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 the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

After you intended to make a perfectly circular pancake, the batter has spread out, filling every last corner of your square pan. (It is unclear why you were using a square pan in the first place.)

To salvage your breakfast, you plan to slice off the corners of your square pancake, giving you something closer to a circle. The image below shows one such slice you might make. Each slice must be straight, and no slice can pass through the inscribed blue circle that represents your original desired pancake.

Of course, there’s a catch. You can make at most five slices. If the blue circle has a radius of 1 unit, what is the minimum possible area your pancake can have after five slices?

(Kudos to anyone who actually constructs an edible pancake with this minimum area.)

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

## Riddler Classic

From Aaron Wilkowski comes a rather enigmatic election:

Riddler Nation’s neighbor to the west, Enigmerica, is holding an election between two candidates, A and B. Assume every person in Enigmerica votes randomly and independently, and that the number of voters is very, very large. Moreover, due to health precautions, 20 percent of the population decides to vote early by mail.

On election night, the results of the 80 percent who voted on Election Day are reported out. Over the next several days, the remaining 20 percent of the votes are then tallied.

What is the probability that the candidate who had fewer votes tallied on election night ultimately wins the race?

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

## Solution to last week’s Riddler Express

Congratulations to ÐÐ¯Ð¡Ð Mark Jackson ÐÐ¯Ð¡Ð of Rochester, New York, winner of last week’s Riddler Express.

Last week, you were creating a variation of a Romulan pixmit deck. Each card was an equilateral triangle, with one of the digits 0 through 9 (written in Romulan, of course) at the base of each side of the card. No number appeared more than once on each card. Furthermore, every card in the deck was unique, meaning no card could be rotated so that it matched (i.e., could be superimposed on) any other card.

What was the greatest number of cards your pixmit deck could have?

The numbers here were small enough that you could have written out every last card. A more efficient way to count them was to use combinatorics. To construct a unique card, you first had to pick three digits from 0 through 9. There were precisely 10 choose 3 — or 120 — ways to do this.

But 120 was not the answer. As noted by solver Yolanda Chang, there were *two* distinct ways to assign any three digits to the three sides of a card. For example, if the three digits were 1, 2 and 3, then one way to arrange them was 1-2-3 as you moved clockwise around the triangle. This card was then equivalent to 2-3-1 and 3-1-2. Alternatively, you could arrange them 3-2-1 as you moved clockwise — an arrangement that was equivalent to 2-1-3 and 1-3-2.

Another way to think about this was that, given three digits to place on the card, the two arrangements were reflections of each other. As observed by Reece Goiffon, pairs of cards connected via reflection were analogous to enantiomers in organic chemistry. Neat!

That meant the greatest number of pixmit cards was twice 120, or **240**.

For extra credit, numbers were allowed to appear two or three times on a given card. Once again, no card could be rotated so that it matched any other card. Now what was the greatest number of cards your pixmit deck could have?

In addition to the 240 cards with three distinct digits we just counted, there were 10 cards that had three of the same digit appear on them — one card for each digit. Finally, you had to count cards with two distinct digits, in which case one digit appeared twice on the card, while another digit appeared once. There were 10 digits that could appear twice, and for each of these digits you then had nine remaining digits to choose from for the digit that appeared once. Multiplying these together, there were 90 unique cards with two digits. Putting this all together, your new pixmit deck had 240 + 10 + 90, or **340** cards.

Anyway, the next time you play pixmit, don’t read too much into the cards. It’s not like the fate of the galaxy depends on it.

## Solution to last week’s Riddler Classic

Congratulations to ÐÐ¯Ð¡Ð Rolfe Petschek ÐÐ¯Ð¡Ð of Kennebunkport, Maine, winner of last week’s Riddler Classic.

Last week marked the beginning of Ramadan in the U.S. The month-long observance traditionally begins with the sighting of a crescent moon, or hilal, after a new moon. For last week’s Classic, you took a closer look at a waxing crescent moon.

After a new moon, the crescent appears to grow slowly at first. At some point, the moon will be one-sixth full by area, then one-quarter full, and so on. Eventually, it becomes a half-moon, at which point its growth begins to slow down. The animation below provided some insight into what was happening here:

How many times faster was the area of the illuminated moon growing when it was a half-moon versus a one-sixth moon?

To keep the puzzle *relatively* simple, you were allowed to make a few assumptions, including:

- The moon is a perfect sphere.
- The moon’s orbit around Earth is a perfect circle.
- The moon orbits the Earth much faster than the Earth orbits the sun.
- The sun is very, very far away.

With these simplifying assumptions — and several more that Rolfe kindly listed but which I won’t get into here — you could think of the moon as having an illuminated hemisphere that was rotating with a constant angular speed. You were a distant observer who happened to be in position to see the full moon, new moon and all the phases in between.

From here, one approach to solving was to find an expression for the illuminated area *A* as a function of time *t*. You could then use calculus to find the derivative, or instantaneous rate of change, in *A*(*t*) when the moon was half-full and one-sixth full.

Here’s a helpful diagram from Thomas Stone, looking down at the moon from atop its axis of rotation and looking at the moon from Earth:

For an angle ÐÐÐ¬Ð¤*t* from a new moon, where ÐÐÐ¬Ð¤ was the angular velocity of the lunar terminator, the area of the crescent was the area of the half-moon — ÐÐÐ¬Ð*R*^{2}/2 — minus the area of a semi-ellipse. The semi-major axis of the ellipse was *R*, while the semi-minor axis was *R*cos(ÐÐÐ¬Ð¤*t*). The semi-ellipse therefore had an area of ÐÐÐ¬Ð*R*^{2}cos(ÐÐÐ¬Ð¤*t*)/2. That meant the crescent, which was the difference between the semicircle and the semi-ellipse, had area *A*(*t*) = ÐÐÐ¬Ð*R*^{2}[1−cos(ÐÐÐ¬Ð¤*t*)]/2.

By the way, since the answer worked for any value of *R* and ÐÐÐ¬Ð¤, you could simplify this expression by choosing an *R* that canceled the ÐÐÐ¬Ð/2 and setting ÐÐÐ¬Ð¤ equal to 1. This gave you the simpler function *A*(*t*) = 1−cos(*t*).

Now that you had a formula for the area, you just needed to know for *which* values of *t* the moon was one-sixth full and half-full. Our simplified function *A*(*t*) oscillated between 0 and 2, meaning the moon was half-full when *A*(*t*) = 1, and one-sixth full when *A*(*t*) = 1/3. Setting 1−cos(*t*) equal to 1 meant that *t* was 90 degrees, or ÐÐÐ¬Ð/2 radians. And setting 1−cos(*t*) equal to 1/3 meant that cos(*t*) was equal to 2/3, or that *t* was cos^{−1}(2/3), or about 48.2 degrees.

Finally, to compare the *rates* at which the crescent was growing, you needed the derivative of *A*(*t*), which was *A*’(*t*) = sin(*t*). When *t* was ÐÐÐ¬Ð/2 radians for a half-moon, *A*’(*t*) was 1. (Note that the moon’s illuminated area grows the fastest at the half-moon!) Meanwhile, when *t* was cos^{−1}(2/3), *A*’(*t*) was sin[cos^{−1}(2/3)]. With the help of some trigonometry — that is, drawing a right triangle with one leg of length 2 and a hypotenuse of 3, whose *other* leg was then √5 — you were able to simplify this expression to √5/3. That meant the half-moon was growing **3/√5** — or about 1.34 — times faster than the one-sixth moon.

For those of you observing, Ramadan Mubarak!

## 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.