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4D Equable Shapes – can you help?

October 7, 2015

In year 8 (ages 12 and 13), we have a project called Equable Shapes which starts off with trying to find a rectangle where the area and perimeter have the same value (ignoring units). Students often stumble across some through trying various shapes and are more successful if they are systematic in their approach. Usually, someone eventual points out that just trying rectangles isn’t very efficient or, more bluntly, calls it tedious. This is the perfect cue for “You’re right, there is a quicker way…” and demonstrating an algebraic approach. It’s really nice to get to the point where students are actually asking if they can use algebra to solve something.

Then, it can go in various ways:

  • Compound rectangles,
  • Triangles (with pythagoras),
  • Some other interesting 2d shapes,
  • 3D – surface area = volume.

Now, I have a couple of students who noticed that in 2D, a 4×4 shape is equable and in 3D, a 6x6x6 shape is equable and so, they suggested that in 4D, an 8x8x8x8 shape should be equable with volume = to hypervolume?

This seems a reasonable prediction. What I’d like help with is how to determine if that’s true. I was able to describe that the 4D shape in question would be bounded (somehow) by 8 cubes of size 8x8x8, which I hope is actually true. What is 4D space called? It’s fair to say that my 4D shape work has been somewhat lacking in the past and I feel that I’m now at a stage in my life where I’m happy to delve into 4D. Can anyone help? Oh, and if you could do it in such a way that a 12 year could understand that’d be great.


From → Uncategorized

  1. I’m no expert, but am interested…

    I believe the shape you are referring to is a hypercube or tesseract. The 4-dimensional volume of this would be 8^4. The 3-dimensional surface-volume (akin to surface-area) would indeed be 8 x 8^3, so this would be an equable 4d shape.

    The number of (n-1)-dimensional boundary elements of an n-dimensional cube is 2n.
    a 2d square has 4 x 1d lines surrounding
    a 3d cube has 6 x 2d squares surrounding
    a 4d hypercube has 8 x 3d cubes surrounding
    …and so on
    (I can’t prove this, but a little internet research reveals it!)

    Generally, an n-dimensional cube with side lengths s has:
    n-dimensional volume = s^n
    (n-1) dimensional surface-volume = (2n) x s^(n-1)

    When s = 2n (side length = 2 x number of dimensions) these formulae will yield equal results.

    You may notice the (n-1)-dimensional surface-volume is the derivative of the n-dimensional volume. This is always the case for the boundary of a shape in any dimension – consider the circle, C = 2(pi)r is the derivative of A = (pi)r^2, or the sphere volume & surface area formulae.

  2. David Radcliffe permalink

    Four-dimensional space is called 4-space, and a four-dimensional cube is called a 4-cube. The hypervolume of an n-cube with sides of length s is s^n.

    An n-cube has 2n faces, because it has n axes, and for each axis there is a front face and a back face. Each face is an (n-1)-cube with sides of length s. So the equation to be solved is s^n = 2n * s^(n-1), which yields s = 2n.

  3. David Radcliffe permalink

    The Wikipedia article for “Tesseract” has some pictures. (Tesseract is an older word for a four-dimensional cube.)

    The unit 4-cube is the set of all quadruples (w, x, y, z) where the variables range between 0 and 1. To find a hyperface of the 4-cube, we pick one of the four variables and set it equal to 0 or 1. There are 4 × 2 = 8 possible choices. It’s hard to visualize high dimensional objects, but we can reason about them algebraically.

  4. I thought of a 4D cube differently – imagine a 2D picture of a cube pulled out of the paper into 3D.

    Are we always looking at perimeter and area, or area and volume too? (and whatever that is in 4D).

    With some made up maths on my 4D cube I think the pattern goes 4, 6 then 12 (rather than 8) but will look at the wiki link above!

    • I’m afraid that I can’t imagine what you’ve suggested! I’ll keep trying.
      We were using area and perimeter for 2d but surface area and volume in 3d. There’s probably something in looking at total edge lengths of cuboids too.

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