1. At the same time, a car departs from point $A$ to point $B$ at a speed of $90 \mathrm{~km} / \mathrm{h}$, and from point $B$ to point $C$ - a train at a speed of $60 \mathrm{~km} /$ h. After 2 hours of travel, they found themselves at the minimum distance from each other. Find the distance between the points, if all three points are equidistant from each other and are connected by straight roads. (8 points)
Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 90t)^{2} + (60t)^{2} - 60t(S - 90t)$. To find the time at which the distance between the car and the train

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-15.jpg?height=428&width=482&top_left_y=503&top_left_x=1438)
was the smallest, we will compute the derivative of the function $r^{2} = r^{2}(t):$

$\left(r^{2}\right)^{\prime} = -2 \cdot 90(S - 90t) + 2 \cdot 3600t - 60(S - 90t) + 90 \cdot 60t = -240S + 2(8100 + 3600 + 5400)t = 0$.

Since the function $r^{2} = r^{2}(t)$ takes its minimum value during the motion at $t = 2$, then $-120S + (8100 + 3600 + 5400) \cdot 2 = 0$, and $S = 285$.

Answer: 285 km.