On calculating a "Mathematically Correct Breakfast"

A friend of mine recently sent me a link to this video by George W. Hart in which he cuts a bagel into two interlocking rings.

After toasting the bagel rings George wonders about how much surface area he has available for cream cheese.  Is there more or less compared to the usual method of just slicing the bagel straight in half?

To answer this question we first need to visualize exactly how George is cutting the bagel.  Let's assume the bagel is perfectly round -- let's assume it's a torus, to be precise.  We'll also assume he's cutting the bagel perfectly evenly.  If we looked closely, we'd find that he's actually cutting the bagel in half along a two-twist Möbius strip.  This cut is shown in the image below.

Calculating the area of the two-twist Möbius strip

We'll need some equations to describe the bagel.  The Wikipedia article for the torus provides us with a parameterization, given by
\begin{align*} & x = (R + r\cos \varphi) \cos \theta \\ & y = (R + r\cos \varphi) \sin \theta \\ & z = r \sin \varphi, \end{align*} where $R$ is the distance from the center of the torus to the center of the tube and $r$ is the radius of the tube. Each point on the torus $(x,y,z)$ is represented by a single combination of the parameters $\theta$ and $\varphi$, each of which lie between $0$ to $2\pi$.  In order for this to represent a physical bagel we can't allow any self-intersections, so we'll also require the diameter of the tube to be less than half the diameter of the torus, which is the same as requiring $0 < r \leq R$.

Using this parameterization for the torus we can obtain a parameterization for the möbius strip inside it.  In the formula above we'll replace $r$ with $\rho$ and let $\rho$ to vary from $-r$ to $r$, allowing us to cross from one side of the tube to the other through its center.  Furthermore, the cut completes a full twist before returning to its starting point, so to capture this we'll also set $\varphi = \theta$.

The (vector form) parameterization of the two-twist Möbius strip is thus $\gamma(\rho,\theta) = \Bigl\langle (R + \rho \cos\theta) \cos\theta,\ (R + \rho \cos\theta) \sin\theta,\ \rho \sin\theta \Bigr\rangle.$ Wikipedia provides us with the formula to calculate the surface area of a parameterized surface, and in our case we can express the surface area of the two-twist Möbius strip as $A_\text{Möb}(R,r) = \int_0^{2\pi} \!\int_{-r}^r \,\left|\frac{\partial\gamma}{\partial\rho} \times \frac{\partial\gamma}{\partial\theta} \right|\,d\rho d\theta.$ After some work we calculate $\left|\frac{\partial\gamma}{\partial\rho} \times \frac{\partial\gamma}{\partial\theta} \right|^2 = R^2 + 2R\rho\cos\theta + \frac{1}{2} \rho^2 (3 + \cos 2\theta),$ so that $A_\text{Möb}(R,r) = \int_0^{2\pi} \!\int_{-r}^r \,\sqrt{R^2 + 2R\rho\cos\theta + \frac{1}{2} \rho^2 (3 + \cos 2\theta)} \,d\rho d\theta. \tag{1}$ This integral we've arrived at seems fairly intractable (Mathematica throws up its hands in frustration) but there is still much we can learn from it.

How much more surface area can we get?

If you recall, our goal in all this is to compare the amount of cream cheeseable surface area revealed by cutting along the two-twist Möbius strip to the amount revealed by cutting the bagel straight in half.  We've obtained an expression for the former amount, and of course we'll need to calculate the latter amount as well.  Luckily this task is much simpler than the work we did above.

The surface exposed by cutting a bagel straight in half is a disk with a hole in the center.  In the notation from the previous section, the outer ring of the disk has radius $R+r$ and the inner ring has radius $R - r$, so all together the surface area is the difference of the areas of these two circles, $A_\text{flat}(R,r) = \pi(R+r)^2 - \pi(R-r)^2 = 4\pi Rr.$ We can now write down the ratio of these two surface areas. Using the expression from equaiton $(1)$ we have \begin{align*} \frac{A_\text{Möb}(R,r)}{A_\text{flat}(R,r)} &= \frac{1}{4\pi Rr} \int_0^{2\pi} \!\int_{-r}^r\,\sqrt{R^2 + 2R\rho\cos\theta + \frac{1}{2} \rho^2 (3 + \cos 2\theta)} \,d\rho d\theta \\ &= \frac{1}{4\pi r} \int_0^{2\pi} \!\int_{-r}^r\,\sqrt{1 + 2\frac{\rho}{R}\cos\theta + \frac{1}{2} \frac{\rho^2}{R^2} (3 + \cos 2\theta)} \,d\rho d\theta \\ &= \frac{1}{4\pi} \int_0^{2\pi} \!\int_{-1}^1\,\sqrt{1 + 2\frac{r}{R}\sigma\cos\theta + \frac{1}{2} \left(\frac{r}{R}\right)^2 \sigma^2 (3 + \cos 2\theta)} \,d\sigma d\theta \tag{2}, \end{align*} where in the second line we factored an $R$ out of the square root and in the third line we made the change of variables $\rho = r\sigma$ in the inner integral.

This expression is quite interesting. For starters it tells us the following fact.

Proposition 1. The ratio of the surface areas $A_\text{Möb}(R,r)/A_\text{flat}(R,r)$ depends only on the ratio of the radii $r/R$.

In particular, as long as the ratio $r/R$ is held constant, it doesn't matter how big the bagel is, the ratio of the surface areas will not change.

Now, if we let $t = r/R$ then we can define a new function $f$ by $f(t) = \frac{A_\text{Möb}(R,r)}{A_\text{flat}(R,r)} = \frac{1}{4\pi} \int_0^{2\pi} \!\int_{-1}^1\,\sqrt{1 + 2t\sigma\cos\theta + \frac{1}{2} t^2 \sigma^2 (3 + \cos 2\theta)} \,d\sigma d\theta. \tag{3}$ Below is a plot of $f(t)$. Since $0 < r \leq R$ our new parameter $t$ ranges from $0$ to $1$.

From this plot it certainly looks like $f(t)$ is an increasing function, so let's try to prove this. We begin by calculating $f'(t) = \frac{1}{4\pi} \int_0^{2\pi} \!\int_{-1}^1\,\frac{2\sigma\cos\theta + t\sigma^2 (3 + \cos 2\theta)}{2\sqrt{1 + 2t\sigma\cos\theta + \frac{1}{2} t^2 \sigma^2 (3 + \cos 2\theta)}} \,d\sigma d\theta.$ From this we find that $f'(0) = 0$, so if we can prove that $f''(t) > 0$ we'll be able to conclude that $f'(t) > 0$ for $t > 0$. Indeed, $f''(t) = \frac{1}{4\pi} \int_0^{2\pi} \!\int_{-1}^1\, \frac{\sigma^2}{\left( 1 + 2t\sigma\cos\theta + \frac{1}{2}t^2\sigma^2 (3+\cos 2\theta)\right)^{3/2}} \,d\sigma d\theta,$ and since the integrand is nonnegative and not identically zero we may conclude that $f''(t) > 0$ for all $t$. As stated above, this implies that $f(t)$ is strictly increasing in $t$.

From this we conclude the following fact.

Proposition 2. The ratio of the surface areas $A_\text{Möb}(R,r)/A_\text{flat}(R,r)$ is a strictly increasing function of $r/R$. Since $0 < r \leq R$ we have $1 < \frac{A_\text{Möb}(R,r)}{A_\text{flat}(R,r)} \leq \frac{A_\text{Möb}(1,1)}{A_\text{flat}(1,1)} \approx 1.1739\ 7932\ 4030.$

In essence what this says is that, for a given size of bagel, cutting the bagel along the two-twist Möbius strip yields the largest gains in surface area when the hole in the center of the bagel is the smallest.   The maximum amount that the surface area can increase when compared with cutting the bagel straight in half is about $17.4\%$.

The amount that the surface area can increase will be less than this for most bagels.  For example, for a bagel $2$ inches thick and $5$ inches wide, cutting along the two-twist Möbius strip yields an increase in cream cheesable surface area of about $7.72\%$.  Depending on how much you enjoy cream cheese this may or may not be worth it.

Approximating the ratio of the surface areas

If we're in the middle of making breakfast then we don't really have time to start up our computer just to evaluate the integral in equation $(2)$. What would be really handy is an accurate approximation. And really, the assumption that our bagels are perfect torii is already an approximation, so we shouldn't feel too bad about the loss of accuracy.

In this problem the path to an approximation is delightfully short. In the definition of the function $f(t)$ in equation $(3)$ the parameter $t$ lies between $0$ and $1$, and in the integrand it is multiplied by bounded functions, so it seems appropriate to expand the integrand in Taylor series about the point $t=0$. We find that $\sqrt{1 + 2t\sigma\cos\theta + \frac{1}{2} t^2 \sigma^2 (3 + \cos 2\theta)} = 1 + t\sigma\cos\theta + \frac{1}{2}t^2\sigma^2 + O\left(t^3\right),$ and if we replace the integrand by these first three terms of its Taylor series we obtain the approximation \begin{align} \frac{A_\text{Möb}(R,r)}{A_\text{flat}(R,r)} &\approx \frac{1}{4\pi} \int_0^{2\pi} \!\int_{-1}^1\, 1 + t\sigma\cos\theta + \frac{1}{2}t^2\sigma^2 \,d\sigma d\theta \\ &= 1 + \frac{1}{6}t^2 \\ &= 1 + \frac{1}{6} \left(\frac{r}{R}\right)^2. \end{align} This approximation produces reasonably good results. It's most accurate when the ratio $r/R$ is small.

Some might say that measuring $R$ and $r$ quickly isn't as easy as just measuring the width and the height of the bagel. Of course we can express these quantities in terms of the others as $R = \frac{\text{width} - \text{height}}{2}$ and $r = \frac{\text{height}}{2},$ so we'll conclude the page with the following statement for practical use.

Proposition 3. The ratio of the surface area exposed by cutting along the two-twist Möbius strip to the area exposed by cutting the bagel straight in half is approximately $1 + \frac{1}{6} \left(\frac{\text{height}}{\text{width} - \text{height}}\right)^2.$