We divided a chord $AB$ of a circle, which has the length of the radius, and the shorter arc $AB$ into three equal parts each. The line connecting the division points closer to $A$ intersects the line connecting the division points closer to $B$ at point $P$. What is the measure of angle $APB$?
I. solution. Let the points dividing the chord into thirds be $C$ and $D$, and the points on the arc be $C^{\prime}$ and $D^{\prime}$. Let $G$ be the orthogonal projection of $C^{\prime}$ onto $A B$, and use the additional notations from the figure.

$$
\text { 1986-09-257-1.eps }
$$

1. figure

Let the radius of the circle be one unit. Due to the obvious symmetry, it is sufficient to determine half of the desired angle. Its tangent can be calculated based on $A F=\frac{1}{2}$ and the knowledge of $F P$. To determine $F P$, notice that $\Delta G C C^{\prime} \sim \Delta F C P$, since their corresponding sides are parallel. Therefore,

$$
\frac{F P}{F C}=\frac{G C^{\prime}}{G C}, \quad \text { that is } \quad F P=F C \cdot \frac{G C^{\prime}}{G C}
$$

Among the segments involved, $F C=\frac{1}{6}$, and the other two can be expressed using trigonometric functions:

$$
\begin{gathered}
G C^{\prime}=F H=O H-O F=\cos 10^{\circ}-\cos 30^{\circ}=2 \sin 10^{\circ} \sin 20^{\circ} \\
G C=G F-C F=C^{\prime} H-C F=\sin 10^{\circ}-\frac{1}{6}
\end{gathered}
$$

Since $\frac{1}{6}=\frac{1}{3} \cdot \sin 30^{\circ}$, we have

$$
G C=\frac{1}{3}\left[2 \sin 10^{\circ}-\left(\sin 30^{\circ}-\sin 10^{\circ}\right)\right]=\frac{2 \sin 10^{\circ}}{3}\left(1-\cos 20^{\circ}\right)
$$

From this,

$$
\operatorname{tg} \angle A P F=\frac{A F}{F P}=\frac{A F}{F C} \cdot \frac{G C}{G C^{\prime}}=\frac{3 G C}{G C^{\prime}}=\frac{1-\cos 20^{\circ}}{\sin 20^{\circ}}=\operatorname{tg} 10^{\circ}
$$

Considering that $\angle A P F$ is certainly an acute angle, $\angle A P F=10^{\circ}$, and thus $\angle A P B=20^{\circ}$.

II. solution. Let the central angle corresponding to the chord $A B$ be $\alpha$. Slightly generalizing the statement of the problem, we will show that for any chord $A B$, $\angle A P B=\frac{\alpha}{3}$.

$1986-09-258-1 . \mathrm{eps}$

$$
\text { 2. figure }
$$

Project the points $A, B$ from $P$ onto the line $C^{\prime} D^{\prime}$. Then $A^{\prime} C^{\prime}=C^{\prime} D^{\prime}=D^{\prime} B^{\prime}$, so in the triangle $A D^{\prime} A^{\prime}$, $C^{\prime}$ is the midpoint of $A^{\prime} D^{\prime}$. The radius $O C^{\prime}$ bisects the chord $A D^{\prime}$, since $C^{\prime}$ is the midpoint of the arc $A D^{\prime}$. Therefore, $O C^{\prime}$ is the axis of the triangle $O A D^{\prime}$, bisects $A D^{\prime}$, and thus is the midline of the triangle $A D^{\prime} A$, and is parallel to $A^{\prime} P$. Similarly, we find that $O D^{\prime}$ is parallel to $B^{\prime} P$. Therefore, as corresponding angles of parallel lines, $\angle A P B=\angle C^{\prime} O D^{\prime}=\frac{\alpha}{3}$.