All-Soviet-Union Mathematical Olympiad

Solution:
As usual let $a$, $b$, $c$ be the lengths of $BC$, $CA$, $AB$ respectively and let $A$, $B$, $C$ denote the angles $BAC$, $ABC$, $BCA$ respectively. We use trigonometry and try to express the quantities of interest in terms of $a$, $b$ and $C$.

a.
Since $AH$ is the longest altitude, $BC$ must be the shortest side (use area $=$ side $\times$ altitude$/2$). So $b^2 \geq a^2$, and $c^2 \geq a^2$. Using the formula $c^2 = a^2 + b^2 - 2ab\cos C$, we deduce that $b^2 \geq 2ab\cos C$. Hence $2b^2 \geq a^2 + 2ab\cos C$. After a little manipulation this gives: $a^2 + b^2 - 2ab\cos C \geq \frac{4}{3}(a^2 + \frac{b^2}{4} - ab\cos C)$ or $c^2 \geq \frac{4}{3}BM^2$. But we are given that $BM \geq AH = b \sin C$, so $\frac{b^2\sin^2 C}{c^2} \leq \frac{3}{4}$. But the sine formula gives $\sin B = \frac{b\sin C}{c}$, so $\sin^2 C \leq \frac{3}{4}$. The triangle is acute-angled, hence $B \leq 60^\circ$.

b.
The angle bisector theorem gives $\frac{AD}{BD} = \frac{b}{a}$, hence $\frac{AD}{AB} = \frac{b}{a + b}$, so $AD = \frac{bc}{a + b}$. Hence, using the sine formula, $\frac{CD}{\sin A} = \frac{AD}{\sin \frac{C}{2}}$. So $CD = \frac{bc \sin A}{(a + b) \sin \frac{C}{2}} = \frac{ba \sin C}{(a + b) \sin \frac{C}{2}}$, using the sine formula again. But we are given that $CD \geq AH = b \sin C$, so $\frac{a}{a + b} \sin \frac{C}{2} \geq 1$. But $a$ is the shortest side, so $\frac{a}{a + b} \leq \frac{1}{2}$ and hence $\sin \frac{C}{2} < \frac{1}{2}$. The triangle is acute-angled, so $\frac{C}{2} \leq 30^\circ$, and $C \leq 60^\circ$. $BC$ is the shortest side, so $A$ is the smallest angle and hence $A \leq 60^\circ$. Also since $AH \leq BM$, $B \leq 60^\circ$. But the angles sum to $180^\circ$, so they must all be $60^\circ$ and hence the triangle is equilateral.