## Distance Between Two Points

If two points are selected at random from the interval $\left [0,1\right ]$, what is the probability that the distance between them is less than one fourth?
Source: NCTM Mathematics Teacher, February 2006

SOLUTION

Consider two points $A(0,0)$ and $B(1/4,0)$ on the $\left [0,1\right ]$ interval. Create an isosceles right triangle $ABC$ by drawing a perpendicular $\overline{BC}$ of length $1/4$ to $\overline{AB}$.

Keeping the length of the segment $\overline{AB}$ at a fixed value of $1/4$, move the segment $\overline{AB}$ from left to right until $A$ is at $(3/4,0)$ and $B$ at $(1,0)$. As the segment $\overline{AB}$ moves from left to right the isosceles right triangle $ABC$ moves within the confines of the trapezoid $ABEF$ from the lower left corner to the upper right hand corner ending at the location of triangle $DEF$.

What we have done is convert the problem of two points $A$ and $B$ on the interval $\left [0,1\right ]$ to the problem of an isosceles right triangle $ABC$ moving inside the trapezoid $ABEF$. The probability that the distance between $A$ and $B$ is less than one fourth equals the probability that the isosceles right triangle $ABC$ stays within the confines of the trapezoid $ABEF$.

The sample space is the isosceles right triangle of side length $1$. Its area equals $(1/2)1^2=1/2$.

The figure shows that the trapezoid is made up of $7$ isosceles triangles $ABC$.
Area of trapezoid = $7(1/2)(1/4)^2 = 7/32$
Probability that the distance between $A$ and $B$ is less than $1/4$ equals
$7/32\div 1/2=7/16$

Answer: $7/16$

Alternative solution
Let $A(x_A,0)$ and $B(x_B,0)$ two points on the interval $\left [0,1\right]$. We want $|x_A-x_B|<1/4$.

The diagonal from $(0,0)$ to $(1,1)$ in the above figure is the set of points $A$ and $B$ such that $x_A=x_B$. The line above the diagonal is the set of points $x_B-x_A=1/4$ and the line under it is the set of points $x_B-x_A=\textrm{-}1/4$. As long as the coordinates $(x_A,x_B)$ stay inside the region delineated by these two lines, $|x_A-x_B|<1/4$.
The square of side $1$ represents the sample space; its area equals $1^2=1$.
Area of the region = area of square – area of two isosceles right triangles of side length $3/4$
$=1-2(1/2)(3/4)^2= 7/16$
Probability that the distance between $A$ and $B$ is less than $1/4$ equals
$7/16\div 1=7/16$