# Diameter of Three Inscribed Circles

What is the diameter of a circle in which are inscribed three smaller identical circles, two of which are on one side of a chord, the third on the other side? This problem came up when cutting a log into billets for turning table legs. I tried including a diagram but the reputation Nazis won't let me.

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Note that $DEA$ are all on one line, tangent to circles $F,H$ and perpendicular to BC. Because of the right angles, the tangents from $E$ to $F,H$ are the radius of the small circles. Chord $BC$ is four times the radius of the small circles. –  Ross Millikan Feb 24 '13 at 22:16
None of the above relies on the tangency at $G$, which will define everything. –  Ross Millikan Feb 24 '13 at 22:23
@Blue: You are right about $FH$, but the more I look at it, $BC$ must be shorter than four small radii. Note that $E$ is toward $D$ from $A$, so the angle between BC and the circle on the side of $FH$ is greater than $\frac \pi 2$ –  Ross Millikan Feb 24 '13 at 22:43
@RossMillikan: My thinking was a little messed-up. I've removed my comment. –  Blue Feb 24 '13 at 22:46

Let $r$ be the radius of the small circle, $s$ the distance from the center of the large circle to the chord, and $t$ the distance from the center of the large circle to the center of one of the two small tangent circles:

Clearly $t = r + s$, and also $t^2 = r^2 + \left(r-s\right)^2$, and we have

$$\left( r + s \right)^2 = r^2 + \left( r - s \right)^2$$

so that, after a tiny bit of algebra,

$$4 s = r$$

The radius of the large circle is therefore $2r+s = \frac{9}{4}r$.

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Let the small radius be $r$ and the distance from the center of the large circle to $BC$ be $x$. Positions the center of the large circle at the origin with BC horizontal and below the $x$ axis. Then the center of the upper right small circle is $(1,1-x)$. Draw the line from $A$ though this point and it will meet the large circle at the point of tangency. The radius of the large circle is $2+x$, looking at the bottom circle. The point of tangency is then $(\frac {2+x}{\sqrt{2-2x+x^2}},\frac {(2+x)(1-x)}{\sqrt{2-2x+x^2}})$. Demanding this be at distance $1$ from the center of the small circle gives $(\frac {2+x}{\sqrt{2-2x+x^2}}-1)^2+(\frac {(2+x)(1-x)}{\sqrt{2-2x+x^2}}-(1-h))^2=1$ which Alpha solves with $x=\frac 14$. This says the radius of the large circle is $\frac 94$ the radius of the smaller.

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