Question 55
Assume that triangle $ABC$ is not equilateral and that both $\beta = \angle ABC$ and $\gamma = \angle ACB$ are larger than $30^\circ$. Let $O$ be the circumcentre of triangle $ABC$. Let the triangles $ACB'$ and $ABC'$ be equilateral with $B$ and $B'$ on opposite sides of $AC$ and $C$ and $C'$ on opposite sides of $AB$. Let $B''$ and $C''$ be such interior points of the segments $BB'$ and $CC'$ that
$$BB'' = \frac{1}{2} \left(1 - \frac{\tan(90^\circ - \beta)}{\tan 60^\circ}\right) BB' \text{ and } CC'' = \frac{1}{2} \left(1 - \frac{\tan(90^\circ - \gamma)}{\tan 60^\circ}\right) CC'$$
Prove $\angle B''OC'' = 120^\circ$.
$$BB'' = \frac{1}{2} \left(1 - \frac{\tan(90^\circ - \beta)}{\tan 60^\circ}\right) BB' \text{ and } CC'' = \frac{1}{2} \left(1 - \frac{\tan(90^\circ - \gamma)}{\tan 60^\circ}\right) CC'$$
Prove $\angle B''OC'' = 120^\circ$.