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Difference between revisions of "2003 AMC 10A Problems/Problem 12"

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The line <math>x=y</math> intersects the rectangle at <math>(0,0)</math> and <math>(1,1)</math>.  
 
The line <math>x=y</math> intersects the rectangle at <math>(0,0)</math> and <math>(1,1)</math>.  
  
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The area which <math>x>y</math> is the right isosceles triangle with side length <math>1</math> that has vertices at <math>(0,0)</math>, <math>(1,1)</math>, and <math>(0,1)</math>.   
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The area which <math>x<y</math> is the right isosceles triangle with side length <math>1</math> that has vertices at <math>(0,0)</math>, <math>(1,1)</math>, and <math>(0,1)</math>.   
  
 
The area of this triangle is <math>\frac{1}{2}\cdot1^{2}=\frac{1}{2}</math>  
 
The area of this triangle is <math>\frac{1}{2}\cdot1^{2}=\frac{1}{2}</math>  
  
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Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A</math>
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Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \boxed{\mathrm{(A)}\ \frac{1}{8}}</math>
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==Video Solution by WhyMath==
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https://youtu.be/JueaMdaNRrs
 +
 
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~savannahsolver
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==Video Solution==
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https://www.youtube.com/watch?v=WdpGXsAYxTQ  ~David
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== See Also ==
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{{AMC10 box|year=2003|ab=A|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 20:58, 19 July 2023

Problem

A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$?

$\mathrm{(A) \ } \frac{1}{8}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{3}{8}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{3}{4}$

Solution

The rectangle has a width of $4$ and a height of $1$.

The area of this rectangle is $4\cdot1=4$.

The line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$.

The area which $x<y$ is the right isosceles triangle with side length $1$ that has vertices at $(0,0)$, $(1,1)$, and $(0,1)$.

The area of this triangle is $\frac{1}{2}\cdot1^{2}=\frac{1}{2}$

Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \boxed{\mathrm{(A)}\ \frac{1}{8}}$

Video Solution by WhyMath

https://youtu.be/JueaMdaNRrs

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=WdpGXsAYxTQ ~David

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 10 Problems and Solutions

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