2. A little goat is tied to a stake with a 4.7 m long rope, located on a flat meadow next to an old tree. The trunk of the tree has the shape of a circular cylinder with a radius of 0.5 m, and the shortest distance from the stake to the surface of the tree is $1 \mathrm{m}$.

Can the goat walk around the tree and approach the stake so closely that the end of the rope coincides with its beginning? Justify your answer.

Assume that the rope, when taut, lies in a horizontal plane.
Answer: No. Solution. When the kid goes around the tree and approaches the stake, the rope will partially lie on the surface of the tree, and the remaining part will consist of two straight segments. The minimum necessary length is the sum of twice the length of the tangent segment drawn from the stake to the tree and the length of the arc between the points of tangency (see figure). Let the angle between the tangents be $2 \alpha$, then

$$
\sin \alpha=\frac{R}{R+d}=\frac{1}{3}
$$

In this case, $L_{\min }=2 R \operatorname{ctg} \alpha+R(\pi+2 \alpha)=2 \sqrt{2}+\frac{\pi}{2}+\arcsin \frac{1}{3}$.

This number is approximately 4.739, but calculating this and comparing it with 4.7 without computational tools is quite difficult.

If we replace the arc $A B$ and the tangent $B K$ with a shorter segment $A K$, we can make such an estimate: $L_{\min }>L_{-}=\pi R+2 \sqrt{R^{2}+(R+d)^{2}}$. In numbers, this gives

$$
L_{\min }>L_{-}=\frac{\pi}{2}+\sqrt{10}>\frac{3.14}{2}+3.15>4.7 \text{. Therefore, the rope will not be enough. }
$$

In variants 2 and 3, the rope is sufficient. This is proven using a different substitution: we replace the arc $A B$ and the tangent $B K$ with the segment $K C$, which is longer because $C B=R \rtimes \operatorname{tg} \alpha>R$. Then

$$
L_{\min }<L_{+}=\pi R+\frac{2(R+d)^{2}}{\sqrt{d(2 R+d)}} . \text{ For example, in variant } 2 \text{ we get } L_{\min }<L_{+}=\frac{\pi}{2}+\frac{9 \sqrt{2}}{4}
$$

$<\frac{3.142}{2}+\frac{9 \rtimes, 42}{4}<4.8$

Answer to variant 192: Yes.

Answer to variant 193: Yes.

Answer to variant 194: No.