B - Points on Cycle

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

先给出自己的证明显然这三个点是要构成一个内接三角形的：

可设内接三角形三点连接圆点得到的3个角a,b,c。


周长C = ans * 2 * r

= 2 * r * (cos(a) + cos(b) + cos(c))

即求MAX(ans)

a + b + c = PI / 2

0 < a,b,c < PI / 2

a = x,b = y,c = PI/2 - x - y,0 < x,y < PI/2

ans =

cos(x) + cos(y) + cos(PI/2 - x - y)

ans =

cos(x)+cos(y)+sin(x+y)

ans =

cos(x)+cos(y)+sin(x)cos(y)+sin(y)cos(x)

ans = f(x,y) =

cos(x)+cos(y)+sin(x)cos(y)+sin(y)cos(x)

求导:

f’(x,y)|x =

-sin(x) + cos(y)cos(x) - sin(y)sin(x) = 0

f’(x,y)|y =

-sin(y) + cos(x)cos(y) - sin(x)sin(y) = 0

->-sin(x) + cos(x+y) = -sin(y) + cos(x+y) = 0

->sin(x) = sin(y) = cos(x + y)

->x = y,1 - 2sin^2(x) = sin(x)

t = sin(x)

->2t^2 + t - 1 = 0

(2t - 1)(t + 1) = 0

t = 1/2

sin(x) = 1/2 -> x = PI/6

x = y = PI / 6

f’(x,y)|xx = -cos(x)-cos(y)sin(x)-sin(y)cos(x)

f’(x,y)|xy = -cos(x)sin(y) - sin(x)cos(y)

f’(x,y)|yy = -cos(y)-cos(x)sin(y)-sin(x)cos(y)

A = -sqrt(3)r,B = -sqrt(3) / 2r,C = -sqrt(3)r

AC - B^2 = 3r^2 - 3/4r^2 > 0,A < 0,取极大值

a = b = c = PI / 6

故该内接三角形应为等边三角形

设三点分别为A,B,C，A已知。

由cos

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <list>
#include <queue>
#include <cstring>
#include <set>
#include <map>
#include <string>
#include <cmath>
using namespace std;
#define INF 0x3f3f3f3f
struct point {
double x;
double y;
};
double dis(double a, double b, double c, double d) {
return sqrt((a - c) * (a - c) + (b - d)*(b - d));
}
int main() {
int T;
point a, b, c;
cin >> T;
while (T--) {
cin >> a.x >> a.y;
double r = sqrt(a.x * a.x + a.y * a.y);
double s = r * sqrt(3);
double t = sqrt(3) * a.y;
b.x = (t - a.x) / 2.0;
if (fabs(a.y - 0) > 1e-7)
b.y = (-r * r / 2.0 - a.x * b.x) / a.y;
else
b.y = sqrt(r * r - b.x * b.x);
c.x = (-t - a.x) / 2.0;
if (fabs(a.y - 0) > 1e-7)
c.y = (-r * r / 2.0 - a.x * c.x) / a.y;
else
c.y = -sqrt(r * r - c.x * c.x);
if (fabs(c.y - b.y) < 0.0005) {
if (c.x < b.x) {
printf("%.3f %.3f %.3f %.3f", c.x, c.y, b.x, b.y);
}
else {
printf("%.3f %.3f %.3f %.3f", b.x, b.y, c.x, c.y);
}
}
else {
if (c.y < b.y) {
printf("%.3f %.3f %.3f %.3f", c.x, c.y, b.x, b.y);
}
else {
printf("%.3f %.3f %.3f %.3f", b.x, b.y, c.x, c.y);
}
}
putchar('\n');
}
return 0;
}