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[几何] 求线段长度

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lrh2006 posted 2021-1-11 19:53 |Read mode
3079FC1B-4D97-43b3-858B-F2668591438F.png
3F95D5F9-4DAC-4024-8A95-9E9CC7B2E507.png
请教各位,谢谢!

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色k posted 2021-1-11 20:17
很简单啊,设BC与圆的另一个交点为P,则BP=MA,所以AP∥MB,故MC=BC

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original poster lrh2006 posted 2021-1-11 21:16
回复 2# 色k

BP=MA是为何?有什么定理吗?还有AP//MB,没反应过来

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力工 posted 2021-1-11 21:29
回复 3# lrh2006


    圆内接四边形的对角互补,角NPB=角NAB啊。

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original poster lrh2006 posted 2021-1-11 22:19
回复 4# 力工


    噢噢明白了,那AP//MB是根据什么?

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TSC999 posted 2021-1-11 23:22
Last edited by TSC999 2021-1-11 23:44注意, 圆半径只有很窄的范围。 sqrt(5)/4=0.559017>R>0.5,并且只有确定了 R,才能确定 x,y 的长度。
例如 R=0.52, 则 x = 0.2856571371417142; y = 0.19171408473721677; 此时 MC=2.04435,不等于 BC,因为 BC=2。
求线段长.png

注: 最后那个方程是 AC · AM=DC · DB。
列出了四个方程都搞不定三个未知数,可见方程中有同解方程。

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色k posted 2021-1-11 23:27
回复 6# TSC999

你是不是看漏了AM=AN的条件?

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TSC999 posted 2021-1-11 23:45
回复 7# 色k

噢,还真是的。漏了这个 AM=AN。

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TSC999 posted 2021-1-11 23:50
求线段长.png

如此就有 MC=BC=2。

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