UVA 11178 Morley's Theorem 计算几何

计算几何： 最基本的计算几何，差积  旋转

                                                                                                         Morley's Theorem

Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

Submit Status

Description

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. Trisectorlike 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

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

 

Source

Root :: Prominent Problemsetters :: Shahriar Manzoor

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 4. Geometry :: Geometric Computations in 2D :: Examples

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const double eps=1e-8;int dcmp(double x){    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){}};Point operator+(Point A,Point B) {return Point(A.x+B.x,A.y+B.y);}Point operator-(Point A,Point B) {return Point(A.x-B.x,A.y-B.y);}Point operator*(Point A,double p) {return Point(A.x*p,A.y*p);}Point operator/(Point A,double p) {return Point(A.x/p,A.y/p);}Point A,B,C;double Dot(Point A,Point B) {return A.x*B.x+A.y*B.y;}double Length(Point A) {return sqrt(Dot(A,A));}double Angle(Point A,Point B) {return acos(Dot(A,B)/Length(A)/Length(B));}double angle(Point v) {return atan2(v.y,v.x);}double Cross(Point A,Point B) {return A.x*B.y-A.y*B.x;}Point Rotate(Point A,double rad){    return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Point GetLineIntersection(Point p,Point v,Point q,Point w){    Point u=p-q;    double t=Cross(w,u)/Cross(v,w);    return p+v*t;}Point getD(Point A,Point B,Point C){    Point v1=C-B;    double a1=Angle(A-B,v1);    v1=Rotate(v1,a1/3);    Point v2=B-C;    double a2=Angle(A-C,v2);    v2=Rotate(v2,-a2/3);    return GetLineIntersection(B,v1,C,v2);}int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        for(int i=0;i<3;i++)        {            double x,y;            scanf("%lf%lf",&x,&y);            if(i==0) A=(Point){x,y};            else if(i==1) B=(Point){x,y};            else if(i==2) C=(Point){x,y};        }        Point D=getD(A,B,C);        Point E=getD(B,C,A);        Point 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;}