The side length of the equilateral triangle $ABC$ is $6 \, \text{cm}$. Starting from vertex $C$, a bug moves uniformly towards vertex $A$ at a speed of $4 \, \text{mm/s}$. At the same time, another bug starts from vertex $B$ and moves towards vertex $C$ at a speed of $3 \, \text{mm/s}$. How much time after their start will they be closest to each other, and what will this distance be?
Solution. Let's examine the position of the beetles $t$ seconds after the start.

![](https://cdn.mathpix.com/cropped/2024_05_02_444748d234814fa01e10g-1.jpg?height=423&width=528&top_left_y=225&top_left_x=793)

The beetle starting from vertex $C$ has traveled $4t$, and the beetle starting from vertex $B$ has traveled $3t$. We need to find the minimum of segment $PQ$. We can calculate $PQ$ using the cosine rule:

$$
PQ^2 = CP^2 + CQ^2 - 2 \cdot CP \cdot CQ \cdot \cos \alpha
$$

Using the fact that $CP = 4t$, $QB = 3t$, and therefore $CQ = 60 - 3t$:

$$
PQ^2 = (4t)^2 + (60 - 3t)^2 - 2 \cdot 4t \cdot (60 - 3t) \cdot 0.5
$$

since $\cos 60^\circ = 0.5$. After expanding the right-hand side and completing the square:

$$
PQ^2 = 37t^2 - 600t + 3600 = 37\left[\left(t - \frac{300}{37}\right)^2 + \frac{43200}{1369}\right]
$$

$PQ^2$ has its minimum where $PQ$ does, since the square function is strictly increasing for positive numbers. Thus, the desired minimum occurs at $t = \frac{300}{37}$, at which time

$$
PQ = \sqrt{\frac{43200}{37}} \approx 34.17
$$

Therefore, after 8.11 seconds, they will be closest to each other, with a distance of $34.17 \text{ mm}$.

Remark. Many solvers made statements about the location of the extremum "intuitively" without any proof attempt, which are not true. (The most common ones: the triangle $APQ$ is isosceles, or the desired segment is parallel to $AB$, or the desired segment is perpendicular to, for example, side $AC$.)

Some misunderstood the problem, thinking that the beetles were moving around the perimeter of the triangle until the faster one caught up with the slower one.