1. At the same time, a cyclist leaves point $A$ for point $B$ at a speed of $15 \mathrm{km} / \mathrm{u}$, and a tourist leaves point $B$ for point $C$ at a speed of 5 km/h. After 1 hour and 24 minutes from the start of their journey, they were at the minimum distance from each other. Find the distance between the points if all three points are equidistant and connected by straight roads. (8 points)
Solution: Let $AB = BC = AC = S$. Denote the distance between the cyclist and the tourist as $r = r(t)$, where $t$ is the time from the start of the movement. Then, by the cosine theorem, we have: $r^{2} = (S - 15t)^{2} + (5t)^{2} - 5t(S - 15t)$. To find the time when the distance between the cyclist and the tourist was the smallest, we compute the derivative of the function $r^{2} = r^{2}(t)$:

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-01.jpg?height=397&width=460&top_left_y=664&top_left_x=1255)
between the start and the movement.

$\left(r^{2}\right)^{\prime} = -2 \cdot 15(S - 15t) + 2 \cdot 25t - 5(S - 15t) + 15 \cdot 5t = -35S + 2(225 + 25 + 75)t = 0$.

Since the minimum value of the function $r^{2} = r^{2}(t)$ during the movement is achieved at $t = 1.4$, then $-35S + (225 + 25 + 75) \cdot 2.8 = 0$, and $S = 26$.

Answer: 26 km.