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.
* *
* * * *
* * * *
* * * *
* * *
* * * *
* * * *
* * * *
* * *
* * * * * *
* * * * * *
* * * * * *
* A * * *
* * * * * *
* * * * * *
* * * * * *
* * *
Let B and C be the centers of the next two circles in the bottom row and D and E the centers of the two circles in the above row. Let M be the intersection between DC and BE. M is the simetry point of the figure formed by the four circles (of centers B, C, D, E) and the zone between them, and A is the simetry point of the first circle in the bottom row. So, the line AM divides the five circles into two equal areas.

Posted by Stefan
on 20060826 12:55:56 