Hamiltonian Puzzle #2 in 3D (version 3)

I was hoping to create nodes that would self-organize.

Unknown Puzzle: verticality

Below is the Unknown puzzle in its original form along with a doctored version. Does "D" seem a more appropriate answer in the doctored version?

ORIGINAL:


DOCTORED:

The shape that comes next has three qualities: 9; vertical mirror (as shown by "1"); style. Both A and D fit two of these three criteria, but the criteria are hierarchical, and only D meets the first two. The second image attempts to reinforce this hierarchy.

Unknown Puzzle: other manipulation

The rounded nine (answer A) remains the apparent correct response in most manipulations.

TierneyLab commentary

additional rationale for Kevin M

Rabbit traces a path that keeps agent at a tangent until center of lake reaches hypotenuse of right triangle created by tangent. Rabbit then follows trajectory to shore.

Kevin M's solution

path to 4.6 an exponential logarithmic spiral?

Why isn't the spiral a bit of an S?

It seems the rabbit would have to begin and end the mad dash at 90º angles. At the beginning the rabbit would have to take an initial stroke directly away from the agent so the agent doesn't change direction. The very end of the spiral would run parallel to the shore so it seems the very last stroke of the rabbit would be directly into shore.

A spiral shape must take care of this but I don't know how to show it.

(π+1) is too slow

The perpendicular line (which generates the (π+1) solution) is not optimal: if the rabbit instead takes off at a tangent it travels an additional unit while the agent must travel an additional five units (approx).

(Am using 3 instead of π in this sketch.)

Agent is (π+1) times as fast as rabbit

Pond circumference takes the same time to travel as the circumference of the rabbit's last circle, so the ratio of speeds in the same as the ratio if radii, or (π+1)/1.

Mad dash must equal pond semi-circle

If Rabbit left any earlier the agent would turn around. (This assumes mad dash is straight line direct to shore. Am not sure how to prove this.. but it does seem apparent line must leave inner circle at a right angle and intersect the shore at a right angle.)

Rabbit can only stay opposite agent for so long

As some point the rabbit reaches a circle that takes as long to circumnavigate as it takes the agent to travel around the pond. At this point the agent will no longer turn back once the rabbit makes a move away from the center.

Rabbit can stay opposite the agent while backing up

The agent, if logical, will reverse direction if that reversal will reduce the distance between the agent and where a line through the agent and rabbit hits the shore opposite the agent. For example, if the agent runs 4x as fast as the rabbit, the rabbit can swim directly away from the agent then return to a point directly opposite the agent just as the agent completes a 60º arc of the circle. At this point, the rabbit crosses this 60º line, so the agent reverses direction. This move can be repeated but is less effective each time.