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

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\qquad\mathrm{(E)}\ 1</math>
 
\qquad\mathrm{(E)}\ 1</math>
 
== Solution ==
 
== Solution ==
 +
 +
  
 
=== Solution 1 ===
 
=== Solution 1 ===
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[[Image:2002_12B_AMC-18.png]]
 
  
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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>.
+
Assume that the point <math>P</math> is randomly chosen within the rectangle with vertices <math>(0,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(0,1)</math>. In this case, the region for <math>P</math> to be closer to the origin than to point <math>(3,1)</math> occupies exactly <math>\frac{1}{2}</math> of the area of the rectangle, or <math>1.5</math> square units.
 +
 
 +
 
 +
If <math>P</math> is chosen within the square with vertices <math>(2,0)</math>, <math>(3,0)</math>, <math>(3,1)</math>, <math>(2,1)</math> which has area <math>1</math> square unit, it is for sure closer to <math>(3,1)</math>.
 +
 
 +
 
 +
Now if <math>P</math> can only be chosen within the rectangle with vertices <math>(0,0)</math>, <math>(2,0)</math>, <math>(2,1)</math>, <math>(0,1)</math>, then the square region is removed and the area for <math>P</math> to be closer to <math>(3,1)</math> is then decreased by <math>1</math> square unit, left with only <math>0.5</math> square unit.
 +
 
 +
 
 +
Thus the probability that <math>P</math> is closer to <math>(3.1)</math> is <math>\frac{0.5}{2}=\frac{1}{4}</math> and that of <math>P</math> is closer to the origin is <math>1-\frac{1}{4}=\frac{3}{4}</math>. <math>\mathrm{(C)}</math>
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 +
 
 +
~ Nafer
 +
 
 +
=== Solution 2 (slightly more calculation but easy:)===
 +
First, join points <math>(0,0)</math> and <math>(3,1)</math>. This line <math>l 1</math> has equation <math>y = \frac{1}{3}x</math>.
 +
 
 +
 
 +
Next, consider the perpendicular bisector <math>l 2</math> of line <math>l 1</math>, any point on the perpendicular bisector is equidistance from points <math>(0,0)</math> and <math>(3,1)</math>.
 +
 
 +
 
 +
If the point <math>P</math> is chosen on the left of <math>l 2</math> , the point is closer to the origin. <math>l 2</math> will cut the rectangle region twice, dividing the region into two smaller trapezoids. The trapezoid on the left side is the area we want.  
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=== Solution 2 ===
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So, first, find the equation of <math>l 2</math>, using the midpoint of <math>(0,0)</math> and <math>(3,1)</math>, which is <math>(1.5,0.5)</math>, the equation <math>y = -3x + 5</math> can be easily derived. Substituting <math> y = 0 </math> and <math> y = 1 </math> to solve for <math>x</math>. For the former, <math>x = \frac{5}{3}</math>, for the latter, <math>x = \frac{4}{3}</math>.
  
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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)--(-1,3.5));
 
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draw((5,0)--(6.5,0));
 
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label("(0,0)",(-1,-0.5));
 
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label("(5,1)",(5.5,2));
 
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label("$A$",(0.5,2.5),W);
 
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label("$B$",(3.5,0.75),W);
 
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label("$C$",(1,1),W);
 
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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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Thus, the area of the trapezoid we want is <math>\frac {(\frac{5}{3} + \frac{4}{3})\cdot1}{2} = \frac{3}{2}</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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Therefore, the probability the question asks equals to <math>\frac{\frac{3}{2}}{2} =\boxed{\textbf{(C) }\frac{3}{4}}</math>
  
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If <math>P</math> is chosen within the square with vertices
+
~Yohann
  
 
== See also ==
 
== See also ==

Latest revision as of 06:44, 12 October 2025

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

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