Problem 3. Let $ABC$ be a triangle where $m(\angle ABC) = 45^{\circ}$ and $m(\angle BAC) > 90^{\circ}$. Let $O$ be the midpoint of side $[BC]$. Consider the point $M \in (AC)$ such that $m(\angle COM) = m(\angle CAB)$. The perpendicular from $M$ to $AC$ intersects line $AB$ at point $P$.

a) Find the measure of the angle $\angle BCP$.

b) Show that if $m(\angle BAC) = 105^{\circ}$, then $PB = 2MO$.
## Solution and grading criteria:

a) Let $P^{\prime}$ be the point where the perpendicular bisector of side $[B C]$ intersects line $A B$, and $M^{\prime}$ the projection of point $P^{\prime}$ onto line $A C$. Then triangle $P^{\prime} B C$ is a right isosceles triangle. From the leg theorem, we obtain that $P^{\prime} C^{2}=C M^{\prime} \cdot C A$ and $P^{\prime} C^{2}=C O \cdot C B$.

Thus, $\frac{C M^{\prime}}{C B}=\frac{C O}{C A}$ and $\angle O C M^{\prime}=\angle A C B$, so triangles $O C M^{\prime}$ and $A C B$ are similar. We deduce that $\angle M^{\prime} O C \equiv \angle B A C \equiv \angle M O C$, which shows that points $M$ and $M^{\prime}$ coincide. Then points $P$ and $P^{\prime}$ also coincide, so $m(\angle B C P)=m\left(\angle B C P^{\prime}\right)=45^{\circ}$. . . 5p

b) Let $D$ be the midpoint of segment $[P C]$. If $m(\angle B A C)=105^{\circ}$, then we successively obtain: $m(\angle A C B)=30^{\circ}, m(\angle M C P)=15^{\circ}, m(\angle M D P)=30^{\circ}$, and $m(\angle M D O)=60^{\circ}$. Furthermore, $M D=O D=C D$, so triangle $M D O$ is equilateral, from which $M O=M D=C D=$ $\frac{P B}{2}$. . . 2p

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