I have asked this question differently and gotten no answer. I have done the wor

Question

I have asked this question differently and gotten no answer. I have done the work, but my answer doesn't match what the professor said it should be. He said that the volume should be V=s^3, which matches the volume of a cube. The problem, and my solution (so far) works like this. You start with a regular tetrahedron, and then do Koch division on it (like a Koch snowflake). I need to find the volume of the resulting 3-dimensional shape. Here is my work so far, if Chegg lets me post it from Word properly... A pyramid with an equilateral triangle as the base has a volume defined by the formula: V = s^3/(6?2), where s is the length of a side of the tetrahedron In the first iteration, the volume would be V1 = s^3/(6?2) Note that there are 4 sides to a regular tetrahedron, so the volume would be the volume of the first one, plus the volume of each of the 4 added tetrahedrons, which are each 1/2 the height of the original one

Explanation / Answer

Interesting problem... and of course you really have to visualize it to see the pattern.

Your algebraic expressions are a little difficult to follow, but I know you did the best you could with the formatting available.

Just thinking about this for a moment, and picturing the progression, I think your statement

" Now with each iteration, each side has 6 times as many tetrahedrons being added,"   

might be wrong. It seems like you went through the first few iterations and tried to establish the pattern. But you might have stopped short, i.e. not enough iterations to clearly see the pattern.

My instinct tells me the pattern is not geometric (i.e. multiplied by a constant factor, like 6, at each step), but more like exponential... you might find that your factor at each iteration is something more like

6^n-1 i.e. 1, 6, 36, 216...

Oh hell, I'll write it out... now I'm thinking about it...

First tetrahedron of height s: one

Second tetrahedron of height s/2: four

Third tetrahedron of height s/4: 24 (this is where you got to, I think)

NOW... fourth tetrahedron of height s/8: I think I counted 108. I think...

Try to count them in levels to see the pattern

Second tetra: four on level 1 (i.e. the side of the first tetra)

Third tetra: 12 on level 1 (3 on each side) and 12 on level 2 (on the sides of the 2nd tetra)

Fourth tetra: 36 on level 1, 36 on level 2, 36 on level 3   

Fifth tetra: 108 on level 1, 108 on level 2, 108 on level 3, 108 on level 4

Sixth tetra: 324 324 324 324 324

So... yeah. The pattern is here...but it's not simply a multiple of 6. It goes   

12 + 12

12*3 + 12*3 + 12*3

12*3*3 + 12*3*3 + 12*3*3

etc etc

Since the pattern starts with the fourth tetra, define n = 4 for the third tetra and you have

number of tetra = (n-1)* 12 * 3^(n-3)

holy smokes! yeah, that gets complicated

in essence, you're going to have to sum, from n = 4 to infinity

(n-1) * 12 * 3^(n-3)

of course you can factor out the 12... but I dont see this converging. OHHHH!!

Of course... you have to include the volume of each iteration. Then you should get something that converges.

See what you can put together from this. You should be able to do

volume of each iteration * number of tetras in each iteration (my expression above) summed from n=4 to infinity.

Of course you also have to add the volumes of the 1st, 2nd and 3rd level tetras when you're done, but those are simple.

Let me know how it goes... I'm very curious now.

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