Euler's formula tells us that the number of vertices, edges and faces of a 3D solid have to satisfy the relationship $V+F=E+2$. How about the converse, if I have a triple of numbers that fulfill this identity, how can I check if such solid (polyhedron) exists?
2026-03-27 07:49:38.1774597778
Euler's Formula and Existence of Solids
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I followed the relevant (linked) resources and found the following.
Every face has at least three sides $2E \ge 3F$, and at least three faces meet at a vertex $2E \ge 3V$. Combining this with the Euler's formula $V+F=E+2$ for convex polyhedra, one has:
$$2F - 4 \ge V \ge F/2 +2$$
And every such pair $(V,F)$ should correspond to at least one convex polyhedron.
I believe this answers your question.
If I'm not mistaken, this result is also stated as a theorem in Eulers Polyederformel, und die Arithmetisierung der Gestalt on page 9., which is a resource linked by Christian Blatter's answer on the related question which was linked in the comments by Shahab.
To see the intuition why it is sufficient, we observe a way to construct examples via truncation and expansion of pyramids and subsequently obtained polygons: (taken and shortened from this article):
Every $n$-sided polygon based pyramid has $V=F=n+1$, $n\ge 3$.
Truncating (a pyramid) at a vertex where three faces meet: $V\to V+2,F\to F+1$.
Expanding on one of the triangular faces (of a pyramid): $V\to V+1,F\to F+2$.
Both operations may be applied infinitely, provided that the polyhedron remains convex. (As for non-convex polyhedra, the Eulers formula might fail $-$ see Euler characteristic.)
Remark; Alternatively, You might be interested in the Steinitz's theorem, cited from Wikipedia:
Which is not needed for this particular question, but should be useful for similar problems.