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2004 AIME I Problems/Problem 4

Revision as of 17:33, 2 December 2006 by 1337h4x (talk | contribs)

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 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, $x^2+y^2=4$. 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 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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