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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