Sticky triangle problem

Say the length of stick is $2 l$ . Let us break it in such a way that $x , y$ and $2l - x - y$ are sides of triangle, then :

$x + y > 2l - x - y \Rightarrow x + y > l$

$x + 2l - x - y > y \Rightarrow y < l$

$y + 2l - x - y > x \Rightarrow x < l$ .

Also, clearly $x + y < 2l$ (as x and y both are part of a stick of length 2l).

Graphically , the situation is represented below. The shaded region shows the favorable region, and $x + y < 2l$ is the sample space.

Clearly, required probability = $\frac{l^2/2}{2l^2} = 0.25$

These type of probability, in most of the cases, is best represented geometrically. Consider the stick is $1$ unit long, broke into segments of $x, y$ and $1-x-y$ . Of course $x+y<1, x>0, y>0$ . These are the conditions for all kinds of possible breakings. A triangle on the Cartesian plane bounded by these lines has area $\frac{1}{2}$ . Now these segments form a triangle if and only if that satisfy all three triangle inequalities used on all three sides, which are

$x+y>1-x-y \iff x+y>\frac{1}{2}$ ,

$x+1-x-y>y \iff y<\frac{1}{2}$ ,

$y+1-x-y>x \iff x<\frac{1}{2}$ ,

and note that these three inequalities, if they all hold true, the previous three inequalities are also true. Therefore, consider the triangle bounded by these lines, it has area $\frac{1}{8}$ , and the probability that results the three segments making a triangle is the probability of choosing a point $(x,y)$ in the triangle that has area $\frac{1}{2}$ , such that it lies in the triangle that has area $\frac{1}{8}$ , which gives the answer $\frac{\frac{1}{8}}{\frac{1}{2}}=0.25$ .

Of course calculus works, but they are basically the same thing, which I think this way is cleaner. Be sure to use this technique when you see this type of problems!

Consider a point $M$ inside an equilateral triangle $ABC$ , as shown. Then $MP+MQ+MR$ is a constant, which is the altitude of the triangle.

This means that for any TWO random points on stick, there is ONE corresponding point M inside an equilateral triangle.

For the three pieces to form a triangle, the longest of these 3 pieces is less than ½ of the original length.

If $M$ is within the pink triangle, then the longest piece is than ½ of the original length. Hence the corresponding 3 pieces will form a triangle if M is within the pink triangle. The probability is $\dfrac{\text{area of pink triangle}}{\text{area of triangle } ABC} = \boxed{\dfrac{1}{4}}$

Discussion can be found in this video .