Compute the probability that one part of a randomly cut equilateral triangle can be made to cover the other (with flipping allowed); see http://math.stackexchange.com/questions/1689806.

Question1689806.java

public class Question1689806 {
	static class Point {
		double x,y;

		public Point (double x,double y) {
			this.x = x;
			this.y = y;
		}
	}
	
	static class Segment {
		Point a,b;
		
		public Segment (Point a,Point b) {
			this.a = a;
			this.b = b;
		}

		public Point midPoint () {
			return new Point ((a.x + b.x) / 2,(a.y + b.y) / 2);
		}
	}

	static Point c = new Point (0,Math.sqrt (3) / 6);
	
	static class Line {
		double theta;
		double d;
		
		public Line (double theta,double d) {
			this.theta = theta;
			this.d = d;
		}
		
		public boolean cuts (Segment s) {
			return sign (s.a) * sign (s.b) < 0;
		}
		
		public double sign (Point p) {
			return Math.sin (theta) * (p.x - c.x) + Math.cos (theta) * (p.y - c.y) - d;
		}
	}
	
	public static void main (String [] args) {
		Point [] points = new Point [] {
				new Point (0,Math.sqrt (3) / 2),
				new Point (-0.5,0),
				new Point (+0.5,0)
		};
		
		Segment [] sides = new Segment [3];

		for (int i = 0;i < 3;i++)
			sides [i] = new Segment (points [i],points [(i + 1) % 3]);
		
		Segment [] altitudes = new Segment [3];
		
		for (int i = 0;i < 3;i++)
			altitudes [i] = new Segment (points [(i + 2) % 3],sides [i].midPoint ());
		
		int count = 0;
		int ntrials = 100000000;
		
		for (int n = 0;n < ntrials;n++) {
			Line line;
			boolean cuts = false;
			
			do {
				line = new Line (Math.random () * 2 * Math.PI,Math.random () / Math.sqrt (3));
				for (Segment side : sides)
					cuts |= line.cuts (side);
			} while (!cuts);
			
			for (Segment altitude : altitudes)
				cuts &= line.cuts (altitude);

			if (!cuts)
				count++;
		}
		
		System.out.println (count / (double) ntrials);
	}
}