Given a circle and a chord inside it, which is equal to the side of a regular triangle inscribed in the circle. The chord rotates $90^{\circ}$ around the center of the circle. What area does the chord sweep during this rotation?
I. solution. Let the center of the circle be $O$, its radius $r$, the endpoints of the chord in its initial position be $A, B$, and its midpoint be $F$; during rotation, let $A$ move along the shorter arc $AB$, and the final positions of $A, B$, and $F$ be $C, D, G$. Thus, $C$ is still on the shorter arc $AB$ because $\angle AOB = 120^\circ$ and $\angle AOC = 90^\circ$; let the intersection of $AB$ and $CD$ be $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_476b9fc09d6ee43ca331g-1.jpg?height=469&width=486&top_left_y=297&top_left_x=814)

The closest point of the moving chord to $O$ is $F$, the farthest points are $A$ and $B$, and all points of the chord are at a distance between $OF = r/2$ and $OA = r$ from $O$. Therefore, the chord sweeps the area bounded by the longer arc $AD$, the segments $AF$, $GD$, and the quarter-circle arc $FG$, except for the area bounded by the arc $BC$ and the segments $EB$, $EC$. By drawing the radii $OF'$ and $OG'$ that bisect $AB$ and $CD$, respectively, the mentioned area is the sum of the areas of the semisegments $AFF'$ and $DGG'$ and the quarter-circular segment $FGG'F'$, minus the area of the lune $BCE$.

The combined area of the semisegments is equal to the area of the segment $ABF'$, which is the difference between the area of the circular sector $OAB$ and the area of the triangle $OAB$. The area of the lune $BCE$ is obtained by subtracting twice the area of the triangle $BOE$ from the area of the circular sector $BOC$. Since $\angle BOC = 30^\circ$ and $BF = r\sqrt{3}/2$, the areas of the mentioned parts and the total area $t$ are:

$$
\begin{gathered}
ABF' = \frac{r^2 \pi}{3} - \frac{r^2 \sqrt{3}}{4}, \quad FFG'G = \frac{\pi}{4}\left(r^2 - \frac{r^2}{4}\right) = \frac{3 \pi r^2}{16} \\
BEC = \frac{r^2 \pi}{12} - BE \cdot OF = \frac{r^2 \pi}{12} - \frac{r^2}{4}(\sqrt{3} - 1) \\
t = r^2 \cdot \frac{7 \pi - 4}{16} \approx r^2 \cdot 1.124
\end{gathered}
$$

Rózália Höss (Makó, József A. g. III. o. t.)

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II. solution (outline). By adding the diameter $AH$, the reflections of $AB$ and $CD$ over $O$, and the circle with radius $OF$ centered at $O$, and by moving the regions $t_i$ to the positions $t_i'$ $(i=1,2,3,4)$ on the diagram, we see that the area swept by the chord is equal to the area bounded by the curve $APFEGKHCA$. The latter is obtained by subtracting the quarter area of the smaller circle (the sum of the 6th and 12th parts) and the square with side $r/2$ from half the area of the larger circle:

$$
t = \frac{r^2 \pi}{2} - \frac{r^2 \pi}{4 \cdot 4} - \frac{r^2}{4}
$$

Further consideration shows that the area in question is equal to the difference between the area of the circular sector $OAS$ and the area of the square $OFEQ$, where $S$ is the first quarter point of the arc $G'H$. Indeed, the area of the circular sector $OGK$ is twice that of the circular sector $OPF$, and four times that of the circular sector $OG'H$, while the area of the circular segment $KGG'H$ is three times as large. Therefore, by adding the circular sector $OPF$ to the area obtained by dissecting, we must subtract two-thirds of the segment. The remaining one-third is equal in area to the circular sector $OGK$, so by taking a circular sector with radius $r$ instead, its arc will be $KG/2 = G'H/4 = G'S$.

Lajos Vigassy