[math-fun] In[569]:= vertexSolidAngles["RhombicTriacontahedron"] // FullSimplify //Tally // tim
During evaluation of In[569]:= 8.60579 secs, 2 components Out[569]= {{π, 12}, {3π/5, 20}} I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π. Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg
<< Question 2: How many vertices, edges, and faces has the Stella Octangula? >> I'll buy it: why isn't the answer just 8, 12, 8 (two tetrahedra) ? WFL On 12/20/17, Bill Gosper <billgosper@gmail.com> wrote:
During evaluation of In[569]:= 8.60579 secs, 2 components
Out[569]= {{π, 12}, {3π/5, 20}}
I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π.
Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-------- Original Message -------- Subject: Re: [math-fun] In[569]:= vertexSolidAngles["RhombicTriacontahedron"] // FullSimplify //Tally // tim Date: 2017-12-20 13:40 From: Fred Lunnon <fred.lunnon@gmail.com> << Question 2: How many vertices, edges, and faces has the Stella Octangula? >> I'll buy it: why isn't the answer just 8, 12, 8 (two tetrahedra) ? WFL Mathematica agrees: In[603]:= PolyhedronData["StellaOctangula", "Vertices"] // Length Out[603]= 8 In[604]:= PolyhedronData["StellaOctangula", "Edges"] Out[604]= {{1, 3}, {1, 6}, {1, 7}, {2, 4}, {2, 5}, {2, 8}, {3, 6}, {3, 7}, {4, 5}, {4, 8}, {5, 8}, {6, 7}} In[592]:= PolyhedronData["StellaOctangula", "Faces"] Out[592]= {{4, 5, 8}, {5, 4, 2}, {8, 2, 4}, {2, 8, 5}, {6, 7, 3}, {7, 6, 1}, {3, 1, 6}, {1, 3, 7}} But that has interior boundaries. If it's really two tetrahedra, why isn't its volume twice the tetrahedron's? What do we call the union of an octahedron and 8 tetrahedra? https://en.wikipedia.org/wiki/Stellated_octahedron equates the two. But they're different. And what are the V and E counts of the stellate? (F=24.) --rwg On 12/20/17, Bill Gosper <billgosper@gmail.com> wrote:
During evaluation of In[569]:= 8.60579 secs, 2 components
Out[569]= {{π, 12}, {3π/5, 20}}
I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π.
Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg
<redundance>
<< What do we call the union of an octahedron and 8 tetrahedra? >> It does at first sight seem reasonable --- on historical grounds, perhaps? --- to define "stella octangula" thus. Unfortunately, an octangle with 14 vertices might rather be reclassified "stella oxymoronic" ... << https://en.wikipedia.org/wiki/Stellated_octahedron equates the two. But they're different. >> I can recall (only just!) being quite concerned about such questions when first encountering polyhedra as a teenager. Traditional discussion of stellated polytopes blithely assumes that they inhabit some kind of exotic branched space where interior regions have multiply-weighted content, without ever bothering to define explicitly an appropriate topological structure! Point made. WFL On 12/21/17, Bill Gosper <billgosper@gmail.com> wrote:
-------- Original Message -------- Subject: Re: [math-fun] In[569]:= vertexSolidAngles["RhombicTriacontahedron"] // FullSimplify //Tally // tim Date: 2017-12-20 13:40 From: Fred Lunnon <fred.lunnon@gmail.com>
<< Question 2: How many vertices, edges, and faces has the Stella Octangula? >>
I'll buy it: why isn't the answer just 8, 12, 8 (two tetrahedra) ? WFL
Mathematica agrees:
In[603]:= PolyhedronData["StellaOctangula", "Vertices"] // Length
Out[603]= 8
In[604]:= PolyhedronData["StellaOctangula", "Edges"]
Out[604]= {{1, 3}, {1, 6}, {1, 7}, {2, 4}, {2, 5}, {2, 8}, {3, 6}, {3, 7}, {4, 5}, {4, 8}, {5, 8}, {6, 7}}
In[592]:= PolyhedronData["StellaOctangula", "Faces"]
Out[592]= {{4, 5, 8}, {5, 4, 2}, {8, 2, 4}, {2, 8, 5}, {6, 7, 3}, {7, 6, 1}, {3, 1, 6}, {1, 3, 7}}
But that has interior boundaries. If it's really two tetrahedra, why isn't its volume twice the tetrahedron's?
What do we call the union of an octahedron and 8 tetrahedra?
https://en.wikipedia.org/wiki/Stellated_octahedron equates the two. But they're different.
And what are the V and E counts of the stellate? (F=24.) --rwg
On 12/20/17, Bill Gosper <billgosper@gmail.com> wrote:
During evaluation of In[569]:= 8.60579 secs, 2 components
Out[569]= {{π, 12}, {3π/5, 20}}
I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π.
Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg
<redundance> _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Has anyone seen a treatment of stellated polyhedra that accurately describes their structured as branched spaces? Jim Propp On Wed, Dec 20, 2017 at 10:19 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< What do we call the union of an octahedron and 8 tetrahedra? >>
It does at first sight seem reasonable --- on historical grounds, perhaps? --- to define "stella octangula" thus. Unfortunately, an octangle with 14 vertices might rather be reclassified "stella oxymoronic" ...
<< https://en.wikipedia.org/wiki/Stellated_octahedron equates the two. But they're different. >>
I can recall (only just!) being quite concerned about such questions when first encountering polyhedra as a teenager.
Traditional discussion of stellated polytopes blithely assumes that they inhabit some kind of exotic branched space where interior regions have multiply-weighted content, without ever bothering to define explicitly an appropriate topological structure!
Point made. WFL
On 12/21/17, Bill Gosper <billgosper@gmail.com> wrote:
-------- Original Message -------- Subject: Re: [math-fun] In[569]:= vertexSolidAngles["RhombicTriacontahedron"] // FullSimplify //Tally // tim Date: 2017-12-20 13:40 From: Fred Lunnon <fred.lunnon@gmail.com>
<< Question 2: How many vertices, edges, and faces has the Stella Octangula? >>
I'll buy it: why isn't the answer just 8, 12, 8 (two tetrahedra) ? WFL
Mathematica agrees:
In[603]:= PolyhedronData["StellaOctangula", "Vertices"] // Length
Out[603]= 8
In[604]:= PolyhedronData["StellaOctangula", "Edges"]
Out[604]= {{1, 3}, {1, 6}, {1, 7}, {2, 4}, {2, 5}, {2, 8}, {3, 6}, {3, 7}, {4, 5}, {4, 8}, {5, 8}, {6, 7}}
In[592]:= PolyhedronData["StellaOctangula", "Faces"]
Out[592]= {{4, 5, 8}, {5, 4, 2}, {8, 2, 4}, {2, 8, 5}, {6, 7, 3}, {7, 6, 1}, {3, 1, 6}, {1, 3, 7}}
But that has interior boundaries. If it's really two tetrahedra, why isn't its volume twice the tetrahedron's?
What do we call the union of an octahedron and 8 tetrahedra?
https://en.wikipedia.org/wiki/Stellated_octahedron equates the two. But they're different.
And what are the V and E counts of the stellate? (F=24.) --rwg
On 12/20/17, Bill Gosper <billgosper@gmail.com> wrote:
During evaluation of In[569]:= 8.60579 secs, 2 components
Out[569]= {{π, 12}, {3π/5, 20}}
I.e., 12 vertices have solid angle π. True or false: Four RTs with disjoint interiors can intersect at a point and fill a neighborhood of it. Can five with disjoint interiors intersect at one point? Six?? 6×⅗π < 4π.
Question 2: How many vertices, edges, and faces has the Stella Octangula? --rwg
<redundance> _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
-
Bill Gosper -
Fred Lunnon -
James Propp