codeforces 1C Ancient Berland Circus

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 
     
       Copy 
     
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000

      output 
    
1.00000000
题意：一个正多边形，只知道其中三个顶点的坐标，求包含这三个顶点的最小的正多边形的面积
步骤：1 ：由海伦公式求出三角形的面积：S=sqrt（p*（p-a）*（p-b）*（p-c）），p=(a+b+c)/2；
 2：外接圆半径R=a*b*c/(4*S);
 3:根据余弦定理求出三条边各自所对应的外接圆圆心角的最大公约数t，（注意求出两个圆心角后第三个圆心角的度数等于（2*PI-前两圆心角和），则多边形边数n=2PI/t；

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 4：多边形的其中一条边所对应的最小单位三角形的面积为：s'1/2*R*R*sint；
 5：输出n*s'即可；

#include <bits/stdc++.h >
const double ff=0.001;
const double PI=acos(-1.0);
using namespace std;
struct pot
{
	double x, y;
}a[3];
double Count(pot n, pot m)
{
	double m1=m.x-n.x;
	double m2=m.y-n.y;
	return sqrt(m1*m1+m2*m2);
}
double fgcd(double a,double b)     
{
	if(a<ff)
	return b;
	if(b<ff)
	return a;
	return fgcd(b,fmod(a,b));
}
int main()
{
	int i;
	for(i=0; i<3; i++)
	{
		cin>>a[i].x>>a[i].y;
	}
	double c,d,e;
	c=Count(a[0],a[1]);
	d=Count(a[0],a[2]);
	e=Count(a[1],a[2]);
	double f=(c+d+e)/2;
	double ss=sqrt(f*(f-c)*(f-d)*(f-e));      //三角形面积 
	double R=c*d*e/(4*ss);
	double angle[3];
	angle[0]=acos(1-c*c/(2*R*R));
	angle[1]=acos(1-d*d/(2*R*R));
	angle[2]=2*PI-angle[0]-angle[1];
	double q=angle[0];
	for(i=1; i<3; i++)
	{
		q=fgcd(q,angle[i]);
	}
	printf("%.6lf\n",(PI*sin(q)*R*R)/q);      //  2PI/q*1/2*R*R*sinq
	return 0;
}

codeforces 1C Ancient Berland Circus

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