1月21日是世界拥抱日，那么一个圆和三角形紧紧拥抱在一起会有什么结果呢？

Problem C: Circle and Triangle
Time Limit: 1 sec
Description
LMY and YY are geometry lovers. They enjoy challenging each other with interesting geometry problems. One day, LMY puts a circle and a triangle on a plane. Then YY moves them. They find that the overlapping area of the circle and the triangle varies as the relative position between the two changes. LMY and YY want to find out the largest common area.
The Input
Input consists of one or more lines. For each line, there are four integers describing one test case: the lengths of three sides of a triangle a, b, c; and the radius of a circle r; where 1≤a≤b≤c≤100, 1≤r≤100, and a+b>c.End of input is indicated by a line consisting four zeros.
The Output
For each test case, output a single line showing the largest overlapping area of the circle and the triangle. The precision should be 10-2.
Sample Input
3 4 5 1 （分别为三角形 三边 和半径）
5 5 8 4
0 0 0 0
Sample Output
3.14
12.00

hujunhua为这个题目配上如下图片并且从力学角度对题目进行分析:

wayne记三角形三边长分别为a,b,c，圆的半径为r,并且设圆心到三边距离分别为x,y,z,得出圆和三角形相交情况的数学形式为:

$\mathrm{arccos}\frac{x}{r}-\sqrt{1-\left(\frac{x}{r}{\right)}^{2}}\frac{x}{r}+\mathrm{arccos}\frac{y}{r}-\sqrt{1-\left(\frac{y}{r}{\right)}^{2}}\frac{y}{r}+\mathrm{arccos}\frac{z}{r}-\sqrt{1-\left(\frac{z}{r}{\right)}^{2}}\frac{z}{r}\arccos\frac\left\{x\right\}\left\{r\right\}-\sqrt\left\{1-\left(\frac\left\{x\right\}\left\{r\right\}\right)^2\right\}\frac\left\{x\right\}\left\{r\right\}+\arccos\frac\left\{y\right\}\left\{r\right\}-\sqrt\left\{1-\left(\frac\left\{y\right\}\left\{r\right\}\right)^2\right\}\frac\left\{y\right\}\left\{r\right\}+\arccos\frac\left\{z\right\}\left\{r\right\}-\sqrt\left\{1-\left(\frac\left\{z\right\}\left\{r\right\}\right)^2\right\}\frac\left\{z\right\}\left\{r\right\}$记图中阴影部分面积的最小值。

mathe指出取极值情况很简单，同三边相交的弦长要同对应边长度成正比，即$\frac{a}{\sqrt{{r}^{2}-{x}^{2}}}=\frac{b}{\sqrt{{r}^{2}-{y}^{2}}}=\frac{c}{\sqrt{{r}^{2}-{z}^{2}}}\frac a\left\{\sqrt\left\{r^2-x^2\right\}\right\}=\frac b\left\{\sqrt\left\{r^2-y^2\right\}\right\} =\frac c\left\{\sqrt\left\{r^2-z^2\right\}\right\}$.