Can You Make the Biggest Bread Bowl?

By boring a cylindrical hole in a hemisphere (or sphere) of bread, how much room can you make for some delicious soup?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

From Michael Lester comes a delectable dilemma about soup:

I have a large, hemispherical piece of bread with a radius of 1 foot. I make a bread bowl by boring out a cylindrical hole with radius r, centered at the top of the hemisphere and extending all the way to the flat bottom crust.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

This Week’s Extra Credit

Instead of a hemisphere, now suppose my bread is a sphere with a radius of 1 foot.

Again, I make a bowl by boring out a cylindrical shape with radius r, extending all the way to (but not through) the curved bottom crust of the bread. The central axis of the hole must pass through the center of the sphere.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a problem in graph theory that I only learned about recently:

Suppose you have two identical graphs whose edges are all assigned weights between 0 and 1. Also, corresponding points between these two graphs are further connected by edges that are assigned arbitrary weights (again, between 0 and 1).

Next, you generate a random subgraph by deleting each edge with a probability equal to that edge’s weight.

Pick any point from one of the two identical graphs, and then pick another point from that same graph. In a randomly generated subgraph, is the first point more likely to be connected to the second point, or to the doppelgänger of the second point over in the other identical graph?

You might think the points in the same graph are always at least as likely to be connected as the first point and the corresponding point in the second graph. But not so fast.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jeffry 🎻 from Redwood City, California. I received 31 timely submissions, of which 26 were correct—good for an 84 percent solve rate. This was a tough one!

Last week, you were doing a 30-minute workout on your stationary bike. There was a live leaderboard that tracked your progress, along with the progress of everyone else who was currently riding, measured in units of energy called kilojoules. (For reference, one kilojoule is 1000 Watt-seconds.) Once someone completed their ride, they were removed from the leaderboard.

Suppose many riders were doing the 30-minute workout alongside you, and that they all began at random times, with many starting before you and many starting after. Further suppose that they were burning kilojoules at different constant rates (i.e., everyone was riding at constant power) that were uniformly distributed between 0 and 200 Watts.

Halfway through (i.e., 15 minutes into) your workout, you noticed that you were exactly halfway up the leaderboard. How far up the leaderboard could you expect to be as you were finishing your workout?

First off, let’s think about the entire distribution of riders. There were two random variables at play here:

• How long a rider has been in the workout, which was uniformly distributed between 0 and 30 minutes.

• The rider’s constant power, which was uniformly distributed between 0 and 200 Watts.

The energy a rider had burned was proportional to the product of these two independent values (i.e., energy equals time multiplied by power). If you rescaled the ranges for the riders’ durations and powers so there went from 0 to 1, then you could think of the continuum of riders as being the points inside the unit square, and the energy burned as the product of their x- and y-coordinates. Here’s an illustration of that unit square:

Riders who were at the same position on the leaderboard were at x- and y-coordinates with the same product, which meant they were all on the same curve xy = C (for some value of C), or y = C/x—a hyperbola! Curves with different values of C have the same color in the graph above, and these values for C increased from the bottom left to the top right of the square.

Now, for some hyperbola, half the square’s area was above it, while the other half of the square’s area was below it, as illustrated by the lower red curve below. This curve represented all the riders who were halfway up the leaderboard, just as you were after 15 minutes. Let’s determine the value of C for this particular curve, which we’ll call C₀.

The two endpoints of the curve inside the unit square were (C₀, 1) and (1, C₀). The area above the curve—which we said had to be half the square’s area, or 1/2—could be computed as the following integral:

\(\int_{C_0}^{1}\left(1-\frac{C_0}{x}\right)dx=\frac{1}{2}\)

The integral of 1−C₀/x is x−C₀ln(x), so evaluating the equation above gave you the rather unfriendly transcendental equation C₀(1−ln(C₀)) = 1/2. The solution (between 0 and 1) to this equation was C₀ = 0.18668.

Halfway through your workout, your x-coordinate in this unit square was 0.5, which meant your y-coordinate was C₀ divided by 0.5, or 2C₀. It turned out you were somewhere around the 37th percentile with respect to power, or putting out about 75 Watts. But that wasn’t what you were asked to figure out.

You needed to determine your position on the leaderboard by the end of the workout, when your x-coordinate was 1 and your y-coordinate was still 2C₀ (since your power remained constant throughout your workout). At this point, the product of the coordinates was 2C₀, a new value that could be plugged into the previous integral formula.

The area above the resulting curve—the higher one in the previous graph—was 1−2C₀(1−ln(2C₀)). The area below the resulting curve was 1 minus that, or 2C₀(1−ln(2C₀)), a value that was approximately 0.741. In other words, you were roughly 74.1 percent of the way up the leaderboard by the end of your workout.

That might have sounded too good to be true for someone who was only in the 37th percentile with respect to power output. But remember—by the end of your workout, all the other riders on the leaderboard still had ways to go, and the majority would indeed surpass your energy total by the end of their respective workouts.

Several solvers, including Q P Liu and Eric Widdison, weren’t intimidated by the aforementioned transcendental equation. They found an exact expression for the proportion: 1−2ln(2)·exp(1+W_-1(-1/(2e))), where W is the Lambert W function.

As an added bonus problem (though not quite Extra Credit), you had to determine how high up the leaderboard you could possibly expect to be 15 minutes into your workout.

In this case, you were exerting the maximum power of 200 Watts. Halfway through your workout, your x-coordinate was definitively 0.5, but now your y-coordinate was maxed out at 1, which meant C (the product of your x- and y- coordinates) was 0.5. The area below your curve was 0.5(1−ln(0.5)), which could be written a little more nicely as (1+ln(2))/2. Numerically, this came out to about 84.6 percent.

So if you ever find yourself halfway through your workout and you’re about five-sixths up the leaderboard, take comfort. You’re well on your way to topping the chart by ride’s end.

Here’s a handy chart from which you can read off a rider’s final leaderboard position based on their position halfway through their workout:

Because the highest you could rank at the halfway point was (1+ln(2))/2, the domain of the graph was restricted to 0 ≤ x ≤ (1+ln(2))/2.

Finally, 🎬 Eric Widdison 🎬 plotted your final ranking in the case where you found yourself X percent up the leaderboard after X percent of your workout. (Note that the two “X”s here represented the same value.) For example, when X was 50 percent, the output was a very familiar 74.1 percent. There were some very neat nonlinear behaviors at play here!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Michael Bradley 🎻 from  London, England. I received 22 timely submissions, of which 16 were correct—good for a 73 percent solve rate.

Again, suppose there were many riders starting their 30-minute workouts at random times, and that their powers were uniformly distributed between 0 and 200 Watts. Now, suppose you decided that you too would be pedaling with a random (but constant) power between 0 and 200 Watts.

If you looked down at the leaderboard at a random time during this random workout, how far up the leaderboard could you have expected to be, on average?