由等边三角形的顶点与外围三角形两角的三分之一作为底部向量的偏转角，两偏转向量的交点就是一个所求点
根据对称性原理求出另外两点
#include<cstdio> #include<cstring> #include<cmath> using namespace std; const double eps=1e-10; 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 - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (Vector A,double B){return Vector(A.x*B,A.y*B);} Vector operator / (Vector A,double B){return Vector(A.x/B,A.y/B);} int dcmp(double x){if(fabs(x)<eps)return 0;return (x>0)?1:-1;} bool operator == (Vector A,Vector 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-B.x*A.y;} Vector Rotate(Vector A,double rad){//逆时针 return Vector(A.x*cos(rad)-A.y*sin(rad) , A.x*sin(rad)+A.y*cos(rad)); } 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 GetPoint(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); } Point read_point(){ int a,b; scanf("%d%d",&a,&b); return Point(a,b); } int main(){ int N; scanf("%d",&N); while(N--){ Point A=read_point(); Point B=read_point(); Point C=read_point(); Point D=GetPoint(A,B,C); Point E=GetPoint(B,C,A); Point F=GetPoint(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; }
