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: 3000MS

Memory Limit: Unknown

64bit 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

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;
}