3. Two baby squirrels are located at points $A$ and $B$, and start jumping simultaneously along lines $A O$ and $B O$ towards point $O$ (after passing point $O$, each continues moving along its own line). The distance $A O=120$ meters, $B O=80$ meters, and the angle $A O B=60^{\circ}$. The baby squirrels have a constant and identical speed. What is the minimum distance between the baby squirrels during their movement?
Answer: $20 \sqrt{3}$ meters.

Solution: Let the squirrels travel a distance $d = A E = B F$. Draw a line through point $E$ parallel to $O B$, and mark a segment $E K$ on it, equal to $A E$. Draw the line $A K$, and let it intersect the line $O B$ at point $C$.

Triangle $A E K$ is isosceles with an angle of $60^{\circ}$, so it is equilateral. Therefore, the similar triangle $A O C$ is also equilateral, with $O A = O C = A C = 120$. The quadrilateral $E K B F$ is a parallelogram because $E K = A E = B F$, and $E K \parallel O B$.

Thus, the distance between the squirrels, $E F$, is equal to $B K$, where $K$ is the current point on the line $A C$. The line $A C$ is determined by points $A$ and $C$, which are equidistant from point $O$ at 120 meters. Therefore, we need to find the shortest distance from the fixed point $B$ to the line $A C$. This distance is measured along the perpendicular, meaning the squirrels will be closest to each other when $B K$ is perpendicular to $A C$. In the right triangle $B K C$, the hypotenuse $B C$ is $120 - 80 = 40$ meters, and $\angle B C K = 60^{\circ}$. Thus, the shortest distance is $40 \cdot \frac{\sqrt{3}}{2} = 20 \sqrt{3}$ (meters).

Comment: Correct solution - 20 points. Correct solution but incorrect answer due to a calculation error at the end - 19 points. Three possible positions of the squirrels are considered, the correct one is chosen, but the minimality is not proven - 15 points. There are errors in the solution - 10 points. The solution is not complete, but there is progress - 5 points. The solution is started, but the progress is minimal - 1 point.