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Circles in Circles

Age 16 to 18
Challenge Level

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 $O$ to $V$, the midpoint of $UC$, as the radius is perpendicular to the tangent, we can find $OA = r_1$. Since triangle $TCU$ is equilateral $\angle TCU=60^o$ and $OC$ bisects $\angle TCU$ so $\angle OCV = 30^o$.

Since $OC=OA+AC$ $${CV\over OA+AC}={\sqrt 3 \over 2}$$ As $AC=CV=1$ and $OA=r_1$, $${1\over r_1+1} = {\sqrt 3 \over 2}$$ $$r_1+1 = {2\over \sqrt 3}$$ $$r_1 = {2\over \sqrt 3}-1$$ With $r_1$ we can find $r_2$ and $r_3$. Since $OV=r_2$, $CV=1$ and $\angle OCV=30^0$, $$r_2={1\over \sqrt 3}.$$ Since $OD=OA+AC+CD$, $$r_3={2\over \sqrt 3} - 1 + 1 + 1 = {2\over \sqrt 3} + 1.$$ Now $$r_1r_2 = \left({2\over \sqrt 3 }- 1\right )\left({2\over \sqrt 3} + 1\right )$$ $$\quad = {4\over 3} + {2\over \sqrt 3} - {2\over \sqrt 3} - 1$$ $$\quad = {1\over 3}$$ $$\quad = \left({1\over \sqrt 3}\right)^2$$ $$\quad = r_2^2$$