Chapter 0: EG.1 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Two points are chosen at random on a line AB, each point being chosen according to the uniform distribution on AB, and the choices being made independently of each other. The line AB may now be regarded as divided into three parts. What is the probability that they may be made in to a triangle?

Each edge has to be shorter than the sum of the two other edges. So, each edge has to be smaller than $\tfrac{1}{2}|AB|$ , and if they all are, then we can build a triangle of them. So, when one point is in either half of $AB$ , the other point has to be in the other half of $AB$ at a distance no longer than $\tfrac{1}{2}|AB|$ , and we have $\begin{eqnarray*} 2\int_{0}^{\tfrac{1}{2}}xdx & = & \tfrac{1}{4}. \end{eqnarray*}$

Another way of thinking about the solution. A point $(x,y)$ is randomly chosen in the square $[0,1]\times[0,1]$ . We need $\begin{eqnarray*} \min\{x,y\} & \le & \tfrac{1}{2},\\ \max\{x,y\} & \ge & \tfrac{1}{2},\\ |x-y| & \le & \tfrac{1}{2}. \end{eqnarray*}$ Below the diagonal the first two equations imply $y\le\tfrac{1}{2}\le x$ , and above it $x\le\tfrac{1}{2}\le y$ , while the third equation implies the point must line between $y=x\pm\tfrac{1}{2}$ . The area of the two resulting triangles is $\tfrac{1}{4}$ .

Subpages

Chapter 0: EG.1 Solution

Chapter 0: EG.1 Solution