# Problem 4.

At 9:00 AM, ships "Anin" and "Vinin" departed from port O to port E. At the same moment, ship "Sanin" departed from port E to port O. All three vessels are traveling on the same course ("Sanin" heading towards "Anin" and "Vinin") at constant but different speeds. At 10:00 AM, "Vinin" was at the same distance from "Anin" and "Sanin". At 10:30 AM, "Sanin" was at the same distance from "Anin" and "Vinin". At what moment will "Anin" be exactly in the middle between "Vinin" and "Sanin"?
# Solution.

For simplicity, let's assume the distance between the ports is 1. Let $x$ be the speed of "Anina", $y$ be the speed of "Vanina", and $z$ be the speed of "Sanina". Then at 10 o'clock (i.e., one hour after departure),

$$
2 y = x + 1 - z,
$$

and one and a half hours after departure,

$$
2(1 - 1.5 z) = 1.5 x + 1.5 y.
$$

After time $t$ from departure, the condition should be met:

$$
2 x t = 1 - t z + t y,
$$

which simplifies to

$$
t(2 x + z - y) = 1.
$$

From (1) and (2),

$$
3 z + 1.5 x + 1.5 y = 4 y + 2 z - 2 x
$$

or $z = 2.5 y - 3.5 x$. Substituting this expression into (1), we get $y - x = 2 / 9$. Considering this, from (1) we have $y + z + y - x = y + z + 2 / 9$. Therefore, $t + z = 7 / 9$. Thus, from (3) we get

$$
\begin{gathered}
t(2 x + z - y) = t(z + y - 2 x + 2 y) = 1 \\
t(7 / 9 - 4 / 9) = 1
\end{gathered}
$$

which gives $t = 3$.

Answer: at 12 o'clock noon.