4. Let $\measuredangle X S Y=\alpha<90^{\circ}$ and $k(O, r)$ be a circle that is entirely within $\alpha$. Let $A B C$ be a triangle such that $A$ is on $k$, $B$ is on the line $p \equiv S X$, and $C$ is on the line $q \equiv S Y$. If $s=\overline{S O}$, calculate the smallest value that the perimeter of such a $\triangle A B C$ can achieve.
Solution. Let $A^{\prime}$ be the intersection point of $S O$ with $k$, which is between $S$ and $O$, $k_{p}\left(O_{p}, r\right), k_{q}\left(O_{q}, r\right)$ are circles symmetric to $k$ with respect to $p$ and $q$ respectively, $A_{p}^{\prime}$ and $A_{q}^{\prime}$ are the images of $A^{\prime}$ under the axial symmetries, and $B^{\prime}$ and $C^{\prime}$ are the intersection points of $A^{\prime}_{p} A_{q}^{\prime}$ with $p$ and $q$ respectively. We will prove that $\triangle A^{\prime} B^{\prime} C^{\prime}$ has the smallest perimeter among the required ones. Let $\triangle A B C$ be an arbitrary such triangle and let $A_{p}$ and $A_{q}$ be the images of $A$ under the axial symmetries, then

$$
\overline{A B}+\overline{B C}+\overline{C A}=\overline{A_{p} B}+\overline{B C}+\overline{C A_{q}} \geq \overline{A_{p} A_{q}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_058dca2a4dafda946906g-8.jpg?height=817&width=1213&top_left_y=1380&top_left_x=249)

## On the other hand

$$
\measuredangle A_{p} S A_{q}=\measuredangle A_{p} S A+\measuredangle A S A_{q}=2(\measuredangle X S A+\measuredangle A S Y)=2 \measuredangle X S Y=2 \alpha
$$

so the angle $\measuredangle A_{p} S A_{q}$ does not depend on the choice of points $A, B$ and $C$. Also, from the axial symmetries, it follows that $\overline{S A_{p}}=\overline{S A}=\overline{S A_{q}}$, so $\Delta A_{p} A_{q} S$ is isosceles. From the estimate:

$$
\begin{aligned}
L_{\triangle A B C} & \geq \overline{A_{p} A_{q}}=2 \overline{S A_{p}} \sin \alpha=2 \overline{S A} \sin \alpha=2((\overline{S A}+\overline{A O})-\overline{A O}) \sin \alpha \\
& \geq 2\left(\overline{S O}-\overline{A^{\prime} O}\right) \sin \alpha=2 \overline{S A^{\prime}} \sin \alpha=L_{\triangle A^{\prime} B^{\prime} C^{\prime}}
\end{aligned}
$$

it follows that $\triangle A^{\prime} B^{\prime} C^{\prime}$ achieves the smallest possible perimeter, which is equal to $L_{\triangle A^{\prime} B^{\prime} C^{\prime}}=2\left(\overline{S O}-\overline{A^{\prime} O}\right) \sin \alpha=2(s-r) \sin \alpha$.

Note: From the condition of the problem, it must be that $s>r$.