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A cuboctahedron is an Archimedean solid. It is generated by truncating the vertices of a cube or of an octahedron at 1/2 edge-length. There are 6 square faces on the cuboctahedron, one for each face of the cube. There are 8 equilateral triangular faces, one for each vertex of the cube. We are going to calculate the volume of an octahedron of edge-length 1 starting from the volume of a cube. If a cuboctahedron has edge-length 1, the cube that contains it is:
The volume of this cube is: To calculate the volume of a cuboctahedron we have to subtract from the volume of the cube the volume of the 8 pyramids that we cut off. The volume of each of these 8 pyramids is:
Now, we can calculate the volume of the cuboctahedron (what we subtract, 8 pyramids, may be reasssembled into an octahedron of edge-length 1)
Then the volume of a cuboctahedron of edge-length a is: Now we are going to see one interesting property of the cuboctahedron. If we think in this cube as made for eight small cubes (with the center of the cube as a vertex shared by the small cubes) we can see that the distance from the center of the cuboctahedron (its center of gravity) to any vertex is the edge length (it is equal to a diagonal of a side of one small cube).
Then a cuboctahedron is made of six half octahedra and eight tetrahedra, all of these pyramids sharing one vertex in the center of gravity of the cuboctahedron. The cuboctahedron is the only spatial configuration in wich the length of each polyhedral edge is equal to that the distance from its center of gravity to any vertex. More than that, its 24 edges lies in four hexagons centered in the center of gravity of the cuboctahedron. In this origami cuboctahedron, can you see the hexagon (paralel to the floor)?
I made this origami cuboctahdron following the instructions in Tomoko Fusè's book 'Unit Origami' (Japan Publications, Inc. 1990)
Can you see here four hexagons?
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