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?

Calculus is not necessary to answer this question and I think there are some key insights if you kick it old-style.

Let's start with the easy part: The area of the regular "sandwich style" bagel cut.

The area here is simply the area of the outer circle minus the area of the inner circle.

π(ρ+r)^{2} - π(ρ-r)^{2} = 4πrρ

Now, on to the (Double) Mobius Bagel.

(This image is very confusing--refer to the above 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 π 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 2π 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.)

I'm going to argue that unrolling the bagel didn't change the cut area. If you imagine re-rolling this cylinder into a bagel, you'll see that one side stretches while the other side shrinks. The red and blue lines are symmetrically spaced around the cylinder, and the two edges are the same length in the rolled state, so the stretching and shrinking should be the same for both. But do the stretch and shrink cancel out?

The inner circumference of the re-rolled bagel will be 2π(ρ-r) = 2πρ-2πr.

The outer will be 2π(&rho+r) = 2πρ+2πr.

We get a stretch of 2πr with a shrink of -2πr. Therefore unrolling the bagel doesn't change the cut area.

Another way to think of it is with a trapezoid. If you shorten the top by the same amount that you lengthen the bottom while the height is constant, the area of the trapezoid is the same.

So the area of the cut in cylindrical form is the same as the area of the cut in bagel form. But what is that area? We need to deal with the twist. For this, we unroll the cylinder.

The width of this rectangle is the original cylinder length: 2πρ.

The height of the rectangle is the original circumference of the cylinder: 2πr.

By Pythagoras: b = sqrt((2πρ)^{2} + (2πr)^{2})) = 2π*sqrt(ρ^{2} + r^{2}).

b is the length of the edge of the cut. The "depth" of the cut is 2r (i.e. the diameter of the "tube" of the bagel). Therefore the total area of the cut is 4πr*sqrt(ρ^{2} + r^{2}).

This area is larger than the sandwich-style cut by a factor of ρ/sqrt(ρ^{2} + r^{2}).

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