You are a mathematical underwater salvage expert. A pirate's trunk filled with millions of dollars worth of jewels is buried in the sand on a flat part of the deep ocean floor. Your client has dropped sensors in the area to try to locate the trunk. The sensors can be dropped to a precise location, but their reports of distance to the treasure trunk are accurate only to within 10 percent.

The first sensor reports the treasure to be at a distance of 450 meters (so the real distance from that sensor falls between 405 meters and 495 meters). A second sensor reports the treasure to be at 350 meters (real distance is between 315 meters and 385 meters). Because your suspicious client doesn't want you to hire your own ship and possibly grab the treasure, he tells you only relative positions of the sensors. Specifically, he tells you the first sensor is at position (0,0) and the second is at position (300, 400).

Warm Up:
If the sensor estimates of distance to the treasure were precisely correct (no plus or minus 10 percent), how could you use one more sensor to find the exact position of the treasure?

Solution to Warm Up:
Draw a circle of radius 450 around (0,0) and another circle of radius 350 around (300, 400). The two circles meet at two points, as you can see in this figure.

A slightly involved algebraic solution tells you that the intersection points are at the coordinates (-46.75, 447.56) and (442.75, 80.44).

All you need do is drop a sensor at one of those intersection points. If the treasure isn't there, it's at the other one.

End of Warm Up:
Unfortunately, as noted, the distance between the sensors and the treasure is accurate only to within 10 percent. Fortunately, your client gives you two more sensors (that is, a third and fourth). The only catch is that these two new sensors, which have the same 10 percent accuracy as the first ones, must be dropped simultaneously. Given those constraints, can you fix the treasure's position to within a single rectangular plot whose area is well under 5,000 square meters? How about under 2,000 square meters?

Hint: Use the fact that the first two sensors, even with their inaccuracies, limit the search to two nearly rectangular areas 70 meters long and 90 meters wide. Those plots are centered at the points of exact intersection: (-46.75, 447.56) and (442.75, 80.44).