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May Puzzle Solutions

You might think it would be a good idea to put the two sensors in the centers of the two rectangles. However, a large distance measurement (say, of 75 meters) yields an over-large rectangle to search.

Consider going literally outside the box: place the sensor 25 meters to the right of the right border of each 70-by-90 rectangle (that is, 70 meters from the center of the rectangle), assuming the 90-meter-long borders of the rectangle are horizontal. Any distance from the sensor would then describe an arc through the rectangle.

If that distance were d, then the leftmost point within the rectangle would be (d ¿ 25) to the left of the right border of the rectangle (and would lie on the center line of the rectangle). The rightmost points would be where the arc intersects the top and bottom edges of the rectangle.

There would be two rightmost points, at the upper and lower rectangular edges. Their position can be determined by the Pythagorean theorem. Because a sensor reading of d could mean an actual distance between 0.9d and 1.1d, we will take the leftmost point from the 1.1d reading and the rightmost point from the 0.9d reading. Choosing this position for the sensor gives us a single rectangle to search of area under 1,940 square meters, regardless of the sensor reading.

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