A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

Lightning
Mar 29, 2018

#1**+2 **

Suppose the points where we break the line are x and y, where 0 < x < y < 1. So our "possible" region is the following triangle on the plane:

The three sides have lengths x, y-x, and 1-y. We must satisfy the Triangle Inequality on all three sides, meaning that the sum of any two of these lengths must be greater than the third. This gives us

1. : This gives y>1-y, so y> 1/2.

2. : This gives 1-x> x, so x<1/2.

3. : This gives , so .

Adding these lines to the diagram gives us the darker "success" region.

The darker region makes up 1/4 of the total region, so the probability is 1/4.

MIRB16
Mar 29, 2018

#3**+2 **

A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?

I would do this with a boolean contour map.

The line segment can be any length so let it be 1

Let the segment by cut into segments that are x, 1-y and y-x units long.

(If you add these you can see that the total length is 1 unit.)

No each of these lengths must be between 0 and 1 unit long.

so

\(0 x \quad (3) \qquad and \qquad y

I need to plot these to get the full sample space. The dark triangle is the sample space.

Now in order for the 3 peices to from a triangle, the sum of the two little ones must be greater than the length of the long one.

Remembering that the segments are x, 1-y and y-x units long.

\(x+(1-y)>y-x\\ 2x+1>2y\\ y

AND

\(x+(y-x)>1-y\\ y>1-y\\ 2y>1\\ y>0.5 \qquad(6)\)

AND

\((1-y)+(y-x)>x\\ 1-x>x\\ 1>2x\\ x<0.5 \qquad (7)\)

So a quarter of the sample space is the area of the desired region.

P(the three pieces form a triangle) = 0.25

Melody
Mar 29, 2018