Task B-4.5. Along a straight road in one direction, 4 trees are planted. The first is 6 m from the second, the second is 4 m from the third, and the third is 8 m from the fourth. At what distance from each of the trees is the observer who sees all three distances between the trees at the same angle?
## Solution.

Let's denote the positions of the trees by points $A, B, C$, and $D$, and the position of the observer by point $P$.

![](https://cdn.mathpix.com/cropped/2024_05_30_6bd7d8198e5de61c26bbg-17.jpg?height=672&width=1003&top_left_y=1971&top_left_x=537)

Introduce the following notations:

$$
\begin{aligned}
& \angle A P B = \angle B P C = \angle C P D = \alpha \\
& |A P| = a, |B P| = b, |C P| = c
\end{aligned}
$$

Apply the Law of Sines to triangles $A P B$ and $B P C$:

$$
\frac{6}{\sin \alpha} = \frac{a}{\sin \angle P B A} \text{ and } \frac{4}{\sin \alpha} = \frac{c}{\sin (180^\circ - \angle P B A)}
$$

It follows that

$$
\frac{a}{c} = \frac{3}{2}
$$

Similarly, from triangles $B P C$ and $C P D$ we get

$$
\frac{b}{d} = \frac{1}{2}
$$

Note: These ratios could also be obtained by applying the Angle Bisector Theorem to triangles $A P C$ and $B P D$.

From the Law of Cosines and the given ratios, we have:

$$
\begin{gathered}
\left\{\begin{array}{l}
a^2 + b^2 - 2ab \cos \alpha = 36 \\
b^2 + c^2 - 2bc \cos \alpha = 16 \\
c^2 + d^2 - 2cd \cos \alpha = 64
\end{array}\right. \\
\left\{\begin{array}{l}
9c^2 + 4b^2 - 12bc \cos \alpha = 144 \\
b^2 + c^2 - 2bc \cos \alpha = 16 \\
c^2 + 4b^2 - 4bc \cos \alpha = 64
\end{array}\right.
\end{gathered}
$$

From the second equation, we have

$$
2bc \cos \alpha = b^2 + c^2 - 16
$$

Substituting this into the first and third equations, we get

$$
\left\{\begin{array}{l}
9c^2 + 4b^2 - 6b^2 - 6c^2 + 96 = 144 \\
c^2 + 4b^2 - 2b^2 - 2c^2 + 32 = 64
\end{array}\right.
$$

which simplifies to

$$
\left\{\begin{array}{l}
3c^2 - 2b^2 = 48 \\
-c^2 + 2b^2 = 32
\end{array}\right.
$$

It follows that $c^2 = 40$ and $b^2 = 36$.

Thus, the desired distances are $c = 2\sqrt{10} \text{ m}$, $b = 6 \text{ m}$, $a = \frac{3}{2}c = 3\sqrt{10} \text{ m}$, and $d = 2b = 12 \text{ m}$.