[1]          If existed, it would be obtainable by taking out 2 adjacent triangular faces off from an Icosahedron and "closing" the gap. This, however, would yield  a vertex with 6 faces, namely a “flat” corner, or an hexagonal face.

[2]          There are only two convex Deltahedra with at least one vertex of valence 3: the Tetrahedron and the Triangular Dipyramid . To see it consider a vertex of valence 3. The three faces meeting at this vertex contribute three ‘free’ edges which are coplanar thus forming a fourth triangle which completes the original corner to form a Tetrahedron (D₄).

Next, let us analyze adding more than one triangular face to the given three faces. The number of added triangles must be odd since the total number of faces in any convex Deltahedron must be even, as we have shown above.

Adding 3 new faces, one to each of the three ‘free’ edges, yield a Triangular Dipyramid (D₆). Note that this solid possesses three vertices of valence 4 and two of valence 3.

Adding 5 new faces can be seen as attaching one face of a Tetrahedron to one face of a Triangular Dipyramid, resulting an 8-face solid with two 5-valence vertices. Although this solid is a Deltahedron it is NOT convex, as the dihedral angle at the edge connecting the two 5-valence vertices is larger than 180⁰.

Adding 7 new faces can be seen as attaching one face of a Tetrahedron to one face of an Octahedron. Surprisingly, the resulting solid, made of 10 triangles has only 7 planar faces – three of which are rhombi made of 2 triangles each. Hence this solid does Not count as a Deltahedra.

In general, a 5-valence vertex connected to a 3-valence vertex forms a dihedral angle which is >180⁰, therefore, the only convex Deltahedra with V₃>1 are the Tetrahedron and the Triangular Dipyramid (D₄,D₆).