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

Revision as of 10:25, 4 July 2013

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 $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow A$

2003 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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

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

Revision as of 10:25, 4 July 2013

Problem

Solution

See Also