Can You Squeeze the Bubbles?

You draw seven circles, one at a time. Each new circle’s center can’t be inside any of the previously drawn circles. How small can their combined area be?

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

Draw a unit circle (i.e., a circle with radius 1). Then draw another unit circle whose center is not inside the first one. Then draw a third unit circle whose center is not inside either of the first two. Keep doing this until you have drawn a total of seven circles.

Consider the entire region that is inside at least one of the circles. What is the minimum possible area of this region?

This Week’s Extra Credit

Instead of seven unit circles, now suppose you draw N of them. As before, the center of each new circle you draw cannot be inside any of the previous circles.

Consider the entire region that is in at least one of the circles. As N gets very, very large, what is the minimum possible area of this region?

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 a new Substack that was recently launched by my colleague Fawn Nguyen. I’ve thoroughly enjoyed my conversations about non-routine math problems with Fawn over the years, so I was quite excited to see her pick up the pen (er, keyboard) and broaden her audience!

For my readers, I specifically recommend her post on counting lines and squares in a grid.

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. 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: 🎻 Alain Camus 🎻 from Ath, Belgium. I received 38 timely submissions, of which 35 were correct—good for a 92 percent solve rate.

Last week, a spider was weaving a web within a unit square (i.e., a square with side length 1) in the following haphazard manner:

First, the spider picked two points at random and uniformly inside the square. Next, the spider connected the two points with a strand of silk and extended the strand to two sides of the square. For example, here was a web made of 10 silk strands that had been picked as described:

Within the unit square, which point (or points) was most likely to be on a new strand of silk, whose two defining points had not yet been picked?

(While the probability that any specific point wound up being precisely on the new strand was zero, some points and regions were nevertheless more likely to be on the strand than others.)

Based on the symmetry of a square, the square’s center was a very logical choice. There wasn’t any justification as to why some other point would have been more likely to be on a strand than the center.

A great way to demonstrate this was to try drawing lots of random strands and see what happened. The following diagram shows 20,000 random strands drawn in the square.

It’s a little hard to see, but the density of points is indeed greater in the middle of the square than anywhere else. So sure enough, the center of the square was the likeliest point to be included on a strand. If you provided coordinates like (0.5, 0.5) that clearly indicated you were describing the center, I awarded full credit.

Before moving on, let’s take one more look at that figure above. Points in the middle were more likely to be on a strand, yes. But points along either diagonal also seemed fairly likely to be on the strand. Tracing along any of the square’s four edges, many strands seemed to be orthogonal to that edge. But when you reached a corner, these orthogonal edges crossed each other, giving such points more opportunities to be on a strand. Thus, the points least likely to be on a strand turned out to be near the midpoints of the four edges.

But for a more quantitative exploration of these probabilities, let’s move on the Extra Credit!

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Gary M. Gerken 🎻 from Littleton, Colorado. I received 25 timely submissions, of which 10 were correct—good for a 40 percent solve rate. I do believe this was one of the toughest puzzles I’ve ever run.

From last week’s Fiddler, we already knew there existed a point in the unit square that was more likely than any others to be on the randomly selected silk strand. That point happened to be at the center of the square.

At the same time, there existed a point (or points) in the unit square that were less likely than any others to be on the random strand. As we just said, it appeared that there were four such points, each at the midpoint of one of the square’s four edges.

How much more likely was a most likely point to be on the strand than a least likely point? More specifically, suppose the maximum of the probability density for being on the strand was p_max and the minimum probability density was p_min. What was the ratio p_max/p_min?