Uva 11178 Morley定理

题意： 给你三角形三个点， 定理是 三个内角的三等分线相交得出 DEF三点，

三角新 DFE是等边三角形

然后要你输出 D E F 的坐标

思路 ：

求出三个内角，对于D 相当于 BC向量逆时针旋转， CB向量顺时针旋转 ，相交得到的点；

同理可以求出其他点  (LRJ 模板真强大）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double eps = 1e-;
struct Point {
    double x, y;
    Point(double x = , double y = ) : x(x), y(y) {}
};
typedef Point Vector;

int dcmp(double x) { if(fabs(x) < eps) return  ; else return x <  ? - : ; }

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 dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == ;
}

double Dot (Vector A, Vector B) { return A.x*B.x + A.y*B.y; } ///点积
double Length (Vector A) { return sqrt(Dot(A,A)); }           ///向量长度
double Angle (Vector A, Vector B) { return acos(Dot(A,B) / Length(A) / Length(B)); }  ///角度
double Cross (Vector A, Vector B) { return A.x*B.y - A.y*B.x; }   ///X积
double Area2 (Point A, Point B, Point C) { return Cross(B-A,C-A); } ///面积
Vector Rotate (Vector A, double rad) {                              ///向量旋转 ， 逆时针，顺时针角度为-
    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)) ;
}
Vector Normal (Vector A) { double L = Length(A);  return Vector(-A.y/L, A.x/L); }  ///单位向量
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 v = C-B;
    double rad = Angle(A-B,v);
    v = Rotate(v, rad/);

    Vector vv =  B-C;
    rad = Angle(A-C,vv);
    vv = Rotate(vv, -rad/);

    return GetLineIntersection(B,v, C,vv);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        Point A, B, C, D, E, F;
        cin >> A.x >> A.y >> B.x >>B.y >> C.x >>C.y;
        D = GetPoint(A,B,C);
        E = GetPoint(B,C,A);
        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 ;
}