Belarusian Mathematical Olympiad

Answer: $\angle B=90^{\circ }$.

Let $A_1$, $B_1$ and $C_1$ be the feet of the perpendiculars from $O$ to the sides $BC$, $CA$ and $AB$ respectively. Since $\angle O{A}_{1}B=90^{\circ }=\angle O{C}_{1}B$, the quadrilateral $B{A}_{1}O{C}_1$ is cyclic and $OB$ is the diameter of its circumcircle. From the sine law

for the triangle $A_{1}B{C}_1$ we get $A_1{C}_1=OB\sin \angle B$. Similarly $A_1{B}_1=OC\sin \angle C$ and $B_1{C}_1=OA\sin \angle A$. Since the triangle $A_1{B}_1{C}_1$ is equilateral it follows that $$\frac{\sin \angle B}{\sin \angle C}=\frac{OC}{OB}=\frac{20}{12}=\frac{5}{3},\frac{\sin \angle B}{\sin \angle A}=\frac{OA}{OB}=\frac{15}{12}=\frac{5}{4}.(1)$$ From the sine law for the triangle $ABC$ we obtain $$AB/\sin \angle C=BC/\sin \angle A=AC/\sin \angle B.$$ Therefore $$AB=AC\frac{\sin \angle C}{\sin \angle B}=\frac{3}{5}AC,BC=AC\frac{\sin \angle A}{\sin \angle B}=\frac{4}{5}AC.$$ Note that $A{B}^2+B{C}^2=\frac{9}{25}A{C}^2+\frac{16}{25}A{C}^2=A{C}^2$, hence the angle $ABC$ is right.