4. Let $D$ and $E$ be the midpoints of sides $A C$ and $B C$ of an equilateral triangle $A B C$. The ray $D E$ intersects the circumcircle of triangle $A B C$ at point $F$. In what ratio does point $E$ divide the segment $D F$?
II/4. Let $a=|A B|$ and $x=|E F|$. The midpoint of segment $D E$ is denoted by $G$. Since $D E C$ is an equilateral triangle with side length $\frac{a}{2}$, we have $|G E|=\frac{a}{4}$ and $|G C|=\frac{a \sqrt{3}}{4}$. Let the circumcircle of triangle $A B C$ have radius $r$ and center at $O$. Then $r=\frac{a \sqrt{3}}{3}$, $|O G|=r-|G C|=\frac{a \sqrt{3}}{3}-\frac{a \sqrt{3}}{4}=\frac{a \sqrt{3}}{12}$, and $|G F|=\frac{a}{4}+x$. Triangle $O F G$ is a right triangle with hypotenuse $O F$, so $|O F|^{2}=|O G|^{2}+|G F|^{2}$. Therefore, $|G F|^{2}=|O F|^{2}-|O G|^{2}=\frac{5 a^{2}}{16}$ and $x=-\frac{a}{4}+\frac{a \sqrt{5}}{4}=\frac{a(\sqrt{5}-1)}{4}$. The desired ratio is thus $|D E|:|E F|=\frac{a}{2}: \frac{a(\sqrt{5}-1)}{4}=2:(\sqrt{5}-1)$.

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-08.jpg?height=566&width=643&top_left_y=1576&top_left_x=1232)
(We can also write $2:(\sqrt{5}-1)=(\sqrt{5}+1): 2$.) The lengths of the sides and heights of triangle $D E C$ in terms of the side length of triangle $A B C \ldots 1$ point Length of $|O G|$ (with analogous notation) . ...............................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-08.jpg?height=66&width=1690&top_left_y=2280&top_left_x=186)
Calculation of the ratio .2 points