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Question: The centers of three circles, each with a radius R, are located on a straight line at points A, B and C. It is known that AB=BC=3. A fourth circle touches each of the three circles. Find all possible radii of the fourth circle if

a) R = 1 b) R = 2 c) R = 5

I am a bit confused as to how to begin solving the problem, as this is a trignometry question but I do not know where I can relate the triangle into the problem.

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  • $\begingroup$ Let $P$ and $r$ be the center and radius of the fourth circle. Since the fourth circle is touching others circles with radius $R$, each of $|PA|, |PB|, |PC|$ can take only two possible values $R+r$ and $|R-r|$. There are only a small number of configurations to check. $\endgroup$ – achille hui May 1 '17 at 3:00
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We just need to use Pythagoras' Theorem.

When $R=1$, there are two possible cases, with the radius equals to $\frac{9}{4}$ and $\frac{9}{2}$ respectively.

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When $R=2$, there is one case, with the radius equals to $\frac{9}{8}$. enter image description here

When $R=5$, there are two cases, with the radius equals to $\frac{9}{20}$ and $\frac{9}{10}$ respectively.

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Here's a hint to get you started (this is for the case $r=1$). Discovering how to draw a fourth circle tangent to the first three is crucial before setting up any equations.

enter image description here

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  • $\begingroup$ I would expect that you would get the minimum value of $r$ when the centre of the fourth circle is on the right bisector of $AC$ ( through $B$) $\endgroup$ – WW1 May 1 '17 at 3:20

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