## Task 3 - 260733

Let $ABC$ be an acute-angled triangle; its circumcircle $k$ has the center $M$. The ray from $A$ through $M$ intersects $k$ at $D$, the ray from $B$ through $M$ intersects $k$ at $E$, the ray from $C$ through $M$ intersects $k$ at $F$.

Determine the ratio of the areas of the hexagon $A F B D C E$ and the triangle $A B C$!
Since the triangles $M F A$ and $M C A$ are equal in the side lengths $M F, M C$ (radii of $k$) and in the length of the corresponding height (perpendicular from $A$ to the diameter $F C$), it follows with the notation: If $P Q R$ is a polygon, then $J(P Q R)$ denotes its area: $J(M F A)=J(M C A)$.

Similarly, we obtain $J(M C D)=J(M C A), J(M F B)=$ $J(M C B), J(M C E)=J(M C B), J(M D B)=J(M A B)$, $J(M A E)=J(M A B)$.

By adding these equations, we get $J(A F B D C E)=$ $2 \cdot J(A B C)$. The desired ratio is therefore 2:1.