Does there exist a higher-dimensional 5-sided “tetrahedron + 1”?












3












$begingroup$


The first shape is "0"-sided and is a point. The next shape is a line segment and it's "1"-sided. The next shape is a triangle and it's 3-sided. The next shape is a tetrahedron and it's 4-sided. Can we define some higher-dimensional 5-sided regular shape which is the next shape in this sequence?



You can reach the next shape by placing a point equidistant to all previous points in the next higher dimension.










share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    In general, this is called a "simplex" or "$n$-simplex".
    $endgroup$
    – Blue
    Dec 31 '18 at 19:40








  • 2




    $begingroup$
    @Blue thanks. I will read that
    $endgroup$
    – I Said Roll Up n Smoke Adjoint
    Dec 31 '18 at 19:43






  • 3




    $begingroup$
    In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
    $endgroup$
    – Rahul
    Dec 31 '18 at 20:02






  • 2




    $begingroup$
    The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 23:17
















3












$begingroup$


The first shape is "0"-sided and is a point. The next shape is a line segment and it's "1"-sided. The next shape is a triangle and it's 3-sided. The next shape is a tetrahedron and it's 4-sided. Can we define some higher-dimensional 5-sided regular shape which is the next shape in this sequence?



You can reach the next shape by placing a point equidistant to all previous points in the next higher dimension.










share|cite|improve this question











$endgroup$








  • 8




    $begingroup$
    In general, this is called a "simplex" or "$n$-simplex".
    $endgroup$
    – Blue
    Dec 31 '18 at 19:40








  • 2




    $begingroup$
    @Blue thanks. I will read that
    $endgroup$
    – I Said Roll Up n Smoke Adjoint
    Dec 31 '18 at 19:43






  • 3




    $begingroup$
    In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
    $endgroup$
    – Rahul
    Dec 31 '18 at 20:02






  • 2




    $begingroup$
    The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 23:17














3












3








3





$begingroup$


The first shape is "0"-sided and is a point. The next shape is a line segment and it's "1"-sided. The next shape is a triangle and it's 3-sided. The next shape is a tetrahedron and it's 4-sided. Can we define some higher-dimensional 5-sided regular shape which is the next shape in this sequence?



You can reach the next shape by placing a point equidistant to all previous points in the next higher dimension.










share|cite|improve this question











$endgroup$




The first shape is "0"-sided and is a point. The next shape is a line segment and it's "1"-sided. The next shape is a triangle and it's 3-sided. The next shape is a tetrahedron and it's 4-sided. Can we define some higher-dimensional 5-sided regular shape which is the next shape in this sequence?



You can reach the next shape by placing a point equidistant to all previous points in the next higher dimension.







general-topology polyhedra natural-numbers polytopes simplicial-stuff






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 19:40







I Said Roll Up n Smoke Adjoint

















asked Dec 31 '18 at 19:37









I Said Roll Up n Smoke AdjointI Said Roll Up n Smoke Adjoint

9,33252659




9,33252659








  • 8




    $begingroup$
    In general, this is called a "simplex" or "$n$-simplex".
    $endgroup$
    – Blue
    Dec 31 '18 at 19:40








  • 2




    $begingroup$
    @Blue thanks. I will read that
    $endgroup$
    – I Said Roll Up n Smoke Adjoint
    Dec 31 '18 at 19:43






  • 3




    $begingroup$
    In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
    $endgroup$
    – Rahul
    Dec 31 '18 at 20:02






  • 2




    $begingroup$
    The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 23:17














  • 8




    $begingroup$
    In general, this is called a "simplex" or "$n$-simplex".
    $endgroup$
    – Blue
    Dec 31 '18 at 19:40








  • 2




    $begingroup$
    @Blue thanks. I will read that
    $endgroup$
    – I Said Roll Up n Smoke Adjoint
    Dec 31 '18 at 19:43






  • 3




    $begingroup$
    In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
    $endgroup$
    – Rahul
    Dec 31 '18 at 20:02






  • 2




    $begingroup$
    The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
    $endgroup$
    – Zachary Hunter
    Dec 31 '18 at 23:17








8




8




$begingroup$
In general, this is called a "simplex" or "$n$-simplex".
$endgroup$
– Blue
Dec 31 '18 at 19:40






$begingroup$
In general, this is called a "simplex" or "$n$-simplex".
$endgroup$
– Blue
Dec 31 '18 at 19:40






2




2




$begingroup$
@Blue thanks. I will read that
$endgroup$
– I Said Roll Up n Smoke Adjoint
Dec 31 '18 at 19:43




$begingroup$
@Blue thanks. I will read that
$endgroup$
– I Said Roll Up n Smoke Adjoint
Dec 31 '18 at 19:43




3




3




$begingroup$
In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
$endgroup$
– Rahul
Dec 31 '18 at 20:02




$begingroup$
In particular, see 5-cell. Here the 5 "sides" are tetrahedra, just like the 4 sides of a tetrahedron are triangles, the 3 sides of a triangle are line segments, and the 2 "sides" (or rather, ends) of a line segment are points.
$endgroup$
– Rahul
Dec 31 '18 at 20:02




2




2




$begingroup$
The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
$endgroup$
– Zachary Hunter
Dec 31 '18 at 23:17




$begingroup$
The 4D 24-cell though not a simplex, shares simplex-like features, being a regular polytope which is self dual.
$endgroup$
– Zachary Hunter
Dec 31 '18 at 23:17










1 Answer
1






active

oldest

votes


















3












$begingroup$

The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.



This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.



In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.



The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being
$$r=sqrt{frac{D}{2(D+1)}},$$
thus prooving its existance for every dimension.



--- rk






share|cite|improve this answer









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    3












    $begingroup$

    The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.



    This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.



    In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.



    The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being
    $$r=sqrt{frac{D}{2(D+1)}},$$
    thus prooving its existance for every dimension.



    --- rk






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.



      This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.



      In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.



      The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being
      $$r=sqrt{frac{D}{2(D+1)}},$$
      thus prooving its existance for every dimension.



      --- rk






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.



        This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.



        In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.



        The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being
        $$r=sqrt{frac{D}{2(D+1)}},$$
        thus prooving its existance for every dimension.



        --- rk






        share|cite|improve this answer









        $endgroup$



        The simplex is the easiest polytope, existing within all dimensions. It is nothing but the pyramid on a simplicial base (of one dimension less). In fact, take any simplex, attach to every facet a copy of that simplex again, and fold those copies up into the next dimension, then those will all meet at a single point atop the central simplex, which then will become its base.



        This very construction already shows by induction, that the D-dimensional simplex would have D+1 facets, which all are D-1 dimensional simplices. Thus I can be affirmative, the 4th dimensional simplex is a pentachoron, i.e. it is built from 5 cells, all being tetrahedra.



        In fact, the element count of all subelements of any dimension would be given by the numbers of the Pascal triangle: eg. a triangle has 3 vertices and 3 sides; a tetrahedron has 4 vertices, 6 edges, and 4 faces; a pentachoron has 5 vertices, 10 edges, 10 triangles, and 5 tetrahedra; etc.



        The constructive device given above surely requires that the circumradius of the lower dimensional simplex is less than one edge unit, in order to allow to build an unit-edged pyramid on top. Again by induction, using right that very construction of a D-dimensional simplex, you could derive that crucial circumradius formula being
        $$r=sqrt{frac{D}{2(D+1)}},$$
        thus prooving its existance for every dimension.



        --- rk







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 3:14









        Dr. Richard KlitzingDr. Richard Klitzing

        1,79526




        1,79526






























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