Indian National Mathematical Olympiad

Solution:

Let $\alpha$ denote the equal angles $\angle BAD = \angle DCA$. Using sine rule in triangles $DAB$ and $DAC$, we get
$$\frac{AD}{\sin B} = \frac{BD}{\sin \alpha}, \quad \frac{CD}{\sin 15^{\circ}} = \frac{AD}{\sin \alpha}$$
Eliminating $\alpha$ (using $BD = DC$ and $2\alpha + B + 15^{\circ} = \pi$), we obtain $1 + \cos(B + 15^{\circ}) = 2 \sin B \sin 15^{\circ}$. But we know that $2 \sin B \sin 15^{\circ} = \cos(B - 15^{\circ}) - \cos(B + 15^{\circ})$. Putting $\beta = B - 15^{\circ}$, we get a relation $1 + 2\cos(\beta + 30) = \cos \beta$. We write this in the form
$$(1 - \sqrt{3}) \cos \beta + \sin \beta = 1$$
Since $\sin \beta \leq 1$, it follows that $(1 - \sqrt{3}) \cos \beta \geq 0$. We conclude that $\cos \beta \leq 0$ and hence that $\beta$ is obtuse. So is angle $B$ and hence $\angle ADC$.

We have the relation $(1 - \sqrt{3}) \cos \beta + \sin \beta = 1$. If we set $x = \tan(\beta/2)$, then we get, using $\cos \beta = (1 - x^2)/(1 + x^2)$, $\sin \beta = 2x/(1 + x^2)$,
$$(\sqrt{3} - 2)x^2 + 2x - \sqrt{3} = 0$$
Solving for $x$, we obtain $x = 1$ or $x = \sqrt{3}(2 + \sqrt{3})$. If $x = \sqrt{3}(2 + \sqrt{3})$, then $\tan(\beta/2) > 2 + \sqrt{3} = \tan 75^{\circ}$ giving us $\beta > 150^{\circ}$. This forces that $B > 165^{\circ}$ and hence $B + A > 165^{\circ} + 15^{\circ} = 180^{\circ}$, a contradiction. Thus $x = 1$ giving us $\beta = \pi/2$. This gives $B = 105^{\circ}$ and hence $\alpha = 30^{\circ}$. Thus $\angle DAO = 60^{\circ}$. Since $OA = OD$, the result follows.

Solution 2:
Let $m_a$ denote the median $AD$. Then we can compute
$$\cos \alpha = \frac{c^2 + m_a^2 - (a^2/4)}{2cm_a}, \quad \sin \alpha = \frac{2\Delta}{cm_a}$$
where $\Delta$ denotes the area of triangle $ABC$. These two expressions give
$$\cot \alpha = \frac{c^2 + m_a^2 - (a^2/4)}{4\Delta}$$
Similarly, we obtain
$$\cot \angle CAD = \frac{b^2 + m_a^2 - (a^2/4)}{4\Delta}$$
Thus we get
$$\cot \alpha - \cot 15^{\circ} = \frac{c^2 - a^2}{4\Delta}$$
Similarly we can also obtain
$$\cot B - \cot \alpha = \frac{c^2 - a^2}{4\Delta}$$
giving us the relation
$$\cot B = 2\cot \alpha - \cot 15^{\circ}$$
If $B$ is acute then $2\cot \alpha > \cot 15^{\circ} = 2 + \sqrt{3} > 2\sqrt{3}$. It follows that $\cot \alpha > \sqrt{3}$. This implies that $\alpha < 30^{\circ}$ and hence
$$B = 180^{\circ} - 2\alpha - 15^{\circ} > 105^{\circ}$$
This contradiction forces that angle $B$ is obtuse and consequently $\angle ADC$ is obtuse.

Since $\angle BAD = \alpha = \angle ACD$, the line $AB$ is tangent to the circumcircle $\Gamma$ of $ADC$ at $A$. Hence $OA$ is perpendicular to $AB$. Draw $DE$ and $BF$ perpendicular to $AC$, and join $OD$. Since $\angle DAC = 15^{\circ}$, we see that $\angle DOC = 30^{\circ}$ and hence $DE = OD/2$. But $DE$ is parallel to $BF$ and $BD = DC$ shows that $BF = 2DE$. We conclude that $BF = DO$. But $DO = AO$, both being radii of $\Gamma$. Thus $BF = AO$. Using right triangles $BFO$ and $BAO$, we infer that $AB = OF$. We conclude that $ABFO$ is a rectangle. In particular $\angle AOF = 90^{\circ}$. It follows that
$$\angle AOD = 90^{\circ} - \angle DOC = 90^{\circ} - 30^{\circ} = 60^{\circ}$$
Since $OA = OD$, we conclude that $AOD$ is equilateral.