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Can You Win The Tour de FiveThirtyEight?

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

This week, Riddler Nation needs your help designing its new credit card, appropriately named Riddler Express™ — don’t solve puzzles without it!

The logo consists of two overlapping circles with radius 1 inch, creating three distinct regions: one region that’s shared between the two circles, and two regions that are part of one circle, but not the other.

Because the Riddler Express™ card is for the mathematically inclined, its parent company has issued a corporate mandate regarding symmetry. In particular, the areas of all three regions must be exactly the same. If that’s the case, how far apart must the centers of the two circles be?

Extra Credit: If you symmetrically arrange three circles of radius 1, you have seven distinct regions. How far apart should the centers of the circles be such that the areas of the largest and smallest of the seven regions are as close to equivalent as possible?

## Riddler Classic

This week’s Riddler Classic is a cycling-themed puzzle of teamwork and strategy.

You are the coach for Team Riddler at the Tour de FiveThirtyEight, where there are 20 teams (including yours). Your objective is to win the Team Time Trial race, which has the following rules:

• Each team rides as a group throughout the course at some fixed pace, specified by that team’s coach. Teams that can’t maintain their pace are said to have “cracked,” and don’t finish the course.
• The team that finishes the course with the fastest pace is declared the winner.
• Teams ride the course one at a time. After each team completes its attempt, the next team quickly consults with its coach (who assigns a pace) and then begins its ride. Coaches are aware of the results of all previous teams when choosing their own team’s pace.

Assume that all teams are of equal ability: At any given pace, they have the exact same probability of cracking, and the faster the pace, the greater the probability of cracking. Teams’ chances of cracking are independent, and each team’s coach knows exactly what a team’s chances of cracking are for each pace.

Team Riddler is the first team to attempt the course. To maximize your chances of winning, what’s the probability that your team will finish the course? What’s the probability you’ll ultimately win?

Extra Credit: If Team Riddler is the last team to attempt the course (rather than the first), what are its chances of victory?

## Solution to last week’s Riddler Express

Congratulations to 👏Adam Opalski 👏 of Warsaw, Poland, winner of last week’s Riddler Express.

Last week, Dakota Jones was close to finding the Lost Arc, a geometric antiquity buried deep in the sands of Egypt. Along the way, she discovered a “highly symmetric crystal” that was needed to precisely locate the Arc. Dakota, now missing, had measured the crystal using her laser scanner, which produced the following looping animation:

What sort of three-dimensional shape is the crystal?

Solver Daniel Thompson took an impressively direct approach: he decomposed the animation into more than 300 separate images, then stacked them on top of each other to produce the following three-dimensional view:

That is one cool-looking crystal!

Having consulted with Professor Jones, who recently resurfaced, I can confirm that the highly symmetric crystal is indeed a cube. To help visualize this, here’s the scan shown at a different angle, along with the cube’s outline:

The scan started at one of the corners of the cube and proceeded to the opposite corner along an internal diagonal. The cross-section starts out as a growing equilateral triangle, turns into a hexagon (becoming a regular hexagon halfway through the scan) and then becomes a shrinking equilateral triangle.

There’s an even more precise answer, though. You could get a similar looking scan for something that isn’t a cube, but what’s called a trigonal trapezohedron (of which a cube is a special case where all six faces happen to be squares). If the crystal were a trigonal trapezohedron, but one that was more stretched out than a cube, and the camera were taking pictures that were farther apart, it could result in the same scan.

Solver Tim Maddux was kind enough to create a paper replica of that trigonal trapezohedron for Dakota Jones:

At first, Tim’s replica appears to have 12 triangular faces. But upon careful inspection, pairs of triangles come together to form six rhombic faces — a trigonal trapezohedron. I’m sure Dakota Jones won’t mind using this paper replica instead of some fancy crystal, right?

## Solution to last week’s Riddler Classic

Congratulations to 👏Jake Bryant 👏 of Weymouth, Massachusetts, winner of last week’s Riddler Classic.

You were challenged to find the longest string of letters in which (1) every pair of consecutive letters is a two-letter abbreviation for a state, territory or freely associated state, and (2) no abbreviation occurs more than once. For example, Guam, Utah and Texas could be combined into the valid four-letter string GUTX. Another valid string would be ALAK (Alabama, Louisiana and Alaska), while ALAL (Alabama, Louisiana and Alabama) was invalid because it includes the same state, Alabama, twice.

This problem is, shall we say, challenging to solve by hand, so riddlers around the world turned to their computers for assistance. Our winner used a technique known as recursion, writing code that breaks the problem down into smaller and smaller pieces, solves those mini-problems and then puts them back together again. The longest string turns out to be 31 characters long, and is made up of 30 states and territories. Here’s the string Jake found: FMPWVINVASDCTNMNCOHIALAKSCARIDE. Each consecutive pair of letters is a state or territory, from FM (Federated States of Micronesia) through DE (Delaware).

But there’s more than one string that fits the bill. Solver Richard Guidry
went so far as to find 9,984 distinct solutions. Here are a few more:

FMNVIDCALAKSCOHINCTNMPWVARIASDE
FMNVINCARIDCOHIAKSCTNMPWVALASDE
FMNCTNMPWVINVAKSCARIDCOHIALASDE
FMPWVALAKSCOHIASDCTNCARINMNVIDE
FMPWVARIASDCTNCALAKSCOHINMNVIDE

Guidry even went the extra mile and included the three additional abbreviations used by the Armed Forces: AA, AE and AP. With this change, he found strings that are a whopping 34 characters long, such as FMNVIDCAALAKSCOHINCTNMPWVAPARIASDE. It would seem that all roads lead to Delaware!

Solver Jason Ash, meanwhile, recognized the connection between this problem and graph theory, a branch of mathematics. As he describes, he treats each abbreviation as a node, and if one abbreviation’s second letter is the same as another abbreviation’s first letter, then they’re connected (in a mathematical sense, rather than a geographical sense). Here’s the resulting graph:

He then tried to find the longest path traversing the graph of connected states or territories, which is indeed 30 places, or 31 characters.

Alas, if only it were possible to connect all the states using their abbreviations. Maybe we can convince Kentucky, New York or Arizona to change their abbreviations. You know, for the sake of connected graphs.

## 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.

## Footnotes

1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.