Points on Cycle

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

There is a cycle with its center on the origin.

Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other

you may assume that the radius of the cycle will not exceed 1000.

Input

There are T test cases, in each case there are 2 decimal number representing the coordinate of the given point.

Output

For each testcase you are supposed to output the coordinates of both of the unknow points by 3 decimal places of precision

Alway output the lower one first(with a smaller Y-coordinate value), if they have the same Y value output the one with a smaller X.

NOTE

when output, if the absolute difference between the coordinate values X1 and X2 is smaller than 0.0005, we assume they are equal.

Sample Input

2
1.500 2.000
563.585 1.251

Sample Output

0.982 -2.299 -2.482 0.299
-280.709 -488.704 -282.876 487.453

题意：

	一个以原点为中心的圆，告诉你圆上的一个点，求与另外的两个点组成的三角形的周长最长的两点作标。

思路：

	设P(x,y),一个方程是pow(x,2)+pow(y,2)=pow(r,2);另一个方程是根据向量知识，向量的夹角公式得到方程。

因为圆心角夹角为120度，已知一个向量（即一个点作标），所以COS(2PI/3)=a*b/|a|*|b|;(a,b为向量)；

已知角和a向量，就可求b向量b(x,y).由方程组可求得（x,y）；最后得到的是一元二次方程组，可得到两个解，即为两个点的作标。

代码：

#include <stdio.h>
#include <math.h>
#define PI 3.1415926

int main()
{
	double x,y,x1,y1,x2,y2,cosx,a,b,c,r,delta;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf",&x,&y);
		r=sqrt(x*x+y*y);
		a=r*r;
		b=r*r*y;
		c=r*r*r*r/4-x*x*r*r;
		delta=b*b-4*a*c;
		y1=(-1*b-sqrt(delta))/(2*a);
		y2=(-1*b+sqrt(delta))/(2*a);
		if(x==0)
		{
			x1=-sqrt(r*r-y1*y1);
			x2=sqrt(r*r-y2*y2);
		}
		else
		{
			x1=(-1*r*r/2-y*y1)/x;
			x2=(-1*r*r/2-y*y2)/x;
		}
		printf("%.3lf %.3lf %.3lf %.3lf/n",x1,y1,x2,y2);
	}
	return 0;
}

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