[math-fun] Platonic regular polyhedra with integer vertices ?
Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices? What are the coordinates? (Here I'm talking about normal square Cartesian coordinates.) Obviously the cube has the trivial embedding with the unit cube.
Only the cube, octahedron and tetrahedron. Tetrahedron can be place in the cube. On Sat, May 25, 2013 at 9:21 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices?
What are the coordinates?
(Here I'm talking about normal square Cartesian coordinates.)
Obviously the cube has the trivial embedding with the unit cube.
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On Sat, May 25, 2013 at 10:34 PM, James Buddenhagen <jbuddenh@gmail.com>wrote:
Only the cube, octahedron and tetrahedron. Tetrahedron can be place in the cube.
This result is attributed to Eugene Ehrhart in the paper http://arxiv.org/pdf/1111.1150.pdf.
Octohedron and tetrahedron can also have integer vertices. I'm fairly sure that the dodecahedron and icosahedron cannot. On Sat, May 25, 2013 at 10:21 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices?
What are the coordinates?
(Here I'm talking about normal square Cartesian coordinates.)
Obviously the cube has the trivial embedding with the unit cube.
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Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center. Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix. This excludes the dodecahedron and icosahedron, and the other three are pretty easy to embed with integer vertices. --Dan On 2013-05-25, at 7:21 PM, Henry Baker wrote:
Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices?
What are the coordinates?
(Here I'm talking about normal square Cartesian coordinates.)
Obviously the cube has the trivial embedding with the unit cube.
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Interesting! Would your proof indicate that the dodecahedron _could_ be embedded in 4D? Or would 4D require a different proof of non-embeddability? At 09:05 PM 5/25/2013, Dan Asimov wrote:
Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center.
Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix.
This excludes the dodecahedron and icosahedron, and the other three are pretty easy to embed with integer vertices.
--Dan
On 2013-05-25, at 7:21 PM, Henry Baker wrote:
Which of the Platonic regular polyhedra can be embedded in 3D with integer vertices?
What are the coordinates?
(Here I'm talking about normal square Cartesian coordinates.)
Obviously the cube has the trivial embedding with the unit cube.
participants (5)
-
Allan Wechsler -
Dan Asimov -
Henry Baker -
James Buddenhagen -
W. Edwin Clark