In triangle $ABC$, the altitude and median from vertex $C$ divide the angle $ACB$ into three equal parts. Determine the ratio of the sides of the triangle.
Let the foot of the altitude and the median from vertex $C$ be $T$ and $F$, and the center of the circumcircle of the triangle be $O$. The angles at vertices $A$ and $B$ are acute, because the altitude from $C$ lies inside the triangle. By symmetry, we can assume that $\angle CACB$ follows.

According to the problem statement, $\angle ACT = \angle FCB$, and also $\angle ACT = \angle OCB$ (if the second intersection points of the lines $CT$ and $CO$ with the circumcircle are $T'$ and $O'$, respectively, then by Thales' theorem $\angle CT'O' = 90^\circ$, so $T'O' \parallel AB$, hence the arcs $AT'$ and $O'B$ of the circumcircle are equal, and therefore the corresponding inscribed angles are also equal), which means that the points $C$, $O$, and $F$ lie on the same line. But $O$ lies on the perpendicular to $AB$ at $F$, and since $CA < CB$, $C$ does not lie on this line, so $O$ coincides with $F$. Then, by Thales' theorem, $\angle ACB = 90^\circ$, so $\angle CAB = 90^\circ - \angle ACT = 90^\circ - \frac{1}{3} \angle ACB = 60^\circ$.

Therefore, the only triangle that can satisfy the conditions of the problem is a right triangle with angles of $60^\circ$ and $30^\circ$, and it is easy to see that this is indeed a valid solution. Thus, the ratio of the sides of the triangle is $2: \sqrt{3}: 1$.

Baldvin Kovács (Fazekas M. Főv. Gyak. Gimn., II. o. t.)