Cut one stick twice to turn it to be three sticks, what is the probability that the three sticks form a triangle? [closed]

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Well it is not a program problem. Is there any hint for such quiz? I am thinking about focusing on two random R1, 开发者_开发技巧R2, both of which is in range (0, 1). and supposing R2 > R1 and then fulfill two equation:

R1 + (1 - R2) > R2 - R1 // two sticks sum longer then the rest one
|R1 - (1 - R2)| < R2 - R1 // the difference of these two should be shorter the rest one

but I cannot move further...

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Think of (r1, r2) as a point in the unit square.

• Which part of the unit square is allowed for r2 > r1?

• Which part of that leads to three lengths that can form a triangle?

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The answer is 1/4. Here is the explanation.

Let x is the length of the leftmost stick and y is the length of the rightmost stick. Then the middle stick has length n-x-y, if the original stick's length was n.

The possible values for x,y are those for which:

1. x > 0
2. y > 0
3. x + y < n

In the plane Oxy this is equivalent to say that the point (x,y) lies within the triangle with vertices (0, 0), (n, 0), (0, n).

Now these three numbers (x, y, n-x-y) form a triangle if all of the three are satisfied:

1. x + y > n - x - y <=> x + y < n/2
2. x + (n - x - y) > y <=> y < n/2
3. y + (n - x - y) > x <=> x < n/2

Again in the Oxy plane these are satisfied when the point (x,y) lies within the triangle with vertices (0, n/2), (n/2, n/2), (n/2, 0).

The area of this triangle is a quarter of the area of the (0, 0), (n, 0), (0, n) triangle, since it's the 'middle' triangle (whose vertices are the midpoints) of the bigger one.

Here is a simple C# program to verify the answer:

Random r = new Random();
int count = 0, total = 0, tries = 1000000;
double x, y;

for (int i = 0; i < tries; i++)
{
    x = r.NextDouble();
    y = r.NextDouble();
    if (x + y > 0.5 && x < 0.5 && y < 0.5) ++count;
    if (x + y < 1.0) ++total;
}

Console.WriteLine((double)count / total);

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I have just made a program to verify it, and I found it is 1/4:

class Program
{
    static void Main(string[] args)
    {
        int nIsTriangle = 0;
        Random ran = new Random(0);

        int nTry = 1000000;

        for (int i = 0; i < nTry; i++)
        {
            double r1 = ran.NextDouble();
            double r2 = ran.NextDouble();
            if (Check(r1, r2)) nIsTriangle++;
        }

        Console.WriteLine((double)nIsTriangle / (double)nTry);
        Console.ReadKey();
    }

    static bool Check(double r1, double r2)
    {
        double first = Math.Min(r1, r2);
        double second = Math.Abs(r1 - r2);
        double third = 1 - Math.Max(r1, r2);
        bool conditionA = (first + second) > third;
        bool conditionB = Math.Abs(first - second) < third;
        return conditionA && conditionB;
    }
}