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$O$ is the center of the incircle of $\Delta ABC$. If $AB=6$ and $\angle ACB=120^{\circ}$, what is the radius of the circumscribed circle of $\Delta AOB$?

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  • $\begingroup$ Aren't there $3$ excircles to a triangle? $\endgroup$ – Shraddheya Shendre Nov 25 '16 at 16:25
  • $\begingroup$ by excircle I mean a circle surrounding the triangle $\endgroup$ – Irena Alexieva Nov 25 '16 at 16:32
  • $\begingroup$ That's a circumcircle. Excircle is a completely different thing. $\endgroup$ – Shraddheya Shendre Nov 25 '16 at 16:33
  • $\begingroup$ I see the error in my terminology and have fixed it. English is not my first language. $\endgroup$ – Irena Alexieva Nov 25 '16 at 16:36
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Let $\alpha, \beta, \gamma$ be the interior angles of the triangle ABC in the usual way, so we know $\gamma=120°$. We have $\sphericalangle BAO = \frac{\alpha}{2}$ and $\sphericalangle ABO = \frac{\beta}{2}$ (center of incircle is the intersection of angle bisectors), hence (sum of inner angles in triangle ABO)

$$ \sphericalangle AOB = 180° -\frac{\alpha+\beta}{2} = 150°,$$

as $\alpha+\beta = 60°$ (sum of inner angles of triangle ABC).

The law of sines in triangle ABO says that $\frac{AB}{\sin \sphericalangle AOB} = 2r$, where $r$ is the radius of the circumscribed circle of ABO. We get $r=6$.

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