# A Triangle out of Three Broken Sticks

### Problem

### Solution, Part 1

Assume the sticks are of length $1$ and consider the cube with vertices $A(0,0,0),$ $B(1,0,0),$ $D(0,1,0),$ $E(0,0,1),$ $C(1,1,0),$ $F(1,0,1),$ $H(0,1,1),$ and $G(1,1,1).$

The left pieces of the sticks are defined by their right points and I shall describe their lengths with the real $x,y,z\in(0,1).$ Thus the cube constitutes the sample space for all possible combinations of the breaking points. For $x,y,z$ to form a triangle, we need three inequalities:

$\begin{align} z&\lt x+y\\ y&\lt x+z\\ x&\lt y+z. \end{align}$

Note that, say, line $AF$ has the equation $z=x$ and line $AH$ the equation $z=y,$ so that the plane through $A,F,H$ is described by $z=x+y.$ The plane divides the space into two half-spaces. In the one that contains $G,$ $x+y\gt z.$

Similarly, the plane $AFC$ is described by $x=y+z$ and the plane $ACH$ by $y=x+z.$ The set of points that satisfies the three inequalities belongs to the intersection of the three half-spaces and the cube: that's the figure $V=ACFHG.$ It consists of two pyramids with the base $CFH$ and apices at $A$ and $G.$

$V$ is obtained from the cube by cutting corner pyramids $ACFB,$ $ACHD,$ and $AFHE.$ Each of these has volume $\displaystyle \frac{1}{3}\cdot\frac{1}{2}\cdot 1=\frac{1}{6}.$ It follows that the volume of $V$ is $\displaystyle 1-3\cdot\frac{1}{6}=\frac{1}{2}.$

### Solution, Part 2

If now $x,y,z$ stand for the lengths of the longest pieces than $\displaystyle \frac{1}{2}\lt x,y,z\lt 1.$ Then for example, $\displaystyle x+y\gt\frac{1}{2}+\frac{1}{2}=1\gt z,$ and, similarly, for the other two inequalities. We conclude that the probability in this case is $1.$

### Solution, Part 3

If now $x,y,z$ stand for the lengths of shortest pieces then $0\lt x,y,z\lt\frac{1}{2}$ and each of the three is drawn uniformly randomly from the interval $\displaystyle\left(0,\frac{1}{2}\right).$ The situation is similar to Part 1 so that the probability in this case is $\displaystyle \frac{1}{2}.$

### Solution, Part 4

First I should mention that quite a few people observed that in Part 1 whether we choose left or right pieces because both side lengths of a broken stick have the same probability distribution. This implies that when a random choice is made between the left and right pieces for each stick, the probability of getting a triangle is no different than in Part 1, i.e., $\displaystyle \frac{1}{2}.$

It occurred to me that there might be a difference with the case where the choice is made between the long and the short pieces, rather than between the left and the right ones. Many people insisted their should not be any difference. I made a mistake of thinking that since $\min(x,1-x)$ is distributed on $\displaystyle \left[0,\frac{1}{2}\right]$ while $\max(x,1-x)$ is distributed on $\displaystyle \left[\frac{1}{2},1\right].$ Tangentially, support also came from the difference in probabilities in Parts 2 and 3. That was explained and corrected by Timon Klugge.

where the three numbers $x,y,z$ satisfied $0\le x,y,z\le\frac{1}{2}$ and formed a triangle. Let $\Delta(x,y,z)=1,$ if $x,y,z$ form a triangle, and $0,$ otherwise. We could denote the probability we found as$\displaystyle P\left(\Delta(x,y,z)=1\,|\,0\le x,y,z\le\frac{1}{2}\right)=\frac{1}{2}.$

Now, if $\Delta(x,y,z)=1$ then also $\Delta(2x,2y,2z)=1,$ such that we were able to conclude that in the domain of definition, $[0,1]^3,$ $\Delta(x,y,z)=1$ half the time.

Now, there is a harder road to arrive at the same conclusion. Assume $\displaystyle 0\le x,y,z\le\frac{1}{2},$ what is the probability of $P(\Delta(x,y,z)=1)?$ In other words, what is $\displaystyle P\left(\Delta(x,y,z)=1\text{ and }0\le x,y,z\le\frac{1}{2}\right)?$ The formula for conditional probability gives

$\displaystyle\begin{align}&P\left(\Delta(x,y,z)=1\text{ and }0\le x,y,z\le\frac{1}{2}\right)\\ &\qquad= P\left(\Delta(x,y,z)=1)\,|\,0\le x,y,z\le\frac{1}{2}\right)\cdot P\left(0\le x,y,z\le\frac{1}{2}\right)\\ &\qquad=\frac{1}{2}\cdot\frac{1}{8}=\frac{1}{16}. \end{align}$

Similarly, with the result of Part 2, we conclude that

$\displaystyle\begin{align}&P\left(\Delta(x,y,z)=1\text{ and }\frac{1}{2}\le x,y,z\le 1\right)\\ &\qquad= P\left(\Delta(x,y,z)=1)\,|\,\frac{1}{2}\le x,y,z\le 1\right)\cdot P\left(\frac{1}{2}\le x,y,z\le 1\right)\\ &\qquad=1\cdot\frac{1}{8}=\frac{1}{8}. \end{align}$

Now we have to consider six additional case that come in two groups of three. If $S,L$ stand for "short" and long, respectively and define $S_x=\min(x,1-x)$ and $L_x=\max(x,1-x),$ and similarly for $y$ and $z$ then there are eight combinations $SSS,$ $SSL,$ ..., $LLL.$ The first, $SSS,$ and the last, $LLL,$ have been just treated.

We need to find $P(\Delta(x,y,z)=1\,|\,SSL)$ and then also $P(\Delta(x,y,z)=1\,|\,SLL).$

### SSL

We have three variables $x,y,z$ that satisfy

$\displaystyle \begin{align} x&\lt\frac{1}{2}\\ y&\lt\frac{1}{2}\\ z&\gt\frac{1}{2} \end{align}$

that, in addition, need to satisfy, $x+y\gt z.$ Points $(x,y,z)$ that satisfy the four conditions are located inside the corner pyramid in a small cube:

The volume of this pyramid, relative to the larger cube is $\displaystyle \frac{1}{8}\cdot\frac{1}{6}=\frac{1}{48}.$ Cycling through the three variables gives $\displaystyle 3\cdot\frac{1}{48}=\frac{1}{16}.$

### SLL

In this case we need to satisfy the following inequalities:

$\displaystyle \begin{align} x&\lt\frac{1}{2}\\ y&\gt\frac{1}{2}\\ z&\gt\frac{1}{2}\\ x+y&\gt z\\ x+z&\gt y. \end{align}$

The points $(x,y,z)$ that satisfy all five inequalities lie outside two pyramids in the small cube:

The relative volume of that space is $\displaystyle \frac{1}{8}\left(1-2\cdot\frac{1}{6}\right)=\frac{1}{12}.$ Cycling through the three variables gives $\displaystyle 3\cdot\frac{1}{12}=\frac{1}{4}.$ Putting everything together, denote $\mathcal{S}=\{SSS, SSL,\ldots, SSS\},$ then, the formula of total probability,

$\displaystyle\begin{align}P(\Delta(x,y,z)=1)&=\sum_{\omega\in\mathcal{S}}P(\Delta)x,y,z)=1\,|\,\omega)\\ &=\frac{1}{16}+\frac{1}{16}+\frac{1}{4}+\frac{1}{8}\\ &=\frac{1}{2}. \end{align}$

As Long Huynh Huu has observed that was clear to start with.

### Solution, Part 5

The expectation that $x,y,z\in(0,1)$ form side lengths of a triangle is $\displaystyle E_{x,y,z}=0\cdot\frac{1}{2}+1\cdot\frac{1}{2}.$

The same is true if we replace $x$ with $1-x,$ $y$ with $1-y,$ or $z$ with $1-z.$ There are eight such combinations. The eight events are not mutually exclusive (explaining the difficulty with Part 4), however the individual expectations can be summed up, giving the total expectation of $\displaystyle 8\cdot \frac{1}{2}=4.$

### Solution 2

WLOG, let the length of the sticks be unity. Let the length of the three pieces (sorted) be $x\geq y\geq z$. Let the regions in the $X,Y,Z$ space (and the projections of the space) where the sorting holds and where forming a triangle is possible be termed "feasible region" and "positive region", respectively.

**Left Pieces**

The figure below shows the feasible region in the $YZ$ plane in white with the conditions on $x$. The conditions in black and blue are for the positive and feasible regions, respectively.

The probability of triangulation is given by

$\displaystyle \begin{align} P&=\frac{\int_{z=0}^{1/2}\int_{y=z}^{1-z}\int_{x=y}^{y+z}dxdydz + \int_{y=1/2}^{1}\int_{z=1-y}^{y}\int_{x=y}^{1} dxdzdy} {\int_{z=0}^{1/2}\int_{y=z}^{1-z}\int_{x=y}^{1}dxdydz + \int_{y=1/2}^{1}\int_{z=1-y}^{y}\int_{x=y}^{1} dxdzdy} \\ &=\frac{1/24+1/24}{1/8+1/24}=1/2. \end{align}$

**Longer Pieces**

The figure below corresponds to this case. Clearly, the probability is $1$ in this case.

**Shorter Pieces**

This case is same as the "left pieces" except that $x$, $y$, and $z$ are drawn from $(0,0.5]$ instead of $(0,1.0)$. The probability does not change by this scaling and the probability is $1/2$.

**Random Chioce**

For the case of "left pieces", the length of the stick has a uniform distribution of $(0,1)$. Choosing one side or the other does not change the distribution for the chosen stick as both sides are identically distributed. Thus, the probability for this case remaines $1/2$.

**Expected Number of Triangles**

There are $8$ combinations depending on which side is chosen for each stick. For each combination, the chosen side for a stick can be arbitrarily labelled "left". This approach maps every combination to the "left pieces" case and the probability of forming a triangle is $1/2$. Thus, the expected number of triangles is $4$.

### Acknowledgment

This is an extension of problem 96 (in my Russian edition) from Challenging Mathematical Problems With Elementary Solutions by A. M. Yaglom and I. M. Yaglom.

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