Barney's Wedge

Once upon a MathsJam, Barney Maunder-Taylor showed up with a curious object, a wedge with a circular base. Why?

Well, if you held a light above it, it cast a circular shadow. From one side, the shadow was an equilateral triangle; along the third axis, a rectangle. A lovely thing.

Several challenges presented themselves, including, but not limited to, designing a shape whose cross-sections were a circle, an equilateral triangle and a square - but I digress: the most pressing challenge was, how to create a net for the shape so that other MathsJams could fold their own?

You may wish to have a think about that before I reveal my thinking. Below the line are spoilers.

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The shape has - depending on how you count, three, four, four or five faces. It’s hard to dispute that the circular base is a face; I would count the top as two faces, each with a straight edge and a curved edge, and the vertical surface as a single ‘face’, even where it degenerates into a single point. (We didn’t say the top face had two curved edges now, did we?) However, for the purposes of paper construction, it may be easier to split it in half.

The base is, as I’ve mentioned at least twice, a circle. There is no difficulty there; I will state that its radius is $R$ and leave drawing it as an exercise for the reader.

What shape is the top? My first approach was extremely algebra-heavy; then I noticed I was cutting a diagonal slide through a (semi-)cylinder, so each face would need to be a semi-ellipse. The minor semi-axis is the same as the radius of the circular base, $R$; because the triangular shadow is equilateral, the major semi-axis has to be $2R$. And, rather than drawing two semi-ellipses, we may as well draw one and fold it across the minor axis.

The vertical side is the interesting one, though. For this, I needed some 3-D thinking.

My starting point was “what would I do to make a cylinder?” - simply roll up a piece of paper with the appropriate width, $2\pi R$. That suggested to me that finding the height of the odd-shaped surface in terms of the angle would be the critical move.

Let’s suppose the fold in the top surface(s) is parallel to the $x$-axis. Then one of the semi-ellipses lies in the plane $z=\sqrt{3}(R-x)$ (the other lies in $z=\sqrt{3}(R+x)$, but we’ll ignore that for now. Symmetry will deal with that.)

However, $x=R\cos (\theta )$, which means $z=R\sqrt{3}(1-\cos (\theta ))$ - and suddenly we’re there (at least for this half-ellipse). We can parameterise the curve we need as $(R\theta ,R\sqrt{3}[1-\cos (\theta )])$ as long as $\theta \ge 0$ - that is, for $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$.

The other half will have exactly the same shape - I think replacing $\cos (\theta )$ with $|\cos (\theta )|$ would have the same effect.

Here is my attempt at a net in Desmos. Have fun!