Description

 

Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry  the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.

Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x1,y1) and B(x2,y2). The third coordinate C(x3,y3) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x1,y1) and B(x2,y2), but he doesn’t know where the C(x3,y3) is. The prince need your help. Can you calculate the C(x3,y3) and tell him?

 

Input

 

The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x1,y1) and B(x2,y2), described by four floating-point numbers x1, y1, x2, y2 ( |x1|, |y1|, |x2|, |y2| <= 1000.0). 

        Please notice that A(x1,y1) and B(x2,y2) and C(x3,y3) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x1,y1) and B(x2,y2) are given by anticlockwise.

 

Output

 

For each test case, you should output the coordinate of C(x3,y3), the result should be rounded to 2 decimal places in a line.

 

Sample Input

4-100.00 0.00 0.00 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 0.00 1.866 0.50

Sample Output

(-50.00,86.60)(-86.60,50.00)(-36.60,136.60)(1.00,1.00)

给你等边三角形的两个点A和B，求第三个点C的坐标；

且ABC是逆时针的；

题解1：

因为要求ABC是逆时针的，所以可以直接用B绕A逆时针旋转60°；

这里有个通用的公式，证明稍微复杂，可以加到模板里以备不时之需：

 

点（x1，y1）绕点（x2，y2）逆时针旋转a角度后新的坐标（X，Y）为:

  X=(x1-x2)*cos(a)-(y1-y2)*sin(a)+x2;

  Y=(x1-x2)*sin(a)+(y1-y2)*cos(a)+y2;

 

如果直接按照题意的等边三角形的情况去画图推导也可以推导出来，不过这个公式比较普适。

#include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             using namespace std;int main() { int t; scanf("%d", &t); while(t--){ double x1,x2,x3,y1,y2,y3; scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); double dx=x2-x1,dy=y2-y1; x3=dx/2-dy*sqrt(3.0)/2+x1; y3=dy/2+dx*sqrt(3.0)/2+y1; printf("(%.2lf,%.2lf)\n",x3,y3); } return 0;}
            
           
          
         
        
       

题解2：

AB线段绕A点逆时针旋转60°后B点的位置

用到平面几何求解

x3=x1+L*cos(60°+angle);

y3=y1+L*sin(60°+angle);

angle=atan2(y2-y1,x2-x1);

#include 
       
          #include
        
           #include
         
            using namespace std;  const double PI=acos(-1.0);  int main()  {      int t;      cin>>t;      double  x1,y1,x2,y2,x3,y3,angle,l;      while(t--)      {          scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);          angle=atan2(y2-y1,x2-x1);          l=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));          x3=x1+l*cos(angle+PI/3.0);          y3=y1+l*sin(angle+PI/3.0);          printf("(%.2lf,%.2lf)\n",x3,y3);      }      return 0;  }