Problem:
Problem E: Spherical Mirrors

A long time ago in a galaxy, far, far away, there were N spheres with various radii.
Spheres were mirrors, that is, they had reflective surfaces . . . .

You are standing at the origin of the galaxy (0, 0, 0), and emit a laser ray to the direction
(u,v, w). The ray travels in a straight line.

When the laser ray from I hits the surface of a sphere at Q, let N be a point outside of the
sphere on the line connecting the sphere center and Q. The reflected ray goes to the direction
towards R that satisfies the following conditions: (1) R is on the plane formed by the three
points I, Q and N , (2) ∠ IQN = ∠ NQR, as shown in Figure 1.

Figure 1: Laser ray reflection

After it is reflected several times, finally it goes beyond our observation. Your mission is to write
a program that identifies the last reflection point.

Input

The input consists of multiple datasets, each in the following format.

N
u v w
x1 y1 z1 r1
.
.
.
xN yN zN rN

The first line of a dataset contains a positive integer N which is the number of spheres. The
next line contains three integers u, v and w separated by single spaces, where (u, v, w) is the
direction of the laser ray initially emitted from the origin.

Each of the following N lines contains four integers separated by single spaces. The i-th line
corresponds to the i-th sphere, and the numbers represent the center position (xi, yi, zi ) and the
radius ri .

N, u, v, w, xi, yi, zi and ri satisfy the following conditions.

1 ≤ N ≤ 100

−100 ≤ u, v, w ≤ 100

−100 ≤ xi, yi, zi ≤ 100

5 ≤ ri ≤ 30

u2 + v2 + w2 > 0

You can assume that the distance between the surfaces of any two spheres is no less than 0.1.
You can also assume that the origin (0, 0, 0) is located outside of any sphere, and is at least 0.1
distant from the surface of any sphere.

The ray is known to be reflected by the sphere surfaces at least once, and at most five times.
You can assume that the angle between the ray and the line connecting the sphere center and
the reflection point, which is known as the angle of reflection (i.e. θ in Figure 1), is less than 85
degrees for each point of reflection.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, you should print the x-, y- and z-coordinates of the last reflection
point separated by single spaces in a line. No output line should contain extra characters.

No coordinate values in the output should have an error greater than 0.01.

Sample Input

3
-20 -20 -24
100 100 100 30
10 8 3 5
-70 -70 -84 5
4
0 47 84
-23 41 42 8
45 -10 14 19
-5 28 47 12
-27 68 34 14
0

Output for the Sample Input

79.0940 79.0940 94.9128
-21.8647 54.9770 34.1761

Solution in JAVA (judge verdict: Accepted)

import java.awt.geom.*;
import java.io.*;
import java.util.*;

import javax.management.ReflectionException;

//reference rankalee's code
public class Main {
	double EPS = 1.0e-08;
	
	class P{
		double [] p;
		public P(double[] p) {
			this.p = new double[3];
			for(int i =0; i < 3; i++){
				this.p[i] = p[i];
			}
			this.p = p;
		}
	}
	
	class Line{
		P p1, p2;

		public Line(P p1, P p2) {
			this.p1 = p1;
			this.p2 = p2;
		}
		
	}
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(true){
			int n = sc.nextInt();
			if(n == 0)break;
			double [] input = new double[3];
			for(int i = 0; i < 3; i++) input[i] = sc.nextInt();
			P last = new P(new double[3]);
			P dir = new P(input);
			P [] data = new P[n];
			double [] rlist = new double[n];
			for(int i = 0; i < n; i++){
				double [] temp = new double[3];
				temp[0] = sc.nextInt();
				temp[1] = sc.nextInt();
				temp[2] = sc.nextInt();
				data[i] = new P(temp);
				rlist[i] = sc.nextInt();
			}
			while(true){
				double mn = 1 << 24;
				int mni = -1;
				Line l = new Line(last, dir);
				for(int i = 0; i < n; i++){
					double disL = distanceLP(l, data[i]);
					if(disL >= rlist[i]) continue;
					double lo = 0, mid = -1;
					double hi = dot(sub(data[i], l.p1), sub(l.p2, l.p1)) / norm(sub(l.p2, l.p1));
					for(int it = 0; it < 100; it++){
						mid = (lo + hi) /2;
						if(abs(sub(add(l.p1, mul(mid,(sub(l.p2, l.p1)))), data[i])) < rlist[i]){
							hi = mid;
						}
						else{
							lo = mid;
						}
					}
					if(lo < EPS )continue;
					if(mn > mid){
						mn = mid;
						mni = i;
					}
					

					
				}
				if(mni == -1){
					break;
				}
				P nextdist = add(last, mul(mn, sub(dir, last)));
				Line midLine = new Line(data[mni], nextdist);
				P nextterm = reflection(midLine, last);
				last = new P(nextdist.p.clone());
				dir = new P(nextterm.p.clone());
			}
			System.out.printf("%.4f %.4f %.4f\n",last.p[0],last.p[1],last.p[2] );
		}
	}

	private P reflection(Line l, P p) {
		return add(p, mul(2.0, sub(projection(l,p) , p)));
	}

	private double distance(P p1, P p2) {
		double res = 0;
		for(int i = 0; i < 3; i++){
			res += (p1.p[i] - p2.p[i]) * (p1.p[i] - p2.p[i]);
		}
		return Math.sqrt(res);
	}

	private double distanceLP(Line l, P p) {
		return abs(sub(p, projection(l,p)));
	}

	private double abs(P p) {
		return Math.sqrt(norm(p));
	}

	private double norm(P p) {
		return dot(p,p);
	}

	private double dot(P p1, P p2) {
		double res = 0.0;
		for(int i = 0; i < 3; i++){
			res += p1.p[i] * p2.p[i];
		}
		return res;
	}

	private P sub(P p1, P p2) {
		double [] res = new double[3];
		for(int i = 0; i < 3; i++){
			res[i] = p1.p[i] - p2.p[i];
		}
		return new P(res);
	}

	private P projection(Line l, P p) {
		double t = dot(sub(p, l.p1), sub(l.p1, l.p2)) / norm(sub(l.p1, l.p2));
		return add(l.p1, mul(t, (sub(l.p1, l.p2))));
	}

	private P add(P p1, P p2) {
		double [] res = new double[3];
		for(int i = 0; i < 3; i++){
			res[i] = p1.p[i] + p2.p[i];
		}
		return new P(res);
	}

	private P mul(double t, P p) {
		double [] res = new double[3];
		for(int i = 0; i < 3; i++){
			res[i] = p.p[i] * t;
		}
		return new P(res);
	}

	public static void main(String [] args){
		new Main().doit();
	}
}