4. Let there be a pentagon $A B C D E$ in which $C D=A B+A E$. Let the bisector of $\measuredangle B A E$ intersect $B E$ at point $P$, let $Q$ be a point on $C D$ such that $C Q=A B$, and let $X$ and $Y$ be the midpoints of segments $B C$ and $D E$, respectively. Let $T$ be the centroid of $\triangle A P Q$ and let $Y T$ intersect $A X$ at point $Z$. Find $\frac{A Z}{Z X}$ as a function of $A B$ and $A E$.
4. First solution. Let $A B = C Q = b$ and $A E = D Q = e$. From the known property of the angle bisector, we have $\frac{P B}{P E} = \frac{A B}{A E} = \frac{b}{e} = \frac{Q C}{Q D}$. Let $X Y$ and $P Q$ intersect at point $R$. Then we have:

$$
\begin{aligned}
\overrightarrow{R P} + \overrightarrow{R Q} & = \frac{e \overrightarrow{R B} + b \overrightarrow{R E}}{b + e} + \frac{e \overrightarrow{R C} + b \overrightarrow{R D}}{b + e} = \frac{e (\overrightarrow{R B} + \overrightarrow{R C}) + b (\overrightarrow{R E} + \overrightarrow{R D})}{b + e} \\
& = \frac{2 e \overrightarrow{R X} + 2 b \overrightarrow{R Y}}{b + e}
\end{aligned}
$$

Since the vector on the left side lies on line $P Q$, and the vector on the right side lies on line $X Y$, it follows that both vectors must be $\overrightarrow{0}$; in other words, $\overrightarrow{R P} = -\overrightarrow{R Q}$ and $e \overrightarrow{R X} = -b \overrightarrow{R Y}$, from which it follows that $R$ is the midpoint of segment $P Q$, and the proportion $\frac{R X}{R Y} = \frac{b}{e}$ holds. Now we observe that points $A, T$, and $R$ are collinear and that $\frac{R T}{T A} = \frac{1}{2}$. Applying Menelaus' theorem to $\triangle A X R$ and line $Y Z$, we get:

$$
-1 = \frac{\overrightarrow{A Z}}{\overrightarrow{Z X}} \cdot \frac{\overrightarrow{X Y}}{\overrightarrow{Y R}} \cdot \frac{\overrightarrow{R T}}{\overrightarrow{T A}} = \frac{\overrightarrow{A Z}}{\overrightarrow{Z X}} \cdot \left(-\frac{b + e}{e}\right) \cdot \frac{1}{2}
$$

from which it follows that

$$
\frac{A Z}{Z X} = \frac{2 e}{b + e} = \frac{2 A E}{A B + A E}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_adcebcbe7d41887b6eefg-30.jpg?height=466&width=1334&top_left_y=1214&top_left_x=210)

Op $20194 \mathrm{~A} 4$

Second solution. Let $A B = C Q = b$ and $A E = D Q = e$, and let $S$ be the midpoint of segment $P Q$. Further, let the line through $E$ parallel to $C D$ intersect line $Q Y$ at point $K$, and the line through $B$ parallel to $C D$ intersect line $Q X$ at point $L$. Clearly, $\triangle Y Q D \cong \triangle Y K E$ and $\triangle X Q C \cong \triangle X L B$, from which it follows that $K Y = Y Q$, $L X = X Q$, $K E = Q D = e$, and $L B = Q C = b$. Since $A P$ is the angle bisector of $\measuredangle B A E$, it follows that $\frac{E P}{P B} = \frac{e}{b} = \frac{K E}{L B}$. From this and $K E \parallel C D \parallel L B$, it follows that $\triangle P E K \sim \triangle P B L$, from which we get $\measuredangle K P E = \measuredangle L P B$, i.e., points $K, P$, and $L$ are collinear. Further, $X Y$ is the midline of $\triangle L Q K$, so since $S$ is the midpoint of segment $P Q$, it follows that $S$ lies on line $X Y$ (i.e., $P Q$ and $X Y$ intersect at point $S$). Additionally, from $X Y \parallel L K$, we have $\frac{S X}{S Y} = \frac{P L}{P K} = \frac{L B}{K E} = \frac{b}{e}$.

Now consider $\triangle A Y X$. Let the parallel from $S$ intersect $Y Z$ at $G$. From $\triangle S G T \sim \triangle A Z T$, it follows that $\frac{G S}{Z A} = \frac{S T}{A T} = \frac{1}{2}$, i.e., $G S = \frac{Z A}{2}$. Further, from $\triangle Y G S \sim \triangle Y Z X$, we get $\frac{Z X}{G S} = \frac{X Y}{S Y} = \frac{X S + S Y}{S Y} = \frac{X S}{S Y} + 1 = \frac{b}{e} + 1 = \frac{b + e}{e}$, i.e., $\frac{Z X}{\frac{Z A}{2}} = \frac{b + e}{e}$. Therefore,

$$
\frac{A Z}{Z X} = \frac{2 e}{b + e} = \frac{2 A E}{A B + A E}
$$