NMO Selection Tests for the Junior Balkan Mathematical Olympiad - Second Selection Test

Let lines $CH$ and $PT$ meet at $U$ and denote by $A'$, $B'$, $C'$ the feet of the altitudes of the triangle $ABC$.

$AHIC$ is a cyclic quadrilateral, so $\angle CHI = \angle CAI = \frac{\angle A}{2}$. Since $PT \parallel HI$, it follows that $\angle HUT = \angle CHI = \frac{\angle A}{2}$.
Hence, $\angle CUP = \angle HUT = \frac{\angle A}{2}$. (1)

The quadrilateral $AHIC$ is also cyclic, so $\angle A'HI = \angle ACI = \frac{\angle C}{2}$.
From the triangle $A'HC$ we have $\angle A'CH = 90^\circ - \angle A'HC = 90^\circ - \frac{\angle A+\angle C}{2} = \frac{\angle B}{2}$. From the right-angled triangle $BCC'$ we get $\angle B + \frac{\angle B}{2} = 90^\circ$, so $\angle B = 60^\circ$.
Notice that $\angle ABB' = \angle ACC' = 90^\circ - \angle A$. The quadrilateral $BCTH$ is cyclic, so $\angle B'TB = \angle TCH = 90^\circ - \angle A = \angle ABB'$.
Therefore, $BB'$ is an angle bisector and also an altitude for triangle $ABT$, so $AB = BT$. Since $AHIC$ is cyclic, it follows that $\angle BPI = \angle CAI = \frac{\angle A}{2}$ and $\angle API = \angle ACI = \frac{\angle C}{2}$. Hence, $\angle APB = \frac{\angle A+\angle C}{2} = 60^\circ = \angle B$, so the triangle $ABP$ is equilateral.
Thereby, $BP = AB = BT$, so triangle $BPT$ is isosceles, hence $\angle BPT = \frac{180^\circ - \angle PBT}{2}$. Triangle $ABT$ is also an isosceles one, so $\angle ABT = 180^\circ - 2\angle A$. Noticing that $\angle PBT = \angle B - \angle ABT$, it results that $\angle PBT = 2\angle A - 120^\circ$, which leads to $\angle BPT = 150^\circ - \angle A$ (2).
From the exterior angle theorem we infer that $\angle BPT = \angle CUP + \angle PCU$. Using (1) and (2), we obtain $150^\circ - \angle A = \frac{\angle A}{2} + \frac{\angle B}{2} = \frac{\angle A}{2} + 30^\circ$, so $\angle A = 80^\circ$ and $\angle C = 40^\circ$.