58th Ukrainian National Mathematical Olympiad

The statement of the problem yields $\angle BQC=\angle BTC=\angle BPC=90^{\circ }$, hence, points $Q,T,P$ lie on the circle with diameter $BC$, and in this exact order: $B,Q,T,P,C$ (Fig. 6), since $△ABC$ is acute. Then, $$\angle QTP=180^{\circ }-\angle ABP=180^{\circ }-(90^{\circ }-\angle BAC)=90^{\circ }+\angle BAC.$$ On the other hand, $QT\perp AC$, $PT\perp AB$, $\angle QTP=180^{\circ }-\angle BAC$. Hence, $$90^{\circ }+\angle BAC=180^{\circ }-\angle BAC\Rightarrow \angle BAC=45^{\circ }.$$