Can You Save the Pizza?

What’s the smallest rectangle of foil you need to completely wrap a triangular slice of pizza, both above and below?

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

I have a slice of uneaten pizza that I’d like to wrap up in aluminum foil before storing in the fridge. The bad news is that I’m all out of foil. The good news is that my unusual neighbor has huge sheets of foil, although I’m not sure why.

I knock on my neighbor’s door, and they’re willing to give me some of their foil in exchange for babysitting their pet snake. For every square inch of foil I borrow, I have to babysit this snake an additional minute. Needless to say, I hate snakes, and I want to request the bare minimum of foil that I need.

The slice in question is an equilateral triangle with a side length of 10 inches. I specifically want to request the smallest rectangle of foil (by area) such that I can completely cover the slice above and below. I can’t fold the pizza, but I’m allowed to make as many straight folds in the foil as I want, without tearing it.

What is the area of the smallest such rectangle of foil that would do the trick?

Extra Credit

Instead of one slice of pizza, I now want to wrap two slices of pizza using a single rectangular piece of foil. The two slices may touch, but they can’t overlap in any way (lest the cheese of one slice stick to the other), and their bottoms and tops must be completely covered by the foil.

Again, what is the area of the smallest such rectangle of foil that would do the trick?

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 some recent musings of James Tanton. James has a habit of asking related questions over the course of multiple days, with each question building on the last in terms of both complexity and intrigue.

Earlier this week, he observed the following, which I have lightly edited:

The Pythagorean triple 3-4-5 appears as the exponents in an equation of the form x³ + y⁴ = z⁵,  where x, y, and z are all positive integers. Indeed, this is possible when x, y, and z are powers of 2. For example, 256³ + 64⁴ = 32⁵.

Is it possible to find similar equations when the exponents are other Pythagorean triples, such as 5-12-13 and 7-24-25? Is it possible to find similar equations for every Pythagorean triple?

I was pleasantly surprised by how this question related to number theory, modular arithmetic, and remainders. It appears that James has extended this line of questioning, and he has moved on to “Pythagorean anti-triples.”

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: 🎻 Kevin L 🎻 from Irvine, California. I received 79 timely submissions, of which 67 were correct—good for an 85 percent solve rate.

Last week, you were driving around a circular road, along which travelers could move in either direction. However, there was only one gas station on the loop. Driving the full loop in your car required 40 gallons of gas, but your car’s fuel tank had a maximum capacity of 20 gallons. That said, you aspired to see every last spot along the route.

Fortunately, you were able to call on any number of your Fiddler Nation friends, all of whom happened to have the same make and model car as you, each with a 20-gallon fuel tank and identical fuel efficiency.

Now, all the cars, including yours, had to start and end at the gas station. However, only your car had to cover the entire route. The gas station could be visited (and refueled at) by any car, any number of times. Cars could also transfer fuel from one to another, provided they met up together at a spot along the route.

What was the smallest number of cars (including yours) needed for you to see every spot on the circular route?

If there was one car in total (i.e., yours), then you could only make it halfway around the loop, since 20 gallons was half of 40 gallons. At this point, you were out of gas, leaving half the loop unseen. It was clear that you needed a little help from your friends. 

If there were two cars in total (i.e., yours plus a friend’s car), then it was possible to see more of the loop. Suppose both cars set off from the gas station together. After they each consumed g gallons of gas (meaning both cars had 20−g gallons remaining), your friend decided to give you some of their gas and head back. Since they needed another g gallons for the return trip, they could afford to transfer 20−2g gallons to your car. At this point, you had (20−g) + (20−2g), or 40−3g gallons in your car. Since the capacity of your tank was 20 gallons, you traveled as far as possible when these two expressions were equal: 40−3g = 20. That meant g was 20/3, and that your car consumed a total of 80/3, or about 26.7 gallons on its journey. Congratulations—you made it more than halfway around the loop! But you didn’t make it far enough such that your friend could safely meet up with you on the other side. And that meant you needed more friends.

Instead of one friend accompanying you for g gallons of the journey, what if you had N friends? And after consuming g gallons, they all transferred gas to you. Now, instead of receiving (20−2g) gallons, you received N·(20−2g) gallons, giving you a total of (20−g) + N·(20−2g) gallons. Of course, the capacity of your tank remained 20 gallons, so you were again constrained by the equation (20−g) + N·(20−2g) = 20. Solving this gave you g = 20N/(2N+1). Alas, no matter how large N was, this value could never be greater than 10. And so, by this strategy, you could never make it three-quarters of the way around the loop, which was what was required for those N friends to all meet up with you on the other side and escort you back to the gas station.

No, another strategy was needed—a strategy in which your friends played asymmetric roles of assistance. And it turned out you didn’t need too many friends after all.

The animation below shows how solver Emily Kelly made it all the way around with just two friends and three total cars:

Here’s a summary of Emily’s strategy, assuming you were in Car A and your two friends were in Cars B and C:

• All three cars began at the gas station with full tanks.

• Next, all three cars burned 5 gallons. They all had 15 gallons remaining.

• Car C transferred 5 gallons each to Cars A and B. Cars A and B now had full tanks, while Car C had 5 gallons remaining.

• Car C (barely) returned to the gas station, where it refueled. Cars A and B each burned another 5 gallons, after which they each had 15 gallons remaining.

• Car B transferred 5 gallons to Car A. Car A now had a full tank, while Car B had 10 gallons remaining.

• Car B (barely) returned to the gas station, where it refueled. Car A passed the midpoint of the loop as it burned all 20 gallons. It was now out of gas on the other side of the loop.

Note that as Car A passed through the midpoint, it had half a tank remaining. The symmetry of this situation gave Emily confidence that, by undoing these steps on the other side of the loop, Car A could safely return to the gas station.

Okay, back to the strategy:

• Cars B and C each burned 5 gallons heading toward Car A, which was now on the other side of the loop. They each had 15 gallons remaining.

• Car C transferred 5 gallons to Car B. Car B had a full tank, while Car C had 10 gallons remaining.

• Car B burned 5 gallons to reach Car A. Car B had 15 gallons remaining, while Car A was still out of gas.

• Car B transferred 10 gallons to Car A. Car B had 5 gallons remaining, while Car A now had 10 gallons.

• Cars A and B each burned 5 gallons to return to Car C. Car A had 5 gallons, Car B was out of gas, and Car C had 10 gallons remaining.

• Car C transferred 5 gallons to Car B. Now all three cars had 5 gallons remaining.

• All three cars (barely) returned to the gas station.

There were other strategies for making it around, but three cars was the minimum number required.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jim Gu 🎻 from Beijing, China. I received 50 timely submissions, of which only two were correct. You read that right—this was the trickiest dang puzzle I’ve run to date! Since there were only two solvers, let’s give that other solver a shoutout as well: Benjamin Phillabaum from Lafayette, Indiana.

For Extra Credit, assuming you used three total cars to complete your tour of the entire circular road, what was the minimum amount of gas collectively needed by all the cars for this journey? (Remember, they all had to begin and end at the gas station.)