Euclidea 14.5 解法与证明

Euclidea 14.5

如图，给定三个两两相切的圆，O，O_1， O_2。三个圆心共线。

任务：尺规作出一个圆，与给定的三个圆都相切。

解法：

设圆O与圆O_1的切点为A
连接OA
过O做OA垂线交圆O于D
以D为圆心，DA为半径作圆，交圆O_1和圆O_2于E和F。
作直线EO_1和FO_2交于G
以G为圆心，GE为半径作圆
圆G即为所求圆

简要证明：

如图，设三个圆切点为A，B，C。明显A，B，C，O，O_1，O_2五点共线。
以A为中心作反演变换，取反演圆与圆O_2正交。于是圆O_2经反演变换后不变。B，C互为反演点。
分别过B和C做AB的垂线L_B和L_C。易知圆O1和圆O经反演变换后分别为L_B和L_C。
如图作圆G'同时与圆O_2，L_B和L_C相切于E'和F'。
过E'F'作直线L，易证C在L上，并且∠E'CB为45°。
设L经过反演变换得到的圆为圆D'，考虑圆D'的性质：

直线AD'与直线L垂直
圆D'经过A（因为L不经过A）
圆D'经过B（因为L经过C，而B，C互为反演点）

于是可知圆D就是圆D'，因此E和E', F和F'分别互为反演点。
设圆G'经过反演变换为圆G''。因为圆G'与L_B和圆O_2相切于E'和F'，所以圆G''与圆O_1和圆O_2相切于E和F。
所以G''即是EO_1和FO_2的交点，且圆G''半径是G''E。于是圆G即是圆G''。
因为圆G'与L_B，圆O_2，L_C都相切，所以圆G与三个给定圆都相切。证明完毕。

参考资料：Pappus chain

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