正十二面体变成五边形柱

hbghlyj

在 math.stackexchange.com/q/5040298 的回答中，

如何作出？

hbghlyj

中间的斜的棱长设为 $x\in[1,\sqrt2]$，其余未变的棱长设为1，解不出精确表达式

1. A1={r,0,z};
2. A2=RotationTransform[2 Pi/5,{0,0,1}][A1];
3. B1=RotationTransform[t,{0,0,1}][{r,0,-z}];
4. B2=RotationTransform[2 Pi/5,{0,0,1}][B1];
5. C1={1/(2 Sin[Pi/5]),0,Sqrt[1-(r-1/(2 Sin[Pi/5]))^2]+z};
6. C2=RotationTransform[2 Pi/5,{0,0,1}][C1];
7. D1=RotationTransform[t,{0,0,1}][{1/(2 Sin[Pi/5]),0,-Sqrt[1-(r-1/(2 Sin[Pi/5]))^2]-z}];
8. {r0,t0,z0}={Cot[Pi/5],-\[Pi]/5,Sqrt[(5-Sqrt[5])/10]/2};
9. Export["output.gif",Table[{r0,t0,z0}={r,t,z}/.FindRoot[{Total[(A1-B2)^2]-x^2,Total[(A1-B1)^2]-1,Det[{A1-B2,A2-B2,C1-A1}]},{{r,r0},{t,t0},{z,z0+MachinePrecision}}];
10. Graphics3D[{Opacity[0.5],GraphicsComplex[{RotationTransform[2Pi/5,{0,0,1}][D1],RotationTransform[4 \[Pi]/5,{0,0,1}][C2],RotationTransform[2 \[Pi]/5,{0,0,1}][B2],B1,RotationTransform[2 \[Pi]/5,{0,0,1}][C2],RotationTransform[6 \[Pi]/5,{0,0,1}][C2],A2,A1,RotationTransform[6Pi/5,{0,0,1}][D1],RotationTransform[-2Pi/5,{0,0,1}][D1],C2,C1,RotationTransform[4 \[Pi]/5,{0,0,1}][A2],RotationTransform[4Pi/5,{0,0,1}][D1],D1,B2,RotationTransform[4 \[Pi]/5,{0,0,1}][B2],RotationTransform[6 \[Pi]/5,{0,0,1}][B2],RotationTransform[2 \[Pi]/5,{0,0,1}][A2],RotationTransform[6 \[Pi]/5,{0,0,1}][A2]}/.{r->r0,t->t0,z->z0},Polygon[{{15,10,9,14,1},{2,6,12,11,5},{5,11,7,3,19},{11,12,8,16,7},{12,6,20,4,8},{6,2,13,18,20},{2,5,19,17,13},{4,20,18,10,15},{18,13,17,9,10},{17,19,3,14,9},{3,7,16,1,14},{16,8,4,15,1}}]]},Boxed->False
11. ],{x,1,Sqrt[2],(Sqrt[2]-1)/40}
12. ],"DisplayDurations"->0.1]

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hbghlyj

hbghlyj 发表于 2025-3-18 10:14
在 math.stackexchange.com/q/5040298 的回答中，
I believe the answer lies in how the side rectangles of the prism are attached

具体是怎样的？