Here, a graphical solution is presented for the icosahedron. As suggested in the comments and in the discussion of Ivan Neretin's answer, the icosahedron is not the dual of the tetartoid, nor could it be except for the Platonic case; and unlike the tetrahedron or octahedron there are enough triangular faces at each vertex to allow concave solutions.
Below left is a graphical representation of a triangular icosahedron. On the right the edges have call been labeled $a,b,c$ representing the distinct lengths of the sides of each triangle. The reader can verify that each face in the graph properly has one side of each length.
The existence of the graph does not fully prove the existence of the polyhedron. We must verify that there are ranges of the side lengths that allow the polyhedron to be assembled in real space. For instance, if we try to specify $a=9, b=6, c=4$ we will find that the icosahedron fails to come together.
Let the angles of the triangles be $\alpha$ opposite side $a$, $\beta$ opposite side $b$, and $\gamma$ opposite side $c$. Without loss of generality we may assume $a>b>c$ from which $\alpha>\beta>\gamma$. From the graph we can then identify four vertices having three $\alpha$ angles, one $\beta$ and one $\gamma$; and cyclic permutations thereof for two remaining groups of four vertices.
One criterion for existence is then the largest face angle at any vertex must be less than the sum of the other angles. With $\alpha>\beta>\gamma$ this implies
Since the three angles of any triangle sum to $180°$, we may render this as
The latter form, with $\alpha$ being the larger angle, implies that the icosahedron must exist when this angle is less than $90°$, thus always for acute triangles. In terms of side lengths, this translates to $a^2<b^2+c^2$. For instance, with $a=6,b=5,c=4$ we have correct angles to form an icosahedron with the graphical structure pictured above.
The second necessary condition is that the polygonal angles at each vertex should sum to less than $360°$. With $\alpha>\beta>\gamma$ this implies
Again $\alpha+\beta+\gamma=180°$, so this criterion is also equivalent to the maximum angle $\alpha$ measuring less than $90°$.
Therefore twenty congruent, acute triangles may indeed be assembled to form the faces of an icosahedron using the graphical scheme pictured above.