Cube into a pyramid

A cube is inserted into a four sided (equal sides) pyramid. How to find the ratio between the volumes of the cube and the pyramid?

That is all , no sides or angles given! :(

Look at the picture to understand better. (the cube may not look like a cube but it is one (belive me)). ;)

Volume = AreaBase * Height / 3

I think the above is the formula for the volume of a pyramid with a square base.

Using the above and my MathCad software, I got various values for the volume of a pyramid which enclosed a unit cube. To get a numeric solution, you need more information.

I imagined a plane though the apex of the pyramid, perpendicular to the base, and through the midpoints of opposite base lines. Consider the triangular intersection of this plane with the pyramid.

If Angle is half the apex angle of this triangle, you can derive the following formulae.

PyramidHeight = CubeSide + CubeSide / (2 * Tan(Angle) )

PyramidSide = CubeSide + 2 * CubeSide * tan(Angle)

Volume = (PyramidHeight * PyramidSide^2 ) / 3

If you consider a unit cube (CubeSide = 1), the above are not too difficult to compute, giving you the ratio of the volumes.

Note that CubeSide factors out of the formulae for Pyramid Height & Side. Therefore the ratio of the Pyramid & Cube volumes depends only on the angle (CubeSide^3 is in both formulae).

Do not bet too much on the above without doing some analysis. While I consider myself infallible, others have sometimes found errors in my work.

There is not more data. The volume of a cube is a^3 and of the pyramid is b^2*H/3. The thing that need to be done is to exsprees a through b or vice versa and that is the ratio. So how to expres a through b (ex. a = x*b....etc.).

Guv, how did you get this formula?

Code:

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PyramidHeight = CubeSide + CubeSide / (2 * Tan(Angle) )

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From the triangle i get

Tan(Angle) = PyramidBaseSide / PyramidHeight

Have a look at the picture.

Vlatko: From the looks of your .bmp, I am fairly sure we are both working with the same geometric object.

Like you, I am working with half the apex angle of the triangle. I think you are missing a factor of two in your formula. Using your notation, I think the following are correct.

• PyramindBaseSide / 2 = PyramidHeight * tan(Angle).
• PyramidBaseSide = 2 * PyramidHeight * tan(Angle).
• PyramidHeight = PyramidBaseSide / ( 2 * tan(angle) )

If you study your .bmp, you will notice that there are a lot of similar right triangles. Two on top of the square (cube intersection), two on either side, and two halves of the entire isosceles triangle. A formula similar to the above applies to half the side of the square and the height of the isosceles triangle on top of the square, giving the following.

• LittleHeight = CubeSide / ( 2 * tan(Angle) )
• PyramidHeight = CubeSide + LittleHeight
• PyramidHeight = CubeSide + CubeSide / ( 2 * tan(Angle) )

My formula for the base of the pyramid uses the similar triangles on either side of the square.

Does the above seem correct to you now? If not, post further questions.

To me, the above formulae seem easier to cope with than formulae based on the dimensions of the pyramid, which requires thinking about an inverse tangent function.

If you want to start with pyramid dimensions and find the ratio, you can do the following. Using the formulae from my previous post, cancel out the CubeSide, resulting in a formula expressing the ratio as a function of tan(Angle). Then replace tan(Angle) with the correct ration of Pyramid dimensions, namely the following.

• Tan(Angle) = PyramidBaseSide / (2 * PyramidHeight)

Note the caveat in my previous post. Do not trust me without doing some analysis. When not being paid, I am not meticulous and do not guarantee correct results.

I have been playing with some algebra and probably come up with a much longer way of doing what Guv has already posted. Anyway, I hope this is of some use:

Extend the vertical sides of the cube to cut through the pyramid. The pyramid is now made up of:
A cube
4 x Wedges
4 x Corners
1 x pyramidion ‘the top pointy bit

Volume of the cube = a^3

Volume of each wedge = cross-sectional area x height
=1/2*(b-a)/2*a * a
=(a^2b-a^3)/4
We have four of these so
Volume of the wedges = (a^b-a^3)

Volume of corners, if you push them together you have a small pyramid, height a and square-base sides = (b-a)
Volume = 1/3 * base * height
= 1/3 * (b-a)^2 * a
=(ab^2 – 2a^2b + a^3)/3

Volume of pyramidion
Volume = 1/3 * base * height
Base area=a^2
Using the half angle at the top of the whole pyramid
Let h be the unknown height
Tan(angle) = (a/2)/h ‘ pyramidion
Tan(angle) = (b/2)/(a+h) ‘whole pyramid

So (a/2)/h = (b/2)/(a+h)

(a^2 + ah)/2 = bh/2
2a^2 + 2ah = 2bh
2a^2 = 2bh - 2ah
2a^2 = 2h(b - a)
a^2 = h(b - a)
h = (a^2) / (b - a)

Volume of pyramidion = 1/3 * base * height = 1/3 * a^2 * (a^2) / (b - a)
Volume of pyramidion =(a^4)/3(b-a)

The volume of the whole pyramid as a sum of parts:

= a^3 + (a^b-a^3) + (ab^2 – 2a^2b + a^3)/3 + (a^4)/3(b-a)

= a(b^2 – 2ab + a^2)/3 + a^2b + (a^4)/3(b-a)

= (ab^2)/3 – (2a^2b)/3 + a^3/3 + a^2b + (a^4)/3(b-a)

= (ab^2)/3 + (a^2b)/3 + a^3/3 + (a^4)/3(b-a)

= (ab^3 – a^2b^2 + a^2b^2 - a^3 + a^3b – a^4 + a^4)/3

= (ab^3 – a^3 + a^3b)/3

Divide cube-volume by pyramid-volume

a^3 / (ab^3 – a^3 + a^3b)/3

= 3a^3 / (ab^3 – a^3 + a^3b)

ratio of cube volume to pyramid volume = 3a^2 / (b^3 + a^2b – a^2)