This suggests a variant on Polya's orchard problem. That problem asks1 for which radius $\epsilon$ of trees at each lattice point within a distance $R$ of the origin block all lines of sight to the exterior of the orchard. It was established in the 1980's that rays to infinity are completely blocked iff $\epsilon \ge 1/\sqrt{R^2 + 1}$, when $R$ is an integer.2 E.g., for $R=20$, $\epsilon = 1/\sqrt{401} \approx 0.05$ suffices.

The variant is: Only lattice points both of whose coordinates are prime numbers are fattened to $\epsilon$-trees, and the source of visibility is $(2,2)$ rather than $(0,0)$. For example, the below shows that $\epsilon=0.3$ is insufficient to block visibility for $R=20$:


The basic question is:

Q1. For a given $R$, which $\epsilon$ suffices to block visibility from $(2,2)$ to beyond $R$ ?

As this does not seem easy to answer precisely, let me pose a very specific question:

Q2. As $R \to \infty$, does $\epsilon \to 0$?

It could be that the thinning of the prime trees is rapid enough to require some lower bound on $\epsilon$ to block visibility, even for large $R$.

1Here I am quoting from a previous MO question.
2Thomas T. Allen, "Polya's orchard problem." (Jstor link.) The American Mathematical Monthly 93(2): 98-104 (1986).)

  • 1
    $\begingroup$ Are you trying to block all lines? Or just in the positive quadrant? Also (except for the offset) this is like approximating various reals by p/q where p and q are both primes. Anyway, I expect the answer to differ from that for Polya's problem by a multiplicative constant. Gerhard "Perhaps Pi Squared Over Six?" Paseman, 2015.07.03 $\endgroup$ – Gerhard Paseman Jul 4 '15 at 2:21
  • $\begingroup$ @GerhardPaseman: In the 1st quadrant only, as illustrated. Nice conjecture re multiplicative constant. $\endgroup$ – Joseph O'Rourke Jul 4 '15 at 11:14
  • $\begingroup$ I can see now that the question would be more natural retaining the origin as the source of visibility, and asking for blockage of all but lines of sight skimming near the $x$- and $y$-axis. For then it would retain the approximation-of-reals flavor by $p/q$ with both prime (as Gerhard mentions). $\endgroup$ – Joseph O'Rourke Jul 4 '15 at 17:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.