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2002 AMC 12B Problems/Problem 18

Revision as of 16:23, 2 July 2019 by Nafer (talk | contribs) (→‎Solution 2)

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

2002 12B AMC-18.png


The region containing the points closer to $(0,0)$ than to $(3,1)$ is bounded by the perpendicular bisector of the segment with endpoints $(0,0),(3,1)$. The perpendicular bisector passes through midpoint of $(0,0),(3,1)$, which is $\left(\frac 32, \frac 12\right)$, the center of the unit square with coordinates $(1,0),(2,0),(2,1),(1,1)$. Thus, it cuts the unit square into two equal halves of area $1/2$. The total area of the rectangle is $2$, so the area closer to the origin than to $(3,1)$ and in the rectangle is $2 - \frac 12 = \frac 32$. The probability is $\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}$.

Solution 2

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 $1-\frac{1}{4}=\frac{3}{4}$. $\mathrm{(C)}$

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

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