1. On a meadow, there are 3 anthills: $A, B$, and $C$. The distance between anthills $A$ and $B$ is 260 mm, and the distance between anthills $B$ and $C$ is 1200 mm, with $\measuredangle A B C=60^{\circ}$. An ant $a$ starts from anthill $A$ towards anthill $B$, moving in a straight line at a speed of $1 \frac{\text { mm }}{\mathrm{c}}$. At the same moment, an ant $b$ starts from anthill $B$ towards anthill $C$, also moving in a straight line at a speed of $3 \frac{\text { mm }}{\mathrm{c}}$. At what moment will the distance between ants $a$ and $b$ be minimal?
1. Let $t$ denote the elapsed seconds since ants $a$ and $b$ started their journey. Let $A_{1}$ and $B_{1}$ be the positions of ants $a$ and $b$ after $t$ seconds, respectively. Then, we have $B A_{1}=260-t$ and $B B_{1}=3 t$. Using the cosine theorem, we have

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-24.jpg?height=421&width=406&top_left_y=1073&top_left_x=1133)

Op 2017 2A 1

\[
\begin{aligned}
A_{1} B_{1}^{2} & =B A_{1}^{2}+B B_{1}^{2}-2 B A_{1} \cdot B B_{1} \cos 60^{\circ} \\
& =(260-t)^{2}+9 t^{2}-2(260-t) 3 t \cdot \frac{1}{2}=13 t^{2}-1300 t+260^{2}
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-24.jpg?height=351&width=398&top_left_y=1783&top_left_x=224)

Op 2017 2A 2
Clearly, the distance between the ants is minimal when this value is minimal. Since this is a quadratic function in $t$, it has a minimum at $t=$ $\frac{1300}{2 \cdot 13}=50$. Since the ants continue to move towards their destinations after 50 seconds, the solution to the problem is 50 seconds.