Can You Make the Smallest Largest Triangle?
Place three point-like barriers in a circle. How large of a triangle can you draw around them? And how small can you make this largest triangle?
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. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.
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 🎬.
Important note: After 6 months, 23 columns, and 46 puzzles, I’ve decided to place part of this and future columns behind a paywall, which can be unlocked with a monthly or annual subscription. This week, only the solution to the Extra Credit is behind the paywall, whereas the solution to the Fiddler remains freely available. And of course, I’m committed to the puzzle statements themselves always remaining free. Thank you to everyone who has already subscribed and supported the continuance of this weekly puzzling tradition!
This Week’s Fiddler
From Jason Zimba, who previously contributed an Extra Credit ont polygons and circles, comes another puzzle about—you guessed it—polygons and circles! He came up with the idea for this one back in 2016.
There are countless ways to pick three points within the unit disk (i.e., the points on the unit circle and inside the unit circle). And for each of these ways, there are countless triangles you can draw within the disk that do not strictly contain any of the three points. By “strictly,” I mean that the three points can be on the edges of a triangle, but not inside a triangle.
For any trio of points in the unit disk, consider the largest triangle (by area) within the disk that doesn’t strictly contain any of the three points.
As an illustration, consider the diagram below:
The three red points are all within the unit disk. Various green and red triangles that don’t strictly contain the red points are shown. The largest of these by area is highlighted with a blue edge, and its area is approximately 1.1842.
By positioning the three points more strategically, it’s possible to reduce the area of the largest such triangle. What is the minimum possible area this largest triangle can have?
Extra Credit
Also from Jason Zimba comes some Extra Credit:
Instead of placing three points, suppose you’re placing four points within the unit disk to minimize the area of the largest triangle (by area) that doesn’t strictly contain any of the points.
Once again, what is the minimum possible area this largest triangle can have?
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 is the final week that I’m sharing Mathigon’s 2023 puzzle calendar, curated by my colleague Eda Aydemir. A new puzzle was posted each of the first 24 days of December.
This week, I’m sharing the puzzle from Dec. 22:
The diagram shows a Galton board—a sorting device where a ball descends, bouncing right or left with equal probability every time it encounters a hexagon. The twist here is that one of the decision points is missing: If the ball initially goes left, it must go right upon reaching the next hexagon. After four decision points, there are five possible outcomes, and the puzzle asks for the probability the ball comes out in bucket B, which is second from the left.
Personally, I’d have asked for the probability that the ball will land in bucket C, the middle bucket. But either way, this one is some quick fun!
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Benjamin Dickman 🎻 from New York, New York. I received 58 timely submissions, of which 47 were correct—good for an 81 percent solve rate!
The year 2024 is now upon us! There are several rectangular prisms with integer edge lengths that have an internal diagonal of length 2024.
What was the greatest volume among these prisms?
First off, there was a question in the comments as to what is meant by “rectangular prism.” I considered it an acceptable shorthand for a right rectangular prism, also known as a rectangular cuboid. In other words, every face of the prism was a rectangle.
That said, what did it mean mathematically to say that its internal diagonal (connecting opposite vertices) had a length of 2024? If the prism’s dimensions were a, b, and c, then the internal diagonal had length √(a2 + b2 + c2). Squaring away the square root gave you the problem’s key constraint: a2 + b2 + c2 = 20242. Your task was to find whole numbers a, b, and c that satisfied this constraint while maximizing abc, the volume of the prism.
As an aside, problems that call for maximizing volume, minimizing surface area, or the like, tend to involve either the most symmetric (or sometimes the most asymmetric) shapes. Here, the most symmetric kind of rectangular prism was a cube, where all three dimensions were equal in length. For 2024 to be the internal diagonal of a cube with side length a, we needed 3a2 = 20242, or a2 = 2024/√(3). Such a value of a wasn’t rational, let alone a whole number, which meant the answer didn’t involve a perfect cube. Instead, you wanted values of a, b, and c that were as close together as possible.
Finding all whole number triples (a, b, c) such that a2 + b2 + c2 = 20242 was no small task. It was no surprise then that almost all solvers turned to their computers for assistance. However, there were ways of listing out such “Pythagorean quadruples” in which a, b, and c were relatively prime. One such way was to pick four relatively prime non-negative integers m, n, p, and q such that m + n + p + q was odd. Once you had picked four such numbers, you could set a = m2 + n2 − p2 − q2, b = 2(mq + np), and c = 2(nq − mp). It turned out that a2 + b2 + c2 = (m2 + n2 + p2 + q2)2, which was indeed a perfect square.
But how was any of this helpful? It appears we just turned a problem involving three variables (a, b, and c) into one involving four variables (m, n, p, and q), upping the complexity. As it so happened, there was a relatively small quadruple (m, n, p, q) that resulted in values of a, b, and c that were fairly close to each other. In particular, when (m, n, p, q) = (0, 3, 1, 1), you had (a, b, c) = (7, 6, 6). The complete Pythagorean quadruple was 72 + 62 + 62 = 112. Multiplying this entire equation by square numbers gave you more Pythagorean quadruples for which the internal diagonal was a multiple of 11.
It just so happened that 2024 was one such multiple 11. In fact, 2024 is one less than 452, which meant 2024 = (45+1)(45−1) = 46·44 = 46·4·11 = 184·11. So if we took the equation 72 + 62 + 62 = 112 and multiplied it all by 1842, we got 12882 + 11042 + 11042 = 20242. The triple (a, b, c) = (1288, 1104, 1104) indeed corresponded to an internal diagonal of length 2024, along with an impressive volume of 1,569,835,008.
Of course, this didn’t prove that (1288, 1104, 1104) were the optimal dimensions for the rectangular prism—that required a bunch more casework. However, in the future, should you ever find yourself stranded on a desert island, without computational assistance, and in need of a Pythagorean quadruple whose three dimensions are all close together, just remember that 72 + 62 + 62 = 112.
By the way, if this puzzle hadn’t constrained the prism’s dimensions to be integers, then the side lengths could each have equaled 2024/√(3), in which case the maximal volume would have been approximately 1,595,694,112. With integer side lengths, we were able to achieve a volume that was roughly 98.4 percent of this maximum. That was surprisingly close to optimal, I’d say!
Solver David Gedye made one additional observation. If the problem had involved a four-dimensional tesseract instead of a three-dimensional rectangular prism, its four dimensions could have been (1012, 1012, 1012, 1012), since 10122 + 10122 + 10122 + 10122 = 20242. This is, of course, thanks to the fact that a unit tesseract has an “internal” (i.e., 4-space) diagonal with integer length 2.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 David Ding 🎻 from Natick, Massachusetts. I received 69 timely submissions, of which 69 were correct. As noted by several readers, I probably should have swapped last week’s Fiddler and Extra Credit. Oh well—I’ll take a 100 percent solve rate for the final Extra Credit of 2023, bookending the only other time it happened with the first Extra Credit of 2023.
As noted in last week’s Extra Credit, the Collatz Conjecture asserts that if you start with any positive integer and repeatedly apply a certain operation, the resulting sequence of numbers will always reach the number 1. The operation is: For even numbers divide by 2; for odd numbers, multiply by 3 and add 1.
For example, if you start with 11, the sequence goes: 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1.
The Collatz Conjecture is a well-known unsolved problem in mathematics. No proof has ever been found to show that it is true, nor has any counterexample been found to show that it is false.
For Extra Credit, what was the smallest starting number for which 2024 appeared somewhere in its Collatz sequence?
Keep reading with a 7-day free trial
Subscribe to Fiddler on the Proof to keep reading this post and get 7 days of free access to the full post archives.