题目大意是给出三维坐标系中三角形三个顶点的坐标，求其内接圆与外接圆的面积比。
因为三点确定一个平面，所以实际上本题还是可以转化为平面上的几何计算题。在三维坐标系中两点面积的计算方法是降维计算，即将三维抓化为二维来计算。计算公式是sqrt(sqrt(sqr(x2-x1)+sqr(y2-y1))+sqr(z2-z1))，化简得sqrt(sqr(x2-x1)+sqr(y2-y1)+sqr(z2-z1))。因为内接圆是由以三边为底，半径为高的三个三角形组成的，故s内接圆=r内接圆(a+b+c)/2。r内接圆=s/(a+b+c)。外接圆的半径的计算方法参看我的ZJU/ZOJ 1090的解题报告。因为s=pi*r^2，故s内接圆:s外接圆=r内接圆^2/r外接圆^2。
2998336 2008-07-23 11:10:21 Accepted 1439 FPC 00:00.12 404K IwfWcf@LZOI
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