51st Ukrainian National Mathematical Olympiad, 4th Round

Let $D$ be a point on the half-line $CA$ such that $CA=AD$ (fig. 37).



Then $DM$ and $BA$ are medians of the triangle $BCD$, and so the point of their intersection divides both of them in the ratio $2:1$ from the vertex. This means that they intersect at the point $N$. Hence $\angle DAB=\pi -\angle CAB=\pi -\angle CMN=\angle DMB$, and so the quadrilateral $BMAD$ is cyclic. Since $AM$ is parallel to $BD$ as a midline, we have that $\angle CDB=\angle CAM=\angle CBD$. This implies that $$CD=CB,\text{ therefore }\frac{AC}{BC}=\frac{AC}{CD}=\frac{1}{2}.$$ It is easy to see that the problem condition is satisfied for every triangle with this ratio of the sides.