Question

A point $P$ is randomly selected from the rectangular region with vertices $(0,0),$ $(2,0),$ $(2,1),$ $(0,1).$ What is the probability that $P$ is closer to the origin than it is to the point $(3,1)?$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$1$

Solution:

The points closer to $(0,0)$ than to $(3,1)$ lie on the origin side of the perpendicular bisector of that segment, the line $3x+y=5.$

Within the rectangle, this region is a trapezoid whose parallel sides have lengths $\dfrac53$ (at $y=0$ ) and $\dfrac43$ (at $y=1$ ), so its area is $\dfrac12\left(\dfrac53+\dfrac43\right)=\dfrac32.$ The rectangle has area $2,$ so the probability is $\dfrac{3/2}{2}=\dfrac34.$

Thus, the correct answer is C.

2002 AMC 12B Problem 18

Solution:

Problem 18 in Other Years