In junior high school plane geometry, there is such a problem: "There is a point $P$ inside the acute angle $\angle X O Y$. Try to find points $Q$ and $R$ on $O X$ and $O Y$ respectively, such that the perimeter of $\triangle P Q R$ is the shortest." This problem can be solved using geometric construction methods.
Proof: Let $Q^{\prime} \backslash R^{\prime}$ be any points on $O X$ and $O Y$ respectively, such that at least one of $Q^{\prime}=Q$ and $R^{\prime}=R$ does not hold. Connect $P Q$, $P R$, $P Q^{\prime}$, $P R^{\prime}$, and $Q^{\prime} R^{\prime}$ to form $\triangle P Q R$ and $\angle P Q^{\prime} R^{\prime}$. Then connect $P_1 Q^{\prime}$ and $P_2 R^{\prime}$ to form the broken line $P_1 Q^{\prime} R^{\prime} P_2$.
$\because P$ and $P$ are symmetric about $O X$,
$\therefore P Q = P Q, P Q^{\prime} = P Q^{\{\prime\}}$.
Similarly, $P R = P R, P R^{\prime} = P_2 R^{\prime}$. Thus, the length of segment $P_1 P_2$ is equal to the perimeter of $\triangle P Q R$, and the length of the broken line $P_1 Q^{\prime} P^{\prime} P_2$ is equal to the perimeter of $\triangle P Q^{\prime} R^{\prime}$. Clearly, $P_1 P_2 < P Q^{\prime} + Q^{\prime} R^{\prime} + R^{\prime} P_2$, which means the perimeter of $\triangle P Q R$ is shorter, making $P$, $Q$ the required points.

If the size of the acute angle $\angle X O Y$ is known and the position of point $P$ is determined, how large is the perimeter of $\triangle P Q R$?

Connect $O P$, $O P_1$, and $O P_2$. Let $O P = a$, $\angle X O P = \alpha$, and $\angle P O Y = \beta$, then $\angle X O Y = \alpha + \beta$. By symmetry, we know $O P_1 = O P_2 = a$, $\angle P_1 O X = \alpha$, and $\angle P_2 O Y = \beta$. In $\triangle P_1 O P_2$, by the cosine rule we get
$$
\begin{array}{c}
P_1 P_2^2 = O P_1^2 + O P_2^2 - 2 O P_1 \\
\quad \cdot O P_2 \cos \angle P_1 O P_2 \\
= a^2 + a - 2 a^2 \cos 2(\alpha + \beta) \\
= 2 a^2 [1 \\
- \cos 2(\alpha + \beta)] \\
= 4 a^2 \sin^2 \\
(\alpha + \beta) \\
= 4 a^2 \sin^2
\end{array}
$$
$\therefore \angle X O Y$.
$$
\begin{array}{c}
\therefore P_1 P_2 = \\
2 a \sin \angle X O Y.
\end{array}
$$

Thus, the perimeter $l$ of $\triangle P Q R$ is $2 a \sin \angle X O Y$. Therefore, we conclude: The shortest perimeter of a triangle formed by a point inside an acute angle and points on the two sides of the angle is twice the product of the distance from the point to the vertex of the angle and the sine of the acute angle.