AoPS - Problem

The answer is $\angle A=80^\circ, \angle B=60^\circ, \angle C=40^\circ$. Note that since $AHIC$ is cyclic, $180^\circ-\angle B=\angle AHC=\angle AIC=90^\circ+\dfrac{\angle B}{2},$ and so $\angle B=60^\circ$. Moreover, note that $AP=BT=AB$, since $\angle BTA=180^\circ-\angle BTC=180^\circ-\angle BHC=\angle A$ and $\angle APB=180^\circ-\angle APC=180^\circ-\angle AHC=\angle B$. Therefore, triangle $APB$ must be equilateral. Therefore, $BP=BA=BT$, that is triangle $BPT$ is isosceles. Thus, $180^\circ=2\angle BPT+\angle TBP=2(\angle BPI+\angle IPT)+\angle BTA-\angle C=2(\angle IAC+180^\circ-\angle HIP)+\angle A-\angle C=$ $=\angle A+2\angle HAP+\angle A-\angle C=2\angle A-(120^\circ-\angle A)+\angle BAP=3\angle A-120^\circ+60^\circ,$ and so we obtain that $\angle A=80^\circ,$ and subsequently $\angle C=40^\circ,$ as desired.