[math-fun] Two random intersection puzzles
1. Given a convex quadrilateral, consider the four bounding line segments (i.e. the sides) and the two segments obtained by joining the midpoints of opposite sides — six line segments in all. Let L be a line not parallel to any of the six line segments. Show that a random translate of L, conditioned to intersect at least one of the six segments, can be expected to intersect 3 of them on average. 2. Consider the twelve line segments in a planar projection of the 1-skeleton of a parallelepiped. Let L be a line not parallel to any of the twelve line segments. Show that a random translate of L, conditioned to intersect at least one of the twelve segments, can be expected to intersect 4 of them on average. Jim Propp
I cannot make sense of the first problem in a way that allows me to reach the desired conclusion. Surely the line always intersects either 3 or 4 of the six lines. How can it ever intersect fewer than 3? And if it never intersects fewer than 3, how can the expected number be as low as 3? On Sun, Feb 4, 2018 at 11:30 PM, James Propp <jamespropp@gmail.com> wrote:
1. Given a convex quadrilateral, consider the four bounding line segments (i.e. the sides) and the two segments obtained by joining the midpoints of opposite sides — six line segments in all. Let L be a line not parallel to any of the six line segments. Show that a random translate of L, conditioned to intersect at least one of the six segments, can be expected to intersect 3 of them on average.
2. Consider the twelve line segments in a planar projection of the 1-skeleton of a parallelepiped. Let L be a line not parallel to any of the twelve line segments. Show that a random translate of L, conditioned to intersect at least one of the twelve segments, can be expected to intersect 4 of them on average.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Allan, If the translate of L clips a corner, it can cross just 2 line segments. Perhaps you were picturing a quadrilateral with its two diagonals, instead of a quadrilateral with its two “medians” (what’s the proper name for a line segment joining the midpoints of opposite sides of a quadrilateral?). Jim On Monday, February 5, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
I cannot make sense of the first problem in a way that allows me to reach the desired conclusion. Surely the line always intersects either 3 or 4 of the six lines. How can it ever intersect fewer than 3? And if it never intersects fewer than 3, how can the expected number be as low as 3?
On Sun, Feb 4, 2018 at 11:30 PM, James Propp <jamespropp@gmail.com> wrote:
1. Given a convex quadrilateral, consider the four bounding line segments (i.e. the sides) and the two segments obtained by joining the midpoints of opposite sides — six line segments in all. Let L be a line not parallel to any of the six line segments. Show that a random translate of L, conditioned to intersect at least one of the six segments, can be expected to intersect 3 of them on average.
2. Consider the twelve line segments in a planar projection of the 1-skeleton of a parallelepiped. Let L be a line not parallel to any of the twelve line segments. Show that a random translate of L, conditioned to intersect at least one of the twelve segments, can be expected to intersect 4 of them on average.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Argh! Allan’s instincts were right; claim #1 is false in general. As a counterexample, consider a trapezoid with one base that’s long and another base that almost vanishes to a point, and let L be almost perpendicular to the bases. Then the average number of intersections is about 2.5, not 3. My mistake lay in thinking that |u.(v+w)| = |u.v|+|u.w| for all vectors u,v,w. The proof works for certain choices of L (a positive angle’s worth of choices, for that matter), but fails for others. Claim #2 still stands, though. Jim Propp On Monday, February 5, 2018, James Propp <jamespropp@gmail.com> wrote:
Allan,
If the translate of L clips a corner, it can cross just 2 line segments.
Perhaps you were picturing a quadrilateral with its two diagonals, instead of a quadrilateral with its two “medians” (what’s the proper name for a line segment joining the midpoints of opposite sides of a quadrilateral?).
Jim
On Monday, February 5, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
I cannot make sense of the first problem in a way that allows me to reach the desired conclusion. Surely the line always intersects either 3 or 4 of the six lines. How can it ever intersect fewer than 3? And if it never intersects fewer than 3, how can the expected number be as low as 3?
On Sun, Feb 4, 2018 at 11:30 PM, James Propp <jamespropp@gmail.com> wrote:
1. Given a convex quadrilateral, consider the four bounding line segments (i.e. the sides) and the two segments obtained by joining the midpoints of opposite sides — six line segments in all. Let L be a line not parallel to any of the six line segments. Show that a random translate of L, conditioned to intersect at least one of the six segments, can be expected to intersect 3 of them on average.
2. Consider the twelve line segments in a planar projection of the 1-skeleton of a parallelepiped. Let L be a line not parallel to any of the twelve line segments. Show that a random translate of L, conditioned to intersect at least one of the twelve segments, can be expected to intersect 4 of them on average.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Actually, my instincts had nothing to do with it. Jim was right when he suggested that I simply had the wrong picture in my head -- I was indeed visualizing a complete graph on four vertices. It was a complete coincidence that claim #1 turned out to be false. On Mon, Feb 5, 2018 at 8:19 AM, James Propp <jamespropp@gmail.com> wrote:
Argh! Allan’s instincts were right; claim #1 is false in general.
As a counterexample, consider a trapezoid with one base that’s long and another base that almost vanishes to a point, and let L be almost perpendicular to the bases. Then the average number of intersections is about 2.5, not 3.
My mistake lay in thinking that |u.(v+w)| = |u.v|+|u.w| for all vectors u,v,w. The proof works for certain choices of L (a positive angle’s worth of choices, for that matter), but fails for others.
Claim #2 still stands, though.
Jim Propp
On Monday, February 5, 2018, James Propp <jamespropp@gmail.com> wrote:
Allan,
If the translate of L clips a corner, it can cross just 2 line segments.
Perhaps you were picturing a quadrilateral with its two diagonals, instead of a quadrilateral with its two “medians” (what’s the proper name for a line segment joining the midpoints of opposite sides of a quadrilateral?).
Jim
On Monday, February 5, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
I cannot make sense of the first problem in a way that allows me to reach the desired conclusion. Surely the line always intersects either 3 or 4 of the six lines. How can it ever intersect fewer than 3? And if it never intersects fewer than 3, how can the expected number be as low as 3?
On Sun, Feb 4, 2018 at 11:30 PM, James Propp <jamespropp@gmail.com> wrote:
1. Given a convex quadrilateral, consider the four bounding line segments (i.e. the sides) and the two segments obtained by joining the midpoints of opposite sides — six line segments in all. Let L be a line not parallel to any of the six line segments. Show that a random translate of L, conditioned to intersect at least one of the six segments, can be expected to intersect 3 of them on average.
2. Consider the twelve line segments in a planar projection of the 1-skeleton of a parallelepiped. Let L be a line not parallel to any of the twelve line segments. Show that a random translate of L, conditioned to intersect at least one of the twelve segments, can be expected to intersect 4 of them on average.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Allan Wechsler -
James Propp