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Article by Alan Beardon# Impossible Polyhedra

Imagine making a polyhedron by taking polygons and fixing them together along their edges. We need four triangles to make a tetrahedron, six squares to make a cube and so on. There are five regular polyhedra and each of these is made entirely out of triangular, square or pentagonal faces. But what about irregular polyhedra? Is it possible, for example, to make an irregular polyhedron using only polygons of, say, six, seven and eight sides? The answer (rather surprisingly) is 'no', but how do we prove a statement like this?

We will need Euler's formula for the polyhedron; this says that if the polyhedron has $F$ faces, $E$ edges, and $V$ vertices, then $F-E+V=2$. We shall also need to count the number of triangular faces, say $F_3$, the number of faces with exactly four edges, say $F_4$, and so on. Obviously, \begin{equation}F= F_3 + F_4 + \ldots, \quad (1) \end{equation} and also as every edge lies on the edge of exactly two faces, we have \begin{equation}2E = 3F_3 + 4F_4 + 5F_5 + \ldots . \quad (2) \end{equation} This shows that $2E \geq 3F$ because \begin{equation*}2E = 3F_3 + 4F_4 + 5F_5 + \ldots \geq 3F_3 + 3F_4 + 3F_5 + \ldots = 3F. \end{equation*} Next, let $V_m$ be the number of vertices which have exactly $m$ edges ending at that vertex. Then \begin{equation}V= V_3 + V_4 + \ldots,\quad (3) \end{equation} and also as every edge has two vertices, we have \begin{equation}2E = 3V_3+4V_4+5V_5+\ldots . \quad (4) \end{equation} This gives the inequality $2E \geq 3V$ because \begin{equation*}2E = 3V_3+4V_4+5V_5+\ldots \geq 3V_3+3V_4+3V_5+\ldots = 3V.\end{equation*} We are going to show that \begin{equation}F_3+F_4+F_5 \geq 4, \quad F_3+V_3 \geq 8. \quad (5) \end{equation} This tells us, for example, that not all of $F_3$, $F_4$ and $F_5$ can be zero, so there must be at least one face with exactly three, four of five sides. Thus it is indeed impossible to make a polyhedron with each face having at least six sides. We know that \begin{equation*}F-E+V = 2, \quad 2E \geq 3F, \quad 2E \geq 3V,\end{equation*} so that \begin{equation*}2+E = F+V \leq F + 2E/3.\end{equation*} This shows that $2+E/3 \leq F$ and hence that $12+2E \leq 6F$. If we substitute for $F$ and for $E$ their expressions in terms of the $F_k$ given in (1) and (2), we get \begin{equation*}12 + 3F_3 + 4F_4 + 5F_5 + \ldots \leq 6F_3 + 6F_4 + 6F_5 + \ldots.\end{equation*} This gives \begin{equation*}12+F_7+2F_8+3F_9+\ldots \leq 3F_3+2F_4+F_5\end{equation*} so that, finally, $12 \leq 3F_3 + 2F_4 + F_5$. This is the first inequality in (5).

To prove the second inequality in (5), we start with Euler's formula $F-E+V=2$ and write this as $(F-E/2)+(V-E/2) = 2$. Multiplying throughout by $4$, we now get $(4F-2E)+(4V-2E) = 8$. If we now replace $4F-2E$ by their expressions in terms of the $F_i$ in (1) and (2), and $4V-2E$ by the expressions in (3) and (4), we get \begin{equation*}4(F_3 + F_4+\ldots) -(3F_3+4F_4+\ldots) +4(V_3 + V_4+\ldots) -(3V_3+4V_4+\ldots) =8.\end{equation*}

Simplifying this we see that $F_3+V_3 \geq 8$ which is the second inequality in (5).

For the cube, we have $F_3 = 0$ and $V_3=8$, so that $F_3+V_3=8$. As $F_3+V_3\geq 8$ in all cases, $8$ is the smallest possible value of $F_3+V_3$. This leads to the question of finding other polyhedra that give other values of $F_3$ and $V_3$ with $F_3+V_3=8$ (that is, its smallest value). We leave this as an exercise for the reader, and suggest that you try the regular polyhedra first.

The inequalities (5) give us a lot of information about regular polyhedra. Suppose that we have a regular polyhedron. Because $F_3+F_4+F_5 \geq 4$, there must be at least four faces which are triangular, quadrilateral, or pentagonal. As the polyhedron is regular, all faces have the same number of sides, and so all are triangles, all are squares, or all are pentagons.

In the case, of squares and pentagons, $F_3=0$ so that $V_3\geq 8$. So, by regularity, every vertex has exactly three edges leaving it. These polyhedra are the cube and the dodecahedron.

It remains to consider the cases when every face is trianglular. Suppose that there are $q$ edges that leave each vertex. Then $F-E+V=2$, $2E = 3F$ and $2E =qV$. If we eliminate $F$ and $V$ from these equations we get $(6-q)E= 6q$ so that $q = 3,4,5$ giving the tetrahedron, octahedron and icosahedron, respectively. We have now listed all possible regular polyhedra.

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Age 16 to 18

Published 2001 Revised 2008

Imagine making a polyhedron by taking polygons and fixing them together along their edges. We need four triangles to make a tetrahedron, six squares to make a cube and so on. There are five regular polyhedra and each of these is made entirely out of triangular, square or pentagonal faces. But what about irregular polyhedra? Is it possible, for example, to make an irregular polyhedron using only polygons of, say, six, seven and eight sides? The answer (rather surprisingly) is 'no', but how do we prove a statement like this?

We will need Euler's formula for the polyhedron; this says that if the polyhedron has $F$ faces, $E$ edges, and $V$ vertices, then $F-E+V=2$. We shall also need to count the number of triangular faces, say $F_3$, the number of faces with exactly four edges, say $F_4$, and so on. Obviously, \begin{equation}F= F_3 + F_4 + \ldots, \quad (1) \end{equation} and also as every edge lies on the edge of exactly two faces, we have \begin{equation}2E = 3F_3 + 4F_4 + 5F_5 + \ldots . \quad (2) \end{equation} This shows that $2E \geq 3F$ because \begin{equation*}2E = 3F_3 + 4F_4 + 5F_5 + \ldots \geq 3F_3 + 3F_4 + 3F_5 + \ldots = 3F. \end{equation*} Next, let $V_m$ be the number of vertices which have exactly $m$ edges ending at that vertex. Then \begin{equation}V= V_3 + V_4 + \ldots,\quad (3) \end{equation} and also as every edge has two vertices, we have \begin{equation}2E = 3V_3+4V_4+5V_5+\ldots . \quad (4) \end{equation} This gives the inequality $2E \geq 3V$ because \begin{equation*}2E = 3V_3+4V_4+5V_5+\ldots \geq 3V_3+3V_4+3V_5+\ldots = 3V.\end{equation*} We are going to show that \begin{equation}F_3+F_4+F_5 \geq 4, \quad F_3+V_3 \geq 8. \quad (5) \end{equation} This tells us, for example, that not all of $F_3$, $F_4$ and $F_5$ can be zero, so there must be at least one face with exactly three, four of five sides. Thus it is indeed impossible to make a polyhedron with each face having at least six sides. We know that \begin{equation*}F-E+V = 2, \quad 2E \geq 3F, \quad 2E \geq 3V,\end{equation*} so that \begin{equation*}2+E = F+V \leq F + 2E/3.\end{equation*} This shows that $2+E/3 \leq F$ and hence that $12+2E \leq 6F$. If we substitute for $F$ and for $E$ their expressions in terms of the $F_k$ given in (1) and (2), we get \begin{equation*}12 + 3F_3 + 4F_4 + 5F_5 + \ldots \leq 6F_3 + 6F_4 + 6F_5 + \ldots.\end{equation*} This gives \begin{equation*}12+F_7+2F_8+3F_9+\ldots \leq 3F_3+2F_4+F_5\end{equation*} so that, finally, $12 \leq 3F_3 + 2F_4 + F_5$. This is the first inequality in (5).

To prove the second inequality in (5), we start with Euler's formula $F-E+V=2$ and write this as $(F-E/2)+(V-E/2) = 2$. Multiplying throughout by $4$, we now get $(4F-2E)+(4V-2E) = 8$. If we now replace $4F-2E$ by their expressions in terms of the $F_i$ in (1) and (2), and $4V-2E$ by the expressions in (3) and (4), we get \begin{equation*}4(F_3 + F_4+\ldots) -(3F_3+4F_4+\ldots) +4(V_3 + V_4+\ldots) -(3V_3+4V_4+\ldots) =8.\end{equation*}

Simplifying this we see that $F_3+V_3 \geq 8$ which is the second inequality in (5).

For the cube, we have $F_3 = 0$ and $V_3=8$, so that $F_3+V_3=8$. As $F_3+V_3\geq 8$ in all cases, $8$ is the smallest possible value of $F_3+V_3$. This leads to the question of finding other polyhedra that give other values of $F_3$ and $V_3$ with $F_3+V_3=8$ (that is, its smallest value). We leave this as an exercise for the reader, and suggest that you try the regular polyhedra first.

The inequalities (5) give us a lot of information about regular polyhedra. Suppose that we have a regular polyhedron. Because $F_3+F_4+F_5 \geq 4$, there must be at least four faces which are triangular, quadrilateral, or pentagonal. As the polyhedron is regular, all faces have the same number of sides, and so all are triangles, all are squares, or all are pentagons.

In the case, of squares and pentagons, $F_3=0$ so that $V_3\geq 8$. So, by regularity, every vertex has exactly three edges leaving it. These polyhedra are the cube and the dodecahedron.

It remains to consider the cases when every face is trianglular. Suppose that there are $q$ edges that leave each vertex. Then $F-E+V=2$, $2E = 3F$ and $2E =qV$. If we eliminate $F$ and $V$ from these equations we get $(6-q)E= 6q$ so that $q = 3,4,5$ giving the tetrahedron, octahedron and icosahedron, respectively. We have now listed all possible regular polyhedra.