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k+d=(D-d)*√3/2
D=d+(k+d)*2√3/3
利用余弦定理:a²=b²+c²-2bccosA
(d+d‘)²=(D-d)²+(D-d’)²-(D-d)(D-d’)cos60°
(d+d‘)²-(D-d’)²+(D-d)(D-d’)/2=(D-d)²
2(d+D)(d-D+2d’)+(D-d)(D-d’)=2(D-d)²
2(d+D)(d-D)+2(d+D)*2d’+(D-d)D-(D-d)d’=2(D-d)²
4d’(d+D)-(D-d)d’=2(D-d)²-2(d+D)(d-D)-(D-d)D
d’(5d+3D)=3D(D-d)
d’=3D(D-d)/(5d+3D)
d0=(d+k)*2√3/3-d
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