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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Now the problem asks for a single solution, what I propose to either show that the solution already found is unique or to find all other solutions.
If you doubt that another solution could possibly exist then consider this. Define the circle with center A as region X and define the region with the other four circles Y. Now let L refer to the line thru A. L divides X into two regions, X1 the region above L and X2 the region below L. L also devides region Y into two regions Y1 the region above L and Y2 the region below. Now in order for L to constitute a solution to the problem we must have that X1+Y1=X2+Y2 now one way to achieve this is by way of the current solution where X1=X2 and Y1=Y2, but it should now be obvious that it is not necessarily the only solution.

Posted by Daniel
on 20060826 14:41:44 