A chord of a unit radius circle is equal to the side of an inscribed equilateral triangle. The chord is rotated $90^{\circ}$ around the center. What area does the chord sweep over in the process? - By what angle should the chord be rotated so that it sweeps over exactly half the area of the circle?
a) Let the center of our circle be $O$, the initial position of the chord be $AB$, and the final position be $A_1B_1$ such that $A$ moves ahead, meaning $B_1$ stops before reaching $A$ on the shorter arc $BA$, since its central angle is $120^\circ$. Let the midpoints of $AB$ and $A_1B_1$ (the points closest to $O$) be $F$ and $F_1$, respectively, and their intersection point be $M$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_9f03e15045ea9be21498g-1.jpg?height=643&width=631&top_left_y=291&top_left_x=736)

Thus, during the rotation, the chord continuously touches the quarter arc $FF_1$ of the circle centered at $O$ with radius $OF$, and the boundaries of the worn-out area $S$ are the segments $A_1F_1$, $FB$, $B_1M$, $MA$, and the quarter arcs $BB_1$, $AA_1$, $FF_1$.

Draw the radius of the circle passing through $M$ and the diameter perpendicular to $OM$, let the endpoints be $M'$ and $N'$, $P'$ (on the shorter arcs $BB_1$ and $AA_1$, respectively), and let $ON'$ intersect $AB$ at $N$, and $OP'$ intersect $A_1B_1$ at $P$. By a $90^\circ$ rotation, the triangular areas $BN'N$ and $A_1P'P$ can be mapped to $B_1M'M$ and $AM'M$, respectively, so the area $t$ of $S$ is equal to the area bounded by the line segments $FNN'M'P'PF_1F$. This area is the area of the quarter circle $OFF_1$ minus the areas of the isosceles right triangles $ONF$ and $OPF_1$ and the semicircle $N'M'P'$. Since the triangle $OFA$ is half of a regular triangle with side length 1, $OF = 1/2$, and thus

$$
t = \frac{1}{2} \pi - \frac{1}{4} \pi \left(\frac{1}{2}\right)^2 - 2 \cdot \frac{1}{2} \left(\frac{1}{2}\right)^2 = \frac{7 \pi}{16} - \frac{1}{4} = 1.124
$$

b) The found $t$ is only slightly more than one-third of the circle's area. Moreover, the area of the worn-out region increases monotonically with the increase in the rotation angle, so the rotation angle must be increased. However, a simpler calculation shows that even at a $120^\circ$ rotation, $S$ does not reach half of the circle's area—not even 45% of it. Therefore, the sought angle $\alpha$ must be greater than $\pi / 3$, so the initial examination of the $AMB_1$ triangular part ceases. However, it is somewhat simpler to calculate here that the unworn area is also half of the circle's area (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_9f03e15045ea9be21498g-1.jpg?height=643&width=585&top_left_y=1657&top_left_x=759)

This can be decomposed into the circular segments $OA_1B$ and $OFF_1$ with central angles $\alpha$ and $(4\pi / 3) - \alpha$, respectively, and the triangles $OBF$ and $OA_1F_1$, which can be combined into a regular triangle with side length 1. Thus, from the requirement

$$
t = \frac{1}{2} \left(\frac{4 \pi}{3} - \alpha\right) + \frac{1}{2} \left(\frac{1}{2}\right)^2 \alpha + \frac{\sqrt{3}}{4} = \frac{\pi}{2}
$$

the final rotation angle is

$$
\alpha = \frac{4 \pi + 6 \sqrt{3}}{9} = 2.551 \text{ radians} = 146^\circ 10'
$$