Difference between revisions of "2004 AIME I Problems/Problem 4"
(11 intermediate revisions by 9 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | [[square (geometry) | Square]] <math>ABCD</math> 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 <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. | |
+ | <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=\boxed{86}</math> | ||
+ | ==Solution 2== | ||
+ | <center><asy> | ||
+ | size(100); | ||
+ | pointpen=black;pathpen = black+linewidth(0.7); | ||
+ | draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); | ||
+ | draw(arc((0,2),1,270,360)); | ||
+ | |||
+ | draw((0,1)--(1.7,2)); draw((0,2)--(1.7,1)); draw((0,1)--(1.7,1)--(1.7,2)); | ||
+ | </asy></center> | ||
+ | If we imagine an arbitrary line with length <math>2</math> connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square. | ||
+ | |||
+ | Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always <math>1</math>, which forms a circle with radius <math>1</math> centered at the corner of the square. | ||
+ | |||
+ | The area of the shape then follows from simple calculations. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2004|n=I|num-b=3|num-a=5}} | |
− | |||
− | |||
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 21:59, 18 September 2020
Contents
Problem
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
Solution 2
If we imagine an arbitrary line with length connecting two sides of the square, we can draw the rectangle formed by drawing a perpendicular from where that line touches the square.
Drawing the other diagonal of the rectangle, it also has length two, and it bisects with the original line. Since their intersection is the midpoint of both lines, the distance from the corner to the midpoint is always , which forms a circle with radius centered at the corner of the square.
The area of the shape then follows from simple calculations.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.