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Problem D
Morley’s Theorem
Input: Standard Input

Output: Standard Output

 Morley’s theorem states that that the lines trisecting the angles of an arbitrary plane triangle meet at the vertices of an equilateral triangle. For example in the figure below the tri-sectors of angles A, B and C has intersected and created an equilateral
triangle DEF.

Of course the theorem has various generalizations, in particular if all of the tri-sectors are intersected one obtains four other equilateral triangles. But in the original theorem only tri-sectors nearest to BC are allowed to intersect to get point D, tri-sectors
nearest to CA are allowed to intersect point E and tri-sectors nearest to AB are intersected to get point F. Trisector like BD and CE are not allowed to intersect. So ultimately we get only one equilateral triangle DEF. Now your task is to find the Cartesian
coordinates of D, E and F given the coordinates of A, B, and C.

Input

First line of the input file contains an integer N (0<N<5001) which denotes the number of test cases to follow. Each of the next lines contain six integers
. This six integers actually indicates that the Cartesian coordinates of point A, B and C are
 respectively. You can assume that the area of triangle ABC is not equal to zero,
 and the points A, B and C are in counter clockwise order.

Output

For each line of input you should produce one line of output. This line contains six floating point numbers
 separated by a single space. These six floating-point actually means that the Cartesian coordinates of D, E and F are
 respectively. Errors less than 
 will be accepted.

Sample Input   Output for Sample Input

2

1 1 2 2 1 2

0 0 100 0 50 50

Problemsetters: Shahriar Manzoor

Special Thanks: Joachim Wulff

题意是说作三角形ABC每个内角的三等分线，相交成三角形DEF，则DEF是等边三角形。求出DEF三点坐标。

如果要求D点，则要计算∠ABC的值a，然后把射线BC逆时针旋转a/3，得到直线BD。同理得到直线CD，求交点即可。

#include<stdio.h>
#include<math.h>
using namespace std;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}//构造函数
};
typedef Point Vector;
Point operator-(Point A,Point B)
{
    return Point(A.x-B.x,A.y-B.y);
}

Point operator+(Point A,Point B)
{
    return Point(A.x+B.x,A.y+B.y);
}

Point operator*(Point A,double p)
{
    return Point(A.x*p,A.y*p);
}


double Dot(Point A,Point B)
{
    return A.x*B.x+A.y*B.y;
}

double Lenth(Point A)
{
    return sqrt(Dot(A,A));
}

double Angle(Point A,Point B)
{
    return acos(Dot(A,B)/Lenth(A)/Lenth(B));
}


Point Rotate(Point A,double rad)//rad是弧度
{
    return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}

double Cross(Point A,Point B)
{
    return A.x*B.y-A.y*B.x;
}

//求相交点的坐标
Point  GetLineIntersection(Point P,Point V,Point Q,Point W)
{
    Point u=P-Q;
    double t=Cross(W,u)/Cross(V,W);
    return P+V*t;
}


Point getD(Point A,Point B,Point C)
{
    Vector v1=C-B;
    double a1=Angle(A-B,v1);
    v1=Rotate(v1,a1/3);

    Vector v2=B-C;
    double a2=Angle(A-C,v2);
    v2=Rotate(v2,-a2/3);//负数表示逆时针旋转
    return GetLineIntersection(B,v1,C,v2);
}
int main()
{
    int t;
    Point A,B,C,D,E,F;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf",&A.x,&A.y);
        scanf("%lf%lf",&B.x,&B.y);
        scanf("%lf%lf",&C.x,&C.y);
        D=getD(A,B,C);
        E=getD(B,C,A);
        F=getD(C,A,B);
        printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",D.x,D.y,E.x,E.y,F.x,F.y);
    }
    return 0;

}