Bulgaria National Olympiad Problem 2

The triangles $\Delta BMC$ and $\Delta ANB$ are similar. Let $\angle MBC = \angle NAB = \alpha$ . Angle-chasing shows us that $\angle COB = 120^\circ$ , and hence the triangle $\Delta OMB$ is also similar to the other two. Suppose that the triangle $\Delta ABC$ has side $R$ , and that $AN = BM = x$ .

Applying the Sine Rule to the triangles $\Delta BMC$ and $\Delta OMB$ , we deduce that $CM \; = \; \frac{x \sin 60^\circ}{\sin\alpha} \hspace{2cm} OM \; = \; \frac{x\sin\alpha}{\sin60^\circ}$ and so the lengths of the altitudes from $O$ and $A$ to $BC$ are $OC\sin\alpha = \frac{3-4\sin^2\alpha}{2\sqrt{3}}x \hspace{2cm} R\sin60^\circ = \tfrac{\sqrt{3}}{2}R$ respectively. The condition on the area of the two triangles tells us that the first of these is $\tfrac27$ the size of the second, and hence $\frac{3-4\sin^2\alpha}{2\sqrt{3}}x = \tfrac{\sqrt{3}}{7}R$ Applying the Sine Rule to the triangle $\Delta CMB$ gives $\frac{R}{\sin(120^\circ - \alpha)} \; = \; \frac{x}{\sin\alpha}$ and hence we have the following equation for $\alpha$ : $\frac{\sin(60^\circ + \alpha)}{\sin\alpha} \; = \; \frac{7(3 - 4\sin^2\alpha)}{6}$ After some manipulation, this equation becomes $10\tan^3\alpha + 3\sqrt{3}\tan^2\alpha - 18\tan\alpha + 3\sqrt{3} \; = \; 0$ which has solutions $\tan\alpha = \tfrac12\sqrt{3}, \tfrac15\sqrt{3}, - \sqrt{3}$ . The condition that $AM > MB$ tells us that we want the smaller of the two positive solutions, and hence $\tan\alpha = \tfrac15\sqrt{3}$ , which tells us that $x = \tfrac13R$ , $OM = \tfrac{x}{\sqrt{7}}$ and $CM = x\sqrt{7}$ .

Since triangles $\Delta OMB$ and $\Delta BMC$ are similar, we have $\frac{OB}{R} =\frac{x}{CM}$ and hence $OB = \tfrac{1}{\sqrt{7}}R$ . Applying the Cosine Rule to the triangle $\Delta ANB$ gives $AO = \sqrt{\tfrac37}R$ . Finally applying the Sine Rule to the triangle $\Delta AOB$ , we see that $\frac{\sin \angle AOB}{R} \; = \; \frac{\sin\alpha}{AO}$ and hence $\sin \angle AOB = \tfrac12$ . Since $\angle AOB$ is clearly obtuse, we deduce that $\angle AOB = \boxed{150^\circ}$ .

Let $AO$ intersect $BC$ at $P$ . Let $x = \frac{BM}{MA} = \frac{AN}{NC}$ and let $s$ be the side length of the triangle.

Observe that by Ratio Lemma, $\frac{\sin \angle CBN}{\sin \angle NBA} = \frac{NC}{NA}$ . From Ratio Lemma on $\triangle NBA$ , we have $\begin{aligned} \frac{BN}{AB} \frac{\sin \angle CBN}{\sin \angle NBA} & = \frac{NO}{OA} = \frac{2}{5} \\ \frac{BN}{BC} \frac{CN}{NA} & = \frac{2}{5} \\ \frac{BN}{BC} & = \frac{2x}{5} \end{aligned}$ So, $\frac{CN}{BN} = \frac{5-2x}{2x}$ . From Ceva's Theorem, we have $\frac{(5-2x)x}{2} = 1 \implies x = \frac{1}{2}.$

Then we can easily find that $BN = \frac{3}{5}s, NC = \frac{12}{5}s.$

Observe that $\angle AOB = 60^{\circ} + \angle NBC + \angle CAN$ . From Ratio Lemma, we have that $\frac{\sin \angle CBN}{\sin (60^{\circ} - \angle CBN)} = 2 \implies \tan \angle CBN = \frac{\sqrt{3}}{2}.$

From Ratio Lemma, we also have $\frac{\sin \angle CAN}{\sin (60^{\circ} - \angle CAN)} = 4 \implies \tan \angle CAN = \frac{2}{\sqrt{3}}.$

Since they are reciprocals of one another, the angles must add to $90^{\circ}$ .

Thus, $\angle AOB = 60^{\circ} + (\angle NBC + \angle CAN) = 60^{\circ} + 90^{\circ} = \boxed{150^{\circ}}$ .