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

I recently came across a rather peculiar recipe for something called Babylonian radish pie. Intrigued, I began to follow the directions, which said I could start with any number of cups of flour.

*Any* number? I mean, I had to start with *some* flour, so zero cups wasn’t an option. But according to the recipe, any positive value was fair game. Next, I needed a second amount of flour that was 3 divided by my original number. For example, if I had started with two cups of flour, then the recipe told me I now needed 3 divided by 2, or 1.5, cups at this point.

I was then instructed to combine these amounts of flour and discard half. Apparently, this was my new starting amount of flour. I was to repeat the process, combining this amount with 3 divided by it and then discarding half.

The recipe told me to keep doing this, over and over. Eventually, I’d have the proper number of cups of flour for my radish pie.

How many cups of flour does the recipe ultimately call for?

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

## Riddler Classic

From Rushabh Mehta comes a highly irregular puzzle:

Lately, Rushabh has been thinking about very large regular polygons — that is, a polygon all of whose sides and angles are congruent. His latest construction is a particular regular 1,000-gon, which has 1,000 sides of length 2. Rushabh picks one of its longest diagonals, which connects two opposite vertices.

Now, this 1,000-gon has many diagonals, but only some are *perpendicular* to that first diagonal Rushabh picked. If you were to slice the polygon along all these perpendicular diagonals, you’d break the first diagonal into 500 distinct pieces. Rushabh is curious — what is the product of the lengths of all these pieces?

*Extra credit:* Now suppose you have a regular 1,001-gon, each of whose sides has length 2. You pick a vertex and draw an altitude to the opposite side of the polygon. Again, you slice the polygon along all the perpendicular diagonals, breaking the altitude into 500 distinct pieces. What’s the product of the lengths of all these pieces *this *time?

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

## Solution to last week’s Riddler Express

Congratulations to 👏 Brian Corrigan 👏 of Los Angeles, winner of last week’s Riddler Express.

Earlier this year, Dakota Jones used a crystal key to gain access to a hidden temple, deep in the Riddlerian Jungle. According to an ancient text, the crystal had exactly six edges, five of which were 1 inch long. Also, the key was the *largest* such polyhedron (by volume) with these edge lengths.

However, after consulting an expert, Jones realized she had the wrong translation. Instead of definitively having *five* edges that were 1 inch long, the crystal only needed to have *four* edges that were 1 inch long. In other words, five edges *could* have been 1 inch (or all six for that matter), but the crystal definitely had *at least four* edges that were 1 inch long.

The translator confirmed that the key was indeed the *largest* such polyhedron (by volume) with these edge lengths.

Once again, Jones needed your help. *Now* what was the volume of the crystal key?

Last time around, when there were five edges that were 1 inch long, you had a tetrahedron with two equilateral faces. To maximize the tetrahedron’s volume, you made sure the faces were perpendicular to each other, giving you an overall volume of 0.125 cubic inches. Now, with only four edges that were 1 inch long, the answer was surely greater.

There were two distinct ways to choose four edges from among the six edges of a tetrahedron. One way was to have three of them all on the same face, forming an equilateral triangular base. To maximize the volume, the fourth edge of length 1 was perpendicular to the base, meaning it was the altitude of the tetrahedron. The volume was therefore 1/3 times the area of the base — √3/4 — times the height of 1. Multiplying this together gave you a volume of √3/12, or about 0.144 cubic inches, which was indeed greater than the 0.125 cubic inches of the previous crystal.

However, as noted by solver Ming Chan, there was another case to check — when the four edges of length 1 were *not* on the same face. Solver Andy Quick offered a sketch of both cases:

In the second case, two of the tetrahedron’s faces were congruent isosceles triangles that were perpendicular to each other along their common base. If the length of that base was *x*, Andy found that the volume was *x*/6 − *x*^{3}/24. This expression was maximized when x was 2/√3, in which case the volume was 2/(9√3), or about 0.1283. That was made for a larger crystal than the original puzzle, but still not as large as the first case.

Thanks to Riddler Nation, Jones was able to forge (or rather, synthesize) an updated crystal key. With it, she will be able to access the temple’s inner sanctum. Still no word on what’s inside, but I’ll be sure to report back with the details.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Emily Boyajian 👏 of Minneapolis, Minnesota, winner of last week’s Riddler Classic.

Last week, Phil DeOrsey was playing with his daughter, who loved to cut paper with her safety scissors. She especially liked cutting paper into “strips,” rectangular pieces of paper whose shorter sides were at most 1 inch long.

Whenever Phil gave her a piece of standard printer paper (8.5 inches by 11 inches), she picked one of the four sides at random and then cut a 1-inch wide strip parallel to that side. Next, she discarded the strip and repeated the process, picking another side at random and cutting the strip. Eventually, she was left with nothing but strips.

On average, how many cuts did she make before she was left only with strips?

There was a (relatively) straightforward way to solve this using recursion. Suppose the paper’s current dimensions were *m* by *n*, and let’s call the expected number of remaining cuts the function *E*(*m*, *n*).

From here, there was a 50 percent chance that Phil’s daughter would cut along one of the two edges of length *m*, and a 50 percent chance she’d cut along an edge of length *n*. Mathematically, this could be written as *E*(*m*, *n*) = 1 + *E*(*m*−1, *n*)/2 + *E*(*m*, *n*−1)/2, with the “1” representing the one cut she needed to make to reach either of the next two rectangular dimensions: *m*−1 by *n*, or *m* by *n*−1.

You were also told that when she was left with “nothing but strips,” meaning when either *m* or *n* was less than or equal to 1, she was done cutting, in which case *E*(*m*, *n*) was equal to 0.

This was enough information to find the expected number of cuts from any given starting dimensions. For example, with *E*(0, 1) and *E*(1, 0), you could determine *E*(1, 1). With *E*(1, 1) and *E*(2, 0), you could determine *E*(2, 1) — and so on. The table below shows the average number of cuts when *m* and *n* were at most 12:

##### Can you strip the paper down … to strips?

The expected number of cuts for paper that is *m* by *n* inches, removing 1-inch strips at a time, until there are only strips left

m | |||||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |

1 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | |

2 | 0.0 | 1.0 | 1.5 | 1.8 | 1.9 | 1.9 | 2.0 | 2.0 | 2.0 | 2.0 | 2.0 | 2.0 | |

3 | 0.0 | 1.5 | 2.5 | 3.1 | 3.5 | 3.7 | 3.8 | 3.9 | 4.0 | 4.0 | 4.0 | 4.0 | |

4 | 0.0 | 1.8 | 3.1 | 4.1 | 4.8 | 5.3 | 5.6 | 5.7 | 5.8 | 5.9 | 5.9 | 6.0 | |

5 | 0.0 | 1.9 | 3.5 | 4.8 | 5.8 | 6.5 | 7.0 | 7.4 | 7.6 | 7.8 | 7.9 | 7.9 | |

6 | 0.0 | 1.9 | 3.7 | 5.3 | 6.5 | 7.5 | 8.3 | 8.8 | 9.2 | 9.5 | 9.7 | 9.8 | |

7 | 0.0 | 2.0 | 3.8 | 5.6 | 7.0 | 8.3 | 9.3 | 10.1 | 10.6 | 11.1 | 11.4 | 11.6 | |

8 | 0.0 | 2.0 | 3.9 | 5.7 | 7.4 | 8.8 | 10.1 | 11.1 | 11.9 | 12.5 | 12.9 | 13.3 | |

9 | 0.0 | 2.0 | 4.0 | 5.8 | 7.6 | 9.2 | 10.6 | 11.9 | 12.9 | 13.7 | 14.3 | 14.8 | |

10 | 0.0 | 2.0 | 4.0 | 5.9 | 7.8 | 9.5 | 11.1 | 12.5 | 13.7 | 14.7 | 15.5 | 16.1 | |

11 | 0.0 | 2.0 | 4.0 | 5.9 | 7.9 | 9.7 | 11.4 | 12.9 | 14.3 | 15.5 | 16.5 | 17.3 | |

12 | 0.0 | 2.0 | 4.0 | 6.0 | 7.9 | 9.8 | 11.6 | 13.3 | 14.8 | 16.1 | 17.3 | 18.3 |

The half-inch from the puzzle was a slight wrinkle but was mathematically equivalent to starting with a sheet of paper that was 9 by 11 inches. According to the table, that meant Phil’s daughter needed an average of about **14.3** cuts.

For extra credit, you were asked to tackle the more general case in terms of *m* and *n*. Solver Rajeev Pakalapati was among those who found a closed-form expression for *E*(*m*, *n*) using generating functions, as shown below:

Meanwhile, solver Josh Silverman looked more closely at the square (*m *by *m* inches) case, finding that *E*(*m*, *m*) grew approximately linearly with *m*.

As you can see, there many have been many short cuts, but there were really no shortcuts.

## 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.