Difference between revisions of "2004 AIME I Problems/Problem 4"
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== Solution == | == Solution == | ||
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}= | 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 <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> | + | \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. |
+ | <center><asy> | ||
+ | size(100); | ||
+ | pointpen=black;pathpen = black+linewidth(0.7); | ||
+ | pair A=(0,0),B=(2,0),C=(2,2),D=(0,2); | ||
+ | D(A--B--C--D--A); | ||
+ | picture p; | ||
+ | draw(p,CR(A,1));draw(p,CR(B,1));draw(p,CR(C,1));draw(p,CR(D,1)); | ||
+ | clip(p,A--B--C--D--cycle); | ||
+ | add(p); | ||
+ | </asy></center> | ||
+ | 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 == |
Revision as of 17:38, 7 June 2008
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
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |