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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* A * * *
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(In reply to
re(2): Equation spoiler by Dej Mar)
Dej Mar wrote: "A question could be asked, if point A were at origin (0,0) of circle of radius = 1, what are the two points, B and C, on the circles where the line segment BC separates the five circles into areas of equal halves?"
The equation of the circle to the right of the "A" circle is:
(x - 2)^2 + y^2 =1
So find the intersection between that circle and the line y = x/3
(3y - 2)^2 + y^2 =1
10 y^2 - 12y +3 = 0
y = {12 +/- sqrt(144-120)} / 20
= .6 +/- .1 sqrt(6)
= .845 or .355
So the 2 points of intersection for that circle are:
(1.065, .355) and (2.535, .845)
So the 2 intersections dividing the group of 4 circles BCDE are:
(1.065, .355) and (3.935, 2.645)
Edited on August 26, 2006, 6:09 pm
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Posted by Larry
on 2006-08-26 18:04:15 |
re(3): Equation spoiler
