the sum of the squares formula was used to measure diagonals. Incidentally, I often need to figure out the mileage saved by using a diagonal road to decide if the time spent at a slower speed would make up for the difference in miles and 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. If you take a 90 mile diagonal at surface speeds, you would have to add 30 minutes of travel time to make up for taking two 60 mile legs at highway speeds. It'd take about 100 minutes to take the legs at 70, 120 minutes to take the 90 mile diagonal at 45. 99 minutes at 55. So the difference would be determined by how much you dare to go over the speed limit on the surface roads.

Were the Ptolemys Greek geeks or Egyptian geeks?

But I believe that Pythagoras' is the first record of a formal proof - even though the Egyptians had probably developed one.

What do they say at birthday parties, "Beware of Geeks bearing gifts"?