What’s the Area of a “Pseudo-Square”?

Does it have four equal sides? Check. Does it have four right angles? Check. Is it a square? Kinda…

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 Tom Keith comes an opportunity to “square the circle,” so to speak:

A post from Anthony Bonato recently caught Tom’s attention. Here’s the image from that post:

Sure enough, the shape (or one very much like it) has four “sides” of equal length, with four right angles. However, two of these sides are curved (in particular, they are arcs of circles), and two of the right angles are exterior, meaning the interior angles measure 270 degrees (rather than the usual 90 degrees).

Let’s call shapes like this one “pseudo-squares.” A pseudo-square has the following properties:

• It is a simple, closed curve.

• It has four sides, all the same length.

• Each side is either a straight line segment or the arc of a circle.

• The four sides are joined at four corners, with each corner having an internal angle of 90 degrees or 270 degrees.

The pseudo-square pictured above has two straight sides, which run radially between arcs of two concentric circles.

Assuming this is a unit pseudo-square (i.e., each side has length 1), what is its area?

This Week’s Extra Credit

Hopefully, by this point, you have determined the area of a unit pseudo-square with two curved sides and two straight sides.

Can you find a unit pseudo-square that has three curved sides and just one straight side?

What is the area of your new unit pseudo-square?

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 an algebraic puzzle that was sent my way from Andy Quick:

Find natural numbers x, y, and z that satisfy the following equation:

x/(y+z) + y/(x+z) + z/(x+y) = 4

I will say that the smallest values that satisfy the equation did not match my expectation. Here’s Andy’s write-up of the problem and a solution.

As always, feel free to discuss this puzzle and your approach in the comments below.

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: 🎻 Eli Wolfhagen 🎻 from Brooklyn, New York. I received 52 timely submissions, of which 34 were correct—good for a 65 percent solve rate.

While trick-or-treating, you encountered a mysterious house in your neighborhood. After ringing the doorbell, you were presented with a giant bag from which to pick candy. You were informed that the bag contained exactly three peanut butter cups (your favorite!), while the rest were individual kernels of candy corn (not your favorite!).

You had absolutely no idea how much candy corn was in the bag—any whole number of kernels (including zero) seemed equally possible in this monstrous bag.

You reached in and pulled out a candy at random. That is, each piece of candy was equally likely to be picked, whether it was a peanut butter cup or a kernel of candy corn. You removed your hand from the bag to find that you had picked a peanut butter cup. Huzzah!

You reached in again and pulled a second candy at random. It was another peanut butter cup! You reached in one last time and pulled a third candy at random. It was the third peanut butter cup!

At this point, whatever was left in the bag was just candy corn. How many candy corn kernels did you expect to be in the bag?

At first, the answer here might have seemed unknowable. There could have been any number of kernels in the bag! But if you thought about it, not every number was equally likely. Before randomly selecting any candy, it was just as likely for there to be five kernels as a billion, or a googol for that matter. But if there had really been that many kernels, the probability of randomly selecting a peanut butter cup (let alone three) would have been infinitesimal. The moment you drew that first peanut butter cup, it was immediately more likely for the bag to have fewer kernels than many kernels. And drawing two more peanut butter cups thereafter made it even more likely that there were few kernels.

That was about as far as you could get with common sense. To calculate the expected number of kernels, you needed some math.

First, suppose there were C candy corn kernels in the bag along with the three initial peanut butter cups. At any given point, the probability of randomly picking a peanut butter cup out of the bag was the number of peanut butter cups divided by the total amount of candy (peanut butter cups plus candy corn kernels).

What was the probability that the first three candies you picked would all be peanut butter cups? For the first candy, it was 3/(C+3). Given that the first candy was a peanut butter cup, the probability the second was also a peanut butter cup was 2/(C+2). And given that the first two candies were both peanut butter cups, the probability the third was the final peanut butter cup was 1/(C+1). The probability of all three of these events happening was the product of their respective probabilities, or 6/((C+1)(C+2)(C+3).

Sure enough, when C was very large, this probability was very small—roughly proportional to 1/C³. And when C was small, this probability was much larger, reaching a maximal value of 1 when C was zero. (When the bag contained nothing but three peanut butter cups, you were guaranteed to draw only the three peanut butter cups.)

At this point, given a value of C, you could compute the probability of drawing three peanut butter cups. But to figure out the expected value of C, you needed things the other way around: the probability of C, given that you drew three peanut butter cups.

The key was recognizing that these two probabilities were proportional to each other. In other words, the more likely it was that a value of C resulted in three peanut butter cups being drawn, the more likely it was that there were in fact C candy corn kernels in the bag.

This idea is captured more rigorously by what’s called Bayes’ theorem, which we’ll explore for a moment. Let A and B be two probabilistic events. (Here, we’ll say that A is that there were C candy corn kernels, while B is that you drew three peanut butter cups.) For any events A and B, Bayes’ theorem says that p(A|B) = p(B|A)p(A)/p(B). If you’re not familiar with this notation, “p(A|B)” can be read aloud as “the probability of A occurring, given that B has occurred (or that B is true). For this problem, we wanted p(A|B), the probability that there were C candy corn kernels given that you drew three peanut butter cups. p(A) was initially a uniform distribution, while p(B) had no dependence on C, and so therefore p(A|B) was proportional to P(B|A).

At this point, we have proven that the probability there were C candy corn kernels was proportional to 6/((C+1)(C+2)(C+3)). Perhaps the probability that there were C was precisely 6/((C+1)(C+2)(C+3))?

Not so fast! The probabilities of an exhaustive set of outcomes should always sum to 1. Meanwhile, adding the infinite series 6/((C+1)(C+2)(C+3) from C = 0 to C = ∞ resulted in something decidedly not 1. To determine the sum, you could use partial fraction decomposition to write 6/((C+1)(C+2)(C+3)) as 3/(C+1) − 6/(C+2) + 3/(C+3). From there, the summation became (3/1−6/2+3/3) + (3/2−6/3+3/4) + (3/3−6/4+3/5) + …, what’s known as a telescoping series. The only terms that didn’t eventually cancel here were those with denominators less than 3, leaving you with 3/1 − 6/2 + 3/2, or 3/2.

To ensure the probability distribution for C was normalized (meaning the probabilities summed to 1), you had to divide each probability by 3/2, so that p(C) = 4/((C+1)(C+2)(C+3)). The expected value of C was the sum of each value of C multiplied by its respective probability, p(C). This was yet another infinite series from C = 0 to C = ∞, but this time each term was 4C/((C+1)(C+2)(C+3)). The partial fraction decomposition was -2/(C+1) + 8/(C+2) − 6/(C+3), and the resulting telescoping series was (-2/1+8/2−6/3) + (-2/2+8/3−6/4) +(-2/3+8/4−6/5) …. Once again, the only terms that didn’t cancel were those with denominators less than 3, leaving you with -2/1 + 8/2 − 2/2, which simplified to -2 + 4 − 1, or 1.

You read that right. After all that work with Bayes theorem, partial fraction decomposition, and telescoping series, the expected number of candy corn kernels in the very large bag was … 1. While the most likely single number of kernels was zero, as illustrated below by solver Michael Nolte, the average was indeed 1./

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Bojan Gornik 🎻 from Ljubljana, Slovenia. I received 20 timely submissions, of which 15 were correct—good for a 75 percent solve rate.

Last week, you moved on to the next house in your neighborhood, which was even more mysterious. The owner of this house presented you with another large bag, which they said had exactly N peanut butter cups, where N ≥ 3. The remainder of the bag was filled with some unknown quantity of candy corn kernels, with any amount of them again being equally likely.

You reached into the bag k times, with 3 ≤ k ≤ N, and pulled out a candy at random. Each time, it was a peanut butter cup!

How many candy kernels did you expect to be in the bag?