Can You See Between the Trees?

You are peering between trees positioned on an infinite grid. Where are the largest gaps between them that you can see?

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

This week, we’ll be returning to the forest with trees at integer coordinates.

You are at the point (0, 0) on the coordinate plane. There is a tree at each point with nonnegative integer coordinates, such as (5, 6), (0, 7), and (9, 1). The trees are very thin, so that they only obstruct trees that are directly behind them. For example, the tree at (2, 2) is not visible, as it is obscured by the tree at (1, 1).

Now, you can’t see infinitely far in this forest. Suppose, for example, that the farthest you can see is 4 units. The diagram below shows the trees you would see and the angles between them:

In truth, you can see much farther than 4 units into the forest. You’re not sure exactly how far you can see, but it’s pretty dang far. To be extra clear about this, the diagram above is just an illustration, and you can in fact see much farther than 4 units.

As you look around, you can make out very narrow angular gaps between the trees. The largest gaps are near the positive x-axis and the positive y-axis (similar to the illustrated case above). After those, the next largest pair of gaps are on either side of the tree at (1, 1), 45 degrees up from the x-axis.

Consider the view between 0 and 45 degrees up from the x-axis. The next largest pair of adjacent gaps in this range are on either side of what angle up from the x-axis? (To be clear, you are not considering the gap just above 0 degrees or the gap just below 45 degrees.)

This Week’s Extra Credit

This week’s Fiddler asked to find the largest pair of adjacent gaps between 0 and 45 degrees.

The fifth largest pair of adjacent gaps in this range are on either side of what angle up from the x-axis? (Again, you are not considering the gap just above 0 degrees or the gap just below 45 degrees.)

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 problem from the 2025 AIME. But there’s a story behind this specific problem.

The commute between Boston and New York (which I’ve driven many, many times) can wear me down. When traveling with my family, my wife will graciously entertain me by reading out AIME problems that seem doable … in one’s head! No pencil, no paper, no calculators. Just mental math and plenty of road (depending on the traffic).

On our recent return from cheering on the runners at the 129th Boston Marathon, I was asked this fun little problem (#9 on the recent exam):

The parabola with equation y = x²−4 is rotated 60° counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has y-coordinate (a−√b)/c, where a, b, and c are positive integers, and a and c are relatively prime. Find a+b+c.

There are many ways to attack this problem. But to do this mentally, you need to settle on a strategy that’s efficient and involves minimal computation. If you have the time, I welcome you to think it through in your head.

Feel free to discuss this puzzle and your approach in the comments below!

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: 🎻 Aaron Gilary 🎻 from Alexandria, Virginia. I received 52 timely submissions, of which 48 were correct—good for a 92 percent solve rate.

Last week, you and an opponent were competing in a golf match. On any given hole you played, each of you had a 50 percent chance of winning the hole (and a zero percent chance of tying). That said, scorekeeping in this match worked a little differently from the norm.

Each hole was worth 1 point. Before starting each hole, either you or your opponent could “throw the hammer.” When the hammer was thrown, whoever did not throw the hammer had to either accept the hammer or reject it. If they accepted the hammer, the hole was worth 2 points. If they rejected the hammer, they conceded the hole and its 1 point to their opponent. Both players could throw the hammer on as many holes as they wished. (When both players decided to throw the hammer at the exact same time—something that couldn’t be planned in advance—the hole was worth 2 points.)

The first player to reach 3 points won the match. Suppose all players always made rational decisions to maximize their own chances of winning.

You won the first hole, and now led 1-0. What was your probability of winning the match?

It wasn’t immediately clear what players should have done early on in the golf match. However, things became clearer when either player was closer to winning—in the “endgame,” if you will. Therefore, solvers like Sanket Alekar opted to work backwards, using probabilities of victory from higher scores to compute similar probabilities for lower scores.

We can start at the highest of scores: When you had 3 (or 4) points, your chances of winning were 100 percent, and the same went for your opponent when they had that many points. That computation wasn’t particularly challenging.

Now suppose the match was tied 2-2. It didn’t matter whether the next hole was worth 1 point or 2. Either way, whoever won that hole would have been the first to reach 3 points and win the match. So your chances of winning in this case were 50 percent. In fact, whenever the score was tied (e.g., when it was 1-1 or 0-0), your chances of winning were 50 percent due to the symmetry between you and your opponent—neither of you had an advantage over the other.

Things got more interesting when the score was 2-1. Suppose you were in the lead. All you needed to do was win one more hole, whether it was worth 1 or 2. However, if the next hole was worth 2 points, your opponent had a chance to win. If the hole was worth 1 point, then your opponent needed to win this next hole and another one after that. So you really wanted the next hole to be worth 1 point. By contrast, your opponent preferred it to be worth 2 points, so they threw the hammer. Therefore, this became another winner-take-all hole, and you each had a 50 percent chance of winning. The same was true when you were down 2-1 and your opponent led.

Next, what if you led 2-0? Suppose the next hole was worth 1 point. There was a 50 percent chance you won the hole, in which case you won the match. But there was also a 50 percent chance you lost the hole, in which case the score would be 2-1, at which point we already said you had a 50 percent chance of winning the match. Putting these together, your chances of winning would have been 75 percent.

Now suppose the hole was instead worth 2 points. There was a 50 percent chance you won the hole, in which case you won the match. But again, there was a 50 percent chance you lost the hole, in which case the score would have been 2-2, at which point we again already said you had a 50 percent chance of winning the match. So whether the hole was worth 1 point or 2, your chances of winning were 75 percent. Correspondingly, when you were behind 2-0, your chances of winning were 25 percent.

The only score we haven’t analyzed yet was the score from the puzzle, 1-0. Suppose the next hole was worth 1 point. There was a 50 percent chance the next score would be 2-0 in your favor (at which point you had a 75 percent chance of winning) and a 50 percent chance the next score would be 1-1 (at which point you had a 50 percent chance of winning). Putting these together, your chances of winning were 62.5 percent.

Now suppose the hole was instead worth 2 points. There was a 50 percent chance the next score would be 3-0 in your favor (meaning you won) and a 50 percent chance the next score would be 2-1 in your opponent’s favor (at which point you had a 50 percent chance of winning). Putting these together, your chances of winning were 75 percent.

Because 75 percent was greater than 62.5 percent, it was in your interest to throw the hammer when you led 1-0. Therefore, the solution to the puzzle was 75 percent.

What I found particularly interesting here was when each player was incentivized to throw the hammer. Initially, you might have thought that the leader in the match should always throw—or never throw. But it was more nuanced than that.

When you led 2-0 or 2-1, it was definitively not in your interest to throw the hammer. But when you led 1-0, we saw that it was in your interest to throw the hammer.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Michael Montuori 🎻 from Morris Plains, New Jersey. I received 41 timely submissions, of which 38 were correct—good for a 93 percent solve rate.

Instead of playing to 3 points, now the first player to 5 points won the match.

You won the first hole, and now led 1-0. What was your probability of winning the match?