Here is another probability problem. (Probability is one of my favorite topics and I plan to post more probability problems in my subsequent posts. These problems shall be of use to school students who prepare for competitive exams and math enthusiasts.)

Two points are chosen randomly on a uniform stick. Find probability that three pieces formed by making two cuts across these two points, shall form a triangle? This problem for first time appeared in a Cambridge University entrance examination paper conducted on 18th January , 1854. Below I present three solutions to the problem and the problem is usually called The Broken Stick Problem.

## Solution#1:

Let 's for simplicity assume a stick of unit length. Let the randomly chosen two points be $x$ and $y$ on the stick.

Case#1: Assume $x \gt y$Then the leftmost edge has a length of x , the middle edge has length y - x and the right edge has length 1 - y.

We know that for a triangle to be formed from three pieces, the length of the largest piece should be smaller than the sum of other two smaller pieces or for that matter, length of one side has to be smaller than the length of the other two sides. This is because the other two pieces should be able to touch or reach each other (when they are laid from the ends of a larger piece) for triangle to be formed. This fact is known as triangle inequality.

Let's write the condition for the above three pieces which satisfies triangle inequality.$$ x \lt (y - x) + (1-y) \rightarrow 2x \lt 1 \rightarrow x \lt 0.5\\

(y - x) \lt (1 - y) + x \rightarrow 2(y - x) \lt 1 \rightarrow y - x \lt 0.5 \\

(1 - y) \lt x + (y - x) \rightarrow 2(1 - y) \lt 1 \rightarrow y \gt 0.5 \\

$$

Let's plot graph with X-axis and Y-axis corresponding to x and y values. The above equations constitute correspond to the upper left triangle. Similar by symmetry or we can derive set of equations as above for y \gt x which is case#2. This set of equations corresponds to the lower right triangle (shaded green) shown as in figure below:

Now the desired probability is the sum of the area of the triangles divided by the area of unit square which is $0.25$ or $\frac 1 4$.## Solution#2:

I will give another proof making use of the properties of equilateral triangle.

In the above figure, there is a larger equilateral triangle. Now connect the midpoints of the sides of the equilateral triangle. The result is four interior equilateral triangles as shown in the figure. The area of these four equilateral triangles sum up to the area of a larger equilateral triangle.

Consider any point P in the larger equilateral triangle. If we drop perpendiculars from the sides of the equilateral , then sum of the three perpendiculars from the sides of the equilateral triangle sum up to the altitude of the triangle. If let the altitude of the equilateral triangle to be of unit length, then the

three perpendicular bisectors represent the three pieces of a stick (broken at random by considering two points on the stick) and the number of different points within the equilateral triangle form the possible sample space/event space for the problem in question.

Now for the three pieces (perpendicular bisectors) to from a triangle, the point P should lie anywhere inside the hatched triangle shown in the figure. This condition ensures the side of larger perpendicular bisector is less than the sum of the other two. This is a necessary condition (the triangle inequality) for a triangle to be formed.

Thus the probability is equal to the area of the hatched triangle divided by the area of the larger triangle and this ratio is clearly .25 or $\frac {1} {4}$.

## Solution#3:

There is yet another solution. Connect the two ends of the stick (on which we have chosen two random points) into a circle. Now on this circle, let's choose three random points and then unroll the circle from one chosen point, then this transforms itself into a stick with remaining two random points. All that we need to prove that the none of the pieces doesn't have length greater than $\frac {1} {2}$. This means we need to find the probability that none of the three points lie on the semicircle. This we find by the probability of the complementary, that is, the probability of three points lying in a semicircle.

We consider one chosen point and proceed counterclockwise to chose two other points. In order to ensure that these points lie in a semicircle , it must be chosen such that it is within $180^{\circ}$ from the first chosen designated point. This happens with probability $ \frac {1} {2} \times \frac {1} {2} = \frac {1} {4}$. The designated first point can be chosen in three ways. So the probability is $\frac {3} {4}$. The complementary probability is $\frac {1} {4}$ which is the probability for the three pieces obtained from a stick broken at two randomly chosen locations to form a triangle.

There is yet another way to arrive at an answer here. Let's consider two points which are within $\le 180^{\circ}$ on the circle. From these two points, draw diameters to connect to their antipodes. Now for three points not to lie within a semicircle , the third point must lie on an arc formed by two antipodal points. Thus the probability is length of these antipodal arc divided by $2\pi$ (circle of unit radius). Now the two diameters cut into 4 arcs which are equal in length. Thus the length is proportionate to $\frac {2\pi}{4}$. Thus the probability is $$\frac {\frac {2\pi}{4}} {2\pi} = \frac {1} {4}$$

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