In an equilateral triangle $ABC$, points $M$ and $N$ are given on side $AB$, point $P$ on side $BC$, and point $Q$ on side $CA$, such that

$$
M A + A Q = N B + B P = A B
$$

What angle can the lines $MP$ and $NQ$ form?
Solution. Since $N B + B P = A B$ and $A N + N B = A B$, we have $A N = B P$. Similarly, $M A = Q C$. Rotate the triangle $ABC$ around its centroid by $120^{\circ}$ such that point $A$ moves to point $B$, point $B$ moves to point $C$, and point $C$ moves to point $A$. During this rotation, since $A N = B P$, point $N$ moves to point $P$, and since $M A = Q C$, point $Q$ moves to point $M$. Therefore, the line $N Q$ is mapped to the line $P M$.

![](https://cdn.mathpix.com/cropped/2024_05_02_64580eafdb63f23419d7g-1.jpg?height=455&width=501&top_left_y=350&top_left_x=801)

Since the two lines are mapped to each other by a $120^{\circ}$ rotation, the angle between the lines is

$$
180^{\circ} - 120^{\circ} = 60^{\circ} \text {. }
$$