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2002 AMC 12B Problems/Problem 18

Problem

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 $(3,1)$?

$\mathrm{(A)}\ \frac 12 \qquad\mathrm{(B)}\ \frac 23 \qquad\mathrm{(C)}\ \frac 34 \qquad\mathrm{(D)}\ \frac 45 \qquad\mathrm{(E)}\ 1$

Solution

Solution 1

Assume that the point $P$ is randomly chosen within the rectangle with vertices $(0,0)$, $(3,0)$, $(3,1)$, $(0,1)$. In this case, the region for $P$ to be closer to the origin than to point $(3,1)$ occupies exactly $\frac{1}{2}$ of the area of the rectangle, or $1.5$ square units.


If $P$ is chosen within the square with vertices $(2,0)$, $(3,0)$, $(3,1)$, $(2,1)$ which has area $1$ square unit, it is for sure closer to $(3,1)$.


Now if $P$ can only be chosen within the rectangle with vertices $(0,0)$, $(2,0)$, $(2,1)$, $(0,1)$, then the square region is removed and the area for $P$ to be closer to $(3,1)$ is then decreased by $1$ square unit, left with only $0.5$ square unit.


Thus the probability that $P$ is closer to $(3.1)$ is $\frac{0.5}{2}=\frac{1}{4}$ and that of $P$ is closer to the origin is $1-\frac{1}{4}=\frac{3}{4}$. $\mathrm{(C)}$


~ Nafer

Solution 2 (slightly more calculation but easy:)

First, join points $(0,0)$ and $(3,1)$. This line $l 1$ has equation $y = \frac{1}{3}x$.


Next, consider the perpendicular bisector $l 2$ of line $l 1$, any point on the perpendicular bisector is equidistance from points $(0,0)$ and $(3,1)$.


If the point $P$ is chosen on the left of $l 2$ , the point is closer to the origin. $l 2$ will cut the rectangle region twice, dividing the region into two smaller trapezoids. The trapezoid on the left side is the area we want.


So, first, find the equation of $l 2$, using the midpoint of $(0,0)$ and $(3,1)$, which is $(1.5,0.5)$, the equation $y = -3x + 5$ can be easily derived. Substituting $y = 0$ and $y = 1$ to solve for $x$. For the former, $x = \frac{5}{3}$, for the latter, $x = \frac{4}{3}$.


Thus, the area of the trapezoid we want is $\frac {(\frac{5}{3} + \frac{4}{3})\cdot1}{2} = \frac{3}{2}$

Therefore, the probability the question asks equals to $\frac{\frac{3}{2}}{2} =\boxed{\textbf{(C) }\frac{3}{4}}$

~Yohann

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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