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This is actually a set of two problems: (in one question, I believe it is useful and convenient to analyse them together)

Problem A

$N$ arbitrary points are given in a plane (all different). $N$ arbitrary lines are also given in the same plane (no two of them are parallel). Show that there exist a set of $N$ perpendiculars from points to lines such that there is a single perpendicular from each point, there is a single perpendicular to each line, and no two such perpendiculars intersect.

Problem B

Is the analoguous claim valid if the word "line" is replaced with "circle", and "parallel" with "concentric"?

NOTE: "A perpendicular" in the context of these problems means a segment going from a point to a line/circle (of course perpendicular to the line/circle).

I know the answer to Problem A, but not to Problem B. I am not attaching the answer to Problem A (for the time being), since I do not want to spoil possible different approaches.

I appreciate any hint/insight/idea of yours.

VividD
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    Problem B feels like a 'plane inversion' of Problem A, since inversions preserve angles and map lines to circles. That may not be sufficiently generic, though. – Semiclassical Jul 20 '14 at 15:48
  • How is it possible that "no two such perpendiculars intersect". Are these line segments, or lines? Otherwise, the perpendicular lines are not parallel, so much intersect. – Calvin Lin Jul 20 '14 at 15:51
  • @Calvin, they (perpendiculars) are segments, I am going to update the question to be clear. – VividD Jul 20 '14 at 15:53
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    @Semiclassical Potential gap is that "parallel" doesn't invert to "concentric". Also, you only go from a line to a circle if the center of inversion is on the line. – Calvin Lin Jul 20 '14 at 15:53
  • What if all the circle's centre and all the point are on the same line? Does that count as "intersect" if the line segments drawn share more than 1 point? – Gina Jul 20 '14 at 16:02
  • @Gina This looks like a valid counterexample. – VividD Jul 20 '14 at 16:19
  • Related: http://math.stackexchange.com/questions/557332/partition-the-points?rq=1 – VividD Jul 20 '14 at 16:29

1 Answers1

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Hint for Problem A:

Consider the $N!$ sets of perpendicular line segments where each of the $N$ points are uniquely joined to one of the $N$ lines. Take the set with minimal sum of lengths.

Claim: No two perpendicular line segments intersect.

Hint for Problem B:

This problem is not true. Find a configuration of 2 circles and 2 lines where it must intersect.

However, it is sufficient if no pair of points and pair of centers are collinear.

[There's an error in this argument which reflects the need for the extra condition:

A line segment is perpendicular to a circle only if it passes through the center of the circle.

Consider the $N!$ sets of lines where each of the $N$ points are uniquely joined to one of the $N$ (distinct by assumption) circle centers. Take the set with minimal sum of lengths.

Claim: No two line segments intersect.

Note: This claim is not true. However, it is true if we add the condition that "no pair of points and pair of centers are collinear."]

Calvin Lin
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