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.

 A: 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 $2r+s = \frac{9}{4}r$.
A: 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.
