Dec 15 / David

Pouring a Cone-Shaped 40 For Archimedes Part I

George Hart presents what he says is a calculus problem:

What is the ratio of the surface area of [the double mobius] cut to the surface area of the usual planar bagel slice?

On my old blog, I took a stab at this using non-calculus means but got the wrong answer. I think I now have the right answer, but it’s less clear that I didn’t use calculus to get there. Also, I’m expanding this into two parts so I have more room to wave my hands in lieu of actual proofs.

Why do I care if I used calculus or not? Because I think the point of this question (and many other math questions, especially of the proof or puzzle variety) is not a numerical answer but an insight. A coworker of mine tried to just directly right down an equation from the bagel and mechanically crank through it from there. You don’t learn anything that way! (This is the Get Off My Lawn Theory of Mathematics.)

Let’s start with the easy part: The area of the regular “sandwich style” bagel cut.


(I’ve chosen slightly odd notation here so it will be consistent in later steps where I need it.) The area here is simply the area of the outer circle minus the area of the inner circle.

\pi (\rho + r)^2 - \pi(\rho - r)^2 = 4\pi r\rho

Now, on to the (Double) Mobius Bagel.


(This image is very confusing–refer to the link for a better visualization of the bagel in 3D.) Note that the red and blue edges of the cut must be the same length because they are the same cut, just offset by \pi radians around the bagel. OK, now unroll the bagel into a cylinder.


The double mobius cut is now a slice through the center of a cylinder. If you imagine the knife traveling down the cylinder, it rotates 2\pi radians while traversing the length. (Aside: The solution to the question of how to make a true mobius cut in a bagel should now be easy to visualize.)

In Part I, I’m going to argue that unrolling the bagel didn’t change the cut area.


Unrolled, the bagel’s sandwich cut area is: (2r)(2\pi \rho ) = 4\pi r\rho. This is the same as the rolled area. Why did this happen? Let’s re-roll the bagel.

The inner circumference of the re-rolled bagel will be 2\pi (\rho - r) = 2\pi \rho - 2\pi r.
The outer circumference of the re-rolled bagel will be 2\pi (\rho + r) = 2\pi \rho + 2\pi r.

We get a stretch of 2\pi r with a shrink of -2\pi r. Think of it is as a trapezoid. If you shorten the top by the same amount that you lengthen the bottom, the area of the trapezoid is the same.

OK, that explains why the sandwich cut doesn’t change area when you unroll the bagel, but what about this crazy double mobius cut? This is where I’m going to resort to some handwaving.

The red and blue lines are symmetrically spaced around the cylinder. If a particular area on the bagel has a blue line that experiences a stretch, then on the same spot \pi of the way around the bagel the red line gets that same stretch.

Furthermore, both lines encircle the cylinder a full rotation each. If a particular area of the bagel has a blue line that experiences a stretch, then on the same spot \pi of the way around the “tube” of the bagel the blue line gets a shrink of the same magnitude.

In other words, because of the way the lines “spend equal time” on either side of the \rho line, the total length of each of the blue and red lines is unchanged when unrolling the bagel. And in general, this is true for any line drawn on the cut area, not just for those lines that represent the edges of the cut.

In Part II, I consider what the area actually is.

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