In the convex quadrilateral $A B C D$, $A B+B D+D C$ is at most 2, and the area of the quadrilateral is $1 / 2$. How long can the diagonal $A C$ be?
Let's use the notations shown in Figure 1. According to the first condition of the problem, $a+e+c \leqq 2$, which means

$$
a+c \leqq 2-e
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_985c473610eb7b1cc2a3g-1.jpg?height=542&width=656&top_left_y=339&top_left_x=713)

Clearly, it also holds that

$$
m_{1} \leqq a \quad \text { and } \quad m^{2} \leqq c
$$

Since the area of the quadrilateral is $\frac{1}{2}$,

$$
e\left(m_{1}+m_{2}\right)=1
$$

By combining the inequalities in (2) with (1),

$$
m_{1}+m_{2} \leqq a+c \leqq 2-e
$$

Thus, according to (3),

$$
1=e\left(m_{1}+m_{2}\right) \leqq e(a+c) \leqq e(2-e)
$$

which implies

$$
1 \leqq e(2-e), \quad(e-1)^{2} \leqq 0
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_985c473610eb7b1cc2a3g-1.jpg?height=545&width=588&top_left_y=1679&top_left_x=758)

This is only possible as an equality, so $e=1$. In this case, equality must also hold in (2) and (1), thus $m_{1}=a, m_{2}=c$ and $a+c=2-e=1$, and $\alpha=\gamma=90^{\circ}$. Therefore, based on Figure 2,

$$
\begin{aligned}
A C^{2} & =(a+c)^{2}+e^{2}=2 \\
A C & =\sqrt{2}
\end{aligned}
$$