Two points, $A$ and $B$, lie on two lines that form a $60^{\circ}$ angle with each other. The distance between the points is $31 \mathrm{~m}$. If point $A$ moves $20 \mathrm{~m}$ closer to the intersection point of the lines, then the distance between the points becomes $21 \mathrm{~m}$. How far are the points from the intersection point of the lines?
Let the intersection point of the two lines be $M$, and the second position of point $A$ be $C$, such that $A C=20 \mathrm{~m}$. Drop a perpendicular from $B$ to $M A$, with the foot of the perpendicular being $B_{1}$. Let $B_{1} C=y$, $M B_{1}=x$, then $M B=2 x$ and $B B_{1}=x \sqrt{3}$, because $\triangle M B B_{1}$ is a right triangle.

From triangles $B B_{1} C$ and $B B_{1} A$, we get:

$$
y^{2}+3 x^{2}=21^{2} \text { and }(y+20)^{2}+3 x^{2}=31^{2} \text {, }
$$

from these equations, we find that $y=3$, $x=12$. Thus, $M B=2 x=24 \mathrm{~m}$ and $M A=x+y+20=35 \mathrm{~m}$.

(Schuster György, Budapest, VIII. district.)

The problem was also solved by: Bánó L., Ehrenstein P., Gedliczka H., Heimlich P., Jánosy Gy., Kiss J., Petrik S., Pichler S., Róth A.