Japan Mathematical Olympiad

78°

Triangles $MBA$ and $MAP$ are similar since $\angle MBA=\angle PAM$, thus $MP\cdot MB=M{A}^2$. Moreover, we have $MD=MA=MC$, since $\angle CDA=90^{\circ }$ and $M$ is the midpoint of $AC$. Then triangles $MBD$ and $MDP$ are similar since $MP\cdot MB=M{D}^2$, thus $\angle DBM=\angle MDP$. By $MD=MC$, we obtain $$\begin{array}{rl}\angle AMP& =\angle CBM+\angle MCB\\ & =\angle MDP+\angle CDM\\ & =\angle CDP\\ & =180^{\circ }-\angle PDB\\ & =180^{\circ }-115^{\circ }=65^{\circ }.\end{array}$$ Since $\angle MBA=\angle PAM$ and the sum of the three angles of $BAM$ is equal to $180^{\circ }$, we obtain $\angle PAM=\frac{180^{\circ }-\angle BAP-\angle AMB}{2}=\frac{180^{\circ }-41^{\circ }-65^{\circ }}{2}=37^{\circ }$. Thus the answer is $\angle BAP+\angle PAM=78^{\circ }$.