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 have four equilateral triangles. I place one on the floor. I pick a random edge of this first triangle and attach it to a side of a second triangle. Next, I randomly pick one of the four edges of the resulting rhombus and attach the third triangle. Finally, I randomly pick an edge from along the perimeter of the resulting shape and attach the fourth triangle.

What is the probability that I can create a regular tetrahedron by folding the four triangles along their edges?

*Extra credit:* Instead of using four equilateral triangles to make a tetrahedron, suppose I use six squares to make a cube. What is the probability I can make a cube by randomly attaching the squares, one at a time? (And what are my chances of making any of the three other Platonic solids using their respective faces?)

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

## Riddler Classic

The game of RaceTrack was published by recreational mathematician Martin Gardner in 1973. There have been a few modifications and even some neat digital versions since then, and it’s high time we had a race here on The Riddler.

So without further ado, welcome to the Riddler-opolis 500!

You begin at the midpoint of the starting line at the bottom (the thicker border), and your goal is to circumnavigate the race track shown below in a single counterclockwise loop. You’ll be moving from point to point on the grid, without ever venturing into the gray surrounding square or the central circle.

Each move takes one second. For your first move, you can choose among nine grid points surrounding and including the starting position.

From there, it’s essential to remember that you have *inertia*. So for your second move, you again have nine possible destinations, but they are determined by your current velocity vector. That is, you can maintain your current direction and speed, or you can alter your destination by one point in any direction (horizontally, vertically or diagonally). For example, if your first move was up and to the right, then your possible second moves are shown below, although two of them will cause you to crash into the wall.

So how quickly can you navigate the following track? Fractional times are allowed, so if the finish line bisects your final move, then that move counts as 0.5 seconds toward your total time. And remember, at no time can your path venture into the wall. (Being tangent to the wall is allowed, as is being on a grid point along a wall.)

Finally, submitting your answer can be tricky work. Please be sure to submit both your total time, as well as your sequence of moves. Each move should be assigned a digit from 1 through 9, corresponding to the nine possible destinations of the move:

Your overall path around the course then corresponds to a unique sequence of digits.

Do you have what it takes to win the Riddler-opolis 500? Puzzle enthusiasts, *start your engines!*

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

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Randy Tobe ÑÑâÐ of Lebanon, Ohio, winner of last week’s Riddler Express.

Last week, Riddler Nation was competing against Conundrum Country at an Olympic archery event. Each team shot three arrows toward a circular target 70 meters away. Hitting the bull’s-eye earned a team 10 points, while regions successively farther away from the bull’s-eye were worth fewer and fewer points.

Whichever team had more points after three rounds won. However, if the teams were tied after each team had taken three shots, both sides shot another three arrows. (If they remained tied, they continued shooting three arrows each until the tie was broken.)

For every shot, each archer of Riddler Nation had a one-third chance of hitting the bull’s-eye (i.e., earning 10 points), a one-third chance of earning 9 points and a one-third chance of earning 5 points.

Meanwhile, each archer of Conundrum Country earned 8 points with every arrow.

Which team was favored to win?

At first, you might have thought both teams were equally likely to win. Since Conundrum Country scored 8 points with every arrow, they were guaranteed to score a total of 24 points. Meanwhile, Riddler Nation scored an average of 10+9+5, which was *also* 24 points. With both teams scoring the same amount on average, how could one team possibly be favored over the other? Not surprisingly, most of Riddler Nation was able to figure this out.

The average of a probability distribution is its center of mass. So if a data point skews off in one direction, it pulls the average along with it, sometimes misrepresenting the distribution.

So while Riddler Nation’s average score was 24, what did the *entire* distribution of scores look like? Of the 27 (i.e., 3×3×3) equally likely cases, there was:

- One way to score 30: 10+10+10
- Three ways to score 29: 10+10+9, 10+9+10 and 9+10+10
- Three ways to score 28: 10+9+9, 9+10+9 and 9+9+10
- One way to score 27: 9+9+9
- Three ways to score 25: 10+10+5, 10+5+10 and 5+10+10
- Six ways to score 24: 10+9+5, 10+5+9, 9+10+5, 9+5+10, 5+10+9 and 5+9+10
- Three ways to score 23: 9+9+5, 9+5+9 and 5+9+9
- Three ways to score 20: 10+5+5, 5+10+5 and 5+5+10
- Three ways to score 19: 9+5+5, 5+9+5 and 5+5+9
- One way to score 15: 5+5+5

As we said, the average of all these results is 24. But if you looked closer, like solver Jane Steele of Santa Clara, California, you found this distribution of scores was not symmetric around 24 — it was skewed. There were six ways to score 24, 10 ways to score less than 24 and 11 ways to score more than 24.

And since Riddler Nation was more likely to exceed 24 points rather than fall short, **Riddler Nation** was favored to win!

For extra credit, you had to calculate Riddler Nation’s probability of winning the match. As we just saw, the chances of an outright victory stood at 11/27. The probability of a tie was 6/27, or 2/9, which extended the match to another three arrows. Moreover, you had to account for any number of rounds that ended in a tie. As noted by solver Rajat Jain of Mumbai, India, this meant the overall probability of victory could be represented as a geometric series: 11/27 + 11/27·(2/9) + 11/27·(2/9)^{2} +11/27·(2/9)^{3} + …. The sum of this series was **11/21**. So while Riddler Nation had an edge, it was a slight edge.

Sorry, Conundrum Country. At least you can chalk it up to home field advantage!

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ David Grabiner ÑÑâÐ of Columbia, Maryland, winner of last week’s Riddler Classic.

Last week, you had a chain with infinitely many flat (i.e., one-dimensional) links. The first link had length 1, and the length of each successive link was a fraction *f* of the previous link’s length. As you might have expected, *f* was less than 1, meaning the chain had a finite length. You placed the chain flat on a table and some ink at the very end of the chain (i.e., the end with the infinitesimal links).

Initially, the chain formed a straight line segment, and the longest link was fixed in place. From there, the links were constrained to move in a very specific way: The angle between each chain and the next, smaller link was always the same throughout the chain. For example, if the *N*^{th} link and the *N*+1^{st} link formed a 40 degree clockwise angle, then so did the *N*+1^{st} link and the *N*+2^{nd} link.

After you moved the chain around as much as you could, what shape was drawn by the ink that was at the tail end of the chain?

Let’s set the coordinates of the end of the chain with the longer links at the origin of the plane, and let’s suppose the chain lies on the positive *x*-axis. First off, how long was the chain when fully extended? Adding up the links gave you a geometric series (as in last week’s Express). This sum was 1 + *f* + *f*^{2} + *f*^{3} + …, which added up to 1/(1−*f*).

Presumably, the chain was maximally “coiled” when the angle between consecutive links was 180 degrees. Beyond this angle, the chain would work its way back toward being fully extended. At this point, the length of the chain was 1 − *f* + *f*^{2} − *f** ^{3}*+ …, or 1/(1+

*f*).

Now what about all the angles in between? Before figuring this out analytically, let’s see what a few solvers were able to do computationally and visually.

And if you’d like to see that on a coordinate plane:

It appeared that the answer was a circle! Its diameter was the difference between the maximal and minimal lengths, and so its radius was half that difference, or 2*f*/(1−*f*^{2}).

Unsatisfied with this empirical evidence, solvers like John Yeager of Lawrenceville, New Jersey, set out to prove this result. John’s key insight was to use complex numbers, which can be added like vectors. In the complex plane, the first link (which never rotated) always had length 1. If we represent the second link as the complex number *z*, then the third link was *z*^{2}, the fourth link was *z*^{3}, and so on. The end of the chain was located at 1 + *z* + *z*^{2} + *z*^{3} + …, yet another geometric series. Fortunately, the formula for the sum of a geometric series works just as well for complex numbers as it does for real numbers, meaning the sum was 1/(1−*z*).

To see how this became a circle, solver Michael DeLyser rewrote z as *f*cosÑÑÑÑ + *if*sinÑÑÑÑ. That meant the end of the chain was at 1/(1−*f*cosÑÑÑÑ−*if*sinÑÑÑÑ). Multiplying the numerator and denominator by the denominator’s conjugate gave you:

(1−*f*cosÑÑÑÑ+*if*sinÑÑÑÑ)/(1−2*f*cosÑÑÑÑ+*f*^{2}cos^{2}ÑÑÑÑ+*f*^{2}sin^{2}ÑÑÑÑ)

And combining the last two terms in the denominator simplified this expression slightly:

(1−*f*cosÑÑÑÑ+*if*sinÑÑÑÑ)/(1−2*f*cosÑÑÑÑ+*f*^{2})

As we said, complex numbers behave like vectors, with their real and imaginary parts representing orthogonal directions. So for any angle ÑÑÑÑ, the real part of the curve formed by the ink was (1−*f*cosÑÑÑÑ)/(1−2*f*cosÑÑÑÑ+*f*^{2}), while the imaginary part was *f*sinÑÑÑÑ/(1−2*f*cosÑÑÑÑ+*f*^{2}). These two expressions looked awfully similar. In fact, after translating the real part by 1/(1−2*f*cosÑÑÑÑ+*f*^{2}), you were left with *f*/(1−2*f*cosÑÑÑÑ+*f*^{2}) times -cosÑÑÑÑ for the real (or horizontal) part and times sinÑÑÑÑ for the imaginary (or vertical) part. In other words, it was a **circle**!

Solver Vamshi Jandhyala elected to be *even more* rigorous, turning the above expressions into a single equation for a circle.

By the way, upon seeing the original animation by Iñaki Huarte, the puzzle’s submitter, Eli Luberoff, still had a few more questions. You might have noticed that if you have a finite (rather than infinite) number of links, then the circle was replaced by a “loopy” approximation of a circle. And depending on the number of links and the value of *f*, the loops became cusps and ultimately a half-decent approximation for a circle.

There remains a *series* of mysteries to unpack in this *geometric* riddle.

## 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.