For a given circle, construct a square using the procedure described under (C) on page 227. To what extent does the area of this square approximate that of the circle?[^0]


[^0]:    ${ }^{1}$ The article part $C$ starts on page 225. (Ed.)
Let the side of the inscribed square be the chord $FG$; this is divided into four equal parts by the points $F^{\prime}, E^{\prime}, G^{\prime}$. The lines $O F^{\prime}$ and $O G^{\prime}$ determine the points $A^{\prime}, B^{\prime}$ on the circle. The line $A z A^{\prime} B^{\prime}$ intersects the radii $O F$ and $O G$ at points $A$ and $B$ respectively, and $AB$ is the side of the sought square, with its midpoint being $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_48fd0ab10ea3adf8e205g-1.jpg?height=422&width=420&top_left_y=304&top_left_x=842)

Let $AB = x$; then $OE = EB = \frac{x}{2}$, because $\widehat{EOB} = 45^{\circ}$. Furthermore,

$EB^{\prime} = \frac{x}{4}$, because $E^{\prime}G^{\prime} = \frac{FG}{4}$. Since $OB^{\prime} = r$, in the right triangle $OEB^{\prime}$,

$$
\left(\frac{x}{2}\right)^{2} + \left(\frac{x}{4}\right)^{2} = r^{2}, \quad \text{that is} \quad \frac{5x^{2}}{16} = r^{2}, \quad x^{2} = \frac{16r^{2}}{5}
$$

Thus, the area of the square with side $AB = x$ is:

$$
t = x^{2} = \frac{16}{5} r^{2} = 3.2 r^{2}
$$

The area of the circle is $3.14159 r^{2}$.

The difference is: $r^{2}(3.2 - 3.14159) = 0.05841$. This difference is approximately the 54th part of the area of the circle.

George Bizám (Bolyai 9. VI. o. Bp. V.)