codeforces--Ancient Berland Circus（三點確定最小多邊形）

codeforces--Ancient Berland Circus（三點確定最小多邊形）

 

Ancient Berland Circus Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Appoint description: System Crawler (2015-01-06)

Description

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.

Sample Input

Input

0.000000 0.000000
1.000000 1.000000
0.000000 1.000000

Output

1.00000000

 

題目大意：給出三個點，求出以這三個點為定點的最小正多邊形。

求最小正多邊形，邊數越多，面積越大，所以要是求得的多邊形的邊盡量的小。

由三個點組成的三角形，可以確定一個外接圓，那麼正多邊形的所有的定點應該都在圓上，求出三邊對應的圓心角，找出圓心角的最大公約數，也就得到了多邊形的最小的邊數。

防止鈍角的情況，邊長最長的對應的圓心角 = 2*PI - 其他兩個圓心角。

 

#include 
#include 
#include 
#include 
using namespace std ;
#define PI acos(-1)
#define eqs 0.01
double gcd(double a,double b)
{
    return a < eqs ? b : gcd(fmod(b,a),a);
}
int main()
{
    double x1 , y1 , x2 , y2 , x3 , y3 ;
    double a , b , c , p , s , r , k ;
    double A , B , C ;
    scanf("%lf %lf %lf %lf %lf %lf", &x1, &y1, &x2, &y2, &x3, &y3) ;
    a = sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ) ;
    b = sqrt( (x2-x3)*(x2-x3) + (y2-y3)*(y2-y3) ) ;
    c = sqrt( (x1-x3)*(x1-x3) + (y1-y3)*(y1-y3) ) ;
    p = ( a + b + c ) / 2.0 ;
    s = sqrt( p * (p-a) * (p-b) * (p-c) ) ;
    r = a * b * c / ( 4 * s ) ;
    if( a > c )
    {
        k = a ; a = c ; c = k ;
    }
    if( b > c )
    {
        k = b ; b = c ; c = k ;
    }
    A = 2 * asin(a/(2*r)) ;
    B = 2 * asin(b/(2*r)) ;
    C = 2 * PI - A - B ;
    //printf("%lf %lf %lf\n", A, B, C) ;
    p = gcd(A,B);
    p = gcd(p,C) ;
    //printf("%lf %lf\n", r, p) ;
    printf("%.6lf\n", (PI*r*r*sin(p))/p ) ;
    return 0;
}