40. A tourist got lost in the forest; he knows the shape of the forest, but does not know where he is.

a) If the forest is a circle with diameter $d$, then by moving in a straight line in any fixed direction, the tourist will certainly exit the forest after traveling a distance $d$: any segment of length $d$ starting inside the forest will intersect its boundary (Fig. $9, a$).
![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-025.jpg?height=388&width=834&top_left_y=526&top_left_x=220)

Fig. 9.

b) If the forest is a half-plane and the tourist accidentally remembers that he moved away from the forest boundary by a distance $d$ (but does not know in which direction the boundary is from him), then he can exit the forest by traveling a segment $AB$ of length $d$ in any direction, and then circling around the circle $S$ with center at the initial point $A$ and radius $d$: this path of length $d + 2\pi d (\approx 7.28d)$ will certainly lead him to the boundary of the forest, located somewhere at a distance $d$ from point $A$ (Fig. 9, b).

Can the tourist choose another path that will certainly lead him out of the forest before he travels a distance $d$, respectively $(1 + 2\pi)d$?

Problems 40 a), b) are from the well-known American mathematician R. Bellman (see pages $161-162$ of book [40] or the note with the same number). In the case of problem 40 a), it is not difficult to find the optimal path that ensures the quickest exit from the forest; however, for a forest of a different shape, this is not easy at all. V. A. Zalgaller [41] found the most advantageous path in the case where the forest is a sufficiently elongated rectangle, with the ratio of the longer side to the shorter side $>2.278$ (and for the case where the forest is an infinite strip of width $d$); however, it is still unknown what path should be chosen in the case where the forest has a rectangular shape with a side ratio less than 2.278 (for example, a square shape), or, say, a triangular shape (for example, an equilateral triangle). Regarding problem 40 b), see J. R. Isbell [42].
40. a) It cannot: a straight path $AB$ of length less than $d$ does not guarantee an exit from the forest. Indeed, suppose the tourist's location and the chosen direction of movement at the beginning of the journey are such that the midpoint of the line $AB$ coincides with the center $O$ of the forest. In this case, the distance $OM$ from any point $M$ on the line $AB$ to the center $O$ of the forest will be less than the radius $d/2$ of the forest (Fig. 103) - and the line $AB$ will entirely pass within the forest, so the tourist will not find an exit.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-129.jpg?height=353&width=400&top_left_y=1234&top_left_x=221)

Fig. 103. 

b) It can. For example, we can proceed as follows: starting from point $A$, we walk the straight path $AB = d$; then we describe the arc $BC$ of a circle $S$ with center $A$ and radius $d$, equal to $270^\circ \left(=\frac{3}{2} \pi\right)$; after that, we walk the path $CD = d$ along the tangent to the circle $S$ at point $C$ (Fig. 104, a). It is easy to see that this path has the property that all lines at a distance $d$ from point $A$ (all tangents to the circle $S$ with center $A$ and radius $d$) have common points with it. But the boundary of the forest is one of these latter lines, so by moving along the path $ABCD$ (Fig. 104, a), we will certainly find an exit from the forest. The length of the path $ABCD$ is

$AB + \cup BC + CD = d + \frac{3}{2} \pi d + d = \left(2 + \frac{3}{2} \pi\right) d \approx 6.71 d < (1 + 2 \pi) d$

$$
(\approx 7.28 d)
$$

Note. The specified path $ABCD$ can be replaced by a shorter one. For this, we first walk the path $AB_1 = \frac{2 \sqrt{3}}{3} d$ in any direction; then along the tangent $B_1C_1$ to the circle $S$ with center $A$ and radius $d$, we exit the circle $S$ at such a point $C_1$ that $\vee BC_1 = 30^\circ$, where $B$ is the intersection point of the segment $AB_1$ with the circle; then we walk the arc $C_1D_1$ of the circle $S$, which is $210^\circ \left(\text{or} \frac{7}{6} \pi\right)$ in degree measure; finally, we walk the segment $D_1E = d$ along the tangent to the circle at point $D_1$ (Fig. 104, b). It is easy to see that the path $AB_1C_1D_1E$ also intersects all tangents to the circle $S$ and, therefore, will certainly lead us out of the forest; the length of this path

$$
\begin{aligned}
AB_1 + B_1C_1 + \cup C_1D_1 + D_1E = \frac{2 \sqrt{3}}{3} d & + \frac{\sqrt{3}}{3} d + \frac{7}{6} \pi d + d = \\
& = \left(1 + \sqrt{3} + \frac{7}{6} \pi\right) d \approx 6.40 d
\end{aligned}
$$

is less than the length $\left(2 + \frac{3}{2} \pi\right) d$ of the path $ABCD$ depicted in Fig. 104, a.

J. R. Isbell [42] proved that the path $AB_1C_1D_1E$ depicted in Fig. 104, b is the shortest of all paths that, under the conditions of problem 40 b), guarantee an exit from the forest.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-130.jpg?height=458&width=980&top_left_y=278&top_left_x=104)

Fig. 104.