莫利定理（Morley's theorem），也称为莫雷角三分线定理。将三角形的三个内角三等分，靠近某边的两条三分角线相交得到一个交点，则这样的三个交点可以构成一个正三角形。这个三角形常被称作莫利正三角形。

11178 - Morley's Theorem

Time limit: 3.000 seconds

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

1 1 2 2 1 2

0 0 100 0 50 50

1.316987 1.816987 1.183013 1.683013 1.366025 1.633975

56.698730 25.000000 43.301270 25.000000 50.000000 13.397460

参考《算法竞赛入门经典——训练指南》第四章计算几何

参考代码+部分注释

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <climits>
#define eps 1e-10
using namespace std;
typedef long long ll;
const int INF=INT_MAX;
const int maxn = 110;
int dcmp(double x){//三态函数，克服浮点数精度陷阱，判断x==0?x<0?x>0?if(fabs(x)<eps) return 0;else return x<0?-1:1;
}
struct Point{double x,y;Point(double x=0,double y=0):x(x),y(y){}//构造函数，方便代码编写
};
typedef Point Vector;//Vector是 Point的别名Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A,Vector 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);}
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;}
double Length(Vector A){return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
//rad是弧度不是角度
Vector Rotate(Vector A,double rad){return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
//调用前请确保两条直线P+tv,Q+tw有唯一交点。当且仅当Cross（v,w）非0
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);//正数表示逆时针旋转Vector v2=B-C;double a2=Angle(A-C,v2);v2=Rotate(v2,-a2/3);//负数表示顺时针旋转return GetLineIntersection(B,v1,C,v2);
}
int main()
{// freopen("input.txt","r",stdin);Point A,B,C,D,E,F;int T;cin>>T;while(T--){cin>>A.x>>A.y>>B.x>>B.y>>C.x>>C.y;D=getD(A,B,C);E=getD(B,C,A);//注意方向对应F=getD(C,A,B);printf("%f %f %f %f %f %f\n",D.x,D.y,E.x,E.y,F.x,F.y);}return 0;
}

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