Let $c_1,c_2,c_3$ be three circles of unit radius touching each other externally. The common tangent to each pair of circles are drawn (and extended so that they intersect) and let the triangle formed by the common tangents be $\Delta ABC$. Find the length of each side of $\Delta ABC$.

I don't have a clue how to proceed. Sorry for not providing a sketch.

  • $\begingroup$ do you agree that by symmetry the triangle ABC is equilateral? $\endgroup$ – amakelov Jun 24 '16 at 12:06
  • $\begingroup$ @amakelov I do now. $\endgroup$ – StubbornAtom Jun 24 '16 at 12:08

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The length is $2AD+DE$. Obviously $DE=2$, so we have to find $AD$. Consider the triangle $ADF$. $DF=1,\angle AFD=60^o$, so $AF=\sqrt3DF=\sqrt3$. Hence $AB=2+2\sqrt3$.

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