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 sixintegers . 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

题意：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。给出A、B、C 3个的位置确定D、E、F 3个点的位置。

分析：本题没什么算法可言，只要根据题意计算。考虑到对称性，只需要知道如何求D点即可。首先需要计算角ABC的值a，然后把射线BC逆时针旋转a/3，得到直线BD，同理可以得到直线CD，求出交点即可。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct Point {double x, y;Point(double x = 0, double y = 0) : x(x), y(y) { }
};typedef Point Vector;Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }
bool operator < (const Point& a, const Point& b) {return a.x < b.x || (a.x == b.x && a.y < b.y);
}
const double eps = 1e-10;
int dcmp(double x) {if(fabs(x) < eps) return 0;return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b) {return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}double Dot(Vector A, Vector B) { return A.x * B.x + A.y * B.y; }  //点乘
double Length(Vector A) { return sqrt(Dot(A, A)); } //向量的模
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } //两个向量的夹角
double Cross(Vector A, Vector B) { return A.x * B.y - A.y * B.x; } //叉乘
double Area(Point A, Point B, Point C) { return Cross(B - A, C - A); } //三个点组成的三角形的面积
Vector Rotate(Vector A, double rad) { //向量A逆时针旋转rad弧度后的坐标return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}//求直线交点
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {Vector 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);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%lf%lf%lf%lf",&A.x, &A.y, &B.x, &B.y, &C.x, &C.y);D = getD(A, B, C);E = getD(B, C, A);F = getD(C, A, B);printf("%lf %lf %lf %lf %lf %lf\n", D.x, D.y, E.x, E.y, F.x, F.y);}return 0;
}