This is an old and very natural question. We shall call f(d) the answer to this problem, and record f (1) = 2, which is trivial. For d = 2 the configuration of four triangles in the margin shows f (2) ≥ 4. There is no similar configuration with five triangles, because from this the dual graph construction, which for our example with four triangles yields a planar drawing of K4, would give a planar embedding of K5, which is impossible (see page 67). Thus we have
$$f(2) = 4$$
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- J. Zaks: No Nine Neighborly Tetrahedra Exist, Memoirs Amer. Math. Soc. No. 447, Vol. 91, 1991.Google Scholar
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