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Angles subtended by an edge in a regular dodecahedron?
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If I have a regular dodecahedron and construct lines between the center of the dodecahedron and its vertices. How do I calculate the angle between such lines, subtended by an edge?
This picture can probably better explain what I meant
For now, I am thinking of calculating the radius of the circumscribed sphere and then using cosine law to find out the angle. Is there a more straightforward way?
geometry platonic-solids
asked 4 hours ago
Brandon
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up vote
3
down vote
favorite
If I have a regular dodecahedron and construct lines between the center of the dodecahedron and its vertices. How do I calculate the angle between such lines, subtended by an edge?
This picture can probably better explain what I meant
For now, I am thinking of calculating the radius of the circumscribed sphere and then using cosine law to find out the angle. Is there a more straightforward way?
geometry platonic-solids
asked 4 hours ago
Brandon
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If I have a regular dodecahedron and construct lines between the center of the dodecahedron and its vertices. How do I calculate the angle between such lines, subtended by an edge?
This picture can probably better explain what I meant
For now, I am thinking of calculating the radius of the circumscribed sphere and then using cosine law to find out the angle. Is there a more straightforward way?
geometry platonic-solids
asked 4 hours ago
Brandon
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If I have a regular dodecahedron and construct lines between the center of the dodecahedron and its vertices. How do I calculate the angle between such lines, subtended by an edge?
This picture can probably better explain what I meant
For now, I am thinking of calculating the radius of the circumscribed sphere and then using cosine law to find out the angle. Is there a more straightforward way?
geometry platonic-solids
geometry platonic-solids
asked 4 hours ago
Brandon
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Brandon
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
asked 4 hours ago
Brandon
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Brandon
162
asked 4 hours ago
Brandon
162
162
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Brandon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago
add a comment |Â
2
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago
2
2
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago
add a comment |Â
4 Answers
4
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oldest
votes
up vote
1
down vote
The plane passing through two opposite edges of the icosahedron is a bisector of the angle between two adjacent faces. From here you can begin to work with regular pentagonal pyramid.
answered 4 hours ago
Michael Rozenberg
91.7k1584181
add a comment |Â
up vote
1
down vote
If you are willing to believe Wolfram Cloud Sandbox the following code
With[vc = PolyhedronData["Dodecahedron", "VertexCoordinates"],
ArcCos[Dot[vc[[1]], vc[[14]]]/Norm[vc[[1]]]^2] // FullSimplify]
returns the result
$, textrmarcsec(3/sqrt5) = textrmarcsin(2/3),$ which translates to $,approx 0.729727 approx 41.8103^circ.$
answered 3 hours ago
Somos
12.4k11034
add a comment |Â
up vote
1
down vote
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$l_cube = sqrt1^2 + 1^2 -2cdot 1cdot 1cos108^circ = frac1+sqrt 52$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $sqrt 3$ times the side length of the cube.
answered 4 hours ago
peterwhy
11.8k21127
add a comment |Â
up vote
0
down vote
First off, you drew a dodecahedron. "Hedron" means "face" and the thing with the pentagonal faces has 12 faces and 20 vertices, not the other way around.
Think of the dodecahedron as alprojected radially onto the sphere in which it is inscribed. Let $A$ and $B$ be any two adjacent vertices and draw a spherical triangle whose vertex $C$ is the center of a Pentagon's face having $AB$ as one of itscedges.
Angle $A$ measures half the angle at the vertex, which in the spherical projection means it measures $60°$. Ditto for angle $B$. Angle $C$ measures one fifth of a revolution $=72°$. From the angles of the triangle you can get the arcs using the spherical Laws of Cosines [sic; there are two to choose from] and then arc $AB$ measures the central angle you want.
You may also want to repeat this with $AB$ as a face diagonal instead of an edge, and compare your result with the edge of a cube.
answered 4 hours ago
Oscar Lanzi
10.7k11734
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The plane passing through two opposite edges of the icosahedron is a bisector of the angle between two adjacent faces. From here you can begin to work with regular pentagonal pyramid.
answered 4 hours ago
Michael Rozenberg
91.7k1584181
add a comment |Â
up vote
1
down vote
The plane passing through two opposite edges of the icosahedron is a bisector of the angle between two adjacent faces. From here you can begin to work with regular pentagonal pyramid.
answered 4 hours ago
Michael Rozenberg
91.7k1584181
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The plane passing through two opposite edges of the icosahedron is a bisector of the angle between two adjacent faces. From here you can begin to work with regular pentagonal pyramid.
answered 4 hours ago
Michael Rozenberg
91.7k1584181
The plane passing through two opposite edges of the icosahedron is a bisector of the angle between two adjacent faces. From here you can begin to work with regular pentagonal pyramid.
answered 4 hours ago
Michael Rozenberg
91.7k1584181
answered 4 hours ago
Michael Rozenberg
91.7k1584181
answered 4 hours ago
Michael Rozenberg
91.7k1584181
answered 4 hours ago
Michael Rozenberg
91.7k1584181
91.7k1584181
add a comment |Â
add a comment |Â
up vote
1
down vote
If you are willing to believe Wolfram Cloud Sandbox the following code
With[vc = PolyhedronData["Dodecahedron", "VertexCoordinates"],
ArcCos[Dot[vc[[1]], vc[[14]]]/Norm[vc[[1]]]^2] // FullSimplify]
returns the result
$, textrmarcsec(3/sqrt5) = textrmarcsin(2/3),$ which translates to $,approx 0.729727 approx 41.8103^circ.$
answered 3 hours ago
Somos
12.4k11034
add a comment |Â
up vote
1
down vote
If you are willing to believe Wolfram Cloud Sandbox the following code
With[vc = PolyhedronData["Dodecahedron", "VertexCoordinates"],
ArcCos[Dot[vc[[1]], vc[[14]]]/Norm[vc[[1]]]^2] // FullSimplify]
returns the result
$, textrmarcsec(3/sqrt5) = textrmarcsin(2/3),$ which translates to $,approx 0.729727 approx 41.8103^circ.$
answered 3 hours ago
Somos
12.4k11034
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you are willing to believe Wolfram Cloud Sandbox the following code
With[vc = PolyhedronData["Dodecahedron", "VertexCoordinates"],
ArcCos[Dot[vc[[1]], vc[[14]]]/Norm[vc[[1]]]^2] // FullSimplify]
returns the result
$, textrmarcsec(3/sqrt5) = textrmarcsin(2/3),$ which translates to $,approx 0.729727 approx 41.8103^circ.$
answered 3 hours ago
Somos
12.4k11034
If you are willing to believe Wolfram Cloud Sandbox the following code
With[vc = PolyhedronData["Dodecahedron", "VertexCoordinates"],
ArcCos[Dot[vc[[1]], vc[[14]]]/Norm[vc[[1]]]^2] // FullSimplify]
returns the result
$, textrmarcsec(3/sqrt5) = textrmarcsin(2/3),$ which translates to $,approx 0.729727 approx 41.8103^circ.$
answered 3 hours ago
Somos
12.4k11034
edited 3 hours ago
answered 3 hours ago
Somos
12.4k11034
answered 3 hours ago
Somos
12.4k11034
answered 3 hours ago
Somos
12.4k11034
12.4k11034
add a comment |Â
add a comment |Â
up vote
1
down vote
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$l_cube = sqrt1^2 + 1^2 -2cdot 1cdot 1cos108^circ = frac1+sqrt 52$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $sqrt 3$ times the side length of the cube.
answered 4 hours ago
peterwhy
11.8k21127
add a comment |Â
up vote
1
down vote
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$l_cube = sqrt1^2 + 1^2 -2cdot 1cdot 1cos108^circ = frac1+sqrt 52$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $sqrt 3$ times the side length of the cube.
answered 4 hours ago
peterwhy
11.8k21127
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$l_cube = sqrt1^2 + 1^2 -2cdot 1cdot 1cos108^circ = frac1+sqrt 52$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $sqrt 3$ times the side length of the cube.
answered 4 hours ago
peterwhy
11.8k21127
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$l_cube = sqrt1^2 + 1^2 -2cdot 1cdot 1cos108^circ = frac1+sqrt 52$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $sqrt 3$ times the side length of the cube.
answered 4 hours ago
peterwhy
11.8k21127
edited 2 hours ago
answered 4 hours ago
peterwhy
11.8k21127
answered 4 hours ago
peterwhy
11.8k21127
answered 4 hours ago
peterwhy
11.8k21127
11.8k21127
add a comment |Â
add a comment |Â
up vote
0
down vote
First off, you drew a dodecahedron. "Hedron" means "face" and the thing with the pentagonal faces has 12 faces and 20 vertices, not the other way around.
Think of the dodecahedron as alprojected radially onto the sphere in which it is inscribed. Let $A$ and $B$ be any two adjacent vertices and draw a spherical triangle whose vertex $C$ is the center of a Pentagon's face having $AB$ as one of itscedges.
Angle $A$ measures half the angle at the vertex, which in the spherical projection means it measures $60°$. Ditto for angle $B$. Angle $C$ measures one fifth of a revolution $=72°$. From the angles of the triangle you can get the arcs using the spherical Laws of Cosines [sic; there are two to choose from] and then arc $AB$ measures the central angle you want.
You may also want to repeat this with $AB$ as a face diagonal instead of an edge, and compare your result with the edge of a cube.
answered 4 hours ago
Oscar Lanzi
10.7k11734
add a comment |Â
up vote
0
down vote
First off, you drew a dodecahedron. "Hedron" means "face" and the thing with the pentagonal faces has 12 faces and 20 vertices, not the other way around.
Think of the dodecahedron as alprojected radially onto the sphere in which it is inscribed. Let $A$ and $B$ be any two adjacent vertices and draw a spherical triangle whose vertex $C$ is the center of a Pentagon's face having $AB$ as one of itscedges.
Angle $A$ measures half the angle at the vertex, which in the spherical projection means it measures $60°$. Ditto for angle $B$. Angle $C$ measures one fifth of a revolution $=72°$. From the angles of the triangle you can get the arcs using the spherical Laws of Cosines [sic; there are two to choose from] and then arc $AB$ measures the central angle you want.
You may also want to repeat this with $AB$ as a face diagonal instead of an edge, and compare your result with the edge of a cube.
answered 4 hours ago
Oscar Lanzi
10.7k11734
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First off, you drew a dodecahedron. "Hedron" means "face" and the thing with the pentagonal faces has 12 faces and 20 vertices, not the other way around.
Think of the dodecahedron as alprojected radially onto the sphere in which it is inscribed. Let $A$ and $B$ be any two adjacent vertices and draw a spherical triangle whose vertex $C$ is the center of a Pentagon's face having $AB$ as one of itscedges.
Angle $A$ measures half the angle at the vertex, which in the spherical projection means it measures $60°$. Ditto for angle $B$. Angle $C$ measures one fifth of a revolution $=72°$. From the angles of the triangle you can get the arcs using the spherical Laws of Cosines [sic; there are two to choose from] and then arc $AB$ measures the central angle you want.
You may also want to repeat this with $AB$ as a face diagonal instead of an edge, and compare your result with the edge of a cube.
answered 4 hours ago
Oscar Lanzi
10.7k11734
First off, you drew a dodecahedron. "Hedron" means "face" and the thing with the pentagonal faces has 12 faces and 20 vertices, not the other way around.
Think of the dodecahedron as alprojected radially onto the sphere in which it is inscribed. Let $A$ and $B$ be any two adjacent vertices and draw a spherical triangle whose vertex $C$ is the center of a Pentagon's face having $AB$ as one of itscedges.
Angle $A$ measures half the angle at the vertex, which in the spherical projection means it measures $60°$. Ditto for angle $B$. Angle $C$ measures one fifth of a revolution $=72°$. From the angles of the triangle you can get the arcs using the spherical Laws of Cosines [sic; there are two to choose from] and then arc $AB$ measures the central angle you want.
You may also want to repeat this with $AB$ as a face diagonal instead of an edge, and compare your result with the edge of a cube.
answered 4 hours ago
Oscar Lanzi
10.7k11734
answered 4 hours ago
Oscar Lanzi
10.7k11734
answered 4 hours ago
Oscar Lanzi
10.7k11734
answered 4 hours ago
Oscar Lanzi
10.7k11734
10.7k11734
add a comment |Â
add a comment |Â
Brandon is a new contributor. Be nice, and check out our Code of Conduct.
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2
That's a dodecahedron!
– Lord Shark the Unknown
4 hours ago
Oh right thanks! edited
– Brandon
4 hours ago