Solving the Pyramid Theorem: Explaining Eudoxus' Method

[.hacker//Kazu]

My math teacher handed this out today without explaining, and told us to try and solve it. He gave us the formula to solve the volume of a pyramid (1/3 x area of base x height) and told us to explain WHY it was like that. I already figured out how to explain the surface area of the tetrahedron (he told us to figure out the surface area, without the base, and volume for a square based pyramid first, and then just multiply by two), but I'm a bit stuck on the volume part. He figured this one would stump us since, although we understand how to find a triangle's area, (b x h)/2, we wouldn't understand a pyramid. He also said we should follow the way of Eudoxus who claimed: a pyramid can be approximated to a collection of slabs of reducing size. For example, a square-based pyramid is comparable to pile of square based cuboids. By increasing the number of slabs and reducing their heights, the approximation would improve.

The theorem is v=(Abase x h)/3, where "h" is the perpendicular height.

The proof, or what I think it is, is:
Consider any pyramid, perpendicular height "h" and with area of base "A"

I believe, you then split the pyramid into n layers. Let n be the unknown number of layers and k be the layer you are measuring. So, by similarity the kth layer will have a base with dimension k/n as a fraction of the original base, right? Then the area of the base would be (k/n)squared x A.

The area of the base of each layer will be (1/n)squared x A, (2/n)squared x A,..., (n/n)squared x A.

That's about all I have so far.

I'm not quite sure how that works out; my friend taught me it, so I don't totally understand it. Like what does it mean "by similarity the kth layer will have a base with dimension k/n as a fraction of the original base"?

 

[Hurkyl]

I'm not quite sure how that works out; my friend taught me it, so I don't totally understand it. Like what does it mean "by similarity the kth layer will have a base with dimension k/n as a fraction of the original base"?

Consider a triangle for simplicity; the idea is the same.

Draw a triangle. Choose one side to be the base. Halfway between the base and the opposite vertex, draw a line parallel to the base. How long is the line segment on that line that lies inside the triangle? What if I drew this line one third of the way between the base and the opposite vertex?




On an unrelated note -- if you have exceptionally good spatial visualization skills, (or the right objects to play with), you can actually pack three congruent tetrahedra inside a cube.

 

[.hacker//Kazu]

I do not understand what the point of drawing the triangle with the line is. I wouldn't know how long the line segment is though because the pyramid is to have unknown values.

I can visualize what you mean by the three congruent tetrahedra in a cube though.

 

[e(ho0n3]

I believe, you then split the pyramid into n layers. Let n be the unknown number of layers and k be the layer you are measuring. So, by similarity the kth layer will have a base with dimension k/n as a fraction of the original base, right? Then the area of the base would be (k/n)squared x A.

The area of the base of each layer will be (1/n)squared x A, (2/n)squared x A,..., (n/n)squared x A.

That's about all I have so far.

Now, find the volume of each layer and sum up the volumes. What happens when n is very large (think infinity)?

 

[.hacker//Kazu]

Umm, you then consider each layer as a prism? So, each will be h/n units tall and the volume of the kth slab would be...er, (h/n) x (k/n)squared x A?

 

[e(ho0n3]

Actually, each layer is a parallelepiped. And yes, the volume would be h/n x (k/n)² x A.

 

[.hacker//Kazu]

A parallelepiped?! What's that?

So the volume of the pyramid would be approximately:

V= (h/n)(1/n)squared x A + (h/n)(2/n)squared x A +...+(h/n)(n/n)squared x A.
= (Ah/n cubed)(1squared + 2squared +...+ n squared)?

 

[e(ho0n3]

Correct. Now simplify and take the limit as n goes to infinity.

 

[.hacker//Kazu]

How do I simplify it?

 

[e(ho0n3]

If I remember correctly, 1² + 2² + ... + n² = n (n + 1) (2n + 1) / 6. (Google around for "summation formulas".)

 

[.hacker//Kazu]

If I remember correctly, 1² + 2² + ... + n² = n (n + 1) (2n + 1) / 6. (Google around for "summation formulas".)

Erm, my friend said it was:1² + 2² + ... + n² = 1/6n (n + 1) (2n + 1)

But the problem is I have the formula I just don't know how to explain why the formula is the way it is. I just don't get the 1/6 part.

 

[denverdoc]

Consider a triangle for simplicity; the idea is the same.


On an unrelated note -- if you have exceptionally good spatial visualization skills, (or the right objects to play with), you can actually pack three congruent tetrahedra inside a cube.

Hurkyl, This got me to thinking since the area of a triangle is 1/2b*h and can be demonstrated by constructing a similar triangle as the original from the left over pieces in the rectangle of dimensions b,h and in an analogous fashion packing three pyramids into a rectanglar solid of dims a,a,h whether by extension such a relation may obtain for higher dimensional spaces of N dims where the enclosed space would be 1/N*a^(n-1)*h.

 

[e(ho0n3]

Erm, my friend said it was:1² + 2² + ... + n² = 1/6n (n + 1) (2n + 1)

That is exactly the same thing I wrote down. There are ways of deriving this formula but it's a bit tricky if you ask me. Anywho, just use it, simplify and take the limit.

 

[.hacker//Kazu]

The problem is, my homework requires me to explain the formulas. I understand the rest, just not why there is a 1/6 in the middle of the equation. My teacher requires that we explain why the formula is the way it is; the rest of my class is stumped too...

 

[e(ho0n3]

That's from the derivation of the formula. I'm sure you can find a derivation somewhere after googling a bit.

 

[AKG]

You can prove the formula for the sum of the first n squares by induction.

 

[.hacker//Kazu]

Induction?

 

[denverdoc]

a really nifty and powerful technique of proving things even non math geeks like myself can understand. Look here for more info:
http://www.cut-the-knot.org/induction.shtml

 

[Edgardo]

As for deriving the formula for the sum of the squares
1/6n (n + 1) (2n + 1), maybe
my post #4 in this thread here helps.

 

[.hacker//Kazu]

Mm, I asked my teacher what induction meant. I think I get the theorem now. Thanks a lot, y'all!

 

[.hacker//Kazu]

But just to make sure, induction is the proof of the nth term and (n+1)th term, right?