CodeForces - 1C：Ancient Berland Circus （幾何）

Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples

Input

0.000000 0.000000
1.000000 1.000000
0.000000 1.000000

Output

1.00000000

題意：有一個正多邊形，如今咱們只知道其中3個點的座標，求原多邊形的面積，若是有多個知足，求最小面積。

思路：已知三點，咱們能夠肯定外接圓，而後顯然，咱們須要多邊形的邊長最大(或者對於的圓心角最大)，可是因爲有鈍角或者銳角，求最大邊長可能要討論。因此咱們求最大圓心角。

可能推論：最大圓心角=三角形的三條邊對應的圓心角的gcd。而後就獲得了有2pi/gcd邊。blabla。

（獲得三個角的時候，第三個角=2pi-A-B。而直接求會WA。。。

#include<bits/stdc++.h>
using namespace std; const double eps=1e-4; const double pi=acos(-1.0); double Gcd(double a,double b) { while(fabs(a)>eps&&fabs(b)>eps){ if(a>b) a-=floor(a/b)*b; else b-=floor(b/a)*a; } return a+b; } double x[4],y[4],L1,L2,L3,S,R; double dist(int a,int b){ return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b])); } double area(){ double p=(L1+L2+L3)/2.0; return sqrt(p*(p-L1)*(p-L2)*(p-L3)); } int main() { for(int i=1;i<=3;i++) scanf("%lf%lf",&x[i],&y[i]); L1=dist(1,2); L2=dist(1,3); L3=dist(2,3); S=area(); R=L1*L2*L3/(S*4); double A=acos((R*R+R*R-L3*L3)/(2*R*R)); double B=acos((R*R+R*R-L2*L2)/(2*R*R)); double C=2*pi-A-B; double ang=Gcd(Gcd(A,B),C); double ans=pi/ang*R*R*sin(ang); printf("%.6lf\n",ans); return 0; }