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I know of one with 12 vertices, 20 faces, and 30 edges. Are there any others? Almost sorry I asked the question. The answer is trivial. Use Euler's formula and the unique solution pops out.

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Let $v,e,f$ be the number of vertices, edges and faces. So by your given condition, we have $ e = 5v/2 , f = 5v/3 $. And use this in the euler characteristic formula for sphere $ v-e+f =2$ to get $v (1-5/2+5/3) = 2 \implies v = 12 $. So there are only one such triangulation.

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  • $\begingroup$ Is it possible that there would be two non-isomorphic triangulations with the same number of vertices, edges, and faces? No of course, but is there an easy way of seeing this? $\endgroup$ – Tyler Seacrest Apr 14 '15 at 15:26
  • $\begingroup$ I think it is possible that two non-isomorphic graph has same v, e, f. But I do not know the details. $\endgroup$ – aNumosh Apr 14 '15 at 20:03

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