Squaring the Circle

Squaring the Circle

By Stephen Meskin

As with many mathematical concepts, “squaring the circle” has become a metaphor, in this case for something impossible to do. The ancient Greeks asked if by using a straight-edge and compass one could draw a square with the same area as a given circle in a finite number of steps. 

Because a circle of radius 1 has area π, this would be equivalent to forming a square with sides . Taking square roots is easy, so the problem is equivalent to constructing π. Mathematicians worked on it for thousands of years but couldn’t solve it. Finally, in 1882 squaring the circle was proved to be impossible. (See the interesting article with this same title in Wikipedia and the “Numberphile” video referenced therein.)

So, how can you be challenged with problems about squaring the circle? Of course, a metaphorical problem about squaring the circle. Namely by being asked to create a circular necklace that you want to give to a friend, which, because you are a numbers person, is decorated with numbered beads having the property that the sum of each pair of adjacent beads is a perfect square. 

That sounds easy enough.

But there some rules to make it interesting. Rule One is that the numbers on the beads must be distinct positive integers, although the squares adjacent beads sum to will not be distinct. For example, pictured here is a squared circle necklace with 6 beads that satisfies Rule One.

  • Problem One: As a warmup, create three squared circle necklaces that satisfy Rule One: one with 3 beads, one with 4 beads, and one with 5 beads. 

If there is one rule, there is probably a second. Rule Two is that the numbers on the beads must consist of all and only integers from 1 to N for some positive integer N ≥ 3. Unfortunately, not all N will work. For example, it is clear that N = 3 will not work since a squared circle necklace cannot be made with the integers 1, 2, and 3. 

Note that both rules apply to the next two problems.

  • Problem Two: Prove that N will not work for N ≤ 30. WARNING: Do not try to prove this by trying to exhibit all the circular arrangements of the integers 1 to N for N from to 3 to 30, thereby showing that none of them can be a squared circle necklace. There are too many circular arrangements and they will certainly not be checked. But there is a neat little proof that works.
  • Problem Three: Create a squared circle necklace for N = 32. No proof is required; all that is needed is to present a necklace. However, if you have a slick derivation, it would be nice to see. 

Notice that N = 31 has been left out. It is said that 31 does not work, but the proof is difficult.

Solutions may be emailed to cont.puzzles@gmail.com

In order to make the solver list, your solutions must be received by Sept. 30, 2020. 

Solution to Previous Puzzle:
The Free Throw Challenge

What is the probability of getting at least 90 out of first 100 free throws? We can use the binomial distribution to solve this, where we have 100 attempts and an 80% likelihood of success of each attempt. Summing getting 90, 91, 92, 93, … 100 free throws is a probability of 0.57%.

What is the probability of somewhere in a set of 5,000 free throws that one makes at least 90 out of 100 attempts? This turned out to be trickier than I thought. There are 4,901 possibilities to achieve this (each attempt from 100 all the way through attempt 5,000), but they are NOT uncorrelated, as 99 free throws will be the same in looking at free throws 1–100 versus free throws 2–101. This makes the math very tricky, and all but calls for a Monte Carlo simulation. Fortunately, this is not that difficult to set up in Excel or any programming language, and when you do, the probability turns out to be right around 83%.

If you make 77 of the first 87 free throws, should you start over or attempt to make 13 in a row? The probability of making 13 in a row is simply (0.8)13 = 5.5%. As that is nearly 10 times the answer as in part A, it is clear you should keep going and not start over.

What is the probability of Swatch finishing the challenge? Again, this turned out to be tougher than intended. The first thing you have to do is figure out when you should or should not start over. You should always start over if your probability of success of completing a set is less than that of question 1, or 0.57%. Working backward, the probability of hitting 23 free throws in a row (0.8)23 is 0.59%, and the probability of hitting 24 free throws in a row is 0.47%. This means that you should start over if you have 10 misses in your first 76 attempts. Similarly, the probability of hitting 32 of 33 free throws is 0.59%, while the probability of hitting 33 out of 34 free throws is 0.48%. So you should start over if you have made 57 of 66 free throws. You can continue this process to figure out when you should start over when you have missed 8, 7, 6 … 1 free throw. Now that you know when the optimal time to start over is—admittedly this slightly changes over the last 100 free throws—you can run a simulation similar to question 2 to figure out the overall probability of success, which is right around 70%. This is far higher than the 25% = {1 – (1 – 0.0087)50}, the probability achieved if one never starts over!


Robert Bartholomew, Bob Byrne, Bob Conger, Bill Feldman, Nick Fiechter, Rui Guo, Steve Itelson, Clive Keatinge, David Promislow, Noam Segal, John Snyder, Al Spooner, Ronald Stokes, Jim Walsh. 

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