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

While watching batter spread out on my waffle iron, and thinking back to a recent conversation I had with Friend-of-The-Riddler™ Benjamin Dickman, I noticed the following sequence of numbers:

1, 4, 4, 0, 4, 8, 0, 0, 4, 4, 8, …

Before you ask — yes, you can find this sequence on the On-Line Encyclopedia of Integer Sequences. However, for the full Riddler experience, I urge you to *not* look it up. See if you can find the next few terms in the sequence, as well as the pattern.

Now, for the actual riddle: Once you’ve determined the pattern, can you figure out the *average* value of the entire sequence?

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

## Riddler Classic

You may recall a previous riddle about the new Olympic event, sport climbing. This week, puzzle submitter Andy Esposito takes us back to this exciting event:

The finals of the sport climbing competition has eight climbers, each of whom compete in three different events: speed climbing, bouldering and lead climbing. Based on their time and performance, each of the eight climbers is given a ranking (first through eighth, with no ties allowed) in each event, as well as a corresponding score (1 through 8, respectively).

The three scores each climber earns are then multiplied together to give a final score. For example, a climber who placed second in speed climbing, fifth in bouldering and sixth in lead climbing would receive a score of 2 × 5 × 6, or 60 points. The gold medalist is whoever achieves the lowest final score among the eight finalists.

What is the highest (i.e., worst) score one could achieve in this event and still have a chance of winning (or at least tying for first place overall)?

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

## Solution to last week’s Riddler Express

Congratulations to 👏 Gary M. Gerken 👏 of Littleton, Colorado, winner of last week’s Riddler Express.

Last week, I 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 (positive) number of cups of flour.

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 did the recipe ultimately call for?

First, let’s see what happened to the aforementioned example that started with two cups. After one round of the process, you had the average of 2 and 3/2, or (2 + 3/2)/2, which was 1.75 cups of flour. Repeating the process again, you next had (1.75 + 3/1.75)/2, which was 97/56, or about 1.73214 cups of flour. And after that, you had (97/56 + 3/(97/56))/2, or about 1.732050810.

Those values seemed to be converging — an encouraging sign — but surely this convergence depended on the initial amount of flour you selected, right? If instead you started with 10 cups, then after one round you had (10 + 3/10)/2, or 5.15 cups of flour. After two rounds, you had (5.15 + 3/5.15)/2, or 2.866 cups. The next few values were 1.956, 1.744 and 1.732 — sure enough, you were right back where you were when you started with only two cups of flour.

As noted by solver An Nguyen, one way to find the point of convergence was to assume that you repeated this process so many times that the value you put in was indistinguishable from the value you got out. Mathematically, that meant that if you had *x* cups, then (*x* + 3/*x*)/2 = *x*. Solving this equation gave you x = **√3**, which, sure enough, is approximately 1.732. In fact, it’s 1.732050807…, remarkably close to what you had when you started with two cups.

At this point, you knew that *if* the process converged at all, then it had to converge to √3. But solver Chris Sears made a convincing argument that the process *always* converged to √3, no matter what initial value was chosen.

The vertical line segments in Chris’s animation show the values used in the iterative rounds, while the horizontal segments connect the values from one round to the next. No matter which starting value Chris chose in the animation, the segments invariably bounced their way to the same point on the coordinate plane, representing a value of √3.

By the way, what was up with the “Babylonian radish” pie from this puzzle? Many solvers made the connection between the words *radish* and *root*. Not only is a radish a root vegetable, but the word itself derives from the Latin word for root, *radix*. Furthermore, it has been speculated that the algorithm in this puzzle — also known as Heron’s method — was developed thousands of years ago by Babylonians.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Tim Stevens 👏 of Littleton, Colorado,^{2} winner of last week’s Riddler Classic.

Last week, you looked at a very large regular polygon — that is, a polygon all of whose sides and angles were congruent. You specifically studied a regular 1,000-gon — which had 1,000 sides of length 2 — and you picked one of its longest diagonals, which connected two opposite vertices.

Now, this 1,000-gon had many diagonals, but only some were *perpendicular* to that first diagonal you picked. If you had sliced the polygon along all these perpendicular diagonals, you would have broken the first diagonal into 500 distinct pieces. What was the product of the lengths of all these pieces?

At first, this riddle might have seemed overwhelming. I can’t say I’ve ever drawn a half-decent heptagon by hand, let alone a 1,000-gon. Solver Ian Rhile decided to keep things simple by starting with regular polygons that had fewer sides and looking for patterns. Below are the pieces created in a square, a regular hexagon and a regular octagon, each with side length 2. (Notice that each of these had an even number of sides, like the 1,000-gon. Odd numbers of sides were relevant for the extra credit.)

Multiplying the lengths from the square gave you (√2)^{2}, or 2. Multiplying the lengths from the hexagon gave you (1)^{2}·2, or 2. Multiplying the lengths from the octagon was a little more work. The topmost and bottommost segments were congruent, and their product was 2−√2. Meanwhile, the middle two segments were also congruent, and their product was 2+√2. Multiplying these together gave you (2−√2)(2+√2), which was, once again, 2.

From there, many solvers hypothesized that the product was **2** for any regular polygon with an even number of sides — and they were right!

To demonstrate this, solver Emily Boyajian inscribed a regular polygon with 2*n* sides and side length 2 in a circle with radius *r* = 1/sin(𝜋/2*n*). The lengths of the *n* resulting segments were then *r*cos((*k*−1)𝜋/*n*) − *r*cos(*k*𝜋/*n*), where the index *k* ranged from 1 to *n*. With the help of some trigonometric identities and complex numbers, Emily was able to prove that these lengths always multiplied to 2.

For extra credit, you had to similarly analyze a 1,001-gon, each of whose sides were length 2. This time, you picked a vertex and drew an altitude to the opposite side of the polygon. Again, you sliced the polygon along all the perpendicular diagonals, breaking the altitude into 500 distinct pieces. What was the product of the lengths of all these pieces *this *time?

Solver Jake Gacuan of Manila, Philippines, applied similar trigonometric identities to show that the product for a regular *n*-gon of side length 2 — when *n* was *odd* — was always the square root of *n*. In this case, the product was **√1001**, or about 31.64.

I don’t know about you, but I thought these results — always 2 and always √*n* — were fascinating in their simplicity. But that’s not all. The submitter of this puzzle, Rushabh Mehta, noticed yet another interesting pattern related to the *sum of the squares* of all the lengths. See if you can figure it out!

## 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.