Problem 4. (Option 2)

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

The line $\mathrm{BE}$ intersects $\mathrm{AC}$ and $\mathrm{AD}$ at points F and $\mathrm{M}$, respectively. Consider the closed broken line BKDMAFB.

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

Let $K B=x, F M=y$,

then from the condition of symmetry of points

A, B, C, D, and E, it follows that

$$
\begin{aligned}
& M A=A F=B F=B K=K C=x \\
& K F=F M=y, D K=K A=B M=D M=x+y \\
& \text { i.e. } \frac{x}{x+y} \cdot \frac{y}{x} \cdot \frac{2 x+y}{x}=1 \text { or } x^{2}+x y=2 x y+y^{2}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3fdcc7ef9dfdd3dab8ecg-04.jpg?height=591&width=609&top_left_y=1875&top_left_x=1266)

Solve the obtained homogeneous equation and get $\frac{x}{y}=\frac{1+\sqrt{5}}{2}$.

Express the vector $\overrightarrow{A K}=\overrightarrow{D K}-\overrightarrow{D A}$ in terms of $\vec{a}_{\text {and }}$

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

Since $\overrightarrow{A C}=\frac{2 x+y}{x+y} \overrightarrow{A K}$, we get

$\overrightarrow{A C}=\frac{2(2+\sqrt{5})}{3+\sqrt{5}} \overrightarrow{A K}=\frac{(1+\sqrt{5})(\sqrt{5}-1)}{4} \vec{b}-\frac{(1+\sqrt{5})}{2} \vec{a}=\vec{b}-\frac{(1+\sqrt{5})}{2} \vec{a}$.

Answer: $\overrightarrow{A C}=\vec{b}-\frac{(1+\sqrt{5})}{2} \vec{a}$.