dimbulb

Lets say you have a line of length 1.

Generate 2 points randomly ( uniform distribution) within the line.

Cut the line at these 2 points.
What is the probability that you can form a triangle with the 3 resulting line segments?

TIO

1/2

To make a triangle, you need one cut on one side of the halfway point, and one cut on the other side. There's a 1 in 2 chance of that happening.

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Strewth

lordjeebus

It's not that simple.

What if each cut is .1 away from the ends? so you have two pieces of .1 and one of .8. No triangle there.

I'm thinking about this problem but I don't think I have the brain-energy tonight.

lordjeebus

I believe the answer is 1/4. I will publish my findings shortly.

EDIT: my theory.

so, you're going to get three segments. let's call them x, y, and z.

because the original line had length 1, x+y+z=1.

To form a triangle, no one side can be longer than the other two sides combined. Hence, x+y>z, y+z>x, x+z>y.

Combing the equations x+y+z=1 and x+y>z, we find that x+y = 1-z > z; 1 > 2z; so z<.5. Similarly, y<.5 and x<.5 to form a triangle.

What are the chances of randomly picking x, y, and z, so that all three are less than .5?

The set of all x, y, and z that add up to 1 form a plane. We will only consider the section of that plane in which x, y, and z are positive. This section of the plane is shaped like an equilateral triangle.

By connecting the midpoints of that equilateral triangle, a second triangle can be formed. Within this triangle, and only within this triangle, are the conditions x+y+z=1, x<.5, y<.5, and z<.5 met.

The larger triangle is the set of ways the line could be cut twice. The smaller one is the set of ways that could form a triangle.

The ratio of the smaller triangle to the larger one is 1:4. Hence, given two random cuts of the line, the chance of being able to form a triangle is 1/4.

If there is interest I'll make some illustrations to make things a bit clearer. I'm sure you can solve this without thinking so geometrically, but I find this way easier to understand.

TIO

Well done, Jeebus! Now that you mention it, I feel silly for having made such a simple omission.

I didn't look at the triangles in your solution, but the rest looks good and I have no reason to believe the triangles aren't the way you have them set up.

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Strewth

TheFirstDuffMan

Jeebus got it.

Dilbert1234567

praise jeebus

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