Italian Mathematical Olympiad

Solution:

Suppose that the bisectors of angles $\angle AST$ and $\angle BTS$ meet at a point $P$ of side $AB$.
Since $\frac{AS}{AC} = \frac{BT}{BC}$, by Thales' theorem, $ST$ and $AB$ are parallel segments, hence the angles $\angle PST$ and $\angle SPA$ are equal. Now, since by hypothesis $\angle ASP = \angle SPA$, it follows that triangle $ASP$ is isosceles and therefore $AS = AP$. In an analogous way one derives that the segments $BT$ and $BP$ also have equal lengths. Recalling that:
$$AS = \frac{1}{3} AC \quad BT = \frac{1}{3} BC,$$
one obtains:
$$AB = AS + BT = \frac{1}{3}(AC + BC).$$

From the relation written above one further derives:
$$MA + NB = \frac{1}{2}(AC + BC) = \frac{3}{2} AB$$
On the other hand $MN = \frac{AB}{2}$ and thus one obtains:
$$MN + AB = AM + BN$$
a condition equivalent to saying that the quadrilateral $ABNM$ is circumscribable about a circle.

Suppose now that the quadrilateral $ABNM$ is circumscribable about a circle, hence we have:
$$MN + AB = AM + BN$$
Since the condition $MN = \frac{AB}{2}$ always holds it follows

$AM + BN = 3 \frac{AB}{2}$ and therefore one derives as before:
$$AS + BT = AB.$$
Let now $P_1$ and $P_2$ be the points of intersection with segment $AB$ respectively of the bisectors of $\angle AST$ and $\angle BTS$. In a completely analogous way as before one derives $AS = AP_1$ and $BT = BP_2$. Hence $AP_1 + P_2B = AB$ and therefore necessarily $P_1 = P_2$.

Solution:

Suppose that the bisectors of angles $\angle AST$ and $\angle BTS$ meet at a point $P$ of side $AB$.
Since $\frac{AS}{AC} = \frac{BT}{BC}$, by Thales' theorem, $ST$ and $AB$ are parallel segments, hence the angles $P\widehat{S}T$ and $S\widehat{P}A$ are equal. Now, since by hypothesis $A\widehat{S}P = S\widehat{P}A$, it follows that triangle $ASP$ is isosceles and therefore $AS = AP$. In an analogous way one derives that the segments $BT$ and $BP$ also have equal lengths. Recalling that:
$$AS = \frac{1}{3} AC \quad BT = \frac{1}{3} BC,$$
one obtains:
$$AB = AS + BT = \frac{1}{3}(AC + BC).$$

From the relation written above one further derives:
$$MA + NB = \frac{1}{2}(AC + BC) = \frac{3}{2} AB$$
On the other hand $MN = \frac{AB}{2}$ and thus one obtains:
$$MN + AB = AM + BN$$
a condition equivalent to saying that the quadrilateral $ABNM$ is circumscribable about a circle.

Suppose now that the quadrilateral $ABNM$ is circumscribable about a circle, hence we have:
$$MN + AB = AM + BN$$
Since the condition $MN = \frac{AB}{2}$ always holds it follows

$AM + BN = 3 \frac{AB}{2}$ and therefore one derives as before:
$$AS + BT = AB.$$
Let now $P_1$ and $P_2$ be the points of intersection with segment $AB$ respectively of the bisectors of $\angle AST$ and $\angle BTS$. In a completely analogous way as before one derives $AS = AP_1$ and $BT = BP_2$. Hence $AP_1 + P_2B = AB$ and therefore necessarily $P_1 = P_2$.