![]() | Morphing Polyhedron Applet (below) doesn't work even though you installed Java? |
Run the Morphing Polyhedron Web-Start Application instead. | |
This downloads a jnlp (Java Web Start) file that tells Java how to run the Morphing Polyhedron outside of your browser. | |
See my Java Web Start notes. |
This is a Java 1.1 applet. See my Java page for Browser Requirements.
Initially (or after clicking the Reset button) you see a solid Icosahedron inscribed in a wireframe Octahedron. The vertices of the Icosahedron lie on the edges of the Octahedron.
You can morph the polyhedron by adjusting the slider control.
You can rotate the image by dragging with the mouse.
The "text box" above the slider shows its setting, which varies from 0.0 to 1.0 in steps of 0.001. (The slider is a "native" control, which is known to have peculiar behaviors on certain platforms -- you may not be able to adjust it all the way to 1.0.)
By adjusting the slider, you can see these polyhedra:
Polyhedron | Slider Setting | Exact |
Octahedron | 0.0 | 0 |
Icosahedron | 0.382 | 1/(1+G) |
Cuboctahedron | 0.5 | 1/2 |
Regular Dodecahedron | 0.618 | G/(1+G) |
Rhombic Dodecahedron | 0.667 | 2/3 |
Regular Dodecahedron | 0.724 | (1+G)/(2+G) |
Tiny cube | Approaching 1.0 | Approaching 1.0 |
Where G is the golden ratio: (1+sqrt(5))/2.
On each of the 8 faces of the Octahedron, we connect the three selected edge-points to form an equilateral triangle (green).
Twelve isoceles triangles (blue) are made by connecting edge-points from adjacent Octrahedron faces (two blue triangles under each of the 6 vertices).
The total number of triangles is 20 (8 green equilateral and 12 blue isoceles).
For a particular slider setting (approximately 0.382), the blue triangles become equilateral (and thus equal to the green triangles) and an icosahedron is formed.
At this particular slider setting, the edge is divided into a larger and smaller part whose lengths are related by the Golden Ratio.
This method of constructing the icosahedron is useful because it provides a quick way to calculate
icosahedron vertex coordinates if you know octahedron vertex coordinates.
Octahedron vertex coordinates are easy to write down. Here are vertices for an octahedron of edge length
sqrt(2):
( 1, 0, 0)
(-1, 0, 0)
( 0, 1, 0)
( 0,-1, 0)
( 0, 0, 1)
( 0, 0,-1)
If the slider is set to 0.5, the blue triangles become right-angled, and pairs of them merge to form the six square faces of a Cuboctahedron.
Understanding what is happening here is a little tricky. I have to explain what the applet is actually doing. It uses a method of "slicing" that is a spin-off from my Hyperspace Polytope Slicer. The blue surfaces are made by slicing the Octahedron with infinite planes. These planes are defined by triplets of selected edge-points, as described above.
For slider settings <0.5 (described above), the effect of the slicing is easy to see because the slices create the blue triangles whose corners are the selected edge-points that define the slicing plane. Each pair of slicing planes meets in a raised edge under an Octahedron vertex.
When the slider setting reaches 0.5, that edge is no longer raised. In fact there is no edge because the pair of slicing planes coincide.
As the slider setting is increased above 0.5, the edge where the planes meet is depressed rather than raised. The two planes undercut each other and the solid figure detaches from the Octahedron edges. You can see this most clearly if you rotate the figure so that you are looking directly down on a vertex of the Octahedron.
Note that when the solid figure has detached from the Octahedron edges, the applet draws orange lines on the Octahedron surface to indicate the locations of the selected points.
Coxeter credits P. Schönemann for the method (1873). (Schonemann01)