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

Revision as of 16:54, 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

$[[Image:2002_12B_AMC-18.png]]$

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

$[asy] unitsize(36); draw((-1,0)--(5,0)--(5,2)--(-1,2)--cycle); draw((-1,2)--(-1,3)); draw((5,0)--(6,0)); draw((-1,0)--(-2,0)); draw((-1,0)--(-1,-1)); draw((3,0)--(3,2)); label("(0,0)",(-1.5,-0.5)); label("(3,1)",(5.5,2)); [/asy]$

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
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 11: / Line 11: @@
 === Solution 1 ===
-[[Image:2002_12B_AMC-18.png]]
+<cmath>[[Image:2002_12B_AMC-18.png]]</cmath>
 The region containing the points closer to <math>(0,0)</math> than to <math>(3,1)</math> is bounded by the [[perpendicular bisector]] of the segment with endpoints <math>(0,0),(3,1)</math>. The perpendicular bisector passes through midpoint of <math>(0,0),(3,1)</math>, which is <math>\left(\frac 32, \frac 12\right)</math>, the center of the [[unit square]] with coordinates <math>(1,0),(2,0),(2,1),(1,1)</math>. Thus, it cuts the unit square into two equal halves of area <math>1/2</math>. The total area of the rectangle is <math>2</math>, so the area closer to the origin than to <math>(3,1)</math> and in the rectangle is <math>2 - \frac 12 = \frac 32</math>. The probability is <math>\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}</math>.

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