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

Line 1: | Line 1: | ||

== Problem == | == Problem == | ||

− | A square has sides of length 2. Set <math> S </math> is the set of all line | + | 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 | + | 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> | + | \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 == | ||

Line 12: | Line 12: | ||

* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||

+ | |||

+ | [[Category:Intermediate Geometry Problems]] |

## Revision as of 18:38, 2 December 2006

## Problem

A square has sides of length 2. Set 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 enclose a region whose area to the nearest hundredth is . Find .

## Solution

Without loss of generality, let , , , and be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex . Let the two endpoints of the segment have coordinates and . Because the segment has length 2, . Using the midpoint formula, we find that the midpoint of the segment has coordinates . Let be the distance from to . Using the distance formula we see that . Thus the midpoints lying on the sides determined by vertex 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 to the nearest hundredth. Thus