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Exercise 168
A circle of radius $1$ is tangent to a circle of radius $3$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?
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Solution
  1. $\triangle AOF\sim\triangle APE\sim\triangle ACD$. Let $AO=x$; since $O,P$ are externally tangent, $AP=x+1+3=x+4$, and $AO:AP=OF:PE=1:3$, so $x:(x+4)=1:3$, giving $x=2$.
  2. $AD=AP+PD=(x+4)+3=9$, and $AF=\sqrt{AO^2-OF^2}=\sqrt{2^2-1^2}=\sqrt3$.
  3. $CD:OF=AD:AF$, so $CD=1\cdot\dfrac9{\sqrt3}=3\sqrt3$.
  4. $[ABC]=CD\cdot AD=3\sqrt3\cdot9=\boxed{27\sqrt3}$.
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