Xichaturyan A.V.

Bus stop B is located on a straight highway between stops A and C. At some point after leaving A, the bus found itself at a point on the highway such that the distance from it to one of the three stops is equal to the sum of the distances to the other two. After the same amount of time, the bus was again at a point with this property, and then after another 25 minutes, it arrived at B. How much time does the bus need for the entire journey from A to C, if its speed is constant, and it stops for 5 minutes at stop B?
At both moments in time mentioned in the problem, the sum will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this moment).

![](https://cdn.mathpix.com/cropped/2024_05_06_1aa9346639cd54e59f2ag-16.jpg?height=89&width=728&top_left_y=1219&top_left_x=665)

In the first case, the bus was at point $X$ and the distance to $C$ was equal to the sum of the distances to $A$ and to $B$. But it is also equal to the sum of the distance to $B$ and the distance $BC$. Therefore, the bus had traveled exactly the distance $BC$. On the diagram, we marked equal distances with arcs.

By the second moment, the bus had traveled another distance $BC$ and was at point $Y$. The sum of the distances from it to $B$ and to $C$ is $BC$ plus $YB$, counted twice. According to the problem, this is the distance to $A$, so $YB$ is half the length of $BC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1aa9346639cd54e59f2ag-16.jpg?height=92&width=734&top_left_y=1626&top_left_x=662)

Since $YB$ was traveled by the bus in 25 minutes, $BC$ will be traveled in 50 minutes, and the entire journey will take $3 \cdot 50 + 25 + 5 = 180$ minutes, which is three hours.

## Answer

3 hours.