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 2006-08-26 12:55:56