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

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

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

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

@@ Line 11: / Line 11: @@
 The line <math>x=y</math> intersects the rectangle at <math>(0,0)</math> and <math>(1,1)</math>.
-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 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>
-Therefore, the probability that <math>x<y</math> is <math>\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A</math>
+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>
+==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]]
+{{AMC10 box|year=2003|ab=A|num-b=11|num-a=13}}
-*[[2003 AMC 10A Problems/Problem 11|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 13|Next Problem]]
 [[Category:Introductory Algebra Problems]]
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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