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Mar 11, 2010 at 14:15 comment added damiano @jp: This might be correct, but I do not see the point in editing it in, given the better argument below! d
Mar 11, 2010 at 13:05 comment added j.p. Another idea to leave out (almost) every 2nd vertex of the $3p$-gon: use just the simplices $\{b_i, b_{i+1}, a_{2i+1\pmod 3}\}$, $\{b_i, a_{2i-1\pmod 3}, a_{2i\pmod 3}\}$ and $\{b_i, a_{2i\pmod 3}, a_{2i+1\pmod 3}\}$ to glue the triangle to the "$(3p+1)/2$"-gon. This way we "make two steps" on the triangle for "each step" on the now "$(3p+1)/2$"-gon. (Needs of course a correct treatment of $i=0$ rsp. $i=(3p+1)/2$).
Mar 11, 2010 at 12:44 comment added j.p. @David: You are right, I set the wrong path to zero.
Mar 11, 2010 at 12:36 comment added David E Speyer Rather, don't add $(a_1, a_2, a_3)$ but, instead, triangulate the $3p$-gon without adding a new internal vertex.
Mar 11, 2010 at 12:19 comment added j.p. You can reduce the number of vertices by 1 if you simply add the simplex $\{a_1, a_2, a_3\}$ to your construction instead of "coning it off". Maybe you should also make clear that you start taking only the boundary of the triangle and the $3p$-gon.
Mar 11, 2010 at 10:55 history answered damiano CC BY-SA 2.5