Can You Fling the Fractal Darts?
A dartboard contains infinitely many circles arranged in a fractal pattern. Each circle’s point value is proportional to its area. What’s the highest score you can earn with a single dart?
Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.
Every Friday morning, I present mathematical puzzles intended to challenge and delight you. Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets 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 puzzles are 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 Quowong Liu comes an unusual game of fractal darts:
You are playing darts with your friend, Apollonius, who has brought his own dartboard. However, this dartboard is somewhat … different. Instead of being a circle divided into concentric rings and sectors, the dartboard is a unit circle (i.e., with radius 1) that’s divided via an Apollonian gasket. In particular, this gasket is defined by two horizontally adjacent, congruent circles with radius 1/2. (Note: An Apollonian gasket also includes circles above, below, and around these two circles of radius 1/2. Put simply, you keep drawing circles that are tangent to three other circles wherever you can.)
But that’s not all! Every circle on the dartboard, no matter how small, is also its own Apollonian sub-gasket. Like the larger circle, every gasket, sub-gasket, sub-sub-gasket, etc., is defined by two horizontally adjacent, congruent circles with radii that are half the radius of their outer circle.
Having trouble imagining what this dartboard looks like? You’re not alone! Fortunately, Quowong created an image of the dartboard:
Now, when you throw a dart at this board, your score is the sum of the areas of every circle for which the dart lies inside or on the circumference. (Remember, the entire board is a unit circle.)
What is the most a single dart can score?
This Week’s Extra Credit
Suppose each point on the dartboard is equally likely to be hit by a dart. On average, what score would you expect a single dart to earn?
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, let’s take a trip down memory lane by checking in on a Riddler Classic from my tenure at FiveThirtyEight:
Congratulations, you’ve made it to the finals of the Riddler Ski Federation’s winter championship! There’s just one opponent left to beat, and then the gold medal will be yours.
Each of you will complete two runs down the mountain, and the times of your runs will be added together. Whoever skis in the least overall time is the winner. Also, this being the Riddler Ski Federation, you have been presented detailed data on both you and your opponent. You are evenly matched, and both have the same normal probability distribution of finishing times for each run. And for both of you, your time on the first run is completely independent of your time on the second run.
For the first runs, your opponent goes first. Then, it’s your turn. As you cross the finish line, your coach excitedly signals to you that you were faster than your opponent. Without knowing either exact time, what’s the probability that you will still be ahead after the second run and earn your gold medal?
Extra Credit: Over in the snowboarding championship, there are 30 finalists, including you (apparently, you’re a dual-sport threat!). Again, you are the last one to complete the first run, and your coach signals that you are in the lead. What is the probability that you’ll win gold in snowboarding?
Spoiler alert: The solution to this puzzle was 75 percent, while the solution to the (rather challenging) Extra Credit was between 31.4 and 31.5 percent.
So what’s new here? Friend-of-the-Fiddler Josh Silverman, as promised five years ago, took another crack at the general solution when there were a total of N finalists! Based on numerical results, Josh previously estimated that your probability of winning gold scaled with N-1/3. But with a more detailed analysis, Josh has instead found that the probability scales with (logN)-(√2−1) · N-(3−2√2).
Just goes to show that it’s never too late to take another crack at a Fiddler (or Riddler)!
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 correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. If you think you see a mistake in the standings, kindly let me know.
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Eugene Shlossberg 🎻 from Doylestown, Pennsylvania. I received 65 timely submissions, of which 59 were correct—good for a 91 percent solve rate.
In the YouTube show, “Hot Ones,” guests answer interview questions while consuming 10 hot sauces, one at a time, ranked in increasing spiciness from 1 to 10.
Last week, you were invited on as a guest and wanted to prepare for the show. However, you didn’t feel like purchasing all 10 sauces in advance. Your plan was to purchase fewer sauces, and then to combine sauces together for any you were missing. For example, if you were missing sauce #7, then you could instead simultaneously consume sauces #3 and #4, since 3 + 4 = 7. (I know the spiciness of the sauces isn’t linear, but for the purposes of this puzzle, let’s assume it is.)
After some pencil-and-paper scratch work, you realized you only needed four spices. For example, here’s how you could generate all the values from 1 to 10 using the spices #1, #2, #3, and #4 at most once:
1
2
3
4
5 = 1 + 4
6 = 2 + 4
7 = 3 + 4
8 = 1 + 3 + 4
9 = 2 + 3 + 4
10 = 1 + 2 + 3 + 4
Including this particular set of four spices (i.e., #1 through #4), for how many sets of four spice numbers was it possible to generate all the numbers from 1 to 10 using each spice at most once?
The only way to generate a value of 1 was to include spice #1 in your set. You also needed spice #2 in your set; that was the only way to get a value of 2, since you couldn’t duplicate values (i.e., 2 = 1 + 1 wasn’t allowed).
With #1 and #2 in hand, the next hotter spice couldn’t have been #5, since that would have left you with no way to generate 4. Thus, the third spice was either #3 or #4.
Suppose the third spice was #3. With just the numbers 1, 2, and 3, you could generate six of the 10 total values:
1
2
3
4 = 1 + 3
5 = 2 + 3
6 = 1 + 2 + 3
The fourth number had to be at least 4 (i.e., greater than 3) and at most 7 (otherwise you couldn’t get 7). Thus, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 3, 6}, and {1, 2, 3, 7} were four possible sets.
Next, suppose the third spice was #4. With just the numbers 1, 2, and 4, it was possible to generate seven of the 10 total values:
1
2
3 = 1 + 2
4
5 = 1 + 4
6 = 2 + 4
7 = 1 + 2 + 4
The fourth number had to be at least 5 (i.e., greater than 4) and at most 8 (otherwise you couldn’t get 8). That meant {1, 2, 4, 5}, {1, 2, 4, 6}, {1, 2, 4, 7}, and {1, 2, 4, 8} were another four viable sets.
And that was it! In total, there were eight viable sets of spices.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Adam Richardson 🎻 from Old Hickory, Tennessee. I received 41 timely submissions, of which 31 were correct—good for a 76 percent solve rate.
Next, you were prepping for a new show, “Hotter Ones,” which had spices ranked from 1 to 100. Let N be the minimum number of spices needed to generate all the numbers from 1 to 100.
For how many sets of N spice numbers was it possible to generate all the numbers from 1 to 100 using each spice at most once?
