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[几何] 求作正三角

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isee Posted 2013-8-22 20:53 |Read mode
已给点$P$及线段$x,y,z$,试作一个正三角形$ABC$,使得点$P$在$\triangle ABC$内,并且$$PA=x,PB=y,PC=z.$$问:为了使所述的正三角形存在,$x,y,z$应满足什么条件?

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kuing Posted 2013-8-22 23:49

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地狱的死灵 Posted 2013-8-22 23:49
QQ图片20130822234335.jpg
貌似只要x,y,z可组成三角形即可
作法:先以x,y,z为三边作出△PBD,
不妨设PD为该三角形最长边,
以该边向形外作正三角形PDC,
于是BC为所求作正三角形的边长……

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 Author| isee Posted 2013-8-23 14:28
完美收官,因“先以x,y,z为三边作出△PBD”故x,y,x 要能围成三角形,再由链接的17楼,知,此三角形的“最大边所对的内角比120°小”

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