4. Inside an angle of $30^{\circ}$ with vertex $A$, a point $K$ is chosen, the distances from which to the sides of the angle are 1 and 2. Through point $K$, all possible lines are drawn, intersecting the sides of the angle. Find the minimum area of the triangle cut off by the line from the angle.
Answer: 8.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-36.jpg?height=388&width=1260&top_left_y=500&top_left_x=478)

Solution 1. Choose points $P$ and $Q$ on the sides of the angle such that point $K$ is the midpoint of segment $P Q$. We will show that line $P Q$ cuts off a triangle of the smallest area. Take another line passing through point $K$. Let it intersect the sides of the angle at points $X$ and $Y$. By swapping $P$ with $Q$ and $X$ with $Y$ if necessary, we can assume that $X$ lies on segment $A P$. It is sufficient to prove that $K X < K Y$. Suppose $K X \geqslant K Y$. Mark a point $Z$ on segment $K X$ such that $K Z = K Y$ (see the left figure). Triangles $K Q Y$ and $K P Z$ are equal by two sides and the angle, hence

$$
\angle K Q Y = \angle K P Z \quad \text{and} \quad \angle Z P Q + \angle A Q P = 180^{\circ}.
$$

But this is impossible, since point $Z$ lies inside triangle $A P Q$ and

$$
\angle Z P Q + \angle A Q P \leqslant \angle A P Q + \angle A Q P = 150^{\circ}.
$$

2. The area of triangle $A X Y$ is expressed by the function

$$
S(x) = \frac{x^2}{x-2}, \quad \text{hence} \quad S'(x) = \frac{x(x-4)}{(x-2)^2}
$$

Thus, the derivative $S$ is negative for $x \in (2,4)$ and positive for $x > 4$. Therefore, the minimum value of $S$ is achieved at $x = 4$ and is equal to 8.