## Task 4 - 120834

Given a circle with center $M$. A diameter of this circle is $A B$. Two points $P_{1}$ and $P_{2}$ start moving uniformly from the same time on each of the two semicircular arcs from $A$ to $B$, with the movement of point $P_{1}$ being four times as fast as that of point $P_{2}$.

Is there a time between the start and the arrival of $P_{1}$ (at $B$) when the triangles $\triangle A B P_{1}$ and $\triangle A B P_{2}$ have the same area?

If yes, determine for each such time the size of the angle $\angle A M P_{2}$!
For every position of the point $P_{1}$, on the semicircular arc it traverses, $P_{2}$ is positioned such that $\angle A M P_{2}=\frac{1}{4} \angle A M P_{1} \cdot 90^{\circ}$ is the condition (1) satisfied exactly when

$$
\angle B M P_{1}=\angle Q_{1} M P_{1}=\angle Q_{2} M P_{2}=\angle A M P_{2}
$$

holds; for from (1) follows (2) via the congruence theorem (ssw), and from (2) follows (1) via (sww). Now (2) is equivalent to $180^{\circ}-4 \cdot \angle A M P_{2}=\angle A M P_{2}$ and this to $\angle A M P_{2}=36^{\circ}$.

Therefore, there is exactly one moment with the required property, namely the one at which the angle $\angle A M P_{2}$ has the size $36^{\circ}$.