Connect the Dots: Prove 2 Line Segments from Red/Blue Points

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SUMMARY

The discussion focuses on proving the existence of non-intersecting line segments connecting red and blue points in a plane. Initially, with four points (two red and two blue), a method is presented where segments are drawn from red points to a blue point while considering visibility constraints. The solution is then extended to six points (three red and three blue), challenging participants to prove that three non-intersecting segments can be drawn. The approach emphasizes geometric visibility and strategic segment placement, showcasing an elegant solution by user AKG.

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Jimmy Snyder
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There are 4 points in the plane. Two are red, two are blue. No three points lie on the same line. Prove that you can draw two line segments, each one connecting a red point to a blue point, and such that the two line segments do not intersect.
 
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Pick a blue dot, and first join both red dots to that blue dot. Pick one of the red dots, and consider the region of the plane that the other red dot can't "see" due to the presence of the segment joining the first red dot to the blue dot (i.e. this segment acts like a wall). If the second blue dot is in this hidden area, then remove the wall, and join the first red dot to the second blue dot. If the second blue dot is visible to the second red dot, then leave the wall in place, remove the segment joining the second red dot to the first blue dot, and instead draw a segment from the second red dot to the second blue dot.
 

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Nice solution, AKG, very elegant. But will it scale? Now there are 6 points in the plane. Three are red, three are blue. No three points lie on the same line. Prove that you can draw three line segments, each one connecting a red point to a blue point, and such that no two line segments intersect.
 

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