Volume relation between regular octahedron & tetrahedron: visual proof​
by Sándor Kabai and Dan Suttin​
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Please find related demonstration here​
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This Demonstration shows a visual proof that the volume of the regular octahedron is four times that of the regular tetrahedron through decomposition . The large octahedron has a side that is twice the length of any of the small octahedra . So the volume of the large octahedron is eight times as much as a small one . But the large octahedron is made of six small octahedra and eight tetrahedra . So the eight tetrahedra must have a volume equal to two small octahedra, and the ratio is 4 to 1.
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Briefly: 8 tet + 6 oct = 8 oct, rearrange the equation and we get oct/tet = 4.
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Manipulate[Module[{oct,octv,tet,t0,t4a,t4b,o1,t1,oct2},oct=GraphicsComplex[PolyhedronData["Octahedron","VertexCoordinates"],Polygon[PolyhedronData["Octahedron","FaceIndices"]]];octv=PolyhedronData["Octahedron","VertexCoordinates"];tet=GraphicsComplex[PolyhedronData["Tetrahedron","VertexCoordinates"],Polygon[PolyhedronData["Tetrahedron","FaceIndices"]]];t0=Translate[Rotate[tet,-ArcTan[Sqrt[2]],{0,1,0}],{0,(-dis2)*0.5,dis2*(0.5/Sqrt[2])}];t4a=Table[Rotate[t0,i*(Pi/2)+Pi/4,{0,0,1}],{i,4}];t4b=Rotate[t4a,Pi,{0,1,0}];o1=(Translate[oct,#1]&)/@(dis1*octv);t1=Scale[{t4a,t4b},1*{1,1,1},{0,0,0}];oct2=Scale[oct,0.99*{2,2,2}];​​​​Graphics3D[{If[oct2s,oct2,{}],RGBColor[0,1,1],o1,Yellow,t1},SphericalRegion->True,Boxed->False,ViewAngle->0.205,ViewPoint->{4,-2,1},ImageSize->400]],{{dis1,3,"move octahedra"},1,3,Appearance->"Labeled"},{{dis2,1,"move tetrahedra"},1,3,Appearance->"Labeled"},{{oct2s,False,"show octahedron"},{False,True},ImageSize->Tiny},TrackedSymbols->{dis2,dis1,oct2s},SaveDefinitionsTrue]
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move octahedra
3.
move tetrahedra
1.
show octahedron