August’s puzzle was a three dimensional throwback to Single Cross from three years ago. Careful computation determined that for lengths D <= 1, the probability of a single cross comes to the pleasantly symmetric spherical coordinate double integral. (As seen above.)

This double integral miraculously simplifies to (1/(4π))*D(-16D + 3D^2 + 6π). This function has a local maximum at D = (16 - sqrt(256 - 54π))/9 ~ 0.7452572091, with value ~ 0.5095346021. The argument that there can’t be a second higher local maximum with D > 1 is left to the puzzler.

Congrats to this month’s solvers who successfully completed the optimal length and probability!