Baby Circle
A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?
Problem
A circle with radius 1 and a circle with radius 2 touch at a point. A third circle fits between these two circles so that all three touch each other and all three have a common tangent. What is the radius of the smallest circle?
Getting Started
Student Solutions
Rosalind of Madras College sent in this solution, well done Rosalind.
All three circles touch each other and have a common tangent $DFE$. The radius of the baby circle is $r$ and the radii of the other circles are $BE = 1$ unit and $AD = 2$ units. So $AC = 1$ unit, $AB = 3$ units, $AH = 2 - r$, $AR = 2 +r$, $BR = 1 + r$ and $BG = 1 - r$. Using Pythagoras Theorem: $$\eqalign{ \; RG &=& \sqrt{(1+r)^2 - (1-r)^2} = \sqrt{4r} = 2\sqrt{r} \\ \; HR &=& \sqrt{(2+r)^2 - (2-r)^2} = \sqrt{8r} = 2\sqrt{2r} \\ \; CB &=& 2\sqrt{2} = DE \\ RG + HR = CB &\Rightarrow& 2\sqrt{r} + 2\sqrt{2r} = 2\sqrt{2}.}$$ This gives $\sqrt{r}(1+\sqrt{2}) = \sqrt{2}$ and hence by squaring $$\eqalign{ r &=& \frac{2}{3+2\sqrt{2}} \\ \; &=& \frac{2(3 - 2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})} \\ \; &=& \frac{6 - 4\sqrt{2}}{9 - 4\times2} \\ \; &=& 6 - 4\sqrt{2}}$$ So the radius of the 'baby circle' is $6 - 4\sqrt{2}$.