Let $P$ denote a Platonic solid. Truncating $P$ at a vertex $v$ consists of marking the midpoints of the edges that touch $v$ and then slicing off a corner of $P$ by the plane that passes through all those points. For each Platonic solid $P$, determine the the polyhedron that results from truncating $P$ simultaneously at each of its vertices.

Now, I had a little difficulty imagining the shapes and their truncations, but this is what I came up with. If we truncate at the midpoint of each edge for all vertices of each Platonic solid, we get its dual polyhedron. So, for the tetrahedron, truncating it in this manner results in another tetrahedron. When you truncate the cube and the octahedron, you get the other Platonic solid. Finally, the same goes for the icosahedron and dodecahedron. Since I am having difficulty envisioning the actual results, I was wondering if you guys could help me out. Also, is there a program I can use to manually do this operation? Thanks!

  • $\begingroup$ What exactly are you asking? $\endgroup$
    – charlotte
    Dec 4 '14 at 4:37
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    $\begingroup$ Truncation to the midpoints of edges is called "rectification", and it does not turn a solid into its dual. It's true that the original vertices "become" faces of equal valence, but portions of the original faces remain as faces of the new figure. The rectified tetrahedron is in fact the octahedron. $\endgroup$
    – Blue
    Dec 4 '14 at 4:46
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    $\begingroup$ For icosahedron and dodecahedron, the resulting figure is an Icosidodecahedron which has 12 pentagons and 20 triangles as faces. $\endgroup$ Dec 4 '14 at 5:22
  • $\begingroup$ @Blue Ahh, I can now visualize the process. Thanks! $\endgroup$
    – RXY15
    Dec 4 '14 at 5:25
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    $\begingroup$ Note that if you start with $P$ with $f$ faces and $v$ vertices, you will end up with a polyhedron with $f+v$ faces. So you won't get the dual. $\endgroup$ Dec 4 '14 at 5:46

Here are the 5 Platonic polyhedra together with their truncations. The best polyhedron program that I know is Stella (you can find the site with Google). A trial download is possible. The following images were created by my own program, MathpadDraw which you can download from mathokay.com.

  • $\begingroup$ Your truncated tetrahedron does not seem correct. It should have 4 triangular faces and 4 hexagonal faces. $\endgroup$
    – user141290
    Jan 3 '15 at 21:48

Let there any platonic solid originally having $\color{blue}{V_{o}} \space \color{blue}{\text{no. of vertices}}$, $\color{blue}{E_{o}}\space \color{blue}{\text{ no. of edges}}$ out of which $\color{blue}{e_{o}\space \text{no. of edges meet at each vertex}}$ & $\color{blue}{F_{o}}\space \color{blue}{\text{ no. of faces}}$.These are co-related by Euler's formula as $F_{o}+V_{o}=E_{o}+2$ .

Now, it is truncated at each vertex to the mid-points of its edges to generate a new solid having $\color{blue}{V}$ no. of vertices , $\color{blue}{E}$ no. of edges & $\color{blue}{F}$ no. of faces then in general we have $$\color{blue}{V= (\text{no. of original edges meeting at each vertex})\times (\text{no. of original vertices})-(\text{no. of original edges}) }$$$$\color{blue}{=e_{o} V_{o}-E_{o}}$$
$$\color{blue}{F=(\text{no. of original vertices})+(\text{no. of original faces}) }$$$$\color{blue}{= V_{o}+F_{o}}$$
Now, applying Euler's formula we get $$\color{blue}{E}=F+V-2$$$$=e_{o} V_{o}-E_{o}+V_{o}+F_{o}-2$$ $$\color{blue}{=(e_{o}+1)V_{o}+F_{o}-E_{o}-2}$$
A regular tetrahedron originally has $\color{blue}{V_{o}=4} \space \color{blue}{\text{no. of vertices}}$, $\color{blue}{E_{o}=6}\space \color{blue}{\text{ no. of edges}}$ out of which $\color{blue}{e_{o}=3\space \text{no. of edges meet at each vertex}}$ & $\color{blue}{F_{o}=4}\space \color{blue}{\text{ no. of faces}}$ Now by truncating it to the mid-points of the edges, a new solid is obtained which has $$V=3(4)-6=6$$ $$F=4+4=8$$ $$E=8+6-2=12$$ Thus, the solid obtained has 6 vertices, 12 edges & 8 (triangular) faces. Hence it is an $\color{blue}{\text{Octahedron}}$

Similarly by truncation of regular hexahedron (cube) & octahedron to the mid-points of the edges will generate a solid called $\color{blue}{\text{Cuboctahedron}}$ having 12 vertices, 24 edges and 8 equilateral triangular faces & 6 square faces.

Similarly by truncation of regular dodecahedron & icosahedron to the mid-points of the edges will generate a solid called $\color{blue}{\text{Icosidodecahedron}}$ having 30 vertices, 60 edges and 20 equilateral triangular faces & 12 regular pentagonal faces.


Truncation need not be to the midpoint of an edge. The traditional truncated Platonics have smaller pieces cut off so that the original edges are shortened but remain present. These solids are know simply as the "truncated [named Platonic].

Cutting back to the midpoint as described is known as rectification. The Platonics pair up in the sense that the rectified cube is also the rectified octahedron, similarly with the icosahedron and dodecahedron. These solids are better known as the cuboctahedron and rhombicuboctahedron. The tetrahedron pairs with itself, producing an octahedron (which in this context may be referred to as a tetratetrahedron). These rectified figures have a property known as quasiregularity, which the conventional truncations do not possess.


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