勾股三角形的聚宝盆
A cornucopia of pythagorean triangles
摘要：
考虑两个外切圆，半径为R1和R2，且R1> R2，这两个圆有三条公共切线，其中一条为T1 T2。设M是T1 T2的中点，K是C1 C2和T1 T2的交点，则构成16个直角三角形（C1和C2是两个圆心），见图1。本文第六节给出了使上述16个直角三角形为勾股三角形的两个整数R1和R2必须具有的精确形式。
英文摘要：
Consider two circles, externally tangential,and with integer radii R1, R2; and with R1>R2.The two circles have three tangent lines in common, one of them being T1T2. If M is the midpoint of T1T2, and K the point of intersection of the lines C1C2 and T1T2;then 16 right triangles are formed(C1 and C2 are the two circle centers), see Figure 1.In Section 6 of this paper, we find the precice form the two integers R1 and R2 must have, in order that the sixteen aforementioned right triangles be Pythagorean.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
