# When Is a Triangle Like a Circle?

### There’s a special relationship between a circle’s area and its circumference. Can you find a similar relationship between a polygon’s area and its perimeter?

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 🎬.

## This Week’s Fiddler

One of my favorite facts about circles is the relationship between their area and their circumference. For a circle with radius *r*, its area is 𝜋*r*^{2} and its circumference is 2𝜋*r*. What’s neat here (or rather, *one *thing that’s neat here) is that if you take the derivative of the area formula with respect to *r*, you get the circumference formula! In other words, d(𝜋*r*^{2})/d*r* = 2𝜋*r*. Amazing, right?

(For those of you who are accustomed to using tau rather than pi, this still works. The area of a circle is 𝜏*r*^{2}/2 and its circumference is 𝜏*r*. Once again, d(𝜏*r*^{2}/2)/d*r* = 𝜏*r*.)

Inspired by this fact, let’s define the term “differential radius.” The differential radius *r* of a shape with area *A* and perimeter *P* (both functions of *r*) has the property that d*A*/d*r* = *P*. (Note that *A* always scales with *r*^{2} and *P* always scales with *r*.)

For example, consider a square with side length *s*. Its differential radius is *r* = *s*/2. The square’s area is *s*^{2}, or 4*r*^{2}, and its perimeter is 4*s*, or 8*r*. Sure enough, d*A*/d*r* = d(4*r*^{2})/d*r* = 8*r* = *P*.

What is the differential radius of an equilateral triangle with side length *s*?

## This Week’s Extra Credit

What is the differential radius of a rectangle with sides of length *a* and *b*? Your answer should be in terms of both *a* and *b*. Oh, and kudos if you can illustrate your solution geometrically!

## 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 an update on the AI Mathematical Olympiad, which I first mentioned a little over a month ago. Back then, the top-performing AI had solved 13 out of 50 AIME-style problems on a private test whose problems are not known to the public.

At the time of this writing, the aptly named Team “pls donate gpus” now sits atop the leaderboard, having solved 27 of the 50 problems—more than double the prior mark.

There’s still about a month left for you to pursue the million-dollar prize pool! (Or to at least donate some GPUs?)

But you know what the best part of winning would be? Having an AI that does fairly well on the weekly puzzles here at The Fiddler!

## 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.

## Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Emily Kelly 🎻 from Brookhaven, Georgia. I received 65 timely submissions, of which 51 were correct, good for a 78 percent solve rate.

Last week, you had infinitely many 3-by-3 cm tiles and infinitely many 5-by-5 cm tiles. You wanted to use some of these tiles to precisely cover a square whose side length was a whole number of centimeters. Tiles could not overlap, and they had to completely cover the larger square, without jutting beyond its borders.

What was the smallest side length this larger square could have, such that it could be precisely covered using at least one 3-by-3 tile *and *at least one 5-by-5 tile?

Many solvers busted out their graph paper and started sketching out a bunch of 3-by-3 and 5-by-5 tiles. The smallest square anyone could seem to find was 18-by-18, and here’s the one generated by solver 🎬 Michael Schubmehl 🎬:

Finding the answer was one thing, but *proving* it was the smallest possible square was another. Here, I’ll walk through the proof similar to ones provided by Benjamin Dickman and Michael Montuori.

Let’s put the graph paper aside for a moment, and consider some algebra instead.

Each 3-by-3 tile had an area of 9 cm^{2}, while each 5-by-5 tile had an area of 25 cm^{2}. So if an *N*-by-*N* square was completely covered by *A* 3-by-3 tiles and *B* 5-by-5 tiles, then it had to be true that *N ^{2}* = 9

*A*+ 25

*B*. At the same time, if there were a total of

*C*3-by-3 tiles and

*D*5-by-5 tiles along one row or column within the

*N*-by-

*N*square, then it had to be true that

*N*= 3

*C*+ 5

*D*. So you were immediately restricted to whole numbers

*N*that could be written as 3

*C*+ 5

*D*(for some positive integers

*C*and

*D*) and whose square could be written as 9

*A*+ 25

*B*(for some positive integers

*A*and

*B*).

The first few numbers that could be written as 3*C* + 5*D* were:

8 (3·1+5·1)

11 (3·2+5·1)

13 (3·1+5·2)

14 (3·3+5·1)

16 (3·2+5·2)

17 (3·4+5·1)

18 (3·1+5·3).

Meanwhile, the first few squares that could be written as 9*A* + 25*B* were:

13

^{2}(9·16+25·1)14

^{2}(9·19+25·1)16

^{2}(9·9+25·7)17

^{2}(9·21+25·4)18

^{2}(9·11+25·9)

The shortest side lengths that matched both criteria were 13, 14, 16, 17, and 18. These were good candidates for the smallest possible value of *N*, and could be analyzed case by case.

Could *N* have been 13? Each row or column had to include *two* 5-by-5 tiles. But the area had to be made up of *just one* 5-by-5 tile (and 16 3-by-3 tiles). How could you have two 5-by-5 tiles along each row or column, but only one in total? This didn’t make any sense, and so *N* could not have been 13.

Could *N* have been 14? The area had to be made up of *one* 5-by-5 tile (and 19 3-by-3 tiles). But there also had to be a 5-by-5 tile somewhere in *each* row or column. How could one 5-by-5 tile possibly coincide with all the rows and columns of the larger square? So *N* could not have been 14.

Could *N* have been 16? Benjamin and his Problem Solving & Posing class broke this up into sub-cases. Along a given side of the rectangle, there were several possible ways to order two 3-by-3 tiles and two 5-by-5 tiles:

3-by-3, 3-by-3, 5-by-5, 5-by-5

3-by-3, 5-by-5, 3-by-3, 5-by-5

3-by-3, 5-by-5, 5-by-5, 3-by-3

5-by-5, 3-by-3, 3-by-3, 5-by-5

(Other orderings were just reflections of these.) The class went through each of these sub-cases, demonstrating that you could never continue adding tiles to cover a 16-by-16 square. So *N* could not have been 16.

Could *N* have been 17? Each row or column had to include exactly one 5-by-5 tile, which meant the four 5-by-5 tiles in the covering could never share a common row or column. But that meant *N* had to be at least 5·4, or 20—a contradiction! So *N* could not have been 17.

And that brings us back to **18**, which was indeed the minimum value of *N*.

Of note, this minimum value of *N* turned out to be 3·5+3. More generally, if *a* and *b* are relatively prime, you can use at least one *a*-by-*a* tile and at least one *b*-by-*b* tile to cover an *N*-by-*N* square when *N* = *ab* + min(*a*, *b*). But is this always the *smallest* value of *N* that works? I’ll leave the proof (or counterexample) of this to you, dear reader.

## Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Q P Liu 🎻 from Santa Cruz, California. I received 25 timely submissions, of which 15 were correct—good for a 60 percent solve rate.

Last week’s Extra Credit was a personal favorite of high schooler Linus Tang. Linus was a Finalist of the 2024 Regeneron Science Talent Search, where he presented his research on measures of peripherality (the opposite of centrality) in graphs. He was also a Gold Award winner at the 2024 USA Math Olympiad.

Here’s was Linus’s puzzle:

This time, you had an infinite supply of square tiles for *each* odd whole number side length (as measured in centimeters) greater than 1 cm. In other words, you had infinitely many 3-by-3 cm tiles, infinitely many 5-by-5 cm tiles, infinitely many 7-by-7 cm tiles, and so on.

You wanted to use one or more of these tiles to precisely cover a square whose side length was *N* cm, where *N* was an integer. Once again, tiles could not overlap, and they had to completely cover the larger square without jutting beyond its borders.

What was the *largest* integer *N* for which this task was *not* possible?

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