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Can You Cut The Square … Into More Squares?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Rob Peacock comes a matter of squaring a multitude of squares:

There are many ways to slice a big square into smaller squares (not necessarily of equal size), so that the smaller squares don’t overlap, while still making up the entire area of the big square.

For example, you can slice the big square into four smaller squares, each a quarter of the area of the big square. Or you could slice it into seven squares, if you take one of those four squares and slice it into four yet smaller squares.

What whole numbers of squares can you not slice the big square into?

The solution to this Riddler Express can be found in the following column.

Riddler Classic

Robin of Foxley has entered the FiveThirtyEight archery tournament. Her aim is excellent (relatively speaking), as she is guaranteed to hit the circular target, which has no subdivisions — it’s just one big circle. However, her arrows are equally likely to hit each location within the target.

Her true love, Marian, has issued a challenge. Robin must fire as many arrows as she can, such that each arrow is closer to the center of the target than the previous arrow. For example, if Robin fires three arrows, each closer to the center than the previous, but the fourth arrow is farther than the third, then she is done with the challenge and her score is four.

On average, what score can Robin expect to achieve in this archery challenge?

Extra credit: Marian now uses a target with 10 concentric circles, whose radii are 1, 2, 3, etc., all the way up to 10 — the radius of the entire target. This time, Robin must fire as many arrows as she can, such that each arrow falls within a smaller concentric circle than the previous arrow. On average, what score can Robin expect to achieve in this version of the archery challenge?

The solution to this Riddler Classic can be found in the following column.

Solution to the last Riddler Express

Congratulations to 👏 Eric Thompson-Martin 👏 of Amherst, Massachusetts, winner of the previous Riddler Express.

Last time, I was playing around in my kitchen with a tall glass and a smaller disk, when, to my surprise, I was able to balance the disk neatly atop the rim of the glass:

glass cup with a white disk balancing on its edge

Suppose you had a disk of radius R and a glass with a circular rim of radius 2R. If you randomly placed the disk so that its center lay within the glass’s rim, what was the probability that the disk would balance atop the glass? (Assume the distribution was uniformly spread across the circular area inside the rim.)

At first, this might have sounded like a physics question. However, the physics gave way to mathematics rather quickly. For the disk to balance atop the rim, the disk’s center (i.e., its center of mass) had to lie between the arc of the glass’s rim and the straight line connecting the endpoints. Otherwise, the disk would either fall into or out of the glass. This region was illustrated by Tom Epp, for the case when the disk was on the right edge of the rim:

Small circle balanced on the rim of a larger circle twice the size. The region in which the center of the small circle can be so that it balances is highlighted, lying between an arc and a chord.

But the disk didn’t have to lie on the right edge of the rim — it could have been anywhere around the circumference. That meant the region in which the center would balance was a ring, or annulus. Solver Alex Vornsand approximated the size of this region by picking random points in the circle and seeing how many would have resulted in a balanced disk.

Of the 200 random points, exactly 50 (i.e., one-quarter of them — a nice round number) were in the annulus.

At this point, solvers like Thanh Nguyen used trigonometry to calculate the inner radius of the annulus, which turned out to be R√3. Thar meant the area of the ring was the area of the entire circle — 𝜋(2R)2, or 4𝜋R2 — minus the area of the inner circle — 𝜋(R√3)2, or 3𝜋R2. This difference was simply 𝜋R2.

Finally, the question was asking for the probability that picking a random point would result in a balanced disk. That meant you had to find the ratio of the area of the annulus (where the disk would balance) to the area of the entire rim. This was 𝜋R2/(4𝜋R2), or one-quarter.

When it’s safe to return to bars and restaurants, I’m sure many readers will apply what they’ve learned here by balancing as many coins as they can on the rim of a pint glass. Your friends will be impressed! (And if they’re not, get better friends.)

Solution to the last Riddler Classic

Congratulations to 👏 Peter Flynn 👏 of Thunder Bay, Ontario, Canada, winner of the previous Riddler Classic.

Last time, you were introduced to the two-player Game of Attrition, where each player started with a whole number of “power points.” Players took turns “attacking” each other, which involved subtracting their own number of power points from their opponent’s until one of the players was out of points.

For example, suppose Player A (who went first) started with 5 points, and Player B started with 7 points. After A’s first attack, A still had 5 points, while B had been reduced to 2 points (i.e., 7 minus 5). Now it was B’s turn, who reduced A to 5 minus 2, or 3 points. Finally, on A’s second turn, B was reduced from 2 points to nothing (since 2 minus 3 is −1). Despite starting with fewer points, A won!

Now suppose A went first and started with N points. In terms of N, what was the greatest number of points B could start with so that A still emerged victorious?

First off, if B started with N or fewer points, then A would win with a single attack. So you immediately knew that the answer was greater than N. But beyond this simplest case, the math got hairier.

One way to gain some intuition for what was happening was to make a plot of who won for each combination of points. The following graph shows the results when each player started with anywhere from 1 to 100 points:

Graph showing who wins, based on Player A's and Player B's starting numbers of points. Player A wins when Player B less than phi times as many points.

The black region on the top left shows where A lost, while the colorful bands indicate where A won. The different colors show how many turns it took for A to win.

As we said before, when A started with at least as many points as B (the blue triangle in the bottom right), then A won in a single turn. Looking closely at the graph, you can see that A won in two turns when B started with at most 3/2·N, in three turns when B started with at most 8/5·N, in four turns when B started with at most 21/13·N and in five turns when B started with at most 55/34·N. If these fractions look familiar, that’s because they contain consecutive Fibonacci numbers!

There are countless patterns contained within the Fibonacci sequence, one of which is that the ratio of consecutive numbers approaches the golden ratio, also written as 𝜑, which equals (1+√5)/2, or approximately 1.618. If we kept going, we’d find that A won whenever B started with at most 𝜑·N points. Technically, since B had to have a whole number of points, the solution was ⌊𝜑·N — that is, the “floor” of 𝜑·N, or the greatest integer less than or equal to 𝜑·N.

Another way to arrive as this solution was to look for edge cases, where A and B were headed for a tie — that is, where the ratio of A’s points to B’s points didn’t fluctuate from one round to the next. Suppose B started with k·N points, where k is a constant we’ll be solving for in a minute. After A’s first turn, B had k·NN, or (k−1)·N points. Then, after B’s turn, A had N−(k−1)·N, or (2−kN points. For the ratio to remain the same, we need k·N/N to equal (k−1)·N/((2−kN). Some rearrangement and simplification (like canceling the N’s) led to the quadratic equation k2−k−1=0, whose positive solution was — wait for it — 𝜑. Once again, as long as B started with at most ⌊𝜑·N points, A was destined to win.

So before you play the Game of Attrition, make sure you know a thing or two (or three or five) about the Fibonacci sequence.

Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Want to submit a riddle?

Email Zach Wissner-Gross at


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.