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

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== Problem ==
 
== Problem ==
A square has sides of length 2. Set <math> S  </math> 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 <math> S </math> enclose a region whose area to the nearest hundredth is <math> k. </math> Find <math> 100k. </math>
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A [[square (geometry) | square]] has sides of length 2. [[Set]] <math> S  </math> is the set of all [[line segment]]s that have length 2 and whose [[endpoint]]s are on adjacent sides of the square. The [[midpoint]]s of the line segments in set <math> S </math> enclose a region whose [[area]] to the nearest hundredth is <math>k</math>Find <math> 100k</math>.
  
 
== Solution ==
 
== Solution ==
Without loss of generality, let (0,0), (2,0), (0,2), and (2,2) be the vertices of the square. Suppose the segment lies 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 two, <math>x^2+y^2=4</math>. Using the midpoint formula, we find that the midpoint of the segment has coordinates <math>\left(\frac{x}{2},\frac{y}{2}\right)</math>. Let d be the distance from (0,0) to <math>\left(\frac{x}{2},\frac{y}{2}\right)</math>. Using the distance we see that <math>d=\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2}=
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Without loss of generality, let <math>(0,0)</math>, <math>(2,0)</math>, <math>(0,2)</math>, and <math>(2,2)</math> be the [[vertex | vertices]] of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex <math>(0,0)</math>. Let the two endpoints of the segment have coordinates <math>(x,0)</math> and <math>(0,y)</math>. Because the segment has length 2, <math>x^2+y^2=4</math>. Using the midpoint formula, we find that the midpoint of the segment has coordinates <math>\left(\frac{x}{2},\frac{y}{2}\right)</math>. Let <math>d</math> be the distance from <math>(0,0)</math> to <math>\left(\frac{x}{2},\frac{y}{2}\right)</math>. Using the [[distance formula]] we see that <math>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</math>. 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 <math>4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86</math> to the nearest hundredth. Thus <math>100\cdot k=086</math>
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\sqrt{\frac{1}{4}\left(x^2+y^2\right)}=\sqrt{\frac{1}{4}(4)}=1</math>. Thus the midpoints lying on the sides determined by vertex <math>(0,0)</math> 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 <math>4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86</math> to the nearest hundredth. Thus <math>100\cdot k=086</math>
  
 
== See also ==
 
== See also ==
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* [[2004 AIME I Problems]]
 
* [[2004 AIME I Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 18:38, 2 December 2006

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

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