Difference between revisions of "2004 AIME I Problems/Problem 4"

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== See also ==
 
== See also ==
* [[2004 AIME I Problems/Problem 3| Previous problem]]
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{{AIME box|year=2004|n=I|num-b=3|num-a=5}}
 
 
* [[2004 AIME I Problems/Problem 5| Next problem]]
 
 
 
* [[2004 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 15:00, 27 April 2008

Problem

A square has sides of length 2. Set $S$ is the set of all line segments that have length 2 and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$.

Solution

Without loss of generality, let $(0,0)$, $(2,0)$, $(0,2)$, and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$. Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$. Because the segment has length 2, $x^2+y^2=4$. Using the midpoint formula, we find that the midpoint of the segment has coordinates $\left(\frac{x}{2},\frac{y}{2}\right)$. Let $d$ be the distance from $(0,0)$ to $\left(\frac{x}{2},\frac{y}{2}\right)$. Using the distance formula we see that $d=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2}= \sqrt{\frac{1}{4}\left(x^2+y^2\right)}=\sqrt{\frac{1}{4}(4)}=1$. Thus the midpoints lying on the sides determined by vertex $(0,0)$ form a quarter-circle with radius 1. The set of all midpoints forms a quarter circle at each corner of the square. The area enclosed by all of the midpoints is $4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86$ to the nearest hundredth. Thus $100\cdot k=086$

See also

2004 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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