In the equilateral triangle $ABC$, the side length is $52 \, \text{m}$. From the vertices $A$ and $B$ of the triangle, a point starts moving simultaneously along the side $AC$ at a speed of $3 \, \text{m} / \text{sec}$ and along the side $BC$ at a speed of $4 \, \text{m} / \text{sec}$, respectively, and heads towards $C$. When will the distance between the two moving points be equal to the height of the triangle?
Let $t$ be the time in seconds after which the position of the point starting from $A$ is $P$, and the position of the point starting from $B$ is $Q$. Denote the projection of $Q$ onto the line $AC$ as $R$ (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_e56592ef3a6fdd8b58f1g-1.jpg?height=575&width=572&top_left_y=260&top_left_x=755)

Calculate the legs of the right triangle $PQR$.

$$
Q C=52-4 t
$$

Thus, from the right triangle $CQR$ with a $60^{\circ}$ angle,

$$
C R=\frac{Q C}{2}=26-2 t, \quad Q R=R C \sqrt{3}=\sqrt{3}(26-2 t)
$$

Therefore,

$$
P R=52-3 t-(26-2 t)=26-t
$$

According to the problem, $P Q$ must be $26 \sqrt{3}$ meters, which gives the equation

$$
\begin{gathered}
P Q^{2}=Q R^{2}+P R^{2}=3(26-2 t)^{2}+(26-t)^{2}=3 \cdot 26^{2} \\
13 t^{2}-14 \cdot 26 t+26^{2}=0
\end{gathered}
$$

Dividing by 13,

$$
t^{2}-28 t+52=0
$$

Thus,

$$
t_{1}=2, \quad t_{2}=26
$$

The second root is not valid because after 26 seconds, both points would be moving on the extensions of the respective sides, so the desired position occurs after 2 seconds.

Note: Some contestants did not realize that $C R=\frac{1}{2} C Q$.