Let $ABCD$ be a rhombus, and let $E$ be the point of intersection of the diagonals. Let $F$ be the midpoint of the segment $[BE]$, and $G$ be the midpoint of the segment $[AD]$. Let $I$ be the point of intersection of the lines $(FG)$ and $(AC)$, and let $K$ be the symmetric point of $A$ with respect to point $I$. What is $\frac{EK}{EA}$?
Let us note that by symmetry, $E K=I K-I E=A I-I E=A E-2 I E$. Therefore, it is sufficient to express $I E$ in terms of $E A$ to calculate the ratio $E K / E A$.

Introduce $H$ as the midpoint of the segment $[D E]$. Thus $H E=\frac{D E}{2}=\frac{B E}{2}=E F$, so $E$ is the midpoint of $[H F]$.

The midline theorem tells us that $(H G)$ and $(E A)$ are parallel. Thus $(E I)$ and $(H G)$ are parallel, so since $E$ is the midpoint of $[H F]$, by the midline theorem, $I$ is the midpoint of $[F G]$ and $E I=\frac{H G}{2}$. By the midline theorem in triangle $A E G, G H=\frac{A E}{2}$, so $E I=\frac{A E}{4}$.

Thus $E K=A E-2 I E=\frac{A E}{2}$, so $\frac{A E^{4}}{A K}=\frac{1}{2}$.

Comment from the graders: Most students had a good intuition about the problem and were able to find or observe the result. However, it is important to justify your answers properly: even when lines appear parallel or perpendicular on the figure, this must be justified. This can often be done using Thales' theorem or its refined version, the midline theorem. Some students correctly noted that the problem could be treated analytically. However, one must be very careful because natural coordinate systems may not be orthonormal, and thus the usual formula for norms, which relies on Pythagoras, cannot be used.