489*. Three cyclists set off simultaneously: the first and second from point A, and the third towards them from point B. After 1.5 hours, the first cyclist was at an equal distance from the other two, and after 2 hours from the start, the third cyclist was at an equal distance from the first and second. How many hours after departure was the second cyclist at an equal distance from the first and third?
$\triangle$ Let's take the path AB as a unit. Denote the speeds of the first, second, and third cyclists as $v_{1}, v_{2}$, and $v_{3}$ (in fractions of this path per hour).

We will denote the positions of the first, second, and third cyclists in each of the three specified situations by the letters $\mathbf{M}_{1}, \mathbf{M}_{2}$, and $\mathbf{M}_{3}$, respectively.

After 1.5 hours from the start of the movement, the first cyclist was halfway between the second and the third (Fig. 4), and after 2 hours, the third cyclist was halfway between the first and the second (Fig. 5).

![](https://cdn.mathpix.com/cropped/2024_05_21_88152b3867f80e3be9a3g-169.jpg?height=103&width=514&top_left_y=1902&top_left_x=497)

Fig. 4

![](https://cdn.mathpix.com/cropped/2024_05_21_88152b3867f80e3be9a3g-169.jpg?height=109&width=568&top_left_y=1896&top_left_x=1007)

Fig. 5

Since on Fig. $4 \mathrm{M}_{2} \mathrm{M}_{1}=\mathrm{M}_{1} \mathrm{M}_{3}$, where

$$
\mathbf{M}_{2} \mathbf{M}_{1}=1.5\left(v_{1}-v_{2}\right), \quad \mathbf{M}_{1} \mathbf{M}_{3}=1-1.5 v_{1}-1.5 v_{3},
$$

then

$$
1.5\left(v_{1}-v_{2}\right)=1-1.5\left(v_{1}+v_{3}\right)
$$

Since on Fig. $5 M_{2} M_{3}=M_{3} M_{1}$, where

$$
\mathbf{M}_{2} \mathbf{M}_{3}=1-2\left(v_{2}+v_{3}\right), \quad \mathbf{M}_{3} \mathbf{M}_{1}=2\left(v_{1}+v_{3}\right)-1
$$

then

$$
1-2\left(v_{2}+v_{3}\right)=2\left(v_{1}+v_{3}\right)-1
$$

We obtain the system of equations:

$$
\left\{\begin{array}{l}
1.5\left(v_{1}-v_{2}\right)=1-1.5\left(v_{1}+v_{3}\right) \\
1-2\left(v_{2}+v_{3}\right)=2\left(v_{1}+v_{3}\right)-1
\end{array}\right.
$$

From this, we find that

$$
v_{3}=\frac{1}{2}\left(1-v_{1}-v_{2}\right), \quad v_{1}-v_{2}=\frac{1}{9}
$$

(consider how to derive these equalities).

We are interested in the third position, when the second cyclist is halfway between the first and the third (Fig. 6).

![](https://cdn.mathpix.com/cropped/2024_05_21_88152b3867f80e3be9a3g-170.jpg?height=92&width=546&top_left_y=1276&top_left_x=755)

Fig. 6

Let $t$ hours be the time from the start of the movement to this position. Since $\mathrm{M}_{3} \mathrm{M}_{2}=\mathrm{M}_{2} \mathrm{M}_{1}$, where

$$
\mathrm{M}_{3} \mathrm{M}_{2}=\left(v_{2}+v_{3}\right) t-1, \quad \mathbf{M}_{2} \mathbf{M}_{1}=\left(v_{1}-v_{2}\right) t
$$

then

$$
\left(v_{1}+v_{3}\right) t-1=\left(v_{1}-v_{2}\right) t .
$$

Solve for $t$, and then, using the obtained expressions for $v_{3}$ and $v_{1}-v_{2}$, calculate $t$:

$t=\frac{1}{2 v_{2}+v_{3}-v_{1}}=\frac{1}{2 v_{2}+\frac{1}{2}\left(1-v_{1}-v_{2}\right)-v_{1}}=\frac{2}{1-3 v_{1}+3 v_{2}}=2:\left(1-\frac{3}{9}\right)=3$.

Answer: after 3 hours.