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 perimeter of the triangle cut off by the line from the angle.
Answer: $4(2+\sqrt{3})$.
![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-32.jpg?height=308&width=1266&top_left_y=794&top_left_x=472)

Solution. Draw a circle $\omega$ that is tangent to the sides of the angle at points $M$ and $N$, such that $K$ lies on the minor arc $MN$ (see the left figure). Let $p$ be the perimeter of the triangle cut off from the angle by the tangent to $\omega$ at point $K$. Then,

$$
p=AP+PK+KQ+AQ=AP+PM+QN+AQ=AM+AN=2 \cdot AN
$$

Note that in this reasoning, $K$ can be replaced by any other point lying on the minor arc $MN$.

We will show that the perimeter $p$ is minimal. Let $\ell$ be a secant to $\omega$ passing through point $K$. Draw a line $\ell_1$ parallel to $\ell$ that is tangent to the minor arc $MN$. The lines $\ell$ and $\ell_1$ cut off similar triangles from the angle. The line $\ell_1$ gives a smaller triangle, the perimeter of which is equal to $p$. Therefore, the minimum is not realized on $\ell$.

Let $T$ and $S$ be the projections of point $K$ onto the sides of the angle, where $KT=2$ and $KS=1$. Construct the circle $\omega$. Let $O$ be the intersection of the angle bisector with the line passing through $K$ parallel to $AT$ (see the right figure). Let $M$ and $N$ be the projections of point $O$ onto the rays $AS$ and $AT$, and $L$ be the projection of $K$ onto $OM$. Then,

$$
OM=ON=KT=2, \quad LO=OM-LM=OM-KS=1 \quad \text{and} \quad OK=2,
$$

since $\angle LKO=\angle MAN=30^{\circ}$. Therefore, $O$ is the center of the desired circle $\omega$. By the above proof,

$$
p=2 \cdot AN=2 \cdot ON \cdot \operatorname{ctg} 15^{\circ}=4 \cdot \frac{1+\cos 30^{\circ}}{\sin 30^{\circ}}=4(2+\sqrt{3})
$$