Five circles are arranged in the following fashion ( Two rows of 3 circles in square arrangement with one end circle removed from the top row ). Circles are of same diameter and touching adjoining circles as per the diagram. Draw a line passing through A ( Centre of the first circle on the bottom row ) in such a way that it divides the five circles into two equal areas.
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(In reply to
new question by Daniel)
Pick any angle. Draw a line at that angle. It divides the plane in two portions. Some part of the circles' area is in the first portion (call it F) and some in the second (call it S). Imagine now moving the line parallel to itself. The difference F-S varies continuosly. Depending on the line's position, F-S can be positive or negative, so by Bolzano's theorem, it must be zero somewhere. Thus, there are infinite solutions, since there are infinite angles for the line.
Now, this problem requires a line through A. Imagine now the line through A, but rotating. F-S will be positive or negative as above, so there is at least one zero. And since F-S varies in a monotone way, there is only ONE solution for this particular problem.
Edited on August 26, 2006, 8:16 pm
Infinite solutions
