Happy 100th Fiddler!

It also happens to be the Fourth of July. This week, you’ll be counting the equilateral triangles on a Dozo board and parallelograms on the American flag.

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “Extra Credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

From Samuel Klein comes a triangular tribulation:

Dozo is a strategy game with a rather distinctive board:

The board features 28 holes in which players place markers, with the goal of making an equilateral triangle of any size with one color.

How many distinct equilateral triangles can you find whose vertices are the centers of holes on the board? (If two triangles are congruent but have different vertices, they should still be counted as distinct.)

This Week’s Extra Credit

Happy Fourth of July! In celebration of America’s birthday, let’s count more shapes—not in a board game, but in the American flag:

In particular, consider the centers of the 50 stars depicted on the flag. How many distinct parallelograms can you find whose vertices are all centers of stars? (If two parallelograms are congruent but have different vertices, they should still be counted as distinct.)

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing “Math, Revealed,” a new series of mathematical nuggets from The New York Times, authored by Steven Strogatz and Jens Mortensen.

So far, these nuggets have covered Manhattan distance, the golden ratio, close packing, and telescoping series. The photographs and images that accompany each of these are delightful!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each.

The results from Q2 are in, and the finest of Fiddlers, who each solved all 26 puzzles, are:

• 👑 Ed Foley 👑 from Redmond, Washington

• 👑 Jason Weisman 👑 from Wilton, Connecticut

• 👑 Michael Schubmehl 👑 from Hinsdale, Illinois

Congratulations to all who participated; the final standings are below. (If you think you see a mistake in the standings, kindly let me know.)

I will be reaching out to the top finishers so I can send along a fancy-schmancy t-shirt prize!

Also, this week’s puzzles mark the beginning of the next quarter (Q3). If you’d like your shot at glory (and a t-shirt), become a paid subscriber today!

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 James Pilloff 🎻 from Mountain View, California. I received 69 timely submissions, of which 65 were correct—good for a 94 percent solve rate.

Last week, you were breaking into a vault that contained an ancient Roman treasure. The vault was locked, and could be opened via a modern-day keypad with three numerical inputs: the Roman numerals “I,” “II,” and “III.”

Fortunately, your accomplice was able to access the numerical key code. Earlier in the day, they had handed you this code on a scroll of paper. Once at the keypad, you took the scroll out of your pocket and unfurled it. It read: “IIIIIIIIII.” Ten vertical marks, without any clear spacing between them.

You quickly realized the combination to unlock the door could have been anywhere from four digits long to 10 digits long. (Or was it IV digits to X digits?) How many distinct combinations were possible? If two combinations used the same numbers but in a different order, they were considered distinct.

Here at the Fiddler, we often find ourselves building up from or working back down to a base case. Why should this week be any different? Let’s start with one vertical mark. There was only one possible code that could correspond to: 1.

But with two vertical marks, there were now two possible codes: 11 and 2. And with three vertical marks, there were four possible codes: 111, 21, 12, and 3. One, two, four … surely with four vertical marks there were eight possible codes, since it appeared we had a doubling pattern on our hands.

Alas, the doubling broke down with four vertical marks. There were only seven possible codes: 1111, 211, 121, 31, 112, 22, 13. A code of “4” would have been an eighth possible code, but remember the keypad only had inputs of “I,” “II,” and “III.” If the pattern wasn’t doubling, what was it doing?

To figure this out, you could organize your thinking around the final digit of the code. Suppose there were n_k codes you could make with k vertical marks. One way to generate these codes was to start with the n_k−3 codes that had k−3 vertical marks and add a final digit of 3 (the three characters, “III”). Another way to generate such codes was to start with the n_k−2 codes that had k−2 vertical marks and add a final digit of 2 (the two characters, “II”). The final way to generate such codes was to start with the n_k−1 codes that had k−1 vertical marks and add a final digit of 1 (the one character, “I”).

Combining these results gave you the recurrence relation n_k = n_k−3 + n_k−2 + n_k−1. Starting with the initial values of n₁ = 1, n₂ = 2, and n₃ = 4 allowed you to compute all the way up to n₁₀. Here were all the values in between:

• n₁ = 1

• n₂ = 2

• n₃ = 4

• n₄ = n₁ + n₂ + n₃ = 1 + 2 + 4 = 7

• n₅ = n₂ + n₃ + n₄ = 2 + 4 + 7 = 13

• n₆ = n₃ + n₄ + n₅ = 4 + 7 + 13 = 24

• n₇ = n₄ + n₅ + n₆ = 7 + 13 + 24 = 44

• n₈ = n₅ + n₆ + n₇ = 13 + 24 + 44 = 81

• n₉ = n₆ + n₇ + n₈ = 24 + 44 + 81 = 149

• n₁₀ = n₇ + n₈ + n₉ = 44 + 81 + 149 = 274

In total, there were 274 possible codes the 10 marks could have represented. Of course, I also accepted answers of “CCLXXIV.” If you really wanted to visualize all 274 codes, solver Jason Weisman had you covered.

By the way, the recurrence relation above looked a lot like that of the Fibonacci sequence, except three prior terms were being summed rather than two. That’s why this sequence has appropriately been dubbed the “Tribonacci” sequence. And just as the ratio of consecutive Fibonacci terms approaches a singular value (the golden ratio), there exists a similar constant for the Tribonacci sequence, which is approximately 1.83929.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Tom Singer 🎻 from Melbourne, Florida. I received 52 timely submissions, of which 45 were correct—good for an 87 percent solve rate.

Having successfully hacked your way through the first keypad, the door opened to reveal a second door with yet another keypad, this time with eight numerical inputs: “I,” “II,” “III,” “IV,” “V,” “VI,” “VII,” and “VIII.”

You were expecting this, which was why your accomplice had handed you a second scroll of paper. You unfurled this one as well, hoping they remembered to add spaces between the numbers.

No such luck. This paper read: “IIIVIIIVIIIVIII,” which was 15 characters in total. How many distinct combinations were possible for this second door?