UVALive 4881

//一个小的技巧就是对于三角形的三点 开出一个6点的数组 这样用取模就方便操作了
//再就是用差乘求面积,设定p0原点后按顺时针防线连线求面积
//另外对于方程2,求ans1和ans2的坐标的时候,利用等比公式会方便许多
//degbug的时候没有区分大小写c 耽误了相当长的时间,以后命名变量的时候也要注意
#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;


const double E=1e-12;

struct Point{
    double x,y;
};

int P,T;

Point p[6];

double S,C;

void read(){
    for(int i=0;i<3;i++){
        cin>>p[i].x>>p[i].y;
        p[i+3]=p[i];
    }
    return ;
}


double CrossProduct(const Point &p1,const Point &p2,const Point &p3){
    return (p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x);
}

double dis(const Point &p1,const Point& p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

double area(){
    double s=0;
    Point p0={0,0};

    for(int i=0;i<3;i++)
        s+=CrossProduct(p0,p[i],p[(i+1)%3]);

    return s;
}

double perimeter(){
    double c=0;
    for(int i=0;i<3;i++){
        c+=dis(p[i],p[(i+1)%3]);
    }
    return c;
}

Point Pos(const Point &i,const Point &j,double l){
//    cout<<"LEN: "<<l<<endl;
    Point x={0,0};
    double d=dis(i,j);
    if(i.x-j.x)
        x.x=i.x-l/d*(i.x-j.x);
    else
        x.x=i.x;
    if(i.y-j.y)
        x.y=i.y-l/d*(i.y-j.y);
    else
        x.y=i.y;

    return x;
}

int dbcmp(const double & x){
    if(x>-E&&x<E)
        return 0;
    return x>0?1:-1;
}

int main()
{
    freopen("i.txt","r",stdin);
    double l,l1,l2;
    Point ans1,ans2;

    cin>>P;
    while(P--){
        cin>>T;
        read();
        S=fabs((area()/2)/2);
        C=perimeter()/2;
//        cout<<S<<" "<<C<<endl;
        for(int i=0;i<3;i++){
            double sina=2*(2*S)/dis(p[i],p[(i+1)%3])/dis(p[i],p[(i+2)%3]);

            double b=-C,c=2*S/sina;
//            cout<<"Flag: "<<b<<" "<<c<<endl;

//            cout<<"sin: "<<sina<<endl;
//            cout<<"S: "<<dis(p[i],p[(i+1)%3])<<" "<<dis(p[i],p[(i+2)%3])<<endl;
            if(dbcmp(b*b-4*c-0.0)<0)
                continue;

            l=(-b+sqrt(b*b-4*c))/2;

//            cout<<"L: "<<l<<" "<<C-l<<endl;

            l1=dis(p[i],p[(i+1)%3]);
            l2=dis(p[i],p[(i+2)%3]);
//            cout<<"l1 l2: "<<l1<<" "<<l2<<endl;
//            break;
//            cout<<l2-l<<" "<<l1-(C-l)<<endl;

            if(dbcmp(l1-l)>=0&&dbcmp(l2-(C-l))>=0){
                ans1=Pos(p[i],p[(i+1)%3],l);
                ans2=Pos(p[i],p[(i+2)%3],C-l);
//                cout<<"yes"<<endl;
            }
            else if(dbcmp(l2-l)>=0&&dbcmp(l1-(C-l))>=0){
                ans1=Pos(p[i],p[(i+2)%3],l);
                ans2=Pos(p[i],p[(i+1)%3],C-l);
//                cout<<"yes"<<endl;
            }
            else
                continue;

//            cout<<"POINT: "<<"("<<ans1.x<<","<<ans1.y<<")"<<" "
//                <<"("<<ans2.x<<","<<ans2.y<<")"<<endl;
            {
                // (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
                double A,B,C;
                if(dbcmp(ans1.x-ans2.x)==0){
                    A=1; B=0;
                    C=ans1.x;

                }
                else if(dbcmp(ans1.y-ans2.y)==0){
                    A=1; B=0;
                    C=ans1.y;
                }
                else{
                    double X=ans2.x-ans1.x,Y=ans2.y-ans1.y,K;
                    A=Y; B=-X;
                    C=ans1.x*Y-X*ans1.y;
                    if(dbcmp(A-0.0)<0){
                        A=-A; B=-B;
                        C=-C;
                    }
                    K=sqrt(A*A+B*B);
//                    cout<<"ANS:"<<A<<" "<<B<<" "<<C<<endl;
//                    cout<<"K: "<<K<<endl;
                    A/=K; B/=K;
                    C/=K;
//                    A=(A*100000+0.5)/100000;
//                    C=(C*100000+0.5)/100000;
//                    C=(C*100000+0.5)/100000;
                }
                cout<<T<<" ";
                printf("%.5lf %.5lf %.5lf\n",A,B,C);
                break;
            }
            break;
        }

    }
    return 0;
}