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. 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 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 Chris Payne comes a baffling bowling brainteaser:
In bowling, you roll a large, heavy ball toward a triangular array of 10 pins many feet away. You’d think the objective is to just knock down as many pins as possible. But thanks to the sport’s byzantine scoring system, it’s not that simple.
Bowling consists of 10 “frames,” and in each frame you have up to two chances to knock down the 10 pins. Your score for the frame is however many pins you knock down, but there are bonuses if you knock down all 10 pins:
If you knock down all 10 in one bowl, it’s called a strike. Your score for the frame is 10 plus however many pins you knock down in your next two bowls. Note that these next two bowls might occur in a single frame, or could be across the next two frames (if your very next bowl is again a strike).
If it takes you two bowls to knock down all 10 pins, it’s called a spare. Your score for the frame is 10 plus however many pins you knock down in your next one bowl.
Let’s look at a quick example. Suppose in frames 1 and 2 you bowl strikes, then in frame 3 you bowl a spare (three pins on the first bowl, seven on the second), and in frame 4 you knock down nine pins (five on the first bowl, four on the second). What’s your score?
So your total score on these four frames is 23 + 20 + 15 + 9 = 67. By my standards, that’s a pretty high score!
Something extra may be needed for the 10th and final frame. If you get a strike on the 10th frame, you get two extra rolls to determine your score for that frame. And if you get a spare, you similarly get a final roll to determine your score. The maximum score for a game is 300, earned from 12 consecutive strikes—10 strikes in the 10 frames, and then two additional strikes to compute the score for the 10th (and ninth) frame.
Now that we’ve reviewed how to score a game, let’s get to the actual puzzle!
If you knock down 100 pins over the course of a game, you are guaranteed to have a score that’s at least 100. But what’s the minimum total number of pins you need to knock down such that you can attain a score of at least 100?
You and your opponent each play a single game in which you both knock down the same number of pins. However, your scores are quite different.
Your opponent remarks, “Given only the information that we knocked down the same number of pins in our two games, there’s no way the difference between our scores could have been any greater!”
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 another great geometry puzzle (spoiler alert: answers appear on the linked thread) from Catriona Agg! Here it is, felt-tip drawing and all:
The three rectangles are congruent. What fraction of the design is shaded?
I buy that the answer is independent of the precise aspect ratios of the rectangles, so long as they’re congruent. But the fact that it’s independent of how much left or right you slide the top blue rectangle along the two green rectangles? That seems very counterintuitive (to me, at least)!
You can solve this by choosing a friendly position for the blue rectangle (e.g., flush left with the right green rectangle below it). But see if you can prove the answer in the general case.
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: 🎻 Mike Porter 🎻 from Haddenham, United Kingdom. I received 53 timely submissions, of which 25 were correct—good for a 47 percent solve rate.
Last week, you were introduced to Dozo, a strategy game with a rather distinctive board:
The board featured 28 holes in which players placed markers, with the goal of making an equilateral triangle of any size with one color.
How many distinct equilateral triangles could you find whose vertices were the centers of holes on the board? (If two triangles were congruent but had different vertices, they still counted as distinct.)
For some readers, this problem was particularly imposing because there were so many triangles on the board. There were 28 choose 3, or 3276, ways to pick three points as your vertices, which meant there were a whopping 3276 distinct triangles to consider. Of course, not any three vertices could have formed an equilateral triangle. Once you picked two vertices, there were only two possible locations for the third vertex—on either side of the first two. So a tighter upper bound on the number of triangles was 28 choose 2, doubled, and then multiplied by a third (since each equilateral triangle would have been counted in triplicate), for a total of 252 triangles.
That said, rather than look through all trios of vertices, a more popular strategy was to organize the triangles according to their side lengths and then count them up.
For convenience, let’s say the distance between each hole’s center and its nearest neighbor’s center was 1. The smallest equilateral triangle you could make then had side length 1. Focusing on such triangles that pointed downward, every hole among the bottom six rows could have been the bottommost vertex, for a total of 21 triangles. Focusing on triangles that pointed upward, any point not along the left or right edges of the board could have been the topmost vertex, for another 15 triangles. Altogether, that was 36 triangles with side length 1.
Let’s continue finding triangles that had a whole number side length. With a side length of 2, there were 15 downward facing triangles and six upward facing triangles, for a total of 21. With a side length of 3, there were 10 downward facing triangles and one upward facing triangle, for a total of 11. With side lengths of 4, 5, and 6, there were six, three, and one downward facing triangles, respectively.
All told, there were 78 equilateral triangles with whole number side lengths. Quite a few readers (17 percent) submitted answers of 78. However, there were other equilateral triangles to be found—triangles whose side lengths were not whole numbers. One example is in the image below:
This wasn’t an “upward” or “downward” pointing triangle; rather, it was tilted in a different direction. Each side had a length of √7. Such “tilted” triangles also had to be included in your total.
The smallest tilted triangles had side lengths of √3, and these pointed either rightward or leftward. There were 10 of each, for a total of 20.
The next smallest had side lengths of √7, one of which was in the figure above. Three triangles (including the one above) had the particular orientation shown above, and another three were reflections of these. Meanwhile, the orientation shown below accounted for another six triangles, and their reflections were yet another six. In total, there were 18 triangles with side lengths of √7.
The next largest triangle you could make had a side length of √12 (or 2√3), of which there were two such triangles. Next, there were six triangles with a side length of √13. Finally, there were two triangles of side length √21.
Altogether, there were 78 + 20 + 18 + 2 + 6 + 2 = 126 equilateral triangles. That was exactly half of our upper bound of 252, which was rather neat.
And if you couldn’t find all 126 triangles, check out this animation by 🎬 Sameer Gauria 🎬, which goes through all of them one at a time.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Peter Exterkate 🎻 from Sydney, Australia. I received 31 timely submissions, of which 23 were correct—good for a 74 percent solve rate.
In celebration of America’s 249th birthday last week, you counted more shapes—not in a board game, but in the American flag:
In particular, you were asked to consider the centers of the 50 stars depicted on the flag. How many distinct parallelograms could you find whose vertices were all centers of stars? (If two parallelograms were congruent but had different vertices, they still counted as distinct.)
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