Stick and a triangle

You have a stick.
You snap it in a random place along the stick.
You snap it again in a random place along the stick.

The two snaps are made independent of each other. IE, the first determines nothing about the second.

What's the probability that the three pieces can make up a triangle?

50% (chance of having no section longer than half the stick).

I think it's 75%. I have no proof, i just have enough of one to convince myself.

Originally posted by doodinthemood
You have a stick.
You snap it in a random place along the stick.
You snap it again in a random place along the stick.

The two snaps are made independent of each other. IE, the first determines nothing about the second.

What's the probability that the three pieces can make up a triangle?

As long as the length of two pieces together is more than the lenth of the third piece you will be able to make a triangle.

If you snap the stick in two halves on the first break, there will be no triangle. Fortunately, the chance of this happening is infinitely small.

Suppose you snap the stick at a/b of the total length, with b > 2a.
Now, if you make the second break in the smaller part, you will not be able to make a triangle. The chance to snap the stick in the longer part is 1 - a/b. You will always be able to make a triangle in this case.

To get a numeric value, we need to make an integral:

Suppose we get a stick of length x at the first break, with x < 1/2. The chance to make a traingle after the second break is 1-x.

2 * INTEGRAL ( 0 to 1/2) (1-x dx) should give the probability requested.

Note that the factor 2 comes from symmetry, x < 1/2 or x > 1/2.

Apologies to any mathematician/physicist if above is nonsense, it's been a while since I had to make my own integrals.

Above integral comes out to be 3/4. That I am sure of 🙂

Precicly how I thought about it in my mind. I just wasn't familiar enough with calculus to be sure that it worked that way: I've only just started MM 1+2.

EDIT: Wait a minute, are you saying that if we first snap it at (a/b) < 1/2 we can always make a triangle if the second snap is in the longer part? That's not true... snap at 1/5 and 4/5...

EDIT: I'm confused let me get this straight...

The first cut was made x along the length of the stick, x < 1/2.
There is 1-x remaining. Another cut now must be made. If this cut is made in the smaller part (x), then there is no triangle. So we have to cut 1-x into y and 1-x-y. Then the following must be satisfied:
x+y > 1-x-y or x+y > 1/2
x+(1-x-y) > y or y < 1/2
y+(1-x-y) > x or x < 1/2 (given)
1/2-x < y < 1/2
The range of possible y is clearly x
So the probabilty is the integral of f(x)=x from 0 to 1/2
The same reasoning applies to x from 1/2 to 1 and we get the integral of f(x)=1-x from 1/2 to 1. Adding these we get 25%

Err... am i right?

Originally posted by Dejection
The first cut was made x along the length of the stick, x < 1/2.
There is 1-x remaining. Another cut now must be made. If this cut is made in the smaller part (x), then there is no triangle. So we have to cut 1-x into y and 1-x-y. Then the following must be satisfied:
x+y > 1-x-y or x+y > 1/2
x+(1-x-y) > y or y < 1/2
y+(1-x-y) > x or x < 1/2 (given)
1 ...[text shortened]... and we get the integral of f(x)=1-x from 1/2 to 1. Adding these we get 25%

Err... am i right?

I make it .25 for a similar reason. I first tested it using a simulation though so knew the answer I was aiming for!

Put me down as another one for 0.25.

If the cuts are made at x and y: w.l.o.g. we can assume x < 0.5.

Then the fragments will make a triangle if 0.5 < y < 0.5 + x - a range of x.

Congratulations people who got 25%

And well done to those who didn't; your effort is appreciated and your courage to post a wrong answer in front of millions online is respectable.

The cuts are made INDEPENDENTLY of each other, so there's no need to say you have 1-x of a stick left, as you can make the second cut on the bit you broke off.

For it to be a triangle, the first cut cannot be more than half way over an allocated side and the second cut cannot be more than half way from the opposit side.

=0.5^2

=0.25

You can make a triangle out of 3 of anything as long as you allow for overlap.

Originally posted by uzless
You can make a triangle out of 3 of anything as long as you allow for overlap.

I don't understand.

Suppose you have one stick that is 8,000,000 meters long. You have two other sticks that are each 1 meter long. You can make a triangle with those three sticks? Maybe, depending on what you mean by "overlap".

Or since it's "3 of anything", I give you 3 chicken eggs. How would you construct your triangle?

Originally posted by LemonJello
I don't understand.

Suppose you have one stick that is 8,000,000 meters long. You have two other sticks that are each 1 meter long. You can make a triangle with those three sticks? Maybe, depending on what you mean by "overlap".

Or since it's "3 of anything", I give you 3 chicken eggs. How would you construct your triangle?

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There is still a triangle on the left [pretend the small space between the lines of type isn't there], even though there is a lot of stick hanging out to the right.

It probably should be stipulated that the sticks must be joined end-to-end.

Originally posted by SwissGambit
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There is still a triangle on the left [pretend the small space between the lines of type isn't there], even though there is a lot of stick hanging out to the right.

It probably should be stipulated that the sticks must be joined end-to-end.

I figured that is what was meant by "overlap".

Originally posted by doodinthemood
Congratulations people who got 25%

And well done to those who didn't; your effort is appreciated and your courage to post a wrong answer in front of millions online is respectable.

The cuts are made INDEPENDENTLY of each other, so there's no need to say you have 1-x of a stick left, as you can make the second cut on the bit you broke off.

For ...[text shortened]... side and the second cut cannot be more than half way from the opposit side.

=0.5^2

=0.25

I think this was the neat solution that had been eluding me.

Originally posted by LemonJello
I figured that is what was meant by "overlap".

Then what didn't you understand about my previous post???