3. In the rectangle $A B C D$, points $P$ and $Q$ are points on sides $A B$ and $B C$ respectively, such that the triangles $A P D, P B Q$ and $Q C D$ have the same area.

Find the ratio $\frac{\overline{A P}}{\overline{P B}}$.
Solution. We introduce notations

$$
\overline{A P}=p_{1}, \overline{P B}=p_{2}, \overline{Q B}=q_{2} \text { and } \overline{Q C}=q_{1}
$$

From the equality

$$
\frac{1}{2}\left(p_{1}+p_{2}\right) \cdot q_{1}=\frac{1}{2}\left(q_{1}+q_{2}\right) p_{1}
$$

we get $\frac{p_{1}}{p_{2}}=\frac{q_{1}}{q_{2}}$, and from

$$
\frac{1}{2}\left(q_{1}+q_{2}\right) p_{1}=\frac{1}{2} q_{2} p_{2}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_f3fdf1bc245e57626c19g-04.jpg?height=317&width=483&top_left_y=1740&top_left_x=979)

we get $\frac{p_{1}}{p_{2}}=\frac{q_{2}}{q_{1}+q_{2}}$.

## Therefore

$$
\frac{p_{1}}{p_{2}}=\frac{q_{2}}{q_{1}+q_{2}}=\frac{1}{\frac{q_{1}}{q_{2}}+1}=\frac{1}{\frac{p_{1}}{p_{2}}+1} \cdot \text { Hence, } \frac{p_{1}}{p_{2}}=\frac{-1+\sqrt{5}}{2} \text {. }
$$