Angels still in fear of treading. Ad.Fa Bf.Ec Cb.De Ae.Bb Cf.Fd Dc.Ea Aa.Cc Df.De Ed.Fb Ab.Dd Ef.Ca Fe.Bc Ac.Ee Ff.Db Ba.Cd Ab.Ji Bj.Ia Cg.Hd Df.Ge Eh.Fc Ac.Ba Cj.Jb Dh.Ie Eg.Hf Fi.Gd Ad.Cb Dj.Bc Ei.Jf Fh.Ig Ga.He Ae.Dc Ej.Cd Fa.Bg Gi.Jh Hb.If Af.Ed Fj.De Gb.Ch Ha.Bi Ic.Jg Ag.Fe Gj.Ef Hc.Di Ib.Ca Jd.Bh Ah.Gf Hj.Fg Id.Ea Jc.Db Be.Ci Ai.Hg Ij.Gh Je.Fb Bd.Ec Cf.Da Aa.Ih Jj.Hi Bf.Gc Ce.Fd Dg.Eb Not very methodical, but the trick is to take a female round-robin and a male one with different fixed points (A and j in the above) and then pair using 2m-2 different chord lengths from among the 2m-1 possible ones, 0, +-1, +-2, ..., +-(m-1), of a (2m-1)-gon. E.g., in the m=5 specimen above, in the polygon with 2m-1 vertices Ja Bb Cc Dd Ee Ff Gg Hh Ii the distances J-i I-a C-g H-d D-f G-e E-h F-c are +1 -1 -4 +4 -2 +2 -3 +3 while the distances A- and -j run through all 2m-1 possibilities. Alice doesn't get to play with jim and Jane doesn't get to play with arthur; otherwise the revulsions are as before No reason why we can't rename a and j, so different fixed points may not be needed. But perhaps the two tournaments can't be isomorphic. A tidier presentation is left to the reader. R.