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The expected distance $d$ of randomly selected points within a unit square to the square's center is

$d = \frac{1}{6} P$

where P is the universal parabolic constant

$P = \sqrt{2} + \ln{\left(1+\sqrt{2}\right)} = 2.2955871 \dots $

see

Is this a mere coincidence or is there an (intuitive) reason why this constant shows up in the solution to this problem?

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2 Answers 2

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There is a geometrical reason for why parabolic constant appears in squares. https://prajwalsouza.github.io/universal-parabolic-constant.html

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  • $\begingroup$ This was really amazing. How did you make this visualization? $\endgroup$
    – Sidharth Ghoshal
    Commented Jul 21, 2023 at 3:32
  • $\begingroup$ Thank you. There were many different attempts and then, this technique of adding and shrinking line segments systematically, led to the parabola. And it was amazing! :D But the visualization is mostly based on an svg plotting library for web : github.com/prajwalsouza/viewX $\endgroup$
    – Prajwal D'Souza
    Commented Jul 22, 2023 at 4:27
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The "reason" that the two given numbers are equal is "write up the integrals, they turn out to be the same integral".

An answer that might satisfy the "intuitive reason" criterion is that the sides of the unit square are given by axis-aligned straight lines; in polar coordinates $r=\frac{1}{cos(t)}$ or $r=\frac{1}{sin(t)}$; and $\frac{1}{cos(t)^2}=\frac{r^2}{r^2cos(t)^2}=\frac{\sqrt{x^2+y^2}}{x^2}$ is the integrand in the integral defining the arc length of the parabolic segment. Truth be told though, this isn't really much more than saying "the integrals turn out to be the same integral", so I'm not sure how much of an "intuitive" explanation this is.

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  • $\begingroup$ I don't see any integral in the definition of the parabolic constant. Which integrals are you talking about? $\endgroup$
    – user4503
    Commented Apr 15, 2015 at 19:59
  • $\begingroup$ The parabolic constant is defined (basically) as the arc length of a certain curve segment, which you find with an integral of the form $\int_a^b\sqrt{1+f'(t)^2}dt$. $\endgroup$
    – Ketil Tveiten
    Commented Apr 16, 2015 at 8:26

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