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Difference between revisions of "2004 AIME I Problems/Problem 4"
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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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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{086}</math>
  
 
== See also ==
 
== See also ==

Revision as of 19:24, 19 July 2017

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.

[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]

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 $100\cdot k=\boxed{086}$

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

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