单交叉 2
Single-Cross 2
题目详情
中文译文待补充。
Consider 3-space (i.e. R3) partitioned into a grid of unit cubes with faces defined by the planes of all points with at least one integer coordinate. For a fixed positive real number D, a random line segment of length D (chosen uniformly in location and orientation) is placed in this cubic lattice.
What length D maximizes the probability that the endpoints of the segment lie in orthogonally adjacent unit cubes (that is, the segment crosses exactly one integer-coordinate plane), and what is this maximal probability? Give your answer as a comma-separated pair of values to 10 significant places (e.g. “1.234567891,0.2468135792”).
解析
中文解析待补充。
Original Explanation
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!