Difference between revisions of "2002 AMC 12B Problems/Problem 18"

Revision as of 15:45, 2 July 2019

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

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

$[asy] unitsize(36); draw((0,0)--(6,0)--(6,2)--(0,2)--cycle); draw((0,2)--(0,3.5)); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label("$a$",(0,1.5),W); label("$c$",(2.5,1),W); label("$A$",(0.5,2.5),W); label("$B$",(3.5,0.75),W); label("$C$",(1,1),W); [/asy]$

Assume that a point $P$ is randomly chosen inside the rectangle with vertices $(0,0)$ , $(3,0)$ , $(3,1)$ , $(0,1)$ .

In this case, the probability that $P$ is closer to the origin than to point $(3,1)$ is $\frac{1}{2}$ .

If $P$ is chosen within the square with vertices

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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