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This is probably an easy question but I don't get it. Please answer in a easy way :P

Imagine you want to paint circles on a linear line. The line is 100cm long. On the right side you have a circle with a diameter of 15 cm. Now I want to draw more circles until the 100cm are full.

So basically this: o o o o o o o but with variable diameters. Right is the biggest and the left the smallest. There must be a space between them.

So since I don't know how I should go about this I also can't define the variables. The right circle, the biggest one, should be variable and all the others should depend on them. Also I'd say I defined the number of circles and the space is "generated".

Anyway, I need a few circles which gets smaller. Whats the best way?

Of course I just could say I make 8 circles and each in is 2cm bigger and try out what goes the best, but somehow I'm also doing this to learn a bit math/geometry etc. :)

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    How many circles? How much space between? How can the line be "full of circles" if there is space between? What do you mean by "best way"?2012-06-07

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Say we want $n$ circles starting from left to right. Let the diameter of the leftmost circle be $a$. Let the difference in diameters between two immediate circles be fixed at $d$. The diameter of $k^{th}$ circle is then $a+(k-1)d$. Hence, the diameter of the $n^{th}$ circle i.e. the rightmost circle is $a +(n-1)d$. We want the rightmost circle to be of diameter $15$. Hence, we get $a + (n-1)d = 15$ Let the "spacing" between two circles be $s$. By "spacing" we mean the shortest distance between the boundary of the two circles. Since the length of the line is $100$, we then need $a + s + (a+d) + s + (a+2d) + s + \cdots + (a+(n-2)d) + s + (a + (n-1)d) = 100$ $na + \dfrac{n(n-1)}{2}d + (n-1)s = 100$ Making use of the fact that $a + (n-1)d = 15$, we get $\dfrac{na}{2} + \dfrac{15n}{2} + (n-1)s = 100$ Hence, you can choose $a$, $n$, $d$ and $s$ satisfying the two equations to get what you want $a + (n-1)d = 15$ $\dfrac{na}{2} + \dfrac{15n}{2} + (n-1)s = 100$

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    That sounds okay. :) Will try it, thanks!2012-06-08