Can It Make A Triangle?

You can also solve it graphically quite easily!

Let the break points be $x$ and $y,$ and by symmetry assume that $y\ge x.$ Our sample space looks like:

The lengths of the segments are $x,\ y-x, \ 1-y.$ The Triangle Inequality should hold for all 3 ordering of lengths:

$x+(y-x) > 1-y \qquad \Rightarrow y>\frac{1}{2}$
$x+(1-y) > y-x \qquad \Rightarrow y < x + \frac{1}{2}$
$(y-x)+(1-y)>x \qquad \Rightarrow x < \frac{1}{2}$

Plotting this "good" region on our outcomes gives:

By Geometric Probability , the probability is thus $\frac{\text{area of good region}}{\text{area of all outcomes region}} = \frac{1/8}{1/2} = \frac{1}{4}=0.25.$

Imagine the rod on a number line so that one end is at $x = 0$ and the other at $x = 1$ . Now cutting the rod into $3$ pieces involves making $2$ cuts, call them $x_{1}$ and $x_{2}$ . By the triangle inequality, the $3$ pieces resulting from the $2$ cuts can form a triangle only if none of the pieces exceeds a length of $\frac{1}{2}$ .

Now suppose the first cut $x_{1}$ lies on the interval $[0, \frac{1}{2}]$ , (which will occur with probability $\frac{1}{2}$ ). Then to ensure that no piece exceeds a length of $\frac{1}{2}$ , the second cut would need to occur somewhere on the interval $[\frac{1}{2}, (x_{1} + \frac{1}{2})]$ , which will occur with probability $(x_{1} + \frac{1}{2}) - \frac{1}{2} = x_{1}$ . Now since the domain of $x_{1}$ is uniform over the interval $[0, \frac{1}{2}]$ , the probability that the pieces can form a triangle will be

$\displaystyle\int_{0}^{\frac{1}{2}} x_{1} dx_{1} = \dfrac{1}{8}$ .

By symmetry, the same result will be found if $x_{1}$ lies in the interval $[\frac{1}{2}, 1]$ , giving a final probability of $2*\dfrac{1}{8} = \dfrac{1}{4} = \boxed{0.25}$ .

A graphical solution: Draw a suare diagram: on X axis is the first cut, on Y axis is the second cut. Hached area is where a triange is possible White are is where a triangle is possible

Answer: 0.25

Say we cut the stick (which will here represent $[0,1]$ ) into three pieces of length $a, b, c$ , and assume that $a$ is the largest.

$a+b+c=1$ and we require $a<b+c$ in order to form a triangle. Since $b+c=1-a$ we have $a<1-a$ so $a<\frac{1}{2}$

Lets assume our first cut is at some point $x$ in the first half of the stick ( $[0,\frac{1}{2}]$ ) for simplicity (w.l.o.g.), and note that the section of the stick where we can safely make the second cut is precisely $[\frac{1}2,x+\frac{1}{2}]$ , whose length is $x$ . Since $x$ varies from $0$ to $\frac{1}{2}$ and is distributed uniformly, its expected value is $\frac{1}{2}=\boxed{0.25}$

Let the stick be broken into pieces of length $p, q, 1-p-q$ . Then by the triangle inequality, we have:

$p + q > 1 - p - q \Rightarrow q > 1/2 - p$ $p + 1 - p - q > q \Rightarrow 1/2 > q$ $q + 1 - p - q > p \Rightarrow 1/2 > p$

which are all the possible orderings. Combining the first and second inequalities, $1/2 - p < q < 1/2$ .

Without loss of generality, pick length $p$ first, as the pieces are still the same length after swapping $p$ and $q$ . We can pick any value of $p$ between $0$ and $1$ , which has probability $1$ in the sample space. Then $q$ is in the region above $q = 1/2 - p$ and below $q = 1/2$ , which has a probability of $1/2$ by graphing (event $A$ ).

Then the remaining length $1-p-q$ must be positive. Adding $p$ to both sides, $1/2 < p + q < 1/2 + p$ , so $-1/2 > -p-q > -1/2 + -p$ as the sign flips when we multiply by a negative number, and adding $1$ to both sides, $1/2 > 1-p-q > 1/2-p$ or $1/2-p < 1-p-q < 1/2$ .

And $1-p-q$ must be above $1/2-p$ but below $1/2$ , which is exactly the same probability as earlier: $1/2$ , (event $B$ ).

The lengths of $q$ and $1-p-q$ are independent as $P(A|B) = P(A|B')$ and $P(B|A) = P(B|A')$ : they are all $\frac{1/2}{1/2} = 1$ as $1 - \frac{1}{2} = \frac{1}{2}$ . Hence we can multiply the probabilities to get $1 \cdot 1/2 \cdot 1/2 = \boxed{\frac{1}{4}}$ .

Answer here: in this link