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On equidistant lines of given line configurations

  • Takayasu Kuwata , Hiroshi Maehara und Horst Martini EMAIL logo
Veröffentlicht/Copyright: 14. August 2019
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Abstract

The equidistant set of a collection F of lines in 3-space is the set of those points whose distances to the lines in F are all equal. We present many examples and results related to the lines possibly contained in the equidistant set of F. In particular, we determine the possible numbers of lines in the equidistant set of a collection of n lines for every n>0. For example, if n=3, then the possible number of such lines is either 4 or 2 or 1 or 0. In a natural way, our results are connected with properties of special types of (ruled) surfaces. For example, we obtain also results on the number of lines in the intersection of quadratic surfaces.


Dedicated to Professor Alexander Kharazishvili on his 70th birthday


A Appendix

Let us present here a proof of Theorem 1.1. First, we prove a lemma.

Lemma A.1.

Let g,ξ be a pair of distinct lines, and let Pξ,Qg (possibly P=Q) be points such that |PQ| attains the minimum distance between g and ξ. Let Xξ and Y,Zg be points such that YXξ and XZg. Then

(A.1)|XY|2=|PQ|2+|PX|2(|XZ|2-|PQ|2|PX|2-|XZ|2+|PQ|2),
(A.2)|XZ|2=|PQ|2+|PX|2(|XY|2-|PQ|2|PX|2+|XY|2-|PQ|2).

Proof.

Let us consider the case PQ. In this case, PQ is perpendicular to both g and ξ. Let g be the line passing through P and parallel to g, and let Y,Zg be such that QPYYZZ; see Figure 7. Then YY and ZZ are both perpendicular to the plane PXY.

Figure 7

Lemma A.1.

Let φ=XPY. Then we have

|PX|2tan2φ+|PQ|2=|XY|2,
|PX|2sin2φ+|PQ|2=|XZ|2.

From these, we have (A.1) and (A.2). ∎

Proof of Theorem 1.1.

Let Pξ, Q1g1 be the points such that |PQ1| attains the minimum distance between ξ and g1. Let us consider the case that PQ1. (The proof of the case P=Q1 is easier.) Let HP be the plane that perpendicularly intersects ξ at P.

First, suppose that all lines in F lie on a surface of revolution with axis ξ. For i=2,3,,n, let Qi be the intersection point of gi and HP. Then |PQ1|==|PQn|. For Xξ, let HX denote the plane that perpendicularly intersects ξ at X, and let {Yi}=HXgi. Since g1,,gn lie on the surface of revolution with axis ξ, we have |XY1|==|XYn|. Hence, by (A.2), we have d(X,g1)==d(X,gn), which implies that ξ is an equidistant line of F.

Next, suppose that ξ is an equidistant line of F. Let be the surface obtained by rotating g1 around ξ. This time, define Qigi as the point such that |PQi|=|PQ1|. Since |PQ1| is the minimum distance between ξ and g1, |PQi| is also the minimum distance between ξ and gi for i=2,,n. For Xξ, define (this time) Yigi as a point such that XYiξ is satisfied. Since d(X,g1)==d(X,gn) and |PQ1|==|PQn|, it follows by (A.1) of Lemma A.1 that |XY1|==|XYn|. Hence Y1,,Yn lie on . Now, moving X on ξ, we can see that the lines g1,,gn all lie on . Thus, all lines in F lie on a surface of revolution with axis ξ. ∎

Acknowledgements

The authors are grateful to the reviewer for careful reading of the manuscript and helpful advises.

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Received: 2018-08-18
Revised: 2019-02-10
Accepted: 2019-03-12
Published Online: 2019-08-14
Published in Print: 2019-12-01

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