A stick is broken into 3 pieces randomly. What are the chances that the 3 pieces will form a triangle ???

Question

length of 3 pieces can be anythng between L and 0.

the point is that if one piece is longer than other two combined, triangle won't be formed....but wot is the probability of this ?????

Tom's explanation looks most logical...lemme explain why....

Lets take a stick of length 10 meters.I break it into two parts(and ignoring physical barriers, I can even get one of the parts like 1 nanometer or smaller).

So now what is the probability that part A will be smaller than part B??.........the answer is impossible to determine....because I can get the length of A to be between 0 and 5 meteres in an infinite number of ways( there are always an infinte numbers between any two continuous quantities.......like 0.000000001,0.000000002 and so on).

So the solution to this problem can be determined in terms of conditions like AB. Thus u can say that probability is 50% for both cases.

If someone disagrees, put forward your point...

Answer 1

Let the 3 sides be x, y and z.
For a triangle to be formed, one of the following triangle inequalities must hold, i.e.

|x+y| > z or
|x+z| > y or
|y+z| > x or

The following equality will also ensure a triangle:
x=y=z (equality)

So, let's consider the favourable possibilities:

x=y=z
x+y > z
x+z > y
y+z > x

Now let's consider unfavourable possibilities:

x+y < z
x+z < y
y+z < x

These are all the possibilities (seven of them).
So the chances of forming a triangle are 4/7 or 57% if the stick is broken randomly.

Answer 2

The basic idea is that the two shortest pieces have to add up to at least a tad more than the longest piece.

But what is the probability that this happens? It's impossible to say... unless you are precise about what you mean by "randomly." There are lots of possibilities. Let's assume one reasonable scenario...

a. Break the stick in two at a point that is equally likely at any place along the stick.

b. Select the longer piece and do the same.

Note that if you break the shorter piece in step b, the answer is zero, since the two shorter pieces will not add up to the longer piece, and you can't make a triangle with those three peices. This is one example of how you need to define "randomly" to get an answer.

Anyway, the probability is a bit tricky to work out, but comes to

.1931471805599453094172321... = ln(2) - 1/2

or approximately 19%.

Taking a different scenario and meaning of "randomly", suppose you "randomly" pick two points to break the stick twice, chosen in advance, again so that every point is equally likely. Then the math is easier and comes out to exactly 1/4, or 25%.

So... it depends! But interestingly, the answer has nothing to do with the length of the stick, L. That's pretty reasonable, when you think about it.

Of course, if an ordinarly person does it with their hands or even an ordinary tool, they are more likely to break the stick somewhere the middle than at the ends. That violates the assumptions of both solutions above, and changes everything. In fact, it's awfully hard to break a stick with your hands close to the end, right-- or even with a pair of vise grips really really close to the end!

But a footnote -- all bets are off if the stick has length 0, as you say is possible. Precisely what ~is~ a stick of length zero? What is a triangle with three sides all of length zero? Anything meaningful?

Answer 3

The only possibility of it not forming a trinagle is when one of the sides is of 0 lenght. However this cannot be possible because then there would only be 2 pieces and not 3.
Hence the probability of the stickes to form a triangle is 100%

Answer 4

It has been kind of answered above but the solution is missing, let me fill in. Let us say the length of the stick is 1m and the first piece broke at the x m mark, second piece broke y m away from that. So that the three pieces of stick you have are of length x, y and 1-(x+y) meters. And of course x+y<=1. Like this

0...........................1
_________________
.............x........y

Now consider the following picture

....| y
....|
....|
..1|\
....|..\
....|....\
....|......\
....|____\______________
............1......................x
Here every point (x,y) you pick inside the isoceles triangle with vertices at (1,0), (0,0) and (0,1) corresponds to a break-up of the stick as we explained above. Now notice that not every break-up will lead to a triangle as the sum of the lengths of two short sides has to be greater than the length of the longest side. That means the following inequalities has to be satisfied:

.................x+y>=1-(x+y)
.....x+(1-(x+y))>=y
.....y+(1-(x+y))>=x
After a little algebra we simplify the above inequalities as:

................x+y>=1/2
.................1/2>=y
.................1/2>=x

Now the conditions above are satisfied for the points that fall into the triangle with vertices (0,1/2), (1/2,1/2), (1/2, 0). Now our event universe corresponds to the big triangle above which has area 1/2*1*1=1/2. In this event universe, we are looking for the events that correspond to the points in the smaller triangle above, which has area 1/2*1/2*1/2=1/8. So the probability that the pieces will form a triangle is
1/8
--- = 1/4
1/2

Answer 5

The first break of the stick will produce 2 parts. The probability that the two parts will be the same length is in effect zero. So we have two parts, one longer than the other. We now have to select one of those to break to end up with 3 parts. The probability of selecting the shortest part is, of course 0.5 or 50%. The probability of selecting the longest part, which would allow a triangle to be formed is also 50%.

So there is your answer 50%.

Answer 6

The answer is 25% probability.

Say you select a stick of length 10" and randomly from a uniform distribution select two lengths between 0 and 10 (x and y) and let the third side by 10-x-y. Eliminating all values that sum up greater than 10, you would get a sample of possible stick lengths. I did this for 40,000 possible breaks.

From this I then looked at the inequalities for a triangle to exist:

ABS(X-Y)<=Z<=(X+Y) or
ABS(X-Z)<=Y<=(X+Z) or
ABS(Y-Z)<=X<=(Y+Z)

When I looked at my results, ~10000 of the 40000 breaks meet one of the three inequalities above and would form a triangle.

Assumptions: Uniform distribution of breaks.
Length of Stick>0.

Look at the link below for an applet to test this out.

Answer 7

All 3 pieces will form a triangle as long as none of them are not zero length

Answer 8

No two sides together can be less than the third. Put the first break wherever you want. The second break must happen in the larger piece. To answer the question specifically, you need to decide whether you break it in two, then pick one or the other, or if you select two break points all at once. This is in the nature of probability problems: the method matters.

Answer 9

It will always be a triangle...it just might not be an equilateral triangle...to answer that question it would be infinent chances to form an equilateral triangle.

Good luck

Answer 10

the three sticks (sides) will always form a triangle.