# Problem 4. (Variant 1)

On a circle, 5 points $A, B, C, D$ and $E$ are located at equal intervals. Given two vectors $\overrightarrow{D A}=\vec{a}, \overrightarrow{D B}=\vec{b}$. Express the vector $\overrightarrow{E C}$ in terms of $\vec{a}$ and $\vec{b}$.
# Solution:

Connect B and E.

Consider the closed broken line BKDMEFB.

By Menelaus' Theorem: $\frac{B K}{K D} \cdot \frac{D M}{M F} \cdot \frac{F E}{E B}=1$.

Let $K D=x, K M=y$,

![](https://cdn.mathpix.com/cropped/2024_05_06_3fdcc7ef9dfdd3dab8ecg-03.jpg?height=397&width=437&top_left_y=2360&top_left_x=1318)
then from the symmetry condition of points A, B, C, D, and E, it follows that

$M D=M E=E F=x, K M=F M=y, B K=K E=B F=x+y$, i.e.,

$\frac{x+y}{x} \cdot \frac{x}{y} \cdot \frac{x}{2 x+y}=1$ or $x^{2}+x y=2 x y+y^{2}$.

Solving the obtained homogeneous equation, we get $\frac{x}{y}=\frac{1+\sqrt{5}}{2}$.

Express the vector $\overrightarrow{M K}=\overrightarrow{D K}-\overrightarrow{D M}$ in terms of $\vec{a}$ and $\vec{b}$:

$\overrightarrow{M K}=\frac{x}{2 x+y} \vec{b}-\frac{x}{2 x+y} \vec{a}=\frac{1+\sqrt{5}}{4+2 \sqrt{5}} \vec{b}-\frac{1+\sqrt{5}}{4+2 \sqrt{5}} \vec{a}=\frac{3-\sqrt{5}}{2} \vec{b}-\frac{3-\sqrt{5}}{2} \vec{a}$

Since $\overrightarrow{E C}=\frac{2 x+y}{y} \overrightarrow{M K}$, we obtain the answer

$\overrightarrow{E C}=(2+\sqrt{5}) \overrightarrow{M K}=\frac{1+\sqrt{5}}{2} \vec{b}-\frac{1+\sqrt{5}}{2} \vec{a}$

Answer: $\overrightarrow{E C}=\frac{1+\sqrt{5}}{2} \vec{b}-\frac{1+\sqrt{5}}{2} \vec{a}$