4. Given a right triangle $ABC$ with legs $\overline{AC}=b$ and $\overline{BC}=b$. Through the vertex $C$, draw a line such that the sum of the distances from the vertices $A$ and $B$ to this line is maximized.
Solution. The line that needs to be drawn through vertex $C$ is fully determined by the angle $\varphi$ that this line makes with the leg $A C$ (see the diagram).

From the right triangles $A C B$, $A A^{\prime} C$, and $B B^{\prime} C$, we get

$$
a=b \operatorname{tg} \alpha, d_{b}=b \sin \varphi \text { and } d_{a}=a \cos \varphi
$$

Therefore, we have:

$$
\begin{aligned}
d_{a}+d_{b} & =a \sin \varphi+b \cos \varphi \\
& =b \operatorname{tg} \alpha \sin \varphi+b \cos \varphi \\
& =b\left(\frac{\sin \alpha}{\cos \alpha} \sin \varphi+\cos \varphi\right) \\
& =b \frac{\cos (\alpha-\varphi)}{\cos \alpha}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_faf4f26986d5dd6dd724g-7.jpg?height=500&width=580&top_left_y=1399&top_left_x=902)

According to this, $d_{a}+d_{b}$ will be the largest if $\cos (\alpha-\varphi)=1$, i.e., for $\phi=\alpha$.

## 4th year