Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

While spending more time at home in recent weeks, I’ve had the chance to revisit one of my favorite video games from recent years — The Legend of Zelda: Breath of the Wild. Within the game, there are hundreds of hidden “Korok Seeds,” which I’m having an increasingly difficult time finding.

Fortunately, there’s a special mask you can acquire in the game that makes a sound any time you’re within a certain distance of a Korok Seed. While playing, I marked nine distinct locations on the game map, forming the 3-by-3 grid shown below:

Each leaf symbol is within range of a Korok Seed, while the point in the middle is *not* within range of a Korok Seed. Given this arrangement, what is the minimum possible number of Korok Seeds I could have detected?

The solution to this Riddler Express can be found in the following week’s column.

## Riddler Classic

From Jim Crimmins comes a puzzle about what would presumably be the largest Zoom meeting of all time:

One Friday morning, suppose everyone in the U.S. (about 330 million people) joins a single Zoom meeting between 8 a.m. and 9 a.m. — to discuss the latest Riddler column, of course. This being a virtual meeting, many people will join late and leave early.

In fact, the attendees all follow the same steps in determining when to join and leave the meeting. Each person independently picks two random times between 8 a.m. and 9 a.m. — not rounded to the nearest minute, mind you, but *any* time within that range. They then join the meeting at the earlier time and leave the meeting at the later time.

What is the probability that at least one attendee is on the call with everyone else (i.e., the attendee’s time on the call overlaps with every other person’s time on the call)?

*Extra credit:* What is the probability that at least *two* attendees are on the call with everyone else?

The solution to this Riddler Classic can be found in the following week’s column.

## Solution to last week’s Riddler Express

Congratulations to ÐÐ¯Ð¡Ð David Goode ÐÐ¯Ð¡Ð of Kula, Hawaii, winner of last week’s Riddler Express.

When cutting a cylindrical muffin into quarters, an “X” pattern would have been sensible. But last week, you were asked to cut a muffin into quarters using an “A” pattern. If you were to produce quarters in this manner, what would have been the ratio of length of the A’s middle bar to the radius of the muffin?

Let’s start with an image of the correct solution, courtesy of Ethan Rubin:

This circle appears to be divided into four equal pieces. But to find the precise dimensions of those pieces required a deeper dive into geometry and trigonometry.

A good first step was to determine the size of the left- and rightmost regions, which we can call arcs of measure *φ*. We can find the areas of these circular segments by taking the area of their corresponding sectors and subtracting isosceles triangles. Assuming the muffin was a unit circle (i.e., it had a radius of 1), the area of each segment turned out to be (*φ* − sin*φ*)/2. Setting this equal to ÐÐÐ¬Ð/4 (since each of these segments had to be one quarter the area of the circle) meant that the arc measure *φ* was about 2.31 radians, or about 132 degrees.

So if the arcs on either side of the A measured 132 degrees, that only left about 96 degrees for the arc at the bottom. The angle at the top of the A, which subtends that bottommost arc, was then half that measure (by the Central Angle Theorem), or about 48 degrees.

At this point, the isosceles triangle that forms the top of the A had to have an area of ÐÐÐ¬Ð/4 and a vertex angle of 48 degrees. That was enough information to nail down the dimensions of this triangle, which was fortunate, because the original question asked for the *base* of this triangle.

If we said the base had a length of 2*x*, then we could split the triangle down the middle to produce two right triangles whose topmost angle was about 24 degrees and whose area was 1/8. In other words, *x*2/(2·tan(24°)) = ÐÐÐ¬Ð/8, which meant 2*x* = √(ÐÐÐ¬Ð·tan(24°)), or about 1.18. Of course, we rounded the angles to the nearest degree along the way. The precise answer, as found by solver Ben Vollmayr-Lee, was closer to **1.177863**.

So that was the solution for an “A” cutting pattern. Solvers James Anderson and Brian Corrigan were further interested in what *other* letters could result in fairly sharing a muffin four ways. James found ways to split the muffin with patterns resembling the letters B, D, H, K and N, where the bumps for B and D were assumed to be semiellipses. Brian further found conditions for M and W cutting patterns.

However, this column wouldn’t be complete without an R (for “Riddler”), which can divide a circle into quarters as follows:

All these muffins made for a rather challenging Riddler Express. From now on, whenever I’m splitting a muffin four ways, I’ll stick to the “X” pattern.

## Solution to last week’s Riddler Classic

Congratulations to ÐÐ¯Ð¡Ð Jason Weill ÐÐ¯Ð¡Ð of Seattle, Washington, winner of last week’s Riddler Classic.

Ohio is the only state whose name doesn’t share any letters with “mackerel.” It’s strange, but it’s true.

But that wasn’t the only pairing of a state and a word you could have said that about — it wasn’t even the only fish! Kentucky had “goldfish” to itself, Montana had “jellyfish,” and Delaware had “monkfish,” just to name a few.

What was the longest “mackerel?” That is, using this word list, what was the longest word that didn’t share any letters with exactly one state?

Most solvers turned to their computers for some assistance. The word list contained a total of 263,533 words to check — a large number, but not prohibitively large. With efficient code (e.g., encoding words and states as arrays of their unique letters, as Josh Silverman and Jason Ash describe), it was possible for a script to pop out the answer in a matter of seconds.

Or in this case, *answers* — there were two, each 23 letters long. Alabama was the only state to have no letters in common with **counterproductivenesses**, while Mississippi was the only state to have no letters in common with **hydrochlorofluorocarbon**.

Of course, the fun didn’t stop there. For extra credit, you had to find the state with the most “mackerels.” In total, across all 50 states, there were 45,385 mackerels. Solver Jenny Mitchell created a map, showing which states had the most mackerels:

And the winning state turned out to be the very first one mentioned in the original puzzle: The great state of **Ohio** had a whopping 11,342 mackerels, or 25 percent of the total. In second place was Alabama, with 8,274 mackerels, and in third place was Utah, with 6,619. Alas, there were also 18 states that were fishless (the lightest blue in Jenny’s map), without a single mackerel to show for themselves.

Finally, solver Michael Branicky decided to make mackerels a global phenomenon. According to Michael, the longest Canadian mackerel (among provinces and territories) was Quebec’s “otorhinolaryngologists.” And among the member states of the United Nations, Fiji claimed the longest mackerel with “electroencephalographers.” Fiji also had the most mackerels, with 18,614.

Go figure that an island nation would take all the glory in a puzzle about fish.

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.