11.1. Mikhail leaves from Berdsk to Cherapanovo at $8:00$ AM; On the same day, at the same time, and on the same road, Khariton and Nikolay leave from Cherapanovo to Berdsk. At 9:30 AM, Khariton was exactly halfway between Mikhail and Nikolay; at 10:00 AM, Mikhail was exactly halfway between Khariton and Nikolay. Determine the time when Mikhail and Khariton met, and the time when Mikhail and Nikolay met, if all of them moved at constant speeds?
Answer. Mikhail and Khariton met at 9:48 AM, Mikhail and Nikolay - at 10:15 AM.

Solution. Let $S$ be the distance between Berdsk and Cherepanovo, and $x, y, z$ be the speeds of Mikhail, Khariton, and Nikolay, respectively. From the fact that at 9:30 AM, Khariton was exactly halfway between Mikhail and Nikolay, it follows that $y > z$ and $\left(S - \frac{3}{2} y\right) - \frac{3}{2} x = \frac{3}{2} y - \frac{3}{2} z$, which is equivalent to $x + 2y - z = \frac{2}{3} S$. From the fact that at 10:00 AM, Mikhail was exactly halfway between Khariton and Nikolay, it follows that $2x - (S - 2y) = S - 2x - 2z$, which is equivalent to $2x + y + z = S$.

The time required for Mikhail and Khariton to meet is $\frac{S}{x + y} = \frac{3S}{(x + 2y - z) + (2x + y + z)} = \frac{3S}{\frac{2S}{3} + S} = \frac{9S}{5S} = \frac{9}{5}$ hours, which is 1 hour and 48 minutes, so this meeting occurred at $9:48$ AM.

The time required for Mikhail and Nikolay to meet is $\frac{S}{x + z} = \frac{3S}{-(x + 2y - z) + 2(2x + y + z)} = \frac{3S}{-\frac{2S}{3} + 2S} = \frac{9S}{4S} = \frac{9}{4}$ hours, which is 2 hours and 15 minutes, so this meeting occurred at $10:15$ AM.

Grading Criteria. ($\cdot$) Correctly setting up the system of equations for $S, x, y, z: 3$ points. ($\cdot$) Calculation of each answer: 2 points each.