# 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.

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 $-\piby 2 \le \theta \le \piby 2$.

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!