Breaking a Stick into a Triangle
A stick of length $L$ is broken at two points chosen independently and uniformly at random along its length. What is the probability that the three resulting pieces can form a triangle?
Hints
- For three pieces summing to a fixed total, the triangle inequality simplifies: no single piece can be more than half the total.
- Set up the two break points as uniform random variables on $[0, 1]^2$ and identify the region where all three pieces are less than $1/2$.
- By symmetry, $P(\text{piece } i > 1/2) = 1/4$ for each piece, and these events are mutually exclusive since the pieces sum to 1.
Worked Solution
How to Think About It: The heuristic is *reduce the condition, then measure area*. Three lengths form a triangle if and only if each one is less than the sum of the other two. Since the three pieces sum to $L$, "piece $<$ sum of the others" is the same as "piece $< L/2$". So the question becomes: if you break a stick at two random points, what is the probability that no piece is $L/2$ or longer? With two uniform break points, that is an area in the unit square.
Quick Estimate: Take $L = 1$ and count the ways to fail. The left piece is too long exactly when both break points land in the right half: $\tfrac12 \cdot \tfrac12 = \tfrac14$. The right piece is the mirror image: $\tfrac14$. The middle piece is too long when the break points are more than $\tfrac12$ apart: two corner triangles of the unit square with legs $\tfrac12$, area $\tfrac18$ each, total $\tfrac14$. At most one piece can be longer than half the stick, so these failures never overlap: $P(\text{fail}) = 3 \cdot \tfrac14 = \tfrac34$, and $P(\text{triangle}) = 1 - \tfrac34 = \tfrac14$.
Approach: Model the break points as a uniform point in the unit square, translate the three conditions into inequalities, and compute the area of the region where all three hold.
Formal Solution:
Set $L = 1$. Let $U_1, U_2 \sim \text{Uniform}(0, 1)$ be the break points, independent, so $(U_1, U_2)$ is uniform on the unit square (area 1) and probability equals area. The pieces are $$A = \min(U_1, U_2), \quad B = |U_1 - U_2|, \quad C = 1 - \max(U_1, U_2).$$ A triangle is possible if and only if $A < \tfrac12$, $B < \tfrac12$ and $C < \tfrac12$. In terms of the break points:
- $A < \tfrac12$: the smaller break point is in the left half.
- $C < \tfrac12$: the larger break point is in the right half.
- $B < \tfrac12$: the break points are less than $\tfrac12$ apart.
*Case $U_1 < U_2$.* The conditions read $U_1 < \tfrac12 < U_2 < U_1 + \tfrac12$. For each $U_1 \in (0, \tfrac12)$, $U_2$ ranges over an interval of length $U_1$, so the area is $\int_0^{1/2} U_1 \, dU_1 = \tfrac18$. This is the triangle with vertices $(0, \tfrac12)$, $(\tfrac12, \tfrac12)$, $(\tfrac12, 1)$.
*Case $U_2 < U_1$.* By symmetry (swap the labels), another area $\tfrac18$: the triangle with vertices $(\tfrac12, 0)$, $(\tfrac12, \tfrac12)$, $(1, \tfrac12)$.
The favorable region is these two triangles, which touch only at the point $(\tfrac12, \tfrac12)$. Total area: $$P(\text{triangle}) = \tfrac18 + \tfrac18 = \boxed{\tfrac14}$$
*Check by complement.* The three failure events $\{A \ge \tfrac12\}$, $\{B \ge \tfrac12\}$, $\{C \ge \tfrac12\}$ are mutually exclusive (two pieces of length at least $\tfrac12$ would leave nothing for the third), each has probability $\tfrac14$ as computed in the Quick Estimate, and $1 - 3 \cdot \tfrac14 = \tfrac14$. A Monte Carlo run with $10^6$ trials gives $0.2497$.
Answer: The probability that the three pieces form a triangle is $\dfrac{1}{4}$.
Intuition
This is a classic geometric probability problem that illustrates a powerful simplification: when pieces sum to a constant, the triangle inequality reduces to a simple "no piece exceeds half the total" condition. The problem then becomes a ratio-of-areas calculation in the space of break points.
The result $1/4$ is surprisingly clean and worth remembering. The same setup appears in many guises: it is equivalent to asking for the probability that three uniformly random points on a circle form a triangle containing the center (all three arcs shorter than half the circumference), or that a uniformly random point in the 2-simplex lies in its middle sub-triangle (the one joining the edge midpoints). These geometric probability arguments (setting up a sample space, identifying the feasible region, and computing its volume) are a staple of quant interviews and show up in problems ranging from order-statistic distributions to random convex hulls.