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Joseph O'Rourke
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Let $S$ be a unit-radius sphere in $\mathbb{R}^3$.

Q0. Where should one place $3$ disjoint lines intersecting $S$ to minimize the maximum distance between any two points in $S$, where distance is measured as follows. Distance off the lines is Euclidean distance, but the distance between any two points on one line is zero.

The lines are like very fast transportation tubes. (I have in mind Star Wars' Coruscant.)

With no tubes, the maximum distance is $2$, realized by any antipodal pair. One tube, or two tubes, do not help reduce the maximum, but three tubes seem to help. In particular, I believe if one selects the three tubes to be the $x,y,z$ axes, slightly displaced to satisfy disjointness, then the maximum distance is $2 \sqrt{\frac{2}{3}} + \epsilon \approx 1.63$, illustrated below.

Q1. Is there a better arrangement of the three tubes?


          [![Tubes][1]][1]
          The red point is the origin/center of the sphere. The point $(1,1,1)/\sqrt{3}$ is $\sqrt{\frac{2}{3}}$ (green) from each axis.
The natural generalization is: $d$ tubes intersecting a unit-radius sphere in $\mathbb{R}^d$.

Let $S$ be a unit-radius sphere in $\mathbb{R}^3$.

Q0. Where should one place $3$ disjoint lines intersecting $S$ to minimize the maximum distance between any two points in $S$, where distance is measured as follows. Distance off the lines is Euclidean distance, but the distance between any two points on one line is zero.

The lines are like very fast transportation tubes.

With no tubes, the maximum distance is $2$, realized by any antipodal pair. One tube, or two tubes, do not help reduce the maximum, but three tubes seem to help. In particular, I believe if one selects the three tubes to be the $x,y,z$ axes, slightly displaced to satisfy disjointness, then the maximum distance is $2 \sqrt{\frac{2}{3}} + \epsilon \approx 1.63$, illustrated below.

Q1. Is there a better arrangement of the three tubes?


          [![Tubes][1]][1]
          The red point is the origin/center of the sphere. The point $(1,1,1)/\sqrt{3}$ is $\sqrt{\frac{2}{3}}$ (green) from each axis.
The natural generalization is: $d$ tubes intersecting a unit-radius sphere in $\mathbb{R}^d$.

Let $S$ be a unit-radius sphere in $\mathbb{R}^3$.

Q0. Where should one place $3$ disjoint lines intersecting $S$ to minimize the maximum distance between any two points in $S$, where distance is measured as follows. Distance off the lines is Euclidean distance, but the distance between any two points on one line is zero.

The lines are like very fast transportation tubes. (I have in mind Star Wars' Coruscant.)

With no tubes, the maximum distance is $2$, realized by any antipodal pair. One tube, or two tubes, do not help reduce the maximum, but three tubes seem to help. In particular, I believe if one selects the three tubes to be the $x,y,z$ axes, slightly displaced to satisfy disjointness, then the maximum distance is $2 \sqrt{\frac{2}{3}} + \epsilon \approx 1.63$, illustrated below.

Q1. Is there a better arrangement of the three tubes?


          [![Tubes][1]][1]
          The red point is the origin/center of the sphere. The point $(1,1,1)/\sqrt{3}$ is $\sqrt{\frac{2}{3}}$ (green) from each axis.
The natural generalization is: $d$ tubes intersecting a unit-radius sphere in $\mathbb{R}^d$.
Source Link
Joseph O'Rourke
  • 150.9k
  • 36
  • 358
  • 958

Transport tubes in a sphere

Let $S$ be a unit-radius sphere in $\mathbb{R}^3$.

Q0. Where should one place $3$ disjoint lines intersecting $S$ to minimize the maximum distance between any two points in $S$, where distance is measured as follows. Distance off the lines is Euclidean distance, but the distance between any two points on one line is zero.

The lines are like very fast transportation tubes.

With no tubes, the maximum distance is $2$, realized by any antipodal pair. One tube, or two tubes, do not help reduce the maximum, but three tubes seem to help. In particular, I believe if one selects the three tubes to be the $x,y,z$ axes, slightly displaced to satisfy disjointness, then the maximum distance is $2 \sqrt{\frac{2}{3}} + \epsilon \approx 1.63$, illustrated below.

Q1. Is there a better arrangement of the three tubes?


          [![Tubes][1]][1]
          The red point is the origin/center of the sphere. The point $(1,1,1)/\sqrt{3}$ is $\sqrt{\frac{2}{3}}$ (green) from each axis.
The natural generalization is: $d$ tubes intersecting a unit-radius sphere in $\mathbb{R}^d$.