A convex polyhedron has exactly two vertices where $10$ edges meet and all the faces of this polyhedron are quadrilateral. Show that this polyhedron has at least $20$ vertices in which three edges meet.
Let's say the polyhedron has $V$ vertices, $E$ edges and $F$ faces in total.
Importantly, all of the faces have $4$ vertices, so they are bounded by $4$ edges: $4F = 2E \iff 2F = E \iff F = \frac{1}{2}E$.
We know that $2$ of the vertices are of degree $10$.
Let's say $V_3$ are vertices of degree 3.
$V′$ are neither of degree $3$ nor $10$. Vertices $V'$ must be of degree $\geq 4$ since we are talking about convex polyhedron (so the minimum degree of every vertice is $3$) and we know that those are not of degree $3$.
Therefore: $V = 2 + V_3 + V′$.
Now, as we look at the relation between number of edges and number of vertices we get: $2E \geq 2 \cdot 10 + 3V_3 + 4V'$
Therefore, from Euler's equation we get that:
$$F + V = E + 2 \iff \frac{1}{2}E + V = E + 2 \iff 4V = 2E + 8 \geq 2 \cdot 10 + 3V_3 + 4V' + 8$$
$$4V \geq 2 \cdot 10 + 3V_3 + 4V' + 8 \iff 4(2 + V_3 + V') \geq 2 \cdot 10 + 3V_3 + 4V' + 8$$
$$8 + 4V_3 + 4V' \geq 2 \cdot 10 + 3V_3 + 4V' + 8 \iff V_3 \geq 20 $$
And from that I get:
$$V_3 \geq 20$$
Is that correct?