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

(→‎Solution 1)
(→‎Solution 2)
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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>.
 
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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=== Solution 2 ===
 
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<center><asy>
 
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unitsize(36);
 
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draw((-1,0)--(5,0)--(5,2)--(-1,2)--cycle);
 
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draw((-1,2)--(-1,3));
 
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draw((5,0)--(6,0));
 
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draw((-1,0)--(-2,0));
 
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draw((-1,0)--(-1,-1));
 
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draw((3,0)--(3,2));
 
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label("(0,0)",(-1.5,-0.5));
 
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label("(3,1)",(5.5,2));
 
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</asy></center>
 
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Assume that a point <math>P</math> is randomly chosen inside the rectangle with vertices <math>(0,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(0,1)</math>.
 
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In this case, the probability that <math>P</math> is closer to the origin than to point <math>(3,1)</math> is <math>\frac{1}{2}</math>.
 
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If <math>P</math> is chosen within the square with vertices
 
  
 
== See also ==
 
== See also ==

Revision as of 15:55, 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

2002 12B AMC-18.png <\center>

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 $\frac{3/2}{2} = \frac 34 \Rightarrow \mathrm{(C)}$.

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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