Smallest polyhedron with an odd number of all n-gon faces

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So this is a question I actually "know" the answer to (in the sense that I know broadly what the answer is, but I'm missing crucial details and expertise to find an exact unique answer).



A long while ago, someone posed this question to me, and I found that the smallest example of a polyhedron with an odd number of faces where each face is an n-gon is an enneahedron with 4-gon faces. I don't remember where I found this particular assertion but I suspect it might have been on OEIS or something.



At the time I had hunted down a page that had a full explanation and image of this object, but unfortunately it is now a 404.



The motivation for this question is I recently got into a discussion about different rhombuses that appear in various geometrical shapes, and I thought I remembered the 4-gons in this object being rhombuses (however, I might be completely mistaken), and I was wondering after their ratio if they were.




polyhedra




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  • Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
    – David K
    Jul 27 at 4:21










  • The requirement was not for congruent n-gons.
    – OmnipotentEntity
    Jul 27 at 4:29










  • That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
    – OmnipotentEntity
    Jul 27 at 4:52






  • 1




    A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
    – David K
    Jul 28 at 14:46










  • I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
    – David K
    Jul 28 at 14:49














up vote
1
down vote

favorite












So this is a question I actually "know" the answer to (in the sense that I know broadly what the answer is, but I'm missing crucial details and expertise to find an exact unique answer).



A long while ago, someone posed this question to me, and I found that the smallest example of a polyhedron with an odd number of faces where each face is an n-gon is an enneahedron with 4-gon faces. I don't remember where I found this particular assertion but I suspect it might have been on OEIS or something.



At the time I had hunted down a page that had a full explanation and image of this object, but unfortunately it is now a 404.



The motivation for this question is I recently got into a discussion about different rhombuses that appear in various geometrical shapes, and I thought I remembered the 4-gons in this object being rhombuses (however, I might be completely mistaken), and I was wondering after their ratio if they were.




polyhedra




share|cite|improve this question





















  • Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
    – David K
    Jul 27 at 4:21










  • The requirement was not for congruent n-gons.
    – OmnipotentEntity
    Jul 27 at 4:29










  • That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
    – OmnipotentEntity
    Jul 27 at 4:52






  • 1




    A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
    – David K
    Jul 28 at 14:46










  • I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
    – David K
    Jul 28 at 14:49












up vote
1
down vote

favorite









up vote
1
down vote

favorite











So this is a question I actually "know" the answer to (in the sense that I know broadly what the answer is, but I'm missing crucial details and expertise to find an exact unique answer).



A long while ago, someone posed this question to me, and I found that the smallest example of a polyhedron with an odd number of faces where each face is an n-gon is an enneahedron with 4-gon faces. I don't remember where I found this particular assertion but I suspect it might have been on OEIS or something.



At the time I had hunted down a page that had a full explanation and image of this object, but unfortunately it is now a 404.



The motivation for this question is I recently got into a discussion about different rhombuses that appear in various geometrical shapes, and I thought I remembered the 4-gons in this object being rhombuses (however, I might be completely mistaken), and I was wondering after their ratio if they were.




polyhedra




share|cite|improve this question













So this is a question I actually "know" the answer to (in the sense that I know broadly what the answer is, but I'm missing crucial details and expertise to find an exact unique answer).



A long while ago, someone posed this question to me, and I found that the smallest example of a polyhedron with an odd number of faces where each face is an n-gon is an enneahedron with 4-gon faces. I don't remember where I found this particular assertion but I suspect it might have been on OEIS or something.



At the time I had hunted down a page that had a full explanation and image of this object, but unfortunately it is now a 404.



The motivation for this question is I recently got into a discussion about different rhombuses that appear in various geometrical shapes, and I thought I remembered the 4-gons in this object being rhombuses (however, I might be completely mistaken), and I was wondering after their ratio if they were.





polyhedra





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 4:15
























asked Jul 27 at 4:08









OmnipotentEntity

391216




391216











  • Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
    – David K
    Jul 27 at 4:21










  • The requirement was not for congruent n-gons.
    – OmnipotentEntity
    Jul 27 at 4:29










  • That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
    – OmnipotentEntity
    Jul 27 at 4:52






  • 1




    A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
    – David K
    Jul 28 at 14:46










  • I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
    – David K
    Jul 28 at 14:49
















  • Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
    – David K
    Jul 27 at 4:21










  • The requirement was not for congruent n-gons.
    – OmnipotentEntity
    Jul 27 at 4:29










  • That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
    – OmnipotentEntity
    Jul 27 at 4:52






  • 1




    A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
    – David K
    Jul 28 at 14:46










  • I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
    – David K
    Jul 28 at 14:49















Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
– David K
Jul 27 at 4:21




Do the faces all have to be the same $n$-gon? (Same sequence of side lengths and angles, not just same number of sides.)
– David K
Jul 27 at 4:21












The requirement was not for congruent n-gons.
– OmnipotentEntity
Jul 27 at 4:29




The requirement was not for congruent n-gons.
– OmnipotentEntity
Jul 27 at 4:29












That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
– OmnipotentEntity
Jul 27 at 4:52




That being said. I don't recall whether or not they actually were congruent or not. I thought they were, but I might be misremembering.
– OmnipotentEntity
Jul 27 at 4:52




1




1




A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
– David K
Jul 28 at 14:46




A more general question whose answer I think applies here too: math.stackexchange.com/questions/2153505/…
– David K
Jul 28 at 14:46












I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
– David K
Jul 28 at 14:49




I notice, however, that the question I linked does not prove the non-existence of a quadrilateral-faced heptahedron.
– David K
Jul 28 at 14:49










2 Answers
2






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1
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Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
$$2sum e_i=sum n_i,f_i$$
Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.



So indeed, the smallest possible value of $n$ within your quest would be 4.



--- rk






share|cite|improve this answer




























    up vote
    1
    down vote













    A partial answer:

    To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:



    Start with Euler's formula $V+F=E+2$. If there are $k$ faces of $n$ sides each this becomes $V+k=frac nk2+2$ because each edge is counted on two faces. If $k$ is odd $n$ must be even to make the fraction integral.



    The total of all the angles at all the vertices must be $4pi$ less than $2pi V$ to make it close. If the faces are $4$ sided we must have two less faces than vertices. As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $frac pi 2$ angle deficit, so there must be at least $9$ vertices. Finally one has to show that there is no solution with $9$ vertices and $7 4$ sided faces. I don't have an approach for that part.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      up vote
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      down vote



      accepted










      Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
      $$2sum e_i=sum n_i,f_i$$
      Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.



      So indeed, the smallest possible value of $n$ within your quest would be 4.



      --- rk






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
        $$2sum e_i=sum n_i,f_i$$
        Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.



        So indeed, the smallest possible value of $n$ within your quest would be 4.



        --- rk






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
          $$2sum e_i=sum n_i,f_i$$
          Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.



          So indeed, the smallest possible value of $n$ within your quest would be 4.



          --- rk






          share|cite|improve this answer













          Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
          $$2sum e_i=sum n_i,f_i$$
          Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.



          So indeed, the smallest possible value of $n$ within your quest would be 4.



          --- rk







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 27 at 7:22









          Dr. Richard Klitzing

          7286




          7286




















              up vote
              1
              down vote













              A partial answer:

              To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:



              Start with Euler's formula $V+F=E+2$. If there are $k$ faces of $n$ sides each this becomes $V+k=frac nk2+2$ because each edge is counted on two faces. If $k$ is odd $n$ must be even to make the fraction integral.



              The total of all the angles at all the vertices must be $4pi$ less than $2pi V$ to make it close. If the faces are $4$ sided we must have two less faces than vertices. As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $frac pi 2$ angle deficit, so there must be at least $9$ vertices. Finally one has to show that there is no solution with $9$ vertices and $7 4$ sided faces. I don't have an approach for that part.






              share|cite|improve this answer

























                up vote
                1
                down vote













                A partial answer:

                To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:



                Start with Euler's formula $V+F=E+2$. If there are $k$ faces of $n$ sides each this becomes $V+k=frac nk2+2$ because each edge is counted on two faces. If $k$ is odd $n$ must be even to make the fraction integral.



                The total of all the angles at all the vertices must be $4pi$ less than $2pi V$ to make it close. If the faces are $4$ sided we must have two less faces than vertices. As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $frac pi 2$ angle deficit, so there must be at least $9$ vertices. Finally one has to show that there is no solution with $9$ vertices and $7 4$ sided faces. I don't have an approach for that part.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  A partial answer:

                  To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:



                  Start with Euler's formula $V+F=E+2$. If there are $k$ faces of $n$ sides each this becomes $V+k=frac nk2+2$ because each edge is counted on two faces. If $k$ is odd $n$ must be even to make the fraction integral.



                  The total of all the angles at all the vertices must be $4pi$ less than $2pi V$ to make it close. If the faces are $4$ sided we must have two less faces than vertices. As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $frac pi 2$ angle deficit, so there must be at least $9$ vertices. Finally one has to show that there is no solution with $9$ vertices and $7 4$ sided faces. I don't have an approach for that part.






                  share|cite|improve this answer













                  A partial answer:

                  To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:



                  Start with Euler's formula $V+F=E+2$. If there are $k$ faces of $n$ sides each this becomes $V+k=frac nk2+2$ because each edge is counted on two faces. If $k$ is odd $n$ must be even to make the fraction integral.



                  The total of all the angles at all the vertices must be $4pi$ less than $2pi V$ to make it close. If the faces are $4$ sided we must have two less faces than vertices. As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $frac pi 2$ angle deficit, so there must be at least $9$ vertices. Finally one has to show that there is no solution with $9$ vertices and $7 4$ sided faces. I don't have an approach for that part.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 27 at 15:43









                  Ross Millikan

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