Draw five tetrahedral in a drawing of {3,5} I tried to do the following exercise, but something is missing for me. I tried to divide the icosahedron in to 5 tetrehedra but it seems it doesn't work.

The symmetry group of the icosahedron (vertices A,B, consists of 120
  rotations and reflections. This number happens to be 5! (five
  factorial).  one may embed five tetrahedra (partitioning the 20
  vertices) and these are permuted by the 60 rotations. Draw these five
  tetrahedra in a drawing of {3,5}
  

 A: There is a regular chiral compound of 5 tetrahedra. Its covex hull is the dodecahedron, its kernel is the icosahedron. In Coxeter's compound notation this is given as $\{5,3\}[5\{3,3\}]\{3,5\}$. Cf. eg. https://bendwavy.org/klitzing/explain/compound.htm#3d
Thus either you'll have to stellate the icosahedron accordingly (extending its face planes), or you would refer to the duality between icosahedron and dodecahedron by using the face centers of the icosahedron for both, the vertex set of the dual dodecahedron and the therein vertex inscribed compound under question.
--- rk
A: You can partition the 20 vertices of a dodecahedron into 5 sets of 4 that each define a regular tetrahedron.
Since the regular dodecahedron and icosahedron have the same symmetry group, it is likely that the wrong polyhedron in the exercise is a typo.
If the drawing in your question comes from the exercise, you would have to partition the face centers rather than the vertices.
A: Evidently, we need to partition the twenty faces, not vertices, of the icosahedron into five sets of tetrahedral faces.
The procedure for doing this References the picture below. (Please pardon the inaccuracies, which are caused by limited options for insering shapes on my device.)

Begin by selecting any face A. Associated with this face are six that share just a single vertex, which are comprised of a pair at each vertex. One triangle from each pair, colored gold, is displaced clockwise from the median edge feeding into the shared vertex; the other triangle from each pair,colored blue,is displaced counterclockwise from this mediant edge.
Select one of these color-coded faces to identify three faces of the tetrahedron. The fourth face is then_opposite_ the originally selected reference face (A).
To get the remaining ingredients tetrahedra, use different reference faces. Starting from A and rotating around any vertex, use the procedure above starting with face B to get the faces of the second tetrahedron, then start with C for the third tetrahedron, and similarly with D and E for the last two tetrahrdra. Important point: once you select the clockwise or counterclockwise group with A in the first tetrahedron, you must use the corresponding groups around B ,C, D, E to get the first three faces of each tetrahedron. Otherwise you will not get exactly one tetrahedron assigned to each icosahrdral face. Since the reference face at each stage is not incorporated into the tetrahedron ot is used to identify, using previously assigned faces for a reference face is allowed.
Because of the initial choice of a group of three faces in the first stage, there are ultimately two mirror-image solutions for the complete set of fivevtetrahedra. This is due to a subtle difference in the "tetrahedral" symmetry of the regular icosahedron versus an actual regular tetrahedron. In the tetrahedron the twofold rotation axes are upgraded to $D_{2d}$, meaning each is accompanied by an obliquely oriented mirror plane that converts it to an improper fourfold axis. In the regular icosahedron, by contrast, the twofold axes are $D_{2h}$, meaning each one is combined with a perpendicular mirror place to generate a center of inversion (the center of the icosahedron), but there is no improper fourfold rotation. So when we partition the icosahedral faces into five tetrahedral subgroups, there must actually be two sets of tetrahedra, reflected into each other through the $D_{2h}$-associated mirror planes, to incorporate the center of inversion. Equally, each tetrahedron must also be associated with a second icosahedron to accommodate the improper fourfold rotation axes of the tetrahedra (it follows that given a regular tetrahedron, there are two ways to fit four faces of an inscribed regular icosahedron flush against the tetrahedral faces).
