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Printjmc
counting and probability senior
Problem
Rectangle has center and . A point is randomly chosen from the interior of rectangle . What is the probability that it is closer to than to any of the four vertices? 
Solution
The original rectangle may be subdivided into four smaller congruent rectangles, all sharing as a vertex. Each of these rectangles is analogous, so we can consider our random point to be without loss of generality in the smaller rectangle with as a vertex. All points in this smaller rectangle are closer to than they are to , , or , so we just need to determine the probability that . Since a rotation about the center of the smaller rectangle takes to , it takes the shaded region to the unshaded region. Therefore, exactly half the area is shaded, and the overall probability is , independent of .
Final answer
\frac{1}{2}