uva 11178 Morley's Theorem（计算几何-点和直线）

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 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

1.316987 1.816987 1.183013 1.683013 1.366025 1.633975 56.698730 25.000000 43.301270 25.000000 50.000000 13.397460

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Problemsetters: Shahriar Manzoor

Special Thanks: Joachim Wulff

题目大意：

作三角形的每一个角的三等分射线，相交成的三角形DEF为等边三角形。

解题思路：

通过向量的旋转以及直线的相交，求出对应的交点。

解题代码：

刘汝佳就是牛逼。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

struct Point{
    double x,y;
    Point(double x0=0,double y0=0){
        x=x0,y=y0;
    }
    void read(){
        scanf("%lf%lf",&x,&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 p) { return Vector(A.x*p,A.y*p); }
Vector operator / (Vector A,double p) { return Vector(A.x/p,A.y/p); }

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; }
Vector Rotate(Vector A,double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); }//逆时针旋转rad弧度

//必须保证相交，也就是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){
    double a1=Angle(A-B,C-B);
    Vector v1=Rotate(C-B,a1/3.0);
    double a2=Angle(A-C,B-C);
    Vector v2=Rotate(B-C,-a2/3.0);
    return GetLineIntersection(B,v1,C,v2);
}

int main(){
    int T;
    scanf("%d",&T);
    while(T-- >0){
        Point A,B,C,D,E,F;
        A.read();
        B.read();
        C.read();
        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;
}