[math-fun] Totally unsupported polygons and polyhedra
I was thinking sloppily when I wrote (as part of the math-fun thread "5 and 7 outcome fair dice") that the pentagram is optimal. For purposes of this problem, the pentagram should be viewed as a non-convex decagon. And 10 sides isn't optimal (see below). So here are my questions: 1) Call a side of a polygon "supporting" if it's part of a support line of the polygon (that is, the polygon lies in one of the closed half-planes determined by the line) and "unsupporting" otherwise, and call a polygon "totally unsupported" if it has no supporting sides. What is the smallest number of sides a totally unsupported polygon can have? My record is 6 (a hexagon whose sides alternate between concave angles and reflex angles). 2) Call a face of a polyhedron "supporting" if it's part of a support plane of the polyhedron (that is, the polyhedron lies in one of the closed half-spaces determined by the plane) and "unsupporting" otherwise, and call a polyhedron "totally unsupported" if it has no supporting faces. What is the smallest number of faces a totally unsupported polyhedron can have? My record is 12 (see below). Also: 3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide. Jim On Tue, Oct 13, 2015 at 12:07 AM, James Propp <jamespropp@gmail.com> wrote:
I like the idea of a concave polyhedron that cannot rest on any of its faces (stably or unstably), i.e., a polyhedron none of whose faces belong to supporting planes. (Sort of like a holeyhedron, but a whole lot easier to construct.)
How few faces can such a polyhedron have?
Take a solid tetrahedron and excavate triangular-pyramidal holes on each of its four faces; we get a polyhedron with twelve faces that can't rest (stably or unstably) on any of them.
Or: stick two tetrahedra together along a face, but give one of the tetrahedra a 180-degree twist, so that the two abutting triangular faces form a star of David. We again get a polyhedron with twelve faces that can't rest on any of them.
Can anyone do better?
(I can prove by brute force that the pentagram is the optimal solution to the 2D version of the question. Is there a slick way to see that a tetragram can't have the property? This last question smells like an Olympiad problem to me.)
Jim Propp
On Monday, October 12, 2015, Bill Gosper <billgosper@gmail.com> wrote:
Odd barrel dice have the problem that two faces are always uppermost, so how do you label them? Alternative to face-transitivity, there are fair dice which cannot rest on any face, and yet have one face uppermost. Spoiler: gosper.org/5&7dice.png --rwg I assume these are old ideas. Check out Wikipedia if you haven't seen spherical, 6 outcome dice. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I think 8 faces is possible: Connect two opposite edges of a regular tetrahedron ABCD with a segment I, and let the 1/3 and 2/3 points of I be 2 new vertices E and F. Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. —Dan
On Oct 13, 2015, at 8:02 AM, James Propp <jamespropp@gmail.com> wrote:
2) Call a face of a polyhedron "supporting" if it's part of a support plane of the polyhedron (that is, the polyhedron lies in one of the closed half-spaces determined by the plane) and "unsupporting" otherwise, and call a polyhedron "totally unsupported" if it has no supporting faces. What is the smallest number of faces a totally unsupported polyhedron can have? My record is 12 (see below).
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out. Dan wrote: I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I,
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters). and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces.
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? Jim
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD. Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC. Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom. The point E makes a triangle with each undeleted edge AB, BC, CD, DA. These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface. Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD. Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E. The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip. —Dan Jim wrote: Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out. Dan wrote: ----- I think 8 faces is possible: Connect two opposite edges of a regular tetrahedron ABCD with a segment I, ----- I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters). Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F. Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. ----- I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)?
That helps a little, but I'm still not seeing it. Can anyone handy with 3D graphics create a picture of Dan's polyhedron? Jim On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Dan's second description allowed me to draw it by hand. Here is another way to describe it: Start with a regular tetrahedron. (It doesn't have to be regular, but that makes it easier to visualize.) Rest that on an edge say on table top, with opposite edge above the table also horizontal. Now take the midpoint of the top edge E' and draw the lines to the bottom vertices, subdividing two of the tetrahedral faces each into 2 right triangles. Similarly, take the midpoint of the bottom edge F' and join it to the two top vertices, thereby subdividing the other two tetrahedral faces each into 2 right triangles. Now deform this tetrahedron (which now is 8 right triangles, 4 pairs of co-planar right triangles), by moving points E' and F' toward each other, making an non-convex octahedron, with the 8 right triangles each deformed (they can still be congruent) so they are obtuse. That is the object. Seems like it must have a name, but I don't know one. On Tue, Oct 13, 2015 at 12:19 PM, James Propp <jamespropp@gmail.com> wrote:
That helps a little, but I'm still not seeing it.
Can anyone handy with 3D graphics create a picture of Dan's polyhedron?
Jim
On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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OK, just in case this kind of surface has no name yet and just in case I have naming rights, I hereby dub it a "pringle".* —Dan P.S. No, they're not caltrops. You can't eat them. _________________________________________________________________________ * Consistent with the edibility of the other shape-name I came up with, the "bialy" — created by rotating a vertical circle tangent to the z-axis about the z-axis.
On Oct 13, 2015, at 11:07 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Dan's second description allowed me to draw it by hand. Here is another way to describe it:
Start with a regular tetrahedron. (It doesn't have to be regular, but that makes it easier to visualize.) Rest that on an edge say on table top, with opposite edge above the table also horizontal. Now take the midpoint of the top edge E' and draw the lines to the bottom vertices, subdividing two of the tetrahedral faces each into 2 right triangles. Similarly, take the midpoint of the bottom edge F' and join it to the two top vertices, thereby subdividing the other two tetrahedral faces each into 2 right triangles.
Now deform this tetrahedron (which now is 8 right triangles, 4 pairs of co-planar right triangles), by moving points E' and F' toward each other, making an non-convex octahedron, with the 8 right triangles each deformed (they can still be congruent) so they are obtuse. That is the object.
Seems like it must have a name, but I don't know one.
Here is a pic: https://www.dropbox.com/s/0nizg6iy93t03eu/pringle1-3.png?dl=0 On Tue, Oct 13, 2015 at 12:19 PM, James Propp <jamespropp@gmail.com> wrote:
That helps a little, but I'm still not seeing it.
Can anyone handy with 3D graphics create a picture of Dan's polyhedron?
Jim
On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Now I see it! Sorry for being so, er, obtuse. Jim On Tue, Oct 13, 2015 at 2:59 PM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Here is a pic: https://www.dropbox.com/s/0nizg6iy93t03eu/pringle1-3.png?dl=0
On Tue, Oct 13, 2015 at 12:19 PM, James Propp <jamespropp@gmail.com> wrote:
That helps a little, but I'm still not seeing it.
Can anyone handy with 3D graphics create a picture of Dan's polyhedron?
Jim
On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I now have a 3D picture of the pringle made with the (annoyingly anemic) Mac Grapher utility, so IF you have this software (as all Macs have had for years), you can rotate it around. I would be happy to e-mail this to anyone who'd like it. BUT, to use it you'll probably have to override the security warning that I am an unregistered developer, or something like that. —Dan P.S. Soon enough I'll have a bunch of still pictures as well.
On Oct 13, 2015, at 11:59 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Here is a pic: https://www.dropbox.com/s/0nizg6iy93t03eu/pringle1-3.png?dl=0
On Tue, Oct 13, 2015 at 12:19 PM, James Propp <jamespropp@gmail.com> wrote:
That helps a little, but I'm still not seeing it.
Can anyone handy with 3D graphics create a picture of Dan's polyhedron?
Jim
On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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How about a 2-D picture that can be folded into the shape in question? --ms On 13-Oct-15 16:03, Dan Asimov wrote:
I now have a 3D picture of the pringle made with the (annoyingly anemic) Mac Grapher utility, so IF you have this software (as all Macs have had for years), you can rotate it around.
I would be happy to e-mail this to anyone who'd like it.
BUT, to use it you'll probably have to override the security warning that I am an unregistered developer, or something like that.
—Dan
P.S. Soon enough I'll have a bunch of still pictures as well.
On Oct 13, 2015, at 11:59 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Here is a pic: https://www.dropbox.com/s/0nizg6iy93t03eu/pringle1-3.png?dl=0
On Tue, Oct 13, 2015 at 12:19 PM, James Propp <jamespropp@gmail.com> wrote:
That helps a little, but I'm still not seeing it.
Can anyone handy with 3D graphics create a picture of Dan's polyhedron?
Jim
On Tue, Oct 13, 2015 at 12:37 PM, Dan Asimov <asimov@msri.org> wrote:
Imagine the skew quadrilateral created by all but 2 opposite edges AC, BD of a regular tetrahedron ABCD.
Let the point E lie inside the tetrahedron, not far from the midpoint of the (deleted) edge AC.
Think of the edge AC as horizontal and on top, and BD as horizontal and on the bottom.
The point E makes a triangle with each undeleted edge AB, BC, CD, DA.
These 4 triangles form a piecewise-linear approximation to the neighborhood of a saddle point on a surface.
Similarly, choose the point F interior to the tetrahedron, close to the midpoint of the (deleted) edge BD.
Likewise to E, the point F taken together with each of the undeleted edges AB, BC, CD, DA will create 4 triangles that also form a saddle-like surface whose center is F, but below the saddle-like surface whose center is E.
The 8 triangles bound a shape roughly like what you would get from dipping a potato chip in sour cream/onion dip.
—Dan
Jim wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron. But that'll be moot if Dan's 8-face solution checks out.
Dan wrote: ----- I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I, -----
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
Dan wrote: ----- and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces. -----
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On 2015-10-13 09:09, James Propp wrote:
Another 12-face solution is a throwing star made by joining four very acute triangular pyramids so that their equilateral sides form a tetrahedron.
That's a caltrop. Inferior aerodynamics.
But that'll be moot if Dan's 8-face solution checks out.
Dan wrote:
I think 8 faces is possible:
Connect two opposite edges of a regular tetrahedron ABCD with a segment I,
I assume that Dan has in mind that I joins the midpoints of those two edges (not that it matters).
and let the 1/3 and 2/3 points of I be 2 new vertices E and F.
Now throw away all faces and the 2 edges that I connects, and draw the 8 new edges between each of E and F with A, B, C, D. The total of 12 edges define 8 triangular faces.
I am having trouble picturing this. Can Dan or someone else suggest an alternative way to see this polyhedron (say by gluing pieces together)?
Jim
Instead of Jim's excavated tetrahedron, you could make nonstandard endododecahedra by, in the octahedral-tetrahedral tessellation, bulging the octahedra into deltoidal icositetrahedra. -rwg
Assume A, B, C, D are distinct points in the plane such that segments AB and CD intersect (necessarily in their interiors). Then ACD and BCD are triangles sharing the edge CD but otherwise disjoint, since A and B are on opposite sides of the line extending CD. Thus BC and AD are disjoint. —Dan
On Oct 13, 2015, at 8:02 AM, James Propp <jamespropp@gmail.com> wrote:
3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide.
Thanks, Dan. Jim On Tuesday, October 13, 2015, Dan Asimov <asimov@msri.org> wrote:
Assume A, B, C, D are distinct points in the plane such that segments AB and CD intersect (necessarily in their interiors).
Then ACD and BCD are triangles sharing the edge CD but otherwise disjoint, since A and B are on opposite sides of the line extending CD.
Thus BC and AD are disjoint.
—Dan
On Oct 13, 2015, at 8:02 AM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide.
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participants (5)
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Dan Asimov -
James Buddenhagen -
James Propp -
Mike Speciner -
rwg