This solution together with the diagram was sent in by Derek Wan, age 17 of Sha Tin College, Hong Kong

Solution to Circles in Circles

To find

OA = r_{1}

OB = r_{2}

and

OD = r_{3}

we must examine the diagram and make appropriate considerations. By dropping a perpendicular from

O

V

, the midpoint of

UC

, as the radius is perpendicular to the tangent, we can find

OA = r_{1}

. Since triangle

TCU

is equilateral

∠ TCU = 60^{o}

and

OC

bisects

∠ OCV

∠ OCV = 30^{o}

Since $OC = OA + AC$

$\frac{CV}{OA + AC} = \frac{\sqrt{3}}{2}$

As $AC = CV = 1$ and $OA = r_{1}$ ,

$\begin{matrix} \frac{1}{r_{1} + 1} & = \frac{\sqrt{3}}{2} \\ r_{1} + 1 & = \frac{2}{\sqrt{3}} \\ r_{1} & = \frac{2}{\sqrt{3}} - 1 . \end{matrix}$

With $r_{1}$ we can find $r_{2}$ and $r_{3}$ . Since $OV = r_{2}$ , $CV = 1$ and $∠ OCV = 30^{0}$ ,

$r_{2} = \frac{1}{\sqrt{3}} .$

Since $OD = OA + AC + CD$ ,

$r_{3} = \frac{2}{\sqrt{3}} - 1 + 1 + 1 = \frac{2}{\sqrt{3}} + 1 .$

Now

$\begin{matrix} r_{1} r_{2} & = (\frac{2}{\sqrt{3}} - 1) (\frac{2}{\sqrt{3}} + 1) \\ = \frac{4}{3} + \frac{2}{\sqrt{3}} - \frac{2}{\sqrt{3}} - 1 \\ = \frac{1}{3} \\ = {(\frac{1}{\sqrt{3}})}^{2} \\ = r_{2}^{2} . \end{matrix}$