Let's discuss Mathematics


May 9, 2020
I do not really get it, but they have figured out a bit more of Anaxagoras of Clazomenae's problem about squaring the circle?

Around 450 BCE, Anaxagoras of Clazomenae had some time to think. The Greek mathematician was in prison for claiming the sun was not a god, but rather an incandescent rock as big as the Peloponnese peninsula. A philosopher who believed that “reason rules the world,” he used his incarceration to grapple with a now-famous math problem known as squaring the circle: Using a compass and a straightedge, can you produce a square of equal area to a given circle?

The exact question posed by Anaxagoras was answered in 1882, when the German mathematician Ferdinand von Lindemann proved that squaring the circle is impossible with classical tools. He showed that pi — the area of a circle with a radius of 1 — is a special kind of number classified as transcendental (a category that also includes Euler’s number, e). Because a previous result had demonstrated that it’s impossible to use a compass and a straightedge to construct a length equal to a transcendental number, it’s also impossible to square a circle that way.

Tarski's Circle Squaring Problem from 1925 asks whether it is possible to partition a disk in the plane into finitely many pieces and reassemble them via isometries to yield a partition of a square of the same area. It was finally resolved by Laczkovich in 1990 in the affirmative. Recently, several new proofs have emerged which achieve circle squaring with better structured pieces: namely, pieces which are Lebesgue measurable and have the property of Baire (Grabowski-Máthé-Pikhurko) or even are Borel (Marks-Unger).

In this paper, we show that circle squaring is possible with Borel pieces of positive Lebesgue measure whose boundaries have upper Minkowski dimension less than 2 (in particular, each piece is Jordan measurable). We also improve the Borel complexity of the pieces: namely, we show that each piece can be taken to be a Boolean combination of Fσ sets. This is a consequence of our more general result that applies to any two bounded subsets of Rk, k≥1, of equal positive measure whose boundaries have upper Minkowski dimension smaller than k.​

Writeup Paper

They are answering a different question here. The classical question of squaring the circle (which means to construct a square with same area of circle via compass and straight-edge construction) is resolved and is deemed impossible. The question stated in the paper asks if you can cut a disk into finitely many pieces and reassemble them by rotation and translation to get a square of the same area. But cutting up a disk into pieces in mathematics can mean absolutely ridiculous things and not just a naive slicing and dicing. This does not contradict the classical question of squaring a circle since they are fundementally asking about different things.


Oct 24, 2003
Can someone explain this question from the ChatGPT does an MBA paper?


The Pennsylvania Department of State is implementing a new electronic voting system. Voters will now use a very simple self-service computer kiosk for casting their ballots. If that kiosk is busy, voters will patiently queue up and wait until it is there turn.

It is expected that voters will spend on average 5 minutes at the kiosk. This time will vary across voters with a standard deviation of 5 minutes. Voters are expected to arrive at a demand rate of 10 voters per hour. These arrivals will be randomly spread out over the hour (you can assume that the number of voters arriving in any time period follows a Poisson distribution).

What is the average amount of time that a voter will have to wait before casting their vote?


To find the right answer, one must look at a standard equation from queuing theory. The equation for the average waiting time states that:

Average Waiting Time = Average Processing Time x Utilization / (1-Utilization). Plugging in an average processing time of 5 minutes and an average utilization of 5/6, we get:

Average Waiting Time = 5 x (5/6) / (1 - 5/6) = 25 minutes.

So, the correct answer is 25 minutes waiting in line. If we add the 5 minutes at the kiosk, we obtain a total of 30 minutes.

Sniff Test

If you can process 12 an hour and expect 10 an hour I would intuitively expect the mean wait time will be closer to zero than five people in front of you in the queue.

The wait for the first person will always be zero. At a rate of 10/hour surely the time the poll is open will be significant to the overall mean waiting time?

However this is an apparently reputable paper that is ited all over in the discussion about AI and stuff, so one would expect me to be wrong rather than them.
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