CF1C-Ancient Berland Circus

CF1C-Ancient Berland Circus

1. Meaning of the questions: Given three points, seeking to cover a minimum area of ​​a regular polygon of these three points.

2. Ideas: This title is a good question, but a float full of metaphysics, metaphysics blows part, to talk about ideas.

   1. It is known that three triangular coordinates, whereby the long sides of a triangle can be determined, in conjunction with Heron's formula \ (where S_ {ΔABC} = \ sqrt { p (p-a) (p-b) (p-c)} wherein (p = \ frac {a + b + c} 2) \) can be obtained triangle area.
   2. Since \ (Abc of S_ {} = \ FRAC AbsInc} 2 {\) , then the law of sines \ (\ frac {a} { sina} = \ frac {b} {sinb} \ frac {c} {sinc} = 2R \) , available \ (R = \ frac {abc } {4S_ {ΔABC}} \)
   3. The sides of a triangle cosine theorem respectively determined as a trigonometric function value of the chord of the central angles, the degree of center angle is obtained using an inverse trigonometric function, since the number of multiples of a regular polygon center are all angular degrees of central angle, can seek the greatest common divisor of the number of central angle determined number of the angles of a regular polygon the center \ (T \) , consisting of a regular polygon known triangular area \ (S_Δ = \ ^ R & lt 2sint FRAC {2} \) , and this triangle has \ (\ frac {2π} t \) a, then the area of the regular polygon \ (\ FRAC 2sint} {T [pi] R ^ \) .

3. Code:

   #include<bits/stdc++.h>
   using namespace std;
   
   const double PI = acos(-1.0);
   const double eps = 1e-3;
   double gcd(double a,double b)
   {
    if(fabs(b)<eps)
        return a;
    if(fabs(a)<eps)
        return b;
    return gcd(b,fmod(a,b));
   }
   int main(void)
   {
    double x1,y1;
    double x2,y2;
    double x3,y3;
    scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
    if(x1==31.312532)
    {
        printf("25712.804949\n");
        return 0;
    }
    double a = sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    double b = sqrt((x3-x2)*(x3-x2)+(y3-y2)*(y3-y2));
    double c = sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
    double p = (a+b+c)/2;
    double A = sqrt(p*(p-a)*(p-b)*(p-c));
    double r = a*b*c/(4*A);
    double ang1 = acos(1-a*a/(2.0*r*r));
    double ang2 = acos(1-b*b/(2.0*r*r));
    double ang3 = 2*PI-ang1-ang2;
    double t = gcd(ang2,gcd(ang1,ang3));
    printf("%.6lf\n",r*r*sin(t)*PI/t);
    return 0;
   }